mm-1075 >However, I think that the paradox is much deeper than this. Consider >four statements the professor could make: > i) there will be an unexpected test tomorrow > ii) there will be an unexepected test in the next three days > iii) there will be an unexpected test next week > iv) there will be an unexpected test this semester >All four have the same logical structure. However, i) seems absurd, >ii) seems questionable, iii) seems fine, iv) is completely unremarkable. >A full resoution of the paradox must explain the difference. I think that if someone says there will be a surprise test sometime in the next week, he usually only means that there is no way *now* to figure out what day it will occur on. People donÕt usually mean the stronger claim that, even on the day of the test, you wonÕt be able to figure out that there will be a test. Obviously, on the last day of the week, you can reason that if you havenÕt had a test yet, youÕre going to have it today. However, if someone says there will be a surprise test tomorrow, the weak interpretation of surprise is not available. The strong interpretation of surprise is paradoxical, while the weak interpretation is not. As someone pointed out, the strong interpretation is not actually paradoxical, once you realize that just because somebody says something doesnÕt make it true. So, if the teacher announces that there will be a surprise test, one possibility is that he is lying---either there wonÕt be a test, or it wonÕt be a surprise. If you consider the *possibility* that the teacher is lying, then you canÕt logically deduce anything from the fact that he said there will be a surprise test, and so on the day of the test, it *will* be a surprise. -- Daryl McCullough Ithaca, NY === Subject: Re: SmullyanÕs Quiz Problem > I think that if someone says there will be a surprise test sometime > in the next week, he usually only means that there is no way *now* > to figure out what day it will occur on. People donÕt usually mean > the stronger claim that, even on the day of the test, you wonÕt be > able to figure out that there will be a test. Obviously, on the last > day of the week, you can reason that if you havenÕt had a test yet, > youÕre going to have it today. > As someone pointed out, the strong interpretation is not actually > paradoxical, once you realize that just because somebody says something > doesnÕt make it true. So, if the teacher announces that there will > be a surprise test, one possibility is that he is lying---either > there wonÕt be a test, or it wonÕt be a surprise. If you consider > the *possibility* that the teacher is lying, then you canÕt logically > deduce anything from the fact that he said there will be a surprise > test, and so on the day of the test, it *will* be a surprise. I think that you have given the solution to the unexpected exam problem. By Thursday morning, the students will know that either (1) the test will be on that day and it will be a surprise, or that (2) the test will be on Friday and it will not be a surprise, in which case the teacher will have said something that is false when he said that the test will be a surprise. However, there is the possibility that the test will not be given on Friday after all, the teacher might forget, or he might have been lying, or be taken ill, or he might decide on Thursday night to give the test the following week. (Put yourself in the shoes of a student sitting in the class on Friday morning you donÕt know for certain that the test will go ahead that day, its happening is a contingent event, not a logical consequence of what the teacher said last week.) Thus the class does not know for certain that the test will be on Friday. So the test is still a surprise when it is on Friday, although less of a surprise. === Subject: Re: SmullyanÕs Quiz Problem >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? > One of the hypothesis is that an examination will be given during the > week. The argument shows that the hypothesis lead to a contradiction: > that no examination will be given during the week. That is, > H->not(H). But > (H -> not(H)) -> not(H) > is a tautology. That means that the only thing we can deduce from the > professorÕs statements are that his conditions will not be met. > Since we can only conclude that his conditions will not be met, none > of the original argument showing that no test can occur is valid. The > test may occur at any time. Arturo, point is that, ON MONDAY, the professor stated You will not know that there is an examination when the class begins. When the Monday morning class began, the students had no idea that there was going to be a surprise examination that week since the professor had yet to make the announcement. Thus giving the exam on Monday would not contract his statement - it would be a surprise examination and the students wonÕt have known that there is an examination when the Monday morning class began. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124 === Subject: Re: SmullyanÕs Quiz Problem days. My association with the Department is that of an alumnus. >>Raymond Smullyan presents the following (paraphrased) riddle in his >>book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise >> examination some day this week. You will not know that there is an >> examination when the class begins. A student then reasoned, I >> canÕt get the quiz on Friday because if I havenÕt gotten it by >> Friday I will know the quiz must be that day. Similarly he reasoned >> for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >>giving the quiz to the class on Monday? >> One of the hypothesis is that an examination will be given during the >> week. The argument shows that the hypothesis lead to a contradiction: >> that no examination will be given during the week. That is, >> H->not(H). But >> (H -> not(H)) -> not(H) >> is a tautology. That means that the only thing we can deduce from the >> professorÕs statements are that his conditions will not be met. >> Since we can only conclude that his conditions will not be met, none >> of the original argument showing that no test can occur is valid. The >> test may occur at any time. >Arturo, >point is that, ON MONDAY, the professor stated You will not know that there is >an examination when the class begins. When the Monday morning class began, the >students had no idea that there was going to be a surprise examination that week >since the professor had yet to make the announcement. Thus giving the exam on >Monday would not contract his statement - it would be a surprise examination >and the students wonÕt have known that there is an examination when the >Monday morning class began. What makes you think that the announcement was made after class that begun? In fact, one might argue that the announcement occurred before class had started, as evidenced by the fact that the student tried to use induction all the way to Monday. Yes, thatÕs being a bit persnickety, but that is math, after all. -- ItÕs not denial. IÕm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: SmullyanÕs Quiz Problem windows-nt) >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? > > One of the hypothesis is that an examination will be given during the > week. The argument shows that the hypothesis lead to a contradiction: > that no examination will be given during the week. That is, > H->not(H). But > > (H -> not(H)) -> not(H) > > is a tautology. That means that the only thing we can deduce from the > professorÕs statements are that his conditions will not be met. > > Since we can only conclude that his conditions will not be met, none > of the original argument showing that no test can occur is valid. The > test may occur at any time. >>Arturo, >>point is that, ON MONDAY, the professor stated You will not know that there is >>an examination when the class begins. When the Monday morning class began, the >>students had no idea that there was going to be a surprise examination that week >>since the professor had yet to make the announcement. Thus giving the exam on >>Monday would not contract his statement - it would be a surprise examination >>and the students wonÕt have known that there is an examination when the >>Monday morning class began. > What makes you think that the announcement was made after class that > begun? In fact, one might argue that the announcement occurred before > class had started, as evidenced by the fact that the student tried to > use induction all the way to Monday. > Yes, thatÕs being a bit persnickety, but that is math, after all. Let me also add (to my adjacent post) that this is a question of the problem definition. The definition of the problem should be clear. If itÕs not, letÕs make define it where it lacks clarity. THEN letÕs work on the solution. I really would like this point to be addressed and resolved before moving forward with discussion of the problem, otherwise we know not what weÕre discussing. -- % Randy Yates % Ticket to the moon, ßight leaves here today %% Fuquay-Varina, NC % from Satellite 2 %%% 919-577-9882 % ŌTicket To The MoonÕ %%%% % *Time*, Electric Light Orchestra http://home.earthlink.net/~yatescr === Subject: Re: SmullyanÕs Quiz Problem days. My association with the Department is that of an alumnus. >Let me also add (to my adjacent post) that this is a question of the >problem definition. The definition of the problem should be clear. If >itÕs not, letÕs make define it where it lacks clarity. THEN letÕs work >on the solution. >I really would like this point to be addressed and resolved before moving >forward with discussion of the problem, otherwise we know not what weÕre >discussing. I think your argument is as follows: let us call the times when the test may be administered the admissible range of the statement by the professor. Let x be the time when the test begins. The admissible range is divided into days; think of them as being integers, so that ßoor(x) represents the beginning of the day at which the test begins. The professorÕs qualifying statement says something about the knowledge of the students at ßoor(x) (the start of the day) where x is the day in which the test will occur. The professor is making a statement at time t. The statement says: (1) x>t, and x is in the admissible range. (2) Something will be true at ßoor(x). Let us denote the admissible range by [Y,Z]. I agree with you that we have two different situations. In the classical Paradox of the Unexpected Hanging/Exam/Egg, we have ßoor(Y)>t. Your interpretation of SmullyanÕs presentation has ßoor(Y)t, all statements made are about future events; since the statements are about knowledge, the information provided at instant t may affect the situation at instant ßoor(Y). That is in fact what the studentÕs argument is. On the other