mm-4179 === Subject: Re: Complete Solutions Manual for College/Grad School Textbooks Bytes: 1882 I'd like to purchase the solution manual for William H. Hoffman, Jr.James E. Smith Eugene Willis West Federal Taxation 2008 - Individual Income Taxes (with RIA Checkpoint and Turbo Tax Premier CD-ROM) 31st Edition ?2008 ISBN:0324380585 Please let me know if you have it. === Subject: Re: Complete Solutions Manual for College/Grad School Textbooks Bytes: 1731 email directly to sbooks4sale[at]hotmail[dot]com === Subject: Re: Software bugs: Carette's law > ...................................................... > ......... Users incorrectly assume that most of the bugs that > they have > encountered are known. [Carette's wording is impressively exact. Same holds > for at > least some other systems. -- VB] ...................................................... > ........ http://www.mapleprimes.com/forum/bug-forum 2007-04-29 10:22 by Jacques Carette http://www.cas.mcmaster.ca/~carette/ I meant to comment on your other point: it is my > personal > experience that a lot of Maple users ``sit'' on their > cache > of (Maple) bugs that they encounter, rarely reporting > them > officially. Whenever they encounter someone somewhat > linked > to Maplesoft (like me), they spill their guts. I got > another > example of this over the last two days at a workshop > I was at. These users incorrectly assume that most of the bugs > that they > have encountered are known: back when I worked at > Maplesoft > (and because of the job I had, had a pretty good idea > of the > known bugs), I almost always learned about new bugs > in these > user-encounters. It really saddened me that these had > not been > reported, because we then had had no chance of fixing > them! ...................................................... > ........ Send Your Fav Maple Bug To Maple Bugs Encyclopaedia http://maple.bug-list.org/sendbug.php ...................................................... > ........ The beta 0.1 of the first world's document written by > a human in a close cooperation with a successor of > the > GEMM machine, our unique VM automated testing expert > system which is a failure prediction oracle http://maple.bug-list.org/maple-crisis.php ...................................................... > ........ like i said many times before: i prefer working with pen and paper. and all your posts are good reason to do so. even for factoring i usually dont use a computer , having tricks of my own... not to mention diophantine equations , integrals and combinatorics. greets tommy1729 === Subject: Re: Software bugs: Carette's law <23685972.1183242836289.JavaMail.jakarta@nitrogen.mathforum.org> Bytes: 1940 >http://www.mapleprimes.com/forum/bug-forum > like i said many times before: > i prefer working with pen and paper. > and all your posts are good reason to do so. > even for factoring i usually dont use a computer , having tricks of my own... > not to mention diophantine equations , integrals and combinatorics. > greets Oh yes? The Ludovicus-Nash theorem says: Given any prime p, for the trinomial T = X^2 + X + A, there are values of A that makes T to have p as minimum divisor. (For some X = 0,1,2,3,4,5.. inf.) Please, with pen and paper give me one A that makes T to have 401 as its smaller divisor. Ludovicus === Subject: Re: Software bugs: Carette's law > On Jun 30, 6:33 pm, tommy1729 i prefer working with pen and paper. > and all your posts are good reason to do so. > even for factoring i usually dont use a computer , > having tricks of my own... > not to mention diophantine equations , integrals > and combinatorics. > greets Oh yes? > The Ludovicus-Nash theorem says: Given any prime p, > for the trinomial > T = X^2 + X + A, there are > values of A that makes T to have p as minimum > divisor. (For some X = > 0,1,2,3,4,5.. inf.) > Please, with pen and paper give me one A that makes T > to have 401 as > its smaller divisor. > Ludovicus > simple A = 401 and X = 0 so T = 401 and 401 is its smallest divisor :-) tommy1729 === Subject: Re: Software bugs: Carette's law Bytes: 2652 .............................................................. http://www.cas.mcmaster.ca/~carette/publications/Testing%20a%20CAS.pdf Testing a Computer Algebra System .............................................................. http://maple.bug-list.org/sendbug.php .............................................................. Do you want to get the bugs you encountered fixed sooner? Send us your Maple bugs. Any version. Any suspicious stuff. If you are not sure this is a bug, all the same send us it. We guarantee you that we will answer you, possibly providing you with some extra details. Within a span of time, we are going to contact Maplesoft and convey its top management your data handled by us, so that hopefully these defects would be fixed. Hesitate not a second. Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director .............................................................. The beta 0.1 of the first world's document written by a human in a close cooperation with a successor of the GEMM machine, our unique VM automated testing expert system which is a failure prediction oracle http://maple.bug-list.org/maple-crisis.php .............................................................. === Subject: Re: Why Null Spaces? Cc: comp.graphics.algorithms Bytes: 2093 needs to be considered especially in the context of a differential equation. The example on page 2 of the paper http://legendre.cse.uiuc.edu/~zvander/fd_null.pdf did a good job of highlighting the non uniqueness of the solution to the ode: u'(t) = 3 solution: u(t) = 3t + c of constant functions, and the basis for this space is the function g(t) = 1. My question now is how is the null space of the biharmonic pde (please refer to page 9 of the slide http://www-cse.ucsd.edu/classes/fa01/cse291/hhyu-presentation.pdf) determined to have the form: a + b.x ... ??? - Olumide === Subject: Re: Why Null Spaces? > My question now is how is the null space of the biharmonic pde (please > refer to page 9 of the slide > http://www-cse.ucsd.edu/classes/fa01/cse291/hhyu-presentation.pdf) > determined to have the form: a + b.x ... ??? Page 9 of this document refers to the null space of phi(f) = integral (f(x)^2) presumably integrating over the real line; it is not a claim about the null space of the biharmonic PDE (which is not finite dimensional). The only way for this integral to be zero is when f(x) = 0 [assuming C^2 functions. Generally, the integrand is zero except possibly on a set of measure zero.] The solution space to f(x) = 0 is f(x) = a + b*x for any constants a and b. -- Dave Eberly http://www.geometrictools.com === Subject: Re: Wanted: All Solutions of 3-Variable Polynomial I would like all integer solutions of the following > polynomial in three variables: 3^8 x^9 - 3^7 (a + b) x^6 + >(3^5 (a + b)^2 - 2^3 3^3 a b) x^3 - (a + b)^3 = 0 if this is an easy question for someone. All or none only please. {(a,b,x) in Z^3: 6561x^9 - 2187(a + b) x^6 + (243(a > + b)^2 - 216a b) > x^3 - (a + b)^3 = 0}. Sorry, I couldn't resist. Note that the set is > infinite, since {(a, -a, > 0): a in N}is a subset thereof. -- > Stephen J. Herschkorn > sjherschko@netscape.net > Math Tutor on the Internet and in Central New Jersey > and Manhattan > .. what method used ? === Subject: Re: Wanted: All Solutions of 3-Variable Polynomial <8863542.1183243390366.JavaMail.jakarta@nitrogen.mathforum.org> Bytes: 1987 I would like all integer solutions of the following > polynomial in three variables: 3^8 x^9 - 3^7 (a + b) x^6 + >(3^5 (a + b)^2 - 2^3 3^3 a b) x^3 - (a + b)^3 = 0 if this is an easy question for someone. All or none only please. {(a,b,x) in Z^3: 6561x^9 - 2187(a + b) x^6 + (243(a + b)^2 - 216a b) > x^3 - (a + b)^3 = 0}. Sorry, I couldn't resist. Note that the set is infinite, since {(a, -a, > 0): a in N}is a subset thereof. > [...] > .. what method used ? Looking for solutions of a special form. Basically, what happens to the equation if you substitute x=0? --- Christopher Heckman === Subject: Re: Catenary at different elevations > bicycle!... or is it a tricycle? wish I knew nore math, but I'd use a different > coordination. also, > don't ride no hands on a your tricycle! > AHA YOU SEE THAT ! brian admits he doesnt know all about math. so why the hell JSH do you consider him as the supermathematician the (male) QUEEN of math ?? its an ass thing right ? and since brian wished he knew more , and hes your hero , and you admitted you dont know as much as he does; that proves your a crankpot !!! checkmate !!! so get out of sci.math besised you said you where leaving and distancing the community anyways. and you were bored bored bored. so get out of sci.math hahahahhahhaahaha tommy1729 === Subject: Re: Catenary at different elevations <12350389.1183244886258.JavaMail.jakarta@nitrogen.mathforum.org> Bytes: 2711 he said WHAT -- did your lips move? as I have stated at least pi times, I can only use tripolar coordinates -- sorry! > wish I knew nore math, but I'd use a different > coordination. also, > don't ride no hands on a your CHAIN-DRIVEN tricycle! thus: I said that, once, on r.synergetics.us.dotcombubble.com; Snelson preinvented the bicycle wheel, or, possibly, the pneumatic self-axial hamster wheel (the Dyminium, which is periodically preinvented (tm) .-) then, came the F-ing Saucer. >http://www.alumnos.unican.es/uc1279/Tensegrity_Structures.htm.) >discovering the principle of tensegrity and Snelson credit for the first >My initial harvest of mathematical structures produced by this new >conceptual tool was a family of four Tensegrity masts characterized by >vertical side-faces of three, four, five and six each, respectively. The >three and four sided masts consisted of discontinuous compression >islands of tetrahedronal strut groups mounted only in tension one above >the other, while the five and six sided masts consisted of local islands >of icosahedronal and octahedronal strut groups mounted vertically above >one another, again only by tensional connectors. I only recall one mast by Fuller, that stack of tetrahedral radii which >was an adaptation of Snelson's original X-Piece. I'm curious about the >other three. Dymaxion World only shows one mast. --n~nerfman~n! === Subject: Re: Catenary at different elevations > David, right side of the equation. I just tried the following closed form > approximation for a: a = p + q + q/p + ((2 - q)*q)/(2*p^2) + (q*(6 - 9*q + 2*q^2))/(6*p^3) + > (q*(12- 36*q + 22*q^2 - 3*q^3))/(12*p^4) > where p = log(2x) and q = log(p) We still cannot arrive at a realistic value though. Sorry to hear that. However, I've checked the formula for sag given in my previous post, and it seems to be correct, given, of course, the simplifying assumptions which you were using. > There must be > something fundamentally wrong with our approach. I did know that we are > only approximating the span because of the chain wrap around the sprocket > but, as you said, that shouldn't make the solution very inaccurate. If > you think of anything else, let me know. I had indeed _guessed_ that that simplifying assumption shouldn't make the solution very inaccurate, but my guess must have been wrong. That's the only thing I can think of which would be preventing your getting realistic values for the sag. I should have mentioned earlier that your simplifying assumptions would cause large error in some cases -- for example, if there were a large amount of slack in the chain or if the sprockets were close together. Intuitively, it seems that if there is very little slack in the chain and the sprockets are far apart, your simplifying assumptions should not cause too much error in calculating the sag. You might check to see if that is the case. If you then do get realistic sag values, I think you can conclude that your simplifying assumptions were the culprit because the chain was too slack or the sprockets too close together for such assumptions to be reasonable. David === Subject: 2*k+3=p (prime number) If for all positive integer K>=0, and all odd prime number p, isn't verified the congruence: K = (p^2-3)(mod.2p) then 2k+3 = p (prime number) Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) Bytes: 1904 On Sat, 30 Jun 2007 19:55:07 EDT, Vincenzo Librandi >If for all positive integer K>=0, >and all odd prime number p, >isn't verified the congruence: K = (p^2-3)(mod.2p) then 2k+3 = p (prime number) I know there's a language problem, but the problem you stated above is so badly worded, it's almost a crime. I actually don't actually believe that you even started with anything close to a correctly stated problem in your native language. To prove me wrong, post the problem in Italian just to see. Not that I know any Italian, but perhaps someone else does. parts -- one part English, the other part in Italian (for reference). quasi === Subject: Re: 2*k+3=p (prime number) >I know there's a language problem, but the problem you >stated above is so badly worded, it's almost a crime. I actually don't actually believe that you even started >with anything close to a correctly stated problem in >your native language. To prove me wrong, post the problem in Italian just to >see. Not that I >know any Italian, but perhaps someone else does. > Ok, quasi, you are right. my language!!. Data la matrice triangolare: 3 6,11 9,16,23 12,21,30,39 15,26,37,48,59 18,31,44,57,70,83 and so on Per ogni K intero, che non appartiene (it's not belong) alla matrice, K<=> (p^2-3)/2] mod.p sar.88: 2k+3=p (prime number) K appartiene (it's belong) alla matrice triangolare se, e solo se, (if and only if) K=[(p^2-3)/2] mod.p Il problema consiste nel determinare quando un K appartiene o non appartiene alla forma triangolare. The problem consists in determining when K belongs or it does not belong to the triangular matrix Come dicevo in un precedente post: K=0, 2k+3=3 K=1, 2k+3=4 K=2, 2k+3=7 K=3 it's belong to matrix K=4, 2k+3=11 K=5, 2k+3=13 K=6, it's belong to matrix K=7, 2k+3=17 and so on. Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) >Mutsumi Suzuki >Magic Squares >Vincenzo Librandi's method for sequential primes< He used a representaion of primes; Prime number = A + 2 k where A is odd number and k is integer. For example, >A=3 and k={0,1,2,4,5,7,8,10,13,, } yield prime sequence >{3,5,7,11,13,,17,19,23,,,} He constructed a magic square by using the set k. He, >then, transform the numbers in the the square to primes >by the above equation. The result is a magic square of >primes. The integer set k is closely related to the Librandi's >triangle A(m,n) = 2 m n + m + n - 1 which is a kind of >Eratosthnes' sieve. The following is his mail on Nov. 25, 1999. Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) Mutsumi suzuki: http://mathforum.org/te/exchange/hosted/suzuki/MagicSquare.html Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) Bytes: 3553 On Sun, 01 Jul 2007 03:35:33 EDT, Vincenzo Librandi Data la matrice triangolare: 3 >6,11 >9,16,23 >12,21,30,39 >15,26,37,48,59 >18,31,44,57,70,83 >and so on Per ogni K intero, che non appartiene >(it's not belong) alla matrice, >K<=> (p^2-3)/2] mod.p >sar.88: 2k+3=p (prime number) K appartiene (it's belong) alla matrice >triangolare se, e solo se, (if and only if) K=[(p^2-3)/2] mod.p Il problema consiste nel determinare >quando un K appartiene o non appartiene >alla forma triangolare. The problem consists in determining >when K belongs or it does not belong >to the triangular matrix Come dicevo in un precedente post: K=0, 2k+3=3 >K=1, 2k+3=4 >K=2, 2k+3=7 >K=3 it's belong to matrix >K=4, 2k+3=11 >K=5, 2k+3=13 >K=6, it's belong to matrix >K=7, 2k+3=17 >and so on. Ok, now it makes some sense. Let me restate the part of the problem that I can follow: Consider the triangular array 3 6,11 9,16,23 12,21,30,39 15,26,37,48,59 18,31,44,57,70,83 and so on If k is a nonnegative integer, then k is not in the array iff 2k+3 is prime. It's essentially just elementary algebra. It will be more convenient to prove it in the form: If k is a nonnegative integer, then k is in the array iff 2k+3 is composite. proof: First, the forward direction (very easy) ... Assume k is in the array. Then k must be in row n for some positive integer n. The elements of row n form an arithmetic progression of n terms, with initial term 3n and difference 2n+1. It follows that k = 3n + t*(2n+1) for some integer t, 0 <= t <= n-1. But then 2k+3 = 6n + 2t*(2n+1) + 3 = (2t+3) (2n+1) so 2k+3 must be composite, as claimed. Next, the converse (also easy, but slightly harder) ... Suppose 2k+3 is composite. Let p be the smallest prime factor of 2k+3. Then 2k+3 = p*(2n+1) for some positive integer n. Write p = 2t+3 for some nonnegative integer t. Then 2k+3 = (2t+3) (2n+1) => k = 3n + t*(2n+1). => k is in row n of the array, provided we can show t <= n-1. But p is the smallest prime factor of 2k+3 => p <= 2n+1 => 2t+3 <= 2n+1 => t <= n-1 => k is in the array, as claimed. This completes the proof. quasi === Subject: Re: 2*k+3=p (prime number) >Ok, now it makes some sense. Let me restate the part of the problem that I can >follow: >Consider the triangular array 3 >6,11 >9,16,23 >12,21,30,39 >15,26,37,48,59 >18,31,44,57,70,83 >and so on If k is a nonnegative integer, then k is not in the >array iff 2k+3 is prime. >It's essentially just elementary algebra. It will be more convenient to prove it in the form: If k is a nonnegative integer, then k is in the array >iff 2k+3 is >composite. proof: First, the forward direction (very easy) ... Assume k is in the array. Then k must be in row n for some positive integer n. The elements of row n form an arithmetic progression of >n terms, with >initial term 3n and difference 2n+1. It follows that k = 3n + t*(2n+1) for some integer t, 0 ><= t <= n-1. But then 2k+3 = 6n + 2t*(2n+1) + 3 = (2t+3) (2n+1) so 2k+3 must be composite, as claimed. Next, the converse (also easy, but slightly harder) ... Suppose 2k+3 is composite. Let p be the smallest prime factor of 2k+3. Then 2k+3 = p*(2n+1) for some positive integer n. Write p = 2t+3 for some nonnegative integer t. Then 2k+3 = (2t+3) (2n+1) => k = 3n + t*(2n+1). => k is in row n of the array, provided we can show t <= n-1. But p is the smallest prime factor of 2k+3 => p <= 2n+1 => 2t+3 <= 2n+1 => t <= n-1 => k is in the array, as claimed. This completes the proof. credo di aver capito I thing to understand. Now, to understand better, example: K=483 is belong to matrix, why ? K=22432 isn't belong to matrix, why ? Vincenzo Librandi vincenzo.librandweoz@alice.it === Subject: Re: 2*k+3=p (prime number) Bytes: 3406 On Sun, 01 Jul 2007 06:52:25 EDT, Vincenzo Librandi >Ok, now it makes some sense. >Let me restate the part of the problem that I can >follow: >Consider the triangular array >3 >6,11 >9,16,23 >12,21,30,39 >15,26,37,48,59 >18,31,44,57,70,83 >and so on >If k is a nonnegative integer, then k is not in the >array iff 2k+3 is prime. >It's essentially just elementary algebra. >It will be more convenient to prove it in the form: >If k is a nonnegative integer, then k is in the array >iff 2k+3 is >composite. >proof: >First, the forward direction (very easy) ... >Assume k is in the array. >Then k must be in row n for some positive integer n. >The elements of row n form an arithmetic progression of >n terms, with >initial term 3n and difference 2n+1. >It follows that k = 3n + t*(2n+1) for some integer t, 0 ><= t <= n-1. >But then 2k+3 >= 6n + 2t*(2n+1) + 3 >= (2t+3) (2n+1) >so 2k+3 must be composite, as claimed. >Next, the converse (also easy, but slightly harder) ... >Suppose 2k+3 is composite. >Let p be the smallest prime factor of 2k+3. >Then 2k+3 = p*(2n+1) for some positive integer n. >Write p = 2t+3 for some nonnegative integer t. >Then 2k+3 = (2t+3) (2n+1) >=> k = 3n + t*(2n+1). >=> k is in row n of the array, >provided we can show t <= n-1. >But p is the smallest prime factor of 2k+3 >=> p <= 2n+1 >=> 2t+3 <= 2n+1 >=> t <= n-1 >=> k is in the array, as claimed. >This completes the proof. credo di aver capito >I thing to understand. Now, to understand better, example: K=483 is belong to matrix, why ? K=22432 isn't belong to matrix, why ? Because I _proved_ that a nonnegative integer k belongs to the matrix if and only if 2k+3 is composite. For k=483, 2k+3=969 which is composite, hence k belongs to the matrix. For k=22432, 2k+3=44867 which is prime, hence k doesn't belong to the matrix. There's nothing more to understand except the proof itself. quasi === Subject: Re: 2*k+3=p (prime number) >Because I _proved_ that a nonnegative integer k belongs >to the matrix if and only if 2k+3 is composite. For k=483, 2k+3=969 which is composite, hence k belongs >to the matrix. For k=22432, 2k+3=44867 which is prime, hence k doesn't >belong to the matrix. There's nothing more to understand except the proof >itself. No, quasi, questo non ci serve, a noi serve il contrario. Cio.8f, per determinare se un numero N .8f primo o no, dobbiamo stabilire se K appartiene o non appartiene alla matrice triangolare. No, quasi, this not us servants, to we serves the contrary. That is, in order to determine if a number N is prime or not, we must establish if K belongs or it does not belong to the triangular matrix. example. N=50543 is prime o not ? 2k+3=50543; k=25270 is o not belong to matrix ? If K=[(p^2-3)/2] mod. p, then k belong to matrix If K<=>[(p^2-3)/2] mod. p, then K isn't belong to matrix. You understand ? Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) Bytes: 4299 On Sun, 01 Jul 2007 12:02:16 EDT, Vincenzo Librandi >Because I _proved_ that a nonnegative integer k belongs >to the matrix if and only if 2k+3 is composite. >For k=483, 2k+3=969 which is composite, hence k belongs >to the matrix. >For k=22432, 2k+3=44867 which is prime, hence k doesn't >belong to the matrix. >There's nothing more to understand except the proof >itself. No, quasi, >questo non ci serve, a noi serve il contrario. >Cio.8f, per determinare se un numero N .8f primo o no, dobbiamo stabilire se K appartiene o non appartiene alla matrice triangolare. No, quasi, >this not us servants, to we serves the contrary. >That is, in order to determine if a number N is >prime or not, we must establish if K belongs or >it does not belong to the triangular matrix. example. N=50543 is prime o not ? >2k+3=50543; k=25270 is o not belong to matrix ? If K=[(p^2-3)/2] mod. p, then k belong to matrix If K<=>[(p^2-3)/2] mod. p, then K isn't belong to matrix. You understand ? I think I understand. I think what you are saying is this ... The goal is not to decide whether a given nonnegative integer k is in the array. As the proof I provided shows, k is in the array if and only if 2k+3 is prime. Rather, the goal is determine whether a given odd positive integer p>1 is prime. As my proof also implies, an odd positive integer p>1 is prime if and only if (p-3)/2 is in the array. Thus, the array can be used a kind of sieve. It's easily built since the elements of the i'th row are in arithmetic progression with initial term 3i and difference 2i+1. Thus, a schoolchild could easily build the array and use it to test small numbers for primality. The algorithm can be described as follows: Given an odd positive integer p to be tested for primality ... (1) Let k = (p-3)/2 (2) Build the array, one row at a time. To build row i, use initial term 3i and then successively add 2i+1 to form an arithmetic progression of i terms. (3) If k is in the array then p is composite, otherwise p is prime. At most floor((p-3)/6) rows need be considered. The above is the schoolchild version. Obviously, if p is large, it's too inefficient to be used in the form described above. However, some improvements can be made ... k is in the array at row i iff p is divisible by 2i+1. Similarly k is in the array at column j iff p is divisible by 2j+1. So it's not necessary to build the array at all. Also, the minimum feasible row is i=floor((sqrt(p)-1)/2) and the maximum feasible row is i=ceil((p-3)/6). Still, that's too many rows for an efficient algorithm. If instead, you consider columns, the minimum feasible column is j=1 and the maximum feasible column is j=floor((sqrt(p)-1)/2). That's certainly better, but still too many columns to be of any use. Progressing diagonally doesn't help. As far as I can see, there's no magic here. It's just a variation on the standard method of trying all prime factors up to sqrt(p). quasi === Subject: Re: 2*k+3=p (prime number) I think I understand. I think what you are saying is this ... The goal is not to decide whether a given nonnegative >integer k is in the array. As the proof I provided >shows, k is in the array if and only if 2k+3 is prime. Rather, the goal is determine whether a given odd >positive integer p>1 is prime. As my proof also >implies, an odd positive integer p>1 is >prime if and only if (p-3)/2 is in the array. Thus, the array can be used a kind of sieve. It's >easily built since the elements of the i'th row are in >arithmetic progression with initial term 3i and >difference 2i+1. Thus, a schoolchild could easily >build the array and use it to test small numbers for >primality. The algorithm can be described as follows: Given an odd positive integer p to be tested for >primality ... (1) Let k = (p-3)/2 (2) Build the array, one row at a time. To build row i, >use initial term 3i and then successively add 2i+1 to >form an arithmetic progression of i terms. (3) If k is in the array then p is composite, otherwise >p is prime. At most floor((p-3)/6) rows need be >considered. The above is the schoolchild version. Obviously, if p is large, it's too inefficient to be >used in the form described above. However, some improvements can be made ... k is in the array at row i iff p is divisible by 2i+1. Similarly k is in the array at column j iff p is >divisible by 2j+1. So it's not necessary to build the array at all. Also, the minimum feasible row is i=floor((sqrt(p)->1)/2) and the maximum feasible row is i=ceil((p-3)/6). Still, that's too many rows for an efficient algorithm. If instead, you consider columns, the minimum feasible >column is j=1 and the maximum feasible column is >j=floor((sqrt(p)-1)/2). That's certainly better, but still too many columns to >be of any use. Progressing diagonally doesn't help. As far as I can see, there's no magic here. It's just a >variation on the standard method of trying all prime >factors up to sqrt(p). I know it. It's a variant to k=(p^2-3)/2 mod.p (K is belong to matrix) Infact you ask (if I undertand) In N odd numbers, if n>=0, then k is belong to matrix when k=n mod.2n+3; (with 2n+3<=sqrt(N)) K=0 mod.3 K=1 mod.5 K=2 mod.7 K=3 mod.9 (trivial) K=4 mod.11 k=5 mod.13 K=6 mod.15 (trivial) K=7 mod.17 K=n mod, sqrt(N) You ask: Progressing diagonally doesn't help, but K is belong also to the j.ma diagonal. Occorre mettere in relazione, quindi, non solo le righe e le colonne, ma anche le diagonali per semplificare il metodo. (It is necessary to put in relation, therefore, not only the lines and the columns, but also the diagonals in order to simplify the method). To controll: If D= diagonal's number 2k+D^2-2D+4 is always square. Example: for D=3 (third diagonal), then 2k+D^2-2D+4=square 2k+7=square 9*2+7=25=5^2 21*2+7=49=7^2 37*2+7=81=9^2 and so on for D=10, then 2k+84=square 30*2+84=144=12^2 56*2+84=196=14^2 86*2+84=256=16^2 and so on Si pu.98 mettere in relazione questa circostanza con le righe e le colonne? (Can be put in relation this circumstance with the lines and the columns?) Vincenzo Librandi === Subject: Re: 2*k+3=p (prime number) Let p=19 then k = (19^2-3) mod 38 = 358 mod 38 = 16 2*16+3 = 35 and that is not prime Same for MANY others === Subject: Re: 2*k+3=p (prime number) >Let p=19 >then >k = (19^2-3) mod 38 = 358 mod 38 = 16 2*16+3 = 35 and that is not prime Oops: repeat: K<=> [(p^2-3)/2] mod. p and for K=16 16=11 mod.5 Vincenzo Librandi === Subject: Re: AN ALTERNATE OPEN LETTER TO MR BONDARENKO Bytes: 4130 Specially for Richard J. Fateman, Professor of Computer Science at University of California at Berkeley http://www.cs.berkeley.edu/~fateman/ who is anxious (by unknown and incomprehensible for us reasons) about the dates of the messages expressing numerous views in support of our multi-year large scale activity directed to CAS improvement, we reproduce here the full text of a letter to us written on Jun 21, 2007 by a Maple customer. Enjoy. And please note that in the nature there could be views not coinciding totally with yours. > I very much appreciate what you are doing. Your posts are > informative, interesting and valuable. Clearly companies are not > going to improve the qualilty of their products if their flaws are not > exposed. So you are doing everyone that uses Maple or Mathematica a > great service. Please keep it up. There has been a lot of bolona about how software flaws are inevitable > but there is nothing inevitable about not fixing known flaws and most > certainly known flaws get fixed and fixed more rapidly when there is > exposure and customer pressure to do so. To suggest that you should > put all your results onto a private web page is to sweep those quality > issues under the rug and that is something that everyone including the > people that work for software companies should be opposed to. In > addition anyone who isn't interested in your postings doesn't have to > read them. It would be very valuable if you could report on how rapidly Maple and > Mathematica fix the bugs that you identify and report. Clearly Maple > and Mathematica need to know that customers want them to clean up > their bug backlog because they the customers want a quality product > that works properly and whose answers I can be trusted and ultimately > the company with the best quality product will survive while the other > probably won't. Also, when a customer reports a bug they should be > thanked and provided with an estimated date when the problem will be > fixed or at least worked on. Any reluctance to do so is a reluctance > to take responsibility for fixing a flawed product and that is > unacceptable. There has been a trend in the software industry to release new > products without fixing the old ones and there comes a time when the > issues are so intermingled that they simply cannot be fixed so the > time has come for companies to decide that they are going to fix what > they have and not allow any more problems to get past them, and any > employee that doesn't agree with that management direction, needs to > work elsewhere. of the products that we buy use. === Subject: Re: AN ALTERNATE OPEN LETTER TO MR BONDARENKO Bytes: 3789 AN ALTERNATE OPEN LETTER TO MR BONDARENKO Local: Thurs, Jun 21 2007 8:13 am === Subject: AN ALTERNATE OPEN LETTER TO MR BONDARENKO I very much appreciate what you are doing. Your posts are informative, interesting and valuable. Clearly companies are not going to improve the quality of their products if their flaws are not exposed. So you are doing everyone that uses Maple or Mathematica a great service. Please keep it up. There has been a lot of bolona about how software flaws are inevitable but there is nothing inevitable about not fixing known flaws and most certainly known flaws get fixed and fixed more rapidly when there is exposure and customer pressure to do so. To suggest that you should put all your results onto a private web page is to sweep those quality issues under the rug and that is something that everyone including the people that work for software companies should be opposed to. In addition anyone who isn't interested in your postings doesn't have to read them. It would be very valuable if you could report on how rapidly Maple and Mathematica fix the bugs that you identify and report. Clearly Maple and Mathematica need to know that customers want them to clean up their bug backlog because they the customers want a quality product that works properly and whose answers I can be trusted and ultimately the company with the best quality product will survive while the other probably won't. Also, when a customer reports a bug they should be thanked and provided with an estimated date when the problem will be fixed or at least worked on. Any reluctance to do so is a reluctance to take responsibility for fixing a flawed product and that is unacceptable. There has been a trend in the software industry to release new products without fixing the old ones and there comes a time when the issues are so intermingled that they simply cannot be fixed so the time has come for companies to decide that they are going to fix what they have and not allow any more problems to get past them, and any employee that doesn't agree with that management direction, needs to work elsewhere. quality of the products that we buy use. === Subject: Radius of an open ball? Bytes: 1201 I think I found an equation for the radius of an open ball -- is it true that for any neighborhood of x, the radius is the distance === Subject: Re: Radius of an open ball? Bytes: 3263 On Sat, 30 Jun 2007 18:37:47 -0700, bit188 true that for any neighborhood of x, the radius is the distance A preliminary comment: Try for more precise wording. Before answering what I think is your intended question, let me critique the lack of precision ... (1) I found an equation for the radius of an open ball Your open balls are in what space? In R^n? In an arbitrary metric space? Since you didn't specify, and since the rest of the question possibly makes sense in an arbitrary metric space, I'll assume that's what you meant. But you should have specified. (2) for any neighborhood of x What's x? You never specified x. Once again, in order to make sense of the problem, one might guess that x is the center of the ball. But if that's what you intended, you should have said so. (3) the radius is the distance between x and the supremum of the set What set? Of any neighborhood? Of the set of all neighborhoods? Of the set of all neighborhoods contained in the open ball? But even with the vagueness as to what set, the wording is fatally flawed in a different way ... The phrase the distance between x and the supremum of the set doesn't make sense. Distances are defined between pairs of points of your metric space. Thus, there's no concept of the distance from x to the sup of a set. With the above critique (offered constructively), here's what I _think_ you might have meant ... Question: Let X be a metric space and let B(x,r) be the open ball, centered at x, with radius r. Let S be the set of all neighborhoods of x contained in B(x,r). Define f:S->R by f(Y) = sup {d(x,y) | y is in Y}. Is it true that r = sup {f(Y) | Y is in S} ? Now I'm not sure if my attempt at rewording your question captures your true intent, but at least it's a correctly posed question (I hope). By the way, the answer to the above question is no, not necessarily. Can you find an example? quasi === Subject: Canonic Maple Experts list Bytes: 3730 http://www.math.usf.edu/~eclark/maple_links_experts.html Homepages of a few people with special interest and expertise in Maple. The order in this list reflects more or less the order in which I discovered the webpages. I make no claim for completeness of this list. I'm sure there are many experts in Maple not on this list. To list them all would be a daunting task. Robert Israel http://www.math.ubc.ca/%7Eisrael/ Preben Alsholm http://www2.mat.dtu.dk/people/P.K.Alsholm/ Rafal Ablamowicz http://math.tntech.edu/rafal/ Douglas B. Meade http://www.math.sc.edu/%7Emeade/ William C. Bauldry http://www.mathsci.appstate.edu/%7Ewmcb/ Joe Riel http://www.k-online.com/~joer/ John B. Cosgrove http://www.spd.dcu.ie/johnbcos/ Carl Eberhart http://www.ms.uky.edu/%7Ecarl/ Gaston Gonnet http://www.inf.ethz.ch/personal/gonnet/ Helmut Kahovec http://profiles.yahoo.com/hkahovec Matthias Kawski http://math.la.asu.edu/%7Ekawski/ Richard Kreckel http://www.ginac.de/~kreckel/ George Labahn http://www.scg.uwaterloo.ca/%7Eglabahn/ Ross Taylor http://www.clarkson.edu/%7Echengweb/faculty/taylor/ Bill Scott http://maple.murdoch.edu.au/%7Escott/ W. D. Joyner http://web.usna.navy.mil/%7Ewdj/homepage.html Mike May, S. J. http://euler.slu.edu/Dept/Faculty/may/may.html Edgardo S. Cheb-Terrab http://www.scg.uwaterloo.ca/~ecterrab/ Doron Zeilberger http://www.math.rutgers.edu/%7Ezeilberg/ Harald Pleym http://www.mamut.com/homepages/Norway/1/17/hpleym/ Alec Mihailovs http://www.mihailovs.com/Alec/ Mark van Hoeij http://web.math.fsu.edu/%7Ehoeij/ Francis J. Wright http://centaur.maths.qmw.ac.uk/ Carl Love (previously known as Carl DeVore) http://profiles.yahoo.com/carldevore Peter Luschny http://www.luschny.de/ Gerald A. Edgar http://www.math.ohio-state.edu/~edgar/ Michael Monagan http://oldweb.cecm.sfu.ca/people/Michael_Monagan/ Kelly Roach http://www.kellyroach.com/ Thomas Richard http://www.scientific.de/ Stephen Forrest http://wandership.ca/ Alejandro S. Jakubi http://www.df.uba.ar/users/jakubi/ Roman Pearce http://www.cecm.sfu.ca/~rpearcea/ Manuel Bronstein http://www-sop.inria.fr/cafe/Manuel.Bronstein/ Janos D. Pinter http://www.pinterconsulting.com/ === Subject: Re: Canonic Maple Experts list Bytes: 4141 Vladimir Bondarenko : > http://www.math.usf.edu/~eclark/maple_links_experts.html Homepages of a few people with special interest and expertise > in Maple. The order in this list reflects more or less the > order in which I discovered the webpages. I make no claim for > completeness of this list. I'm sure there are many experts in > Maple not on this list. To list them all would be a daunting > task. Robert Israel http://www.math.ubc.ca/%7Eisrael/ Preben Alsholm http://www2.mat.dtu.dk/people/P.K.Alsholm/ Rafal Ablamowicz http://math.tntech.edu/rafal/ Douglas B. Meade http://www.math.sc.edu/%7Emeade/ William C. Bauldry http://www.mathsci.appstate.edu/%7Ewmcb/ Joe Riel http://www.k-online.com/~joer/ John B. Cosgrove http://www.spd.dcu.ie/johnbcos/ Carl Eberhart http://www.ms.uky.edu/%7Ecarl/ Gaston Gonnet http://www.inf.ethz.ch/personal/gonnet/ Helmut Kahovec http://profiles.yahoo.com/hkahovec Matthias Kawski http://math.la.asu.edu/%7Ekawski/ Richard Kreckel http://www.ginac.de/~kreckel/ George Labahn http://www.scg.uwaterloo.ca/%7Eglabahn/ Ross Taylor http://www.clarkson.edu/%7Echengweb/faculty/taylor/ Bill Scott http://maple.murdoch.edu.au/%7Escott/ W. D. Joyner http://web.usna.navy.mil/%7Ewdj/homepage.html Mike May, S. J. http://euler.slu.edu/Dept/Faculty/may/may.html Edgardo S. Cheb-Terrab http://www.scg.uwaterloo.ca/~ecterrab/ Doron Zeilberger http://www.math.rutgers.edu/%7Ezeilberg/ Harald Pleym http://www.mamut.com/homepages/Norway/1/17/hpleym/ Alec Mihailovs http://www.mihailovs.com/Alec/ Mark van Hoeij http://web.math.fsu.edu/%7Ehoeij/ Francis J. Wright http://centaur.maths.qmw.ac.uk/ Carl Love (previously known as Carl DeVore) > http://profiles.yahoo.com/carldevore Peter Luschny http://www.luschny.de/ Gerald A. Edgar http://www.math.ohio-state.edu/~edgar/ Michael Monagan http://oldweb.cecm.sfu.ca/people/Michael_Monagan/ Kelly Roach http://www.kellyroach.com/ Thomas Richard http://www.scientific.de/ Stephen Forrest http://wandership.ca/ Alejandro S. Jakubi http://www.df.uba.ar/users/jakubi/ Roman Pearce http://www.cecm.sfu.ca/~rpearcea/ Manuel Bronstein http://www-sop.inria.fr/cafe/Manuel.Bronstein/ Janos D. Pinter http://www.pinterconsulting.com/ Nice staff comrade Vladimir. Could you add some similar links for Mathematica? Dimitris === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 2193 We ask you kindly to tell us if in Maple sqrt(z) and z^(1/2) should show the *mathematically* identical behavior, that is the calculations including sqrt(z) and z^(1/2) should yield the outputs that are equivalent mathematically? We mean that z might be an expression. The seconds request is to quote a place in the Maple Help where the customer can find the exact indication of mathematical identity of/difference between sqrt() and ()^(1/2)? We'd much prefer an explanation however short rather than a simple yes/no answer. Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 2746 Could Robert Israel share with us his opinion on account of sqrt() and ()^(1/2) ? Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing We ask you kindly to tell us if in Maple sqrt(z) and z^(1/2) should show the *mathematically* identical > behavior, that is the calculations including sqrt(z) and z^(1/2) should yield the outputs that are equivalent > mathematically? We mean that z might be an > expression. The seconds request is to quote a place in the > Maple Help where the customer can find the exact > indication of mathematical identity of/difference > between sqrt() and ()^(1/2)? We'd much prefer an explanation however > short rather than a simple yes/no answer. > Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/ CAS Testing === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 2107 > Could Robert Israel share with us his opinion > on account of sqrt() and ()^(1/2) ? > OK, since you asked: in Maple, sqrt(z) and z^(1/2) should be mathematically equivalent (though not necessarily presented in the same way). They should both be the principal branch of the square root: for z <> 0 it is exp(ln(z)/2) where -Pi < Im(ln(z)) <= Pi. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 2538 Could Robert Israel share with us his opinion > on account of sqrt() and ()^(1/2) ? OK, since you asked: in Maple, sqrt(z) and z^(1/2) > should be mathematically equivalent (though not necessarily > presented in the same way). They should both be the > principal branch of the square root: for z <> 0 it is > exp(ln(z)/2) where -Pi < Im(ln(z)) <= Pi. I guess I should add that this should is theoretical: for an arbitrary constant expression z it may be impossible to tell which side of the branch cut z is on, and it is entirely possible that sqrt(z) and z^(1/2) will produce different results by attempting to decide this using different methods. Robert Israel israel@math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) See your doctor. E.Z. > Could Robert Israel share with us his opinion > on account of sqrt() and ()^(1/2) ? > Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC > http://maple.bug-list.org/ Maple Bugs Encyclopaedia > http://www.CAS-testing.org/ CAS Testing > We ask you kindly to tell us if in Maple > sqrt(z) and z^(1/2) > should show the *mathematically* identical > behavior, that is the calculations including > sqrt(z) and z^(1/2) > should yield the outputs that are equivalent > mathematically? We mean that z might be an > expression. > The seconds request is to quote a place in the > Maple Help where the customer can find the exact > indication of mathematical identity of/difference > between sqrt() and ()^(1/2)? > We'd much prefer an explanation however > short rather than a simple yes/no answer. > Best wishes, > Vladimir Bondarenko > VM and GEMM architect > Co-founder, CEO, Mathematical Director > http://www.cybertester.com/ Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/ CAS Testing > === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 3529 EZ> See your doctor. Is your piece of advice based on your personal experience? Cheer up, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing > See your doctor. E.Z. Could Robert Israel share with us his opinion > on account of sqrt() and ()^(1/2) ? > Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC >http://maple.bug-list.org/ Maple Bugs Encyclopaedia >http://www.CAS-testing.org/ CAS Testing > We ask you kindly to tell us if in Maple > sqrt(z) and z^(1/2) > should show the *mathematically* identical > behavior, that is the calculations including > sqrt(z) and z^(1/2) > should yield the outputs that are equivalent > mathematically? We mean that z might be an > expression. > The seconds request is to quote a place in the > Maple Help where the customer can find the exact > indication of mathematical identity of/difference > between sqrt() and ()^(1/2)? > We'd much prefer an explanation however > short rather than a simple yes/no answer. > Best wishes, > Vladimir Bondarenko > VM and GEMM architect > Co-founder, CEO, Mathematical Director >http://www.cybertester.com/Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/CAS Testing === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Bytes: 2616 We ask you kindly to tell us if in Maple sqrt(z) and z^(1/2) should show the identical behavior, that > is the calculations including sqrt(z) and z^(1/2) should yield the same output (up to the > terms ordering)? We mean that z might > be an expression. We'd much prefer an explanation however > short rather than a simple yes/no answer. > Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director is(sqrt(z) = z^(1/2)); true solve(y = sqrt(I), y); 1 (1/2) 1 (1/2) - 2 + - I 2 2 2 solve(y = (-1)^(1/4), y); (1/4) (-1) simplify(%); 1 (1/2) 1 (1/2) - 2 + - I 2 2 2 is((1/2)*sqrt(2)+(1/2*I)*sqrt(2) = (-1)^(1/4)); FAIL So sqrt(z) = z^(1/2), except when it fails. === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) Originator: roberson@ibd.nrc-cnrc.gc.ca (Walter Roberson) > is(sqrt(z) = z^(1/2)); > true > solve(y = sqrt(I), y); > 1 (1/2) 1 (1/2) > - 2 + - I 2 > 2 2 > solve(y = (-1)^(1/4), y); > (1/4) > (-1) > simplify(%); > 1 (1/2) 1 (1/2) > - 2 + - I 2 > 2 2 > is((1/2)*sqrt(2)+(1/2*I)*sqrt(2) = (-1)^(1/4)); > FAIL >So sqrt(z) = z^(1/2), except when it fails. You introduced a simplify(). is( x = simplify(x) ) is not certain to succeed, independant of the properties of sqrt() and ^ . -- No one has the right to destroy another person's belief by demanding empirical evidence. -- Ann Landers === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) We ask you kindly to tell us if in Maple sqrt(z) and z^(1/2) should show the identical behavior, that > is the calculations including sqrt(z) and z^(1/2) should yield the same output (up to the > terms ordering)? We mean that z might > be an expression. We'd much prefer an explanation however > short rather than a simple yes/no answer. > Try this... > lprint(sqrt(z)); z^(1/2) So it seems sqrt(z) is stored internally as z^(1/2) ... > z^(1/2); (1/2) z ...that display has a square-root sign in typeset form... However, it does require evaluation: > op(0,sqrt(z)); ^ > op(0,'sqrt(z)'); sqrt > op(0,'z^(1/2)'); ^ > evalb('sqrt(z)'='z^(1/2)'); false > evalb(sqrt(z)=z^(1/2)); true -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) <300620070701098281%edgar@math.ohio-state.edu.invalid> On Jun 30, 4:01 am, G. A. Edgar GAE> lprint(sqrt(z)); GAE> op(0,sqrt(z)); GAE> evalb(sqrt(z)=z^(1/2)); I am not sure about the precise meaning of your answer... So I will try to re-shape a bit my question. If the customer has an expression A containing (one or several) sqrt's and the expression B in which one or several sqrt's is/ are replaced (of course, in a mathematically correct way) by ^(1/2), my question is, Should Maple return the same output in both cases? Can we find in Help any direct indication on what should happen? If yes, could you please quote this passage? Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing On Jun 30, 4:01 am, G. A. Edgar is the calculations including sqrt(z) and z^(1/2) should yield the same output (up to the > terms ordering)? We mean that z might > be an expression. We'd much prefer an explanation however > short rather than a simple yes/no answer. Try this... lprint(sqrt(z)); z^(1/2) So it seems sqrt(z) is stored internally as z^(1/2) ... z^(1/2); (1/2) > z ...that display has a square-root sign in typeset form... However, it does require evaluation: op(0,sqrt(z)); ^> op(0,'sqrt(z)'); sqrt> op(0,'z^(1/2)'); ^> evalb('sqrt(z)'='z^(1/2)'); false> evalb(sqrt(z)=z^(1/2)); true -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) If the customer has an expression A containing (one or several) > sqrt's and the expression B in which one or several sqrt's is/ > are replaced (of course, in a mathematically correct way) by > ^(1/2), my question is, Should Maple return the same output in > both cases? No. For example sqrt(-12) and (-12)^(1/2) return different answers. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: A question to Maple Experts about sqrt(z) and z^(1/2) (we'd like to hear at least 2-3 opinions...) <300620070701098281%edgar@math.ohio-state.edu.invalid> <300620071612042693%edgar@math.ohio-state.edu.invalid> Bytes: 2362 On Jun 30, 1:12 pm, G. A. Edgar sqrt's and the expression B in which one or several sqrt's is/ > are replaced (of course, in a mathematically correct way) by > ^(1/2), my question is, Should Maple return the same output in > both cases? No. For example sqrt(-12) and (-12)^(1/2) return different answers. -- > G. A. Edgar http://www.math.ohio-state.edu/~edgar/ A related question is whether exp(1/3) and e^(1/3) are the same, or does the expression e^(1/3) involve some collection of the 3 roots. In Macsyma, sqrt(x) is simply a syntactic variant of the internal form x^(1/2). This may or may not be what you would like to see. RJF === Subject: (( The Last Breath )) Cc: sci.math Bytes: 1047 http://english.islamway.com/flashes/2/last.swf === Subject: Re: (( The Last Breath )) > http://english.islamway.com/flashes/2/last.swf Congratulations! You finally got it! Yes! Behold June's Top Moron. Keep up the good work - buffoons like you are always entertaining. === Subject: Re: Galilei's Customer Principle Bytes: 1562 Another Galileo's useful maxim reads, Customer, try before you buy. (He invented it having tried some beer in a place he never visited before...) === Subject: join Bytes: 1093 hi friend i'd like to join in ur group === Subject: Re: join Bytes: 1834 > hi friend i'd like to join in ur group You don't need anyone's permission to use an unmoderated news group such as . Just go ahead and post any mathematical message you like. === Subject: Re: join > .com... > hi friend i'd like to join in ur group You want to join a group in an ancient southern > Mesopotamian city? It has really made me smile. Fernando. === Subject: join hi friend i'd like to join in ur group === Subject: What makes Math easy? Bytes: 1724 Questions, What makes some people have math come natural to them? While others struggle with Math as if it is a foreign language to them? Not the native, tongue that they were raised to speak in. Does it mean one that is unable to fully understand Math and enjoy it lacks logic or form of reasoning skills? Or something wrong with them to have a blind Math spot? To be very good in Math what traits does a person need to have? It seems some people try and try but like blinded to grasping qualities one has to appreciate Math? Afterall everything in existance virtually lives off of Math. === Subject: Re: What makes Math easy? Bytes: 2504 Am 01.07.2007 08:01 schrieb mickbmcc@gmail.com: > Questions, What makes some people have math come natural to them? While others > struggle with Math as if it is a foreign language to them? Not the > native, tongue that they were raised to speak in. Does it mean one > that is unable to fully understand Math and enjoy it lacks logic or > form of reasoning skills? Or something wrong with them to have a blind > Math spot? To be very good in Math what traits does a person need to > have? It seems some people try and try but like blinded to grasping > qualities one has to appreciate Math? Afterall everything in existance > virtually lives off of Math. > You may do math like a musisican or artist or like an account-manager. For me, I like finding structures, symmetries, the music in math, and I'm getting better with it by speaking and sometimes even composing it. So what would it mean to have a blind spot in this? Not everyone is a Vincent van Gogh in arts or an Archimedes, Euler or Gauss in math, or a J.S.Bach in music. But also a good garage band enjoys its music, and its fans too - even without being able to play the required three chords. But by continuously speaking math one may aquire experience and virtuosity, in my opinion, for the type of understanding which is most natural for one person. Just my 2 c - Gottfried === Subject: What makes Math easy? Bytes: 1744 Questions, What makes some people have math come natural to them? While others struggle with Math as if it is a foreign language to them? Not the native, tongue that they were raised to speak in. Does it mean one that is unable to fully understand Math and enjoy it lacks logic or form of reasoning skills? Or something wrong with them to have a blind Math spot? To be very good in Math what traits does a person need to have? It seems some people try and try but like blinded to grasping qualities one has to appreciate Math? Afterall everything in existance virtually lives off of Math. === Subject: Re: Selling off math books > My mom died last year and finally we sell off her books. She collected > a lot and was kind a packrat. I dont have a list but she really > collected everything Send me an email and ask elakenad@yahoo.com ps sorry if this not appropiate here, but i couldnt figure ebay If you can get ISBNs, the 10-digit numbers that appear on the title page of most books published in the last 50 years or so, you can use half.com, an ebay subsidiary that is targeted at those buying and selling used books. It is much easier for that purpose than the general ebay auctions. 10-digit ISBNs are often punctuated as n-nnnnn-nnn-x, where the n's are digits and the x is either a digit or the letter X. -- Randy Hudson === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2507 > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe ----------------- i am not sure you saied the above seriously or sarcastically now if you are serious let metell you that :: waht is worse than an idiot ?? the answer is an arrogant idiot and just take my word there ar etoo many of them !! a lot of them bump parasites tha tnot only thet dont contribute anything they are a pain parasites on the neck that prevents advance Y.Porat ---------------- === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2539 Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything EricGisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe ----------------- > i am not sure you saied the above seriously > or sarcastically > now if you are serious > let metell you that :: waht is worse than an idiot ?? the answer is > an arrogant idiot The irony is goddamn BREATHTAKING. [...] === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2092 The chimps are easy to spot. They are so stupid that you have to send them to schools to put at least something between the ears since they can't do it themselves. Then they come out the other side, thinking that they have the smarts. Meanwhile I turn on the TV and wait for that full hour of news, packed with info concerning today's thousands upon thousands upon thousands upon..... of daily breakthroughs made by those endless numbers of brilliant university grads. I'm still waiting. Maybe it's because in truth they are just a bunch of copies. Monkey see, monkey do. ADVANCED BANNANA'S, news at 11:00. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > The chimps are easy to spot. They are so stupid that you have to send them to schools to put at > least something between the ears since they can't do it themselves. Then they come out the other side, thinking that they have the smarts. Meanwhile I turn on the TV and wait for that full hour of news, packed > with info concerning today's thousands upon thousands upon thousands > upon..... of daily breakthroughs made by those endless numbers of > brilliant university grads. I'm still waiting. Maybe it's because in truth they are just a bunch of copies. Monkey > see, monkey do. ADVANCED BANNANA'S, news at 11:00. > Turn off the TV and crack a textbook. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2417 > The chimps are easy to spot. They are so stupid that you have to send them to schools to put at > least something between the ears since they can't do it themselves. Then they come out the other side, thinking that they have the smarts. Meanwhile I turn on the TV and wait for that full hour of news, packed > with info concerning today's thousands upon thousands upon thousands > upon..... of daily breakthroughs made by those endless numbers of > brilliant university grads. I'm still waiting. Maybe it's because in truth they are just a bunch of copies. Monkey > see, monkey do. ADVANCED BANNANA'S, news at 11:00. Turn off the TV and crack a textbook.- Hide quoted text - - Show quoted text - I prefer to crack EGG-heads === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > The chimps are easy to spot. > They are so stupid that you have to send them to schools to put at > least something between the ears since they can't do it themselves. > Then they come out the other side, thinking that they have the smarts. > Meanwhile I turn on the TV and wait for that full hour of news, packed > with info concerning today's thousands upon thousands upon thousands > upon..... of daily breakthroughs made by those endless numbers of > brilliant university grads. > I'm still waiting. > Maybe it's because in truth they are just a bunch of copies. Monkey > see, monkey do. > ADVANCED BANNANA'S, news at 11:00. > Turn off the TV and crack a textbook.- Hide quoted text - > - Show quoted text - > I prefer to crack EGG-heads > Lots of folks that couldn't hack it try to crack it. Welcome to cranksville! === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything > Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe > Hey Eugene--my first degree was a BS Mathematics, but you will be > disappointed in that I was not trained to prove theorems. My education > in mathematics has certainly proved be valuable as a tool in pursuit of > areas of physics, however! I would say a lot of physicists, or those interested in physics, have degrees in math as a their first degree, or as a double degree. Bill > === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe My first degree was an MS in mathematics. > Back to the drawing board, shoob. Dirk Vdm > I also hold an MS in mathematics, and I was indeed trained > in proving theorems. > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) Thinking up your own proofs is very instructive and fun. I whiled away this morning deriving my own proof for Ito's extension theorem needed to define the stochastic integral beyond elementary functions. The one in the text was way too complex, a bit opaque, and skipped important details while referring you to textbooks on functional analysis for others. It started with continuous functions then worked its way up to unbounded ones. Working in Lebesque integrals from the start obviated the need the assumption of continuity. Indeed I have found stochastic calculus, and probability in general, can be simplified by working in Dirac spaces and using generalised functions eg the derivative of Brownian motion exists as a generalised function allowing one to write usual differential equations without resorting to the fiddle of Ito's equation - which is still useful for solving particular problems. It inspired me enough to write a paper about it. Makes you feel great, until you find reading the fine print of the text someone else has already done it. Bill > Looks like Shubee's taxonomy is going to take a little work. A lot. > I see Androcles listed there :-( Dirk Vdm === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2828 > Thinking up your own proofs is very instructive and fun. I whiled away this morning deriving my own proof for Ito's extension > theorem needed to define the stochastic integral beyond elementary functions. The one in the text was way too complex, a bit > opaque, and skipped important details while referring you to textbooks on functional analysis for others. It started with > continuous functions then worked its way up to unbounded ones. Working in Lebesque integrals from the start obviated the need the > assumption of continuity. Indeed I have found stochastic calculus, and probability in general, can be simplified by working in > Dirac spaces and using generalised functions eg the derivative of Brownian motion exists as a generalised function allowing one to > write usual differential equations without resorting to the fiddle of Ito's equation - which is still useful for solving > particular problems. It inspired me enough to write a paper about it. Makes you feel great, until you find reading the fine > print of the text someone else has already done it. hey, but that should make you feel even greater, don't you agree? Dirk Vdm === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > Thinking up your own proofs is very instructive and fun. I whiled away > this morning deriving my own proof for Ito's extension theorem needed to > define the stochastic integral beyond elementary functions. The one in > the text was way too complex, a bit opaque, and skipped important details > while referring you to textbooks on functional analysis for others. It > started with continuous functions then worked its way up to unbounded > ones. Working in Lebesque integrals from the start obviated the need the > assumption of continuity. Indeed I have found stochastic calculus, and > probability in general, can be simplified by working in Dirac spaces and > using generalised functions eg the derivative of Brownian motion exists > as a generalised function allowing one to write usual differential > equations without resorting to the fiddle of Ito's equation - which is > still useful for solving particular problems. It inspired me enough to > write a paper about it. Makes you feel great, until you find reading the > fine print of the text someone else has already done it. hey, but that should make you feel even greater, don't you agree? Well - is your cup half full or half empty? - but I get your point. Bill Dirk Vdm > === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything > Eric Gisse > Androcles > Bilge > Bill Hobba I suppose my degree in math does not count - nor the masters I am doing. > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe My first degree was an MS in mathematics. > Back to the drawing board, shoob. Dirk Vdm Same here. Its obvious Shuberts envy, that he can't understand what nearly everyone who studied it can, has warped him a lot. Bill === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2537 > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe Hey Eugene--my first degree was a BS Mathematics, but you will be > disappointed in that I was not trained to prove theorems. My education > in mathematics has certainly proved be valuable as a tool in pursuit of > areas of physics, however! I got straight As at Physics but quickly realised it was too easy - I studied Engineering instead. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything > Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe > Hey Eugene--my first degree was a BS Mathematics, but you will be > disappointed in that I was not trained to prove theorems. My education > in mathematics has certainly proved be valuable as a tool in pursuit of > areas of physics, however! I got straight As at Physics but quickly realised it was too easy - I > studied Engineering instead. That's true for many an engineer taking freshman physics. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2930 > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything > Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe > Hey Eugene--my first degree was a BS Mathematics, but you will be > disappointed in that I was not trained to prove theorems. My education > in mathematics has certainly proved be valuable as a tool in pursuit of > areas of physics, however! I got straight As at Physics but quickly realised it was too easy - I > studied Engineering instead. That's true for many an engineer taking freshman physics. Yall have an interesting little closed system going here. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 4297 > [...] > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) I was taking a graduate-level class (analysis?) as an undergraduate, and the professor remarked that I was turning in short proofs of the results, which meant that I was either a genius or lazy. 8-) For my first fractal geometry test, there was a question asking for the Hausdorff dimension of the set S = {1, 1/2, 1/3, 1/4, ...}. I got 1/2. Asking around after the exam, I found out that everyone else seemed to have gotten an answer of 0 or 1, including the professor. After reading my response and thinking about it, he changed his mind to 1/2. (While he was thinking it over, he told me that he thought the answer was 0 because S can be covered by finitely many sets with diameter epsilon, for any epsilon. I told him that that was the case for any finite set (instead of S), and that it's the number of smaller sets which counts, not just that there are finitely many.) After recovering from a year off due to a nervous breakdown, I went into my first graduate algorithms class being less than confident. We had learned about how to find a minimum weight spanning tree of a connected graph G in class, which is a spanning tree of G which minimizes the total weights on the edges. On the test, the professor asked for a bottleneck spanning tree; we had to choose a tree whose _maximum_ weight (not the sum of the weights) was minimized. Again, brilliance struck: I showed that a bottleneck spanning tree was a minimum spanning tree, so you could use the same algorithm. I don't know how many other students got it, but the number had to have been low, since he assigned that result as a problem after handing back the test. --- Christopher Heckman === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > [...] > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. > heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) I was taking a graduate-level class (analysis?) as an undergraduate, > and the professor remarked that I was turning in short proofs of the > results, which meant that I was either a genius or lazy. 8-) No - it indicates you are a mathematician, and, despite Shuberts silly claims, most physicists are also mathematicians. I think it was Erdos that said he wanted proofs that are not merely correct - they are from Gods book - meaning they are derived in a few lines in a very neat way. Most of the proofs in Apostle's book on analysis are like that, and those that aren't can often be improved on eg his treatment of Riemann integrals as a special case of the Riemann-Stieltjes integral is neater directly from the Lebesque integral - Riemann-Stieltjes integrals can be handled very easily using generalised functions, as all functions have a derivative. As mentioned in another posit I have discovered a neat treatment of stochastic calculus using Dirac spaces - generalised functions are examples of a Dirac space. Looking for 'God' proofs is a very strong urge in virtually every one bitten by the maths bug. For my first fractal geometry test, there was a question asking for > the Hausdorff dimension of the set S = {1, 1/2, 1/3, 1/4, ...}. I got > 1/2. Asking around after the exam, I found out that everyone else > seemed to have gotten an answer of 0 or 1, including the professor. > After reading my response and thinking about it, he changed his mind > to 1/2. (While he was thinking it over, he told me that he thought the > answer was 0 because S can be covered by finitely many sets with > diameter epsilon, for any epsilon. I told him that that was the case > for any finite set (instead of S), and that it's the number of smaller > sets which counts, not just that there are finitely many.) After recovering from a year off due to a nervous breakdown, I went > into my first graduate algorithms class being less than confident. We > had learned about how to find a minimum weight spanning tree of a > connected graph G in class, which is a spanning tree of G which > minimizes the total weights on the edges. On the test, the professor > asked for a bottleneck spanning tree; we had to choose a tree whose > _maximum_ weight (not the sum of the weights) was minimized. Again, > brilliance struck: I showed that a bottleneck spanning tree was a > minimum spanning tree, so you could use the same algorithm. I don't > know how many other students got it, but the number had to have been > low, since he assigned that result as a problem after handing back the > test. I am not sure that setting difficult problems on exams is a good idea. These are best done in assignments or projects outside an exam IMHO. Exams should be for testing if you have learnt the subject material - not if you can think using it under exam conditions. The best math subject I ever did (applied linear algebra) had a number of assignments - worth 25%, a project - worth 25%, and a final worth 50%, with half the marks on the final being a write up of your project - the other half simply regurgitating problems similar to what was done in tutorials. Bill --- Christopher Heckman > === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 8207 > [...] > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. > heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) I was taking a graduate-level class (analysis?) as an undergraduate, > and the professor remarked that I was turning in short proofs of the > results, which meant that I was either a genius or lazy. 8-) No - it indicates you are a mathematician, and, despite Shuberts silly > claims, most physicists are also mathematicians. I think it was Erdos that > said he wanted proofs that are not merely correct - they are from Gods > book - meaning they are derived in a few lines in a very neat way. My advisor and I found one of the proofs from the Book. (Fortunately, other people have said this as well 8-).) It was published as A New Proof of the Independence Ratio of Triangle-Free Cubic Graphs Discrete Mathematics, Vol 233/1-3, pp. 233-237, available online as gzip'ed Postscript: http://math.asu.edu/~checkman/514.ps.gz . Not only did we get the proof out of it, we found a linear-time algorithm to get what we were looking for, and the cases of equality actually turned out to be easy to characterize. > Most of > the proofs in Apostle's book on analysis are like that, and those that > aren't can often be improved on eg his treatment of Riemann integrals as a > special case of the Riemann-Stieltjes integral is neater directly from the > Lebesque integral - Riemann-Stieltjes integrals can be handled very easily > using generalised functions, as all functions have a derivative. As > mentioned in another posit I have discovered a neat treatment of stochastic > calculus using Dirac spaces - generalised functions are examples of a Dirac > space. Looking for 'God' proofs is a very strong urge in virtually every > one bitten by the maths bug. For my first fractal geometry test, there was a question asking for > the Hausdorff dimension of the set S = {1, 1/2, 1/3, 1/4, ...}. I got > 1/2. Asking around after the exam, I found out that everyone else > seemed to have gotten an answer of 0 or 1, including the professor. > After reading my response and thinking about it, he changed his mind > to 1/2. (While he was thinking it over, he told me that he thought the > answer was 0 because S can be covered by finitely many sets with > diameter epsilon, for any epsilon. I told him that that was the case > for any finite set (instead of S), and that it's the number of smaller > sets which counts, not just that there are finitely many.) After recovering from a year off due to a nervous breakdown, I went > into my first graduate algorithms class being less than confident. We > had learned about how to find a minimum weight spanning tree of a > connected graph G in class, which is a spanning tree of G which > minimizes the total weights on the edges. On the test, the professor > asked for a bottleneck spanning tree; we had to choose a tree whose > _maximum_ weight (not the sum of the weights) was minimized. Again, > brilliance struck: I showed that a bottleneck spanning tree was a > minimum spanning tree, so you could use the same algorithm. I don't > know how many other students got it, but the number had to have been > low, since he assigned that result as a problem after handing back the > test. I am not sure that setting difficult problems on exams is a good idea. > These are best done in assignments or projects outside an exam IMHO. Exams > should be for testing if you have learnt the subject material - not if you > can think using it under exam conditions. This was an actual graduate school class, and the argument wasn't that bad, since the professor had done a similar problem in class. It was basically: Fix an ordering of the edges of G, and let T1 be minimum spanning tree from the usual algorithm. (That is, go from the edges with the smallest weight to the heaviest weight --- this is a function which is also provided --- and choose an edge if it doesn't form a cycle with the edges you've chosen already. This produces a spanning tree, the sum of the weights of edges being as small as possible.) If there is a spanning tree T2 whose maximum weight m is less than the heaviest edge of T1 (whose weight is M), then do the following: Let H consist of the edges of G whose weights are <= m. H is connected, since H contains the spanning tree T2 (maybe with more edges), ordered the same way as in the original order. Now run the MST algorithm; you either don't choose an edge somewhere in the algorithm when you should choose it, or else you end up with a forest, since you only get a connected tree once you include the edges whose weight is > m. In either case, the failure contradicts the fact that the MST algorithm works, which negates the assumption that m < M. It looks horrible when written out, but the idea is a neat one. --- Christopher Heckman > The best math subject I ever did > (applied linear algebra) had a number of assignments - worth 25%, a > project - worth 25%, and a final worth 50%, with half the marks on the final > being a write up of your project - the other half simply regurgitating > problems similar to what was done in tutorials. Bill === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 4478 > [...] > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. > heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) I was taking a graduate-level class (analysis?) as an undergraduate, > and the professor remarked that I was turning in short proofs of the > results, which meant that I was either a genius or lazy. 8-) Is there a difference really? For my first fractal geometry test, there was a question asking for > the Hausdorff dimension of the set S = {1, 1/2, 1/3, 1/4, ...}. I got > 1/2. Asking around after the exam, I found out that everyone else > seemed to have gotten an answer of 0 or 1, including the professor. > After reading my response and thinking about it, he changed his mind > to 1/2. (While he was thinking it over, he told me that he thought the > answer was 0 because S can be covered by finitely many sets with > diameter epsilon, for any epsilon. I told him that that was the case > for any finite set (instead of S), and that it's the number of smaller > sets which counts, not just that there are finitely many.) After recovering from a year off due to a nervous breakdown, I went > into my first graduate algorithms class being less than confident. We > had learned about how to find a minimum weight spanning tree of a > connected graph G in class, which is a spanning tree of G which > minimizes the total weights on the edges. On the test, the professor > asked for a bottleneck spanning tree; we had to choose a tree whose > _maximum_ weight (not the sum of the weights) was minimized. Again, > brilliance struck: I showed that a bottleneck spanning tree was a > minimum spanning tree, so you could use the same algorithm. I don't > know how many other students got it, but the number had to have been > low, since he assigned that result as a problem after handing back the > test. That's great - a real thumbs up there. Weren't you offered an assistent position in the end? Dirk Vdm === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 5514 > [...] > I can't claim any published theorems in my name, but I > did startle my professors with a shortcut in a qualifying > exam proof. > heh - same here on my first linear algebra exam :-) > One of the 4 proofs I could't possibly remember how to > start - so kept it as the last one. > Then found a nice little proof. The prof looked at it, and > what he saw didn't look familiar, so since the other 3 > proofs were perfect, he gave me 15/20. I insisted that he > would take the trouble to look at my proof, since I was > quite sure it was okay. He was a bit reluctant and > hesitated ... and then proposed 18/20, which was okay > with me. That felt really great :-) I was taking a graduate-level class (analysis?) as an undergraduate, > and the professor remarked that I was turning in short proofs of the > results, which meant that I was either a genius or lazy. 8-) Is there a difference really? For my first fractal geometry test, there was a question asking for > the Hausdorff dimension of the set S = {1, 1/2, 1/3, 1/4, ...}. I got > 1/2. Asking around after the exam, I found out that everyone else > seemed to have gotten an answer of 0 or 1, including the professor. > After reading my response and thinking about it, he changed his mind > to 1/2. (While he was thinking it over, he told me that he thought the > answer was 0 because S can be covered by finitely many sets with > diameter epsilon, for any epsilon. I told him that that was the case > for any finite set (instead of S), and that it's the number of smaller > sets which counts, not just that there are finitely many.) After recovering from a year off due to a nervous breakdown, I went > into my first graduate algorithms class being less than confident. We > had learned about how to find a minimum weight spanning tree of a > connected graph G in class, which is a spanning tree of G which > minimizes the total weights on the edges. On the test, the professor > asked for a bottleneck spanning tree; we had to choose a tree whose > _maximum_ weight (not the sum of the weights) was minimized. Again, > brilliance struck: I showed that a bottleneck spanning tree was a > minimum spanning tree, so you could use the same algorithm. I don't > know how many other students got it, but the number had to have been > low, since he assigned that result as a problem after handing back the > test. That's great - a real thumbs up there. > Weren't you offered an assistent position in the end? Grad school was actually smooth sailing after that. (Well, until the comprehensive exam, which was for the Algorithms/Combinatorics/ Optimization program at Georgia Tech: eight hours to solve eight problems from eight different subject areas 8-). My thesis was actually well-written; there was another student who, at his defense, was told that he had to rewrite his thesis from scratch.) I was actually a research assistant before that, as well as after, and I also did some teaching before going on to Arizona State, where my research career has been slowly dying. --- Christopher Heckman === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 2323 > Mathematicians are the highest order of human beings on the planet. > They are trained to think and prove theorems. Physicists are > chimpanzees that are trained to revere dominant alpha male > chimpanzees. Like most humans, chimpanzees have clan wars and throw > on other chimpanzees. I respectfully request that if you are on > my list of the dirtiest and the smelliest of the -throwing > chimpanzees, that you stay away from me. I believe that you're uncivil > and irrational and therefore incapable of discussing anything Do you mean professional mathematicians? That crank/crackpot James Harris and/or Bassam King Karzeddin (two distinct people, according to some) purports to be a mathematician, and yet he/they, as you say, acts like a chimpanzee with clan wars, throwing on other chimpanzees, is uncivil and quite irrational, and is incapable of discussing anything mathematically without reverting to chimp-like behavior. So, why this sudden anger at physicists? === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. Bytes: 1526 [...] So, why this sudden anger at physicists? Flunked out of physics grad school. That, and he hasn't done in the last 20 years and is bitter about it. === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > Mathematicians are the highest order of human beings > on the planet. > They are trained to think and prove theorems. > Physicists are > chimpanzees that are trained to revere dominant alpha > male > chimpanzees. Like most humans, chimpanzees have clan > wars and throw > on other chimpanzees. I respectfully request > that if you are on > my list of the dirtiest and the smelliest of the > -throwing > chimpanzees, that you stay away from me. I believe > that you're uncivil > and irrational and therefore incapable of discussing > anything Eric Gisse > Androcles > Bilge > Bill Hobba > Dirk Van de moortel > YBM > Dono a.k.a. karandash2...@yahoo.com > Sam Wormley > Igor > Randy Poe > if this is a list of people you hate ; i disagree with you. as for my point of view on physics , it is not positive either. if you are ever at a university and you hear to prof discussing physics , note that they dont agree on anything !! electrons are not even understood . but to say something constructive instead of saying they dont know anything and insult them , id rather say : i am a believer in hidden variable theory. look at the advances in physics historicly ?? ( after newton ) 3 SIMPLE examples : einstein added the variable c.(at constant but thats irrelevant ) same for strong nucleair action. which proves my point !! tommy1729 ** not including nutty prof thinking of 177 dimensions or string theory , which cannot be falsified and has no predictive power THEREFORE NO PLACE IN PHYSICS BUT METAPHYSICS ! === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. To the unmoral couple (Tommy 1729 & Neilist) If you think deeply you would find that both of you are the best types of chimpanzees, especially with your long tongue deeds & unity So enjoy it for ever... Who said that monkey is better than a donkey? The problem is that I don't have time for you kids, other wise I can make of your couple the best joke on line.. Indeed you are very lucky I also realized that both of you love to be insulted badly on the net for attention, therefore I am asking others to do it on my behalf Have fun B.Karzeddin === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. <23740279.1183219998017.JavaMail.jakarta@nitrogen.mathforum.org To the unmoral couple (Tommy 1729 & Neilist) If you think deeply you would find that both of you are the best types of chimpanzees, especially with your long tongue deeds & unity So enjoy it for ever... Who said that monkey is better than a donkey? The problem is that I don't have time for you kids, other wise I can make of your couple the best joke on line.. Indeed you are very lucky I also realized that both of you love to be insulted badly on the net for attention, therefore I am asking others to do it on my behalf Have fun B.Karzeddin Hey Harris/Bassam/Quinn's-sperm-receptacle, Quinn Tyler Jackson and you and the other high-IQ phonies at the Mega Foundation only had to pay money to join your closed list of intellectuals. Hahahahahahahahahahahahahahahahahahahahahaha Mega Foundation, huh? Mega-bull Foundation, that is. Any society that would let YOU and Quinn join is clearly a fraud, just like YOU, Bassam King Bull-Artist. Hahahahahahahahahahahahahahahahahahahahahaha Since you are not a professional mathematician, you're a chimp who keeps flinging crap at REAL mathematicians in this newsgroup. SO, GET OUT OF SCI.MATH ... B.Karzeddin Hahahahahahahahahahahahahahahahahahahahahaha I'm laughing at the superior intellect. Hahahahahahahahahahahahahahahahahahahahahaha P.S. Quinn is gay === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > To the unmoral couple (Tommy 1729 & Neilist) If you think deeply you would find that both of you > are the best types of chimpanzees, especially with > your long tongue deeds & unity So enjoy it for ever... Who said that monkey is better than a donkey? The problem is that I don't have time for you kids, > other wise I can make of your couple the best joke on > line.. Indeed you are very lucky I also realized that both of you love to be insulted > badly on the net for attention, therefore I am asking > others to do it on my behalf Have fun B.Karzeddin you are projecting again. confusing us with you. first were not a couple , were not a gay fagot crankpot like you (trying ? ) to get quincy's ass. secondly were friends , something you aint got. third your a monkey and a child. fourth , you/jsh love to be insulted here. fifth , the entire forum finds you a gay crankpot. have fun playing with your ass. and hey dont take it personal , i like you better than some other crankpots here... well i mean your alter ego's actually :p === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. <14848917.1183246709234.JavaMail.jakarta@nitrogen.mathforum.org> Bytes: 2611 > To the unmoral couple (Tommy 1729 & Neilist) If you think deeply you would find that both of you > are the best types of chimpanzees, especially with > your long tongue deeds & unity So enjoy it for ever... Who said that monkey is better than a donkey? The problem is that I don't have time for you kids, > other wise I can make of your couple the best joke on > line.. Indeed you are very lucky I also realized that both of you love to be insulted > badly on the net for attention, therefore I am asking > others to do it on my behalf Have fun B.Karzeddin you are projecting again. confusing us with you. first were not a couple , were not a gay fagot crankpot like you (trying ? ) to get quincy's ass. secondly were friends , something you aint got. third your a monkey and a child. fourth , you/jsh love to be insulted here. fifth , the entire forum finds you a gay crankpot. have fun playing with your ass. and hey dont take it personal , i like you better than some other crankpots here... well i mean your alter ego's actually :p- Hide quoted text - - Show quoted text - Yeah, so there! === Subject: Re: Mathematicians are trained to think and prove theorems. Physicists are trained to imitate chimpanzees. > On Jun 30, 7:37 pm, tommy1729 you > are the best types of chimpanzees, especially > with > your long tongue deeds & unity So enjoy it for ever... Who said that monkey is better than a donkey? The problem is that I don't have time for you > kids, > other wise I can make of your couple the best > joke on > line.. Indeed you are very lucky I also realized that both of you love to be > insulted > badly on the net for attention, therefore I am > asking > others to do it on my behalf Have fun B.Karzeddin you are projecting again. confusing us with you. first were not a couple , were not a gay fagot > crankpot like you (trying ? ) to get quincy's ass. secondly were friends , something you aint got. third your a monkey and a child. fourth , you/jsh love to be insulted here. fifth , the entire forum finds you a gay crankpot. have fun playing with your ass. and hey dont take it personal , i like you better > than some other crankpots here... well i mean your alter ego's actually :p- Hide > quoted text - - Show quoted text - Yeah, so there! > ;-) tommy1729 === Subject: Integral extension of Integral Domains Bytes: 1307 One can show that if A and B are integral domain with B integral over A, then the quotient field of B is integral over the quotient field of A. Now the question is, if A is an integral domain integrally closed in B, then can I say that Quot(A) is integrally closed in Quot(B)? Jose Capco === Subject: Re: Integral extension of Integral Domains Bytes: 1841 On Sun, 01 Jul 2007 01:53:40 -0700, Jose Capco >One can show that if A and B are integral domain with B integral over >A, then the quotient field of B is integral over the quotient field of >A. Now the question is, if A is an integral domain integrally closed >in B, then can I say that Quot(A) is integrally closed in Quot(B)? Yes, it's true, but only in a trivial way. The hypothesis is way too strong. Your hypothesis: A,B are integral domains. B is integral over A. A is integrally closed in B. The above conditions trivially force B=A. So of course Quot(A) is integrally closed in Quot(B) since they are equal. quasi === Subject: Re: Integral extension of Integral Domains Bytes: 1644 > Yes, it's true, but only in a trivial way. The hypothesis is way too strong. Your hypothesis: A,B are integral domains. B is integral over A. A is integrally closed in B. The above conditions trivially force B=A. > No, the second statement is not related to the firs. I didnt required that B be an integral extension of A when I posed the question. Jose Capco === Subject: Re: Integral extension of Integral Domains Bytes: 2058 On Sun, 01 Jul 2007 03:52:58 -0700, Jose Capco > Yes, it's true, but only in a trivial way. > The hypothesis is way too strong. > Your hypothesis: > A,B are integral domains. > B is integral over A. > A is integrally closed in B. > The above conditions trivially force B=A. No, the second statement is not related to the firs. I didnt required >that B be an integral extension of A when I posed the question. In that case, it's false. Here's a counterexample ... Let A=Z, B=Z[x, x*sqrt(2)] It's not hard to verify that A is integrally closed in B. Then Quot(A)=Q but Quot(B) contains sqrt(2), hence Quot(A) is not integrally closed in Quot(B). quasi === Subject: Re: Integral extension of Integral Domains <7i9f83l3avmdca4cnrq0sdnutug0nn9t2c@4ax.com> Bytes: 2281 > On Sun, 01 Jul 2007 03:52:58 -0700, Jose Capco In that case, it's false. Here's a counterexample ... Let A=Z, B=Z[x, x*sqrt(2)] It's not hard to verify that A is integrally closed in B. Then Quot(A)=Q but Quot(B) contains sqrt(2), hence Quot(A) is not integrally closed in Quot(B). quasi Yes indeed.. nice and simple example. But I claim the following: If - A and B are integral domian - A contained in B and integrally closed in it - For every b in B (nonzero) there is an a in A such that a/b is in B (I might rephrase this as : For all nonzero b in B, there is a c in B such that bc is in A, there is actually a name for this.. it is said B is an essential extension of A) Then it is true that Quot(A) is integrally closed in Quot(B). The proof uses the fact that for any element in Quot(B) one can use without loss of generality a denominator that is in A. Jose Capco === Subject: Re: Integral extension of Integral Domains Bytes: 2560 On Sun, 01 Jul 2007 10:06:57 -0700, Jose Capco > On Sun, 01 Jul 2007 03:52:58 -0700, Jose Capco > In that case, it's false. > Here's a counterexample ... > Let A=Z, B=Z[x, x*sqrt(2)] > It's not hard to verify that A is integrally closed in B. > Then Quot(A)=Q but Quot(B) contains sqrt(2), > hence Quot(A) is not integrally closed in Quot(B). > quasi Yes indeed.. nice and simple example. But I claim the following: Ok, but this is a new claim, not the original problem. >If >- A and B are integral domian >- A contained in B and integrally closed in it >- For every b in B (nonzero) there is an a in A such that a/b is in B >(I might rephrase this as : For all nonzero b in B, there is a c in B >such that bc is in A, >there is actually a name for this.. it is said B is an essential >extension of A) Then it is true that Quot(A) is integrally closed in Quot(B). The >proof uses the fact that >for any element in Quot(B) one can use without loss of generality a >denominator that is in A. Yes, the new claim is true and not hard to prove. quasi === Subject: Re: Integral extension of Integral Domains > If > - A and B are integral domain > - A contained in B and integrally closed in it > - For every b in B (nonzero) there is an a in A such that a/b is in B > (I might rephrase this as : For all nonzero b in B, there is a c in B > such that bc is in A, there is actually a name for this.. it is said > B is an essential extension of A) > Then it is true that Quot(A) is integrally closed in Quot(B). > The proof uses the fact that> for any element in Quot(B) one can > use without loss of generality a denominator that is in A. Yes, the new claim is true and not hard to prove. I'm curious to see the details of both of your proofs. Could you please sketch what you _had_ in mind (without being influenced by this request or the other's proof). I ask because I'm very interested to see how students reason with such concepts as they are learning them. --Bill Dubuque === Subject: Re: Integral extension of Integral Domains Bytes: 3537 On 02 Jul 2007 13:27:33 -0400, Bill Dubuque - A and B are integral domain > - A contained in B and integrally closed in it > - For every b in B (nonzero) there is an a in A such that a/b is in B > (I might rephrase this as : For all nonzero b in B, there is a c in B > such that bc is in A, there is actually a name for this.. it is said > B is an essential extension of A) Then it is true that Quot(A) is integrally closed in Quot(B). > The proof uses the fact that> for any element in Quot(B) one can > use without loss of generality a denominator that is in A. > Yes, the new claim is true and not hard to prove. I'm curious to see the details of both of your proofs. >Could you please sketch what you _had_ in mind (without >being influenced by this request or the other's proof). >I ask because I'm very interested to see how students >reason with such concepts as they are learning them. Sure. First the statement of the problem: Let A and B be integral domains such that A is a subring of B A is integrally closed in B For all b in B, there exists nonzero c in B such that b*c is in A Let Quot(A), Quot(B) denote the quotient fields of A,B respectively. Prove that Quot(A) is integrally closed in Quot(B). proof: Let x be in Quot(B) and suppose x is integral over Quot(A). Since x is integral over Quot(B), x is algebraic over A. Write x=b1/b2 where b1,b2 are in B. Choose nonzero c in B such that b2*c=a where a is in A. Then x = b1/b2 = (b1*c)/(b2*c) = (b1*c)/a. Thus b1*c=a*x. Since x is algebraic over A, so is a*x, hence b1*c is algebraic over A. But then a1*(b1*c) is integral over A for some a1 in A. Since a1*(b1*c) is in B, and A is integrally closed in B, it follows that a1*(b1*c) is in A. a1*(b1*c) in A => b1*c in Quot(A) => (b1*c)/a in Quot(A) => x in Quot(A) It follows that Quot(A) is integrally closed in Quot(B), as was to be shown. quasi === Subject: Re: Integral extension of Integral Domains Bytes: 3781 >On 02 Jul 2007 13:27:33 -0400, Bill Dubuque If > - A and B are integral domain > - A contained in B and integrally closed in it > - For every b in B (nonzero) there is an a in A such that a/b is in B > (I might rephrase this as : For all nonzero b in B, there is a c in B > such that bc is in A, there is actually a name for this.. it is said > B is an essential extension of A) > Then it is true that Quot(A) is integrally closed in Quot(B). > The proof uses the fact that> for any element in Quot(B) one can > use without loss of generality a denominator that is in A. Yes, the new claim is true and not hard to prove. >I'm curious to see the details of both of your proofs. >Could you please sketch what you _had_ in mind (without >being influenced by this request or the other's proof). >I ask because I'm very interested to see how students >reason with such concepts as they are learning them. Sure. First the statement of the problem: Let A and B be integral domains such that A is a subring of B > A is integrally closed in B > For all b in B, there exists nonzero c in B such that b*c is in A Let Quot(A), Quot(B) denote the quotient fields of A,B respectively. Prove that Quot(A) is integrally closed in Quot(B). proof: Let x be in Quot(B) and suppose x is integral over Quot(A). Since x is integral over Quot(B), x is algebraic over A. Typo: Since x is integral over Quot(A), x is algebraic over A. >Write x=b1/b2 where b1,b2 are in B. Choose nonzero c in B such that b2*c=a where a is in A. Then x = b1/b2 = (b1*c)/(b2*c) = (b1*c)/a. Thus b1*c=a*x. Since x is algebraic over A, so is a*x, hence b1*c is algebraic over >A. But then a1*(b1*c) is integral over A for some a1 in A. Since a1*(b1*c) is in B, and A is integrally closed in B, it follows >that a1*(b1*c) is in A. a1*(b1*c) in A => b1*c in Quot(A) => (b1*c)/a in Quot(A) => x in >Quot(A) It follows that Quot(A) is integrally closed in Quot(B), as was to be >shown. quasi === Subject: Re: Integral extension of Integral Domains <7i9f83l3avmdca4cnrq0sdnutug0nn9t2c@4ax.com> Bytes: 3807 > I'm curious to see the details of both of your proofs. > Could you please sketch what you _had_ in mind (without > being influenced by this request or the other's proof). > I ask because I'm very interested to see how students > reason with such concepts as they are learning them. --Bill Dubuque With Pleasure... But first, upon examining my original post.. I am thinking that I might need that B essential over A as well for the first statement. The first statement says: If A and B are integral domains. A a subring of B and B an integral extension of A, then Quot(B) is an integral extension of Quot(A).. which .. well, I haven't seen any algebra reference that says this (which is surprising, because this is very natural to think about).. I was thinking in the lines: .. Take the algebraic closure of Quot(A) then B is in it and.... etc., but this is not clear to me.. so I might indeed need that B is essential over A, having essentiality I can prove this. First lets show the proof of the second statement which says: Let A and B be integral domains, B an essential over-ring of A (ie. for all b in B nonzero there is a c in B such that bc is in A and is nonzero) and A integrally closed in B THEN Quot(A) is integrally closed over Quot(B) Proof: Set K=Quot(B) and F=Quot(A) .. so F <= K Let b/c be an element of K, with b,c in B and c nonzero. Without loss of generality we may assume that c is in A, since B is essential over A there is a d in B with cd in A{0} and b/c=bd/cd and we may take cd as our denominator otherwise. Suppose furthermore that b/c is an integral element of F. Then there is a polynomial f(T)=T^n + sum_{i=0}^{n-1} T^i * a_i /x in F[T] with a_i in A and x in A{0} and b/c a zero of f(T). So f(b/c)=b^n/c^n + sum_{i=0}^{n-1} b^i * a_i /xc^i = 0 multiply f(b/c) by (cx)^n (which isnt 0, since c and arent), then c^nx^nf(b/c)= (bx)^n + sum_{i=0}^{n-1} (bx)^i c^(n-i)x^(n-i-1)a_i = 0 and therefore bx in B is a zero of T^n + sum_{i=0}^{n-1} b_i T^i in A[T] where b_i = c^(n-i)x^(n-i-1)a_i (in A) for i=0,...,n-1 but A is integrally closed in B. So there is an x in A{0} such that bx is in A/{0} But this implies that b/a=bx/ax , and since a and x are both in a we have b/a is in Quot(A). q.e.d. === Subject: Re: Integral extension of Integral Domains Bytes: 2902 On Mon, 02 Jul 2007 13:01:42 -0700, Jose Capco >But first, upon examining my original post.. I am >thinking that I might need that B essential over A as well for the >first statement. The first statement says: If A and B are integral domains. A a subring of B and B an integral >extension of A, then Quot(B) is an integral extension of Quot(A).. >which .. well, I haven't seen any algebra reference that says this >(which is surprising, because this is very natural to think about).. I >was thinking in the lines: .. Take the algebraic closure of Quot(A) >then B is in it and.... etc., but this is not clear to me.. so I might >indeed need that B is essential over A, having essentiality I can >prove this. You're making it too complicated. The first question is truly simple ... prove: Let A,B be integral domains with A a subring of B. If B integral over A, then Quot(B) is integral over Quot(A). proof: It's not even necessary to assume B is integral over A. It's sufficient that B is algebraic over A. The set of elements of Quot(B) which are algebraic over A form a subfield F of Quot(B). But since B is algebraic over A, it follows that B is a subset of F. Then, since F is a field, the chain B subset F subset Quot(B) implies F=Quot(B). Thus, all elements of Quot(B) are algebraic over A, and hence integral over Quot(A). quasi === Subject: Re: Integral extension of Integral Domains <7i9f83l3avmdca4cnrq0sdnutug0nn9t2c@4ax.com> On Jul 2, 10:01 pm, Jose Capco was thinking in the lines: .. Take the algebraic closure of Quot(A) > then B is in it and.... etc., but this is not clear to me.. ^^^^^^^^^^^^^^^^^^^^^^^^ What I meant with this is not clear to me is that its not clear to me that B is in the algebraic closure of Quot(A). And I doubt it.. or I think I should :) Jose Capco === Subject: Normalization of a curve Bytes: 1126 Hello everybody! Let X be a curve over a field, has a node at x. Let f: Y ---->X be === Subject: Yet another Maple 11 regression bug the long liver (1997--2007--?): TRIVIAL integral (CauchyPrincipalValue) ................................................................. If the same bugs exist through numerous software releases, I think that is valuable public information. It just should not happen. -- Brad Cooper ................................................................. Hello again from the VM machine... ................................................................. BUG # XXXXX int (1-D): INVALID undefined REGRESSION: YES REPRODUCIBLE: ALWAYS BUG HISTORY: PRESENT Maple 11.00,IBM INTEL NT, Feb 16 2007 Build ID 277223 PRESENT Maple 10.06,IBM INTEL NT, Oct 2 2006 Build ID 255401 PRESENT Maple 8.00, IBM INTEL NT, May 10 2002 Build ID 111221 PRESENT Maple 7.00, IBM INTEL NT, May 28 2001 Build ID 96223 PRESENT Maple 6.01, IBM INTEL NT, Jun 9 2000 Build ID 79514 PRESENT Maple V, Release 5, IBM INTEL NT, Nov 27 1997 UNEVAL Maple V, Release 4, IBM INTEL NT, Dec 15, 1995 UNEVAL Maple V, Release 3, IBM INTEL NT, Jan 10, 1994 UNEVAL Maple V, Release 2, IBM INTEL NT, Jan 10, 1992 DESCRIPTION: Welcome to the continuing Maple development crisis. None of the Maple versions since 1992 to 2007 can calculate this trivial integral correctly. The integrand is a LINEAR function over the reals. TEST CASE: int(ln(-exp(2*z))-I*Pi, z= -infinity..infinity, CauchyPrincipalValue); ACTUAL: undefined EXPECTED: 0 CHECKUP: int((simplify(ln(-exp(2*z))-I*Pi) assuming z::real), z= -infinity..infinity, CauchyPrincipalValue); 0 COMMENT: UNEVAL = int(ln(-exp(2*z))-I*Pi,z = -infinity .. infinity) ................................................................. Man+Machine Review Of Maple Crisis http://maple.bug-list.org/maple-crisis.php Main Maple's Quality Results 1. Maple 9.5 is an unstable, inconsistent, non-linear, non-uniform, randomized, self-incompatible environment where fundamental math properties (uniqueness of the answer for a good-defined problem commutativity and linearity property etc) now hold, now fail making Maple breaking down grotesquely. [ ... ] ................................................................. Best wishes, Vladimir Bondarenko VM and GEMM architect Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC http://maple.bug-list.org/ Maple Bugs Encyclopaedia http://www.CAS-testing.org/ CAS Testing ................................................................. === Subject: Re: Yet another Maple 11 regression bug the long liver (1997--2007--?): TRIVIAL integral (CauchyPrincipalValue) Bytes: 5264 Vladimir Bondarenko : > ................................................................. If the same bugs exist through numerous software releases, > I think that is valuable public information. It just should not happen. -- Brad Cooper ................................................................. Hello again from the VM machine... ................................................................. > BUG # XXXXX int (1-D): INVALID undefined REGRESSION: YES REPRODUCIBLE: ALWAYS BUG HISTORY: PRESENT Maple 11.00,IBM INTEL NT, Feb 16 2007 Build ID 277223 > PRESENT Maple 10.06,IBM INTEL NT, Oct 2 2006 Build ID 255401 > PRESENT Maple 8.00, IBM INTEL NT, May 10 2002 Build ID 111221 > PRESENT Maple 7.00, IBM INTEL NT, May 28 2001 Build ID 96223 > PRESENT Maple 6.01, IBM INTEL NT, Jun 9 2000 Build ID 79514 > PRESENT Maple V, Release 5, IBM INTEL NT, Nov 27 1997 > UNEVAL Maple V, Release 4, IBM INTEL NT, Dec 15, 1995 > UNEVAL Maple V, Release 3, IBM INTEL NT, Jan 10, 1994 > UNEVAL Maple V, Release 2, IBM INTEL NT, Jan 10, 1992 DESCRIPTION: Welcome to the continuing Maple development crisis. None of the Maple versions since 1992 to 2007 can > calculate this trivial integral correctly. The integrand is a LINEAR function over the reals. TEST CASE: int(ln(-exp(2*z))-I*Pi, z= -infinity..infinity, > CauchyPrincipalValue); ACTUAL: undefined EXPECTED: 0 CHECKUP: int((simplify(ln(-exp(2*z))-I*Pi) assuming z::real), > z= -infinity..infinity, CauchyPrincipalValue); 0 COMMENT: UNEVAL = int(ln(-exp(2*z))-I*Pi,z = -infinity .. infinity) ................................................................. Man+Machine Review Of Maple Crisis http://maple.bug-list.org/maple-crisis.php Main Maple's Quality Results 1. Maple 9.5 is an unstable, inconsistent, non-linear, > non-uniform, randomized, self-incompatible environment > where fundamental math properties (uniqueness of the > answer for a good-defined problem commutativity and > linearity property etc) now hold, now fail making Maple > breaking down grotesquely. [ ... ] ................................................................. Best wishes, Vladimir Bondarenko VM and GEMM architect > Co-founder, CEO, Mathematical Director http://www.cybertester.com/ Cyber Tester, LLC > http://maple.bug-list.org/ Maple Bugs Encyclopaedia > http://www.CAS-testing.org/ CAS Testing ................................................................. I think that the setting CauchyPrincipalValue is to treat simple poles. I must admit I didn't check it, but I am based on the similar saetting of Mma: PrincipalValue->True. In[2]:= Integrate[Log[-Exp[2*z]]-I*Pi, {z,- , },PrincipalValue->True] !(* RowBox[{(Integrate::idiv ), ((:)( )), Integrate::idiv, ButtonStyle->RefGuideLinkText, ButtonFrame->None])>}]) Out[2]= Integrate[(-I)*Pi + Log[-E^(2*z)], {z, -Infinity, Infinity}, PrincipalValue -> True] I belive that the setting is not proper for this integral. On the contrary, In[5]:= Integrate[Log[-Exp[2*z]] - I*Pi, {z, -Infinity, Infinity}, GenerateConditions -> False] Out[5]= 0