hand, if ßoor(Y) windows-nt) >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? > > One of the hypothesis is that an examination will be given during the > week. The argument shows that the hypothesis lead to a contradiction: > that no examination will be given during the week. That is, > H->not(H). But > > (H -> not(H)) -> not(H) > > is a tautology. That means that the only thing we can deduce from the > professorÕs statements are that his conditions will not be met. > > Since we can only conclude that his conditions will not be met, none > of the original argument showing that no test can occur is valid. The > test may occur at any time. >>Arturo, >>point is that, ON MONDAY, the professor stated You will not know that there is >>an examination when the class begins. When the Monday morning class began, the >>students had no idea that there was going to be a surprise examination that week >>since the professor had yet to make the announcement. Thus giving the exam on >>Monday would not contract his statement - it would be a surprise examination >>and the students wonÕt have known that there is an examination when the >>Monday morning class began. > What makes you think that the announcement was made after class that > begun? In fact, one might argue that the announcement occurred before > class had started, as evidenced by the fact that the student tried to > use induction all the way to Monday. > Yes, thatÕs being a bit persnickety, but that is math, after all. ThatÕs a possibility. I interpreted the problem statement to reasonably mean that the professor was making the statement during the meeting time of the very class he is proposing to give the quiz in. To be persnicketier, who said the studentÕs reasoning was correct??? :) -- % Randy Yates % I met someone who looks alot like you, %% Fuquay-Varina, NC % she does the things you do, %%% 919-577-9882 % but she is an IBM. %%%% % ŌYours Truly, 2095Õ, *Time*, ELO http://home.earthlink.net/~yatescr === Subject: Re: SmullyanÕs Quiz Problem This is the Swedish civil defense paradox: page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover books,1982 It was analyed by Williard Van Orman Quine in 1953 The analysis was that the 4th logical possiblity is best : The test will occur this week but it will be unexpected Martin Gardner also has a version of this in an Scientific American Bunch concludes that reasoning isnÕt enough to solve the paradox. >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? >Now if the professor had stated to his class on Friday, I will give >you a surprise examiniation some day next week, then there would be >inconsistency problems for all days of the week, but thatÕs not the >way the problem is stated. >Comments anyone? -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: SmullyanÕs Quiz Problem > This is the Swedish civil defense paradox: > page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover books,1982 > It was analyed by Williard Van Orman Quine in 1953 > The analysis was that the 4th logical possiblity is best : > The test will occur this week but it will be unexpected > Martin Gardner also has a version of this in an Scientific American > Bunch concludes that reasoning isnÕt enough to solve the paradox. >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? >Now if the professor had stated to his class on Friday, I will give >you a surprise examiniation some day next week, then there would be >inconsistency problems for all days of the week, but thatÕs not the >way the problem is stated. >Comments anyone? > Agreed, the professor can give the exam on Monday and it will indeed be unexpected. However, that really doesnÕt deal with the paradox. This paradox has been presented under many names: the Swedish civil defense paradox, the prediction paradox, the unexpected exam, the unexpected egg to name a few. It gets my vote as the best paradox. I have not seen a resoltuion that I find satisfactory. My own analysis is that the error is the insistance that the students must have a rational basis for believing that there is or is not going to be an exam. In essence the professor is saying to the students: you have a rational basis for believing there will be an exam if and only if you do not have a rational basis for believing there will be an exam. If the students believe the professor, they must conclude that they do not have a rational reason to either to believe that there will be an exam or to believe that there will not be an exam. The best resolution is that the students should not believe the professor. In real life, of course, the professor can only say There is a very high probability that you will have an unexpected exam next week. This statement does not lead to a paradox, so the impression that a professor can make this kind of statement and communicate information is justified. In other forms of the paradox, where there is no prediction element and hence no probablity, the statement (e.g. there is an unexpected egg) cannot be believably made. -William Hughes === Subject: Re: SmullyanÕs Quiz Problem http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Smullyan.html Looking at his history the exam question isnÕt paradoxical at all, but expected. unexpected has the connotation of Random: If you substitute at random for unexpected , it becomes an probability problem with an expectation distribution , and not a paradox or logical question anymore. It becomes equally probable that he will give the exam on the class days that week ( including the day he announces it). HereÕs an alternative: The Swedish civil defense paradox on the chalk board. He, then, announces he with give an unexpected exam at some time that week. The week passes and he gives no exam. At the first day of the next week when no one has responded to the phrase which he has left on the chalkboard, he announces that they all failed the exam: no one has told him what the The Swedish civil defense paradox is about. Dirty trick? Or a clever way to make the students pay better attention? >>This is the Swedish civil defense paradox: >>page 34Mathematical Fallacies and Paradoxes, Bryan Bunch, Dover books,1982 >>It was analyed by Williard Van Orman Quine in 1953 >>The analysis was that the 4th logical possiblity is best : >>The test will occur this week but it will be unexpected >>Martin Gardner also has a version of this in an Scientific American >>Bunch concludes that reasoning isnÕt enough to solve the paradox. >> >Raymond Smullyan presents the following (paraphrased) riddle in his >book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? >My question is this: Why is there any inconsistency in the professor >giving the quiz to the class on Monday? >Now if the professor had stated to his class on Friday, I will give >you a surprise examiniation some day next week, then there would be >inconsistency problems for all days of the week, but thatÕs not the >way the problem is stated. >Comments anyone? > > >Agreed, the professor can give the exam on Monday and >it will indeed be unexpected. However, that really doesnÕt >deal with the paradox. >This paradox has been presented under many names: >the Swedish civil defense paradox, the prediction paradox, >the unexpected exam, the unexpected egg >to name a few. It gets my vote as the best paradox. >I have not seen a resoltuion that I find satisfactory. >My own analysis is that the error is the insistance that the >students must have a rational basis for believing >that there is or is not going to be an exam. >In essence the professor is saying to the students: >you have a rational basis for believing there >will be an exam if and only if you do not have a rational >basis for believing there will be an exam. If the >students believe the professor, they must conclude >that they do not have a rational reason to either >to believe that there will be an exam or to believe that >there will not be an exam. The best resolution is that >the students should not believe the professor. >In real life, of course, the professor can only >say There is a very high probability that you >will have an unexpected exam next week. This statement >does not lead to a paradox, so the impression that >a professor can make this kind of statement and >communicate information is justified. >In other forms of the paradox, where there >is no prediction element and hence no probablity, >the statement (e.g. there is an unexpected egg) >cannot be believably made. > -William Hughes -- Respectfully, Roger L. Bagula tftn@earthlink.net, 11759Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 : alternative email: rlbtftn@netscape.net URL : http://home.earthlink.net/~tftn === Subject: Re: SmullyanÕs Quiz Problem * Randy Yates > Raymond Smullyan presents the following (paraphrased) riddle in his > book, Forever Undecided: > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? > My question is this: Why is there any inconsistency in the professor > giving the quiz to the class on Monday? > Now if the professor had stated to his class on Friday, I will give > you a surprise examiniation some day next week, then there would be > inconsistency problems for all days of the week, but thatÕs not the > way the problem is stated. Cannot see that there are any differences if the statement is made on Friday from giving it on the Monday. Anyway, this is discussed to death earlier on this group and on rec.puzzles. Basically, such a statement is self-contradictory, as if I say, My name is Jon, but you are not able to figure out my name. Therefore, the professor does not have any real information in his statement. But when the stunned class get the surprise examination on Wednesday, the statement was sensible anyway. -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: SmullyanÕs Quiz Problem > [...] > Basically, such a statement is self-contradictory, as if I say, My > name is Jon, but you are not able to figure out my name. Therefore, > the professor does not have any real information in his statement. Jon, I donÕt think this analogy is appropriate. The problem with SmullyanÕs riddle, as I see it, is that the consistency of the professorÕs statement(s) depends on the intelligence and awareness of the student. If the student never pondered the logic, he would feel that any day is a possible quiz day. --Randy -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124 === Subject: Re: SmullyanÕs Quiz Problem * Randy Yates > [...] > Basically, such a statement is self-contradictory, as if I say, My > name is Jon, but you are not able to figure out my name. Therefore, > the professor does not have any real information in his statement. > I donÕt think this analogy is appropriate. The problem with SmullyanÕs > riddle, as I see it, is that the consistency of the professorÕs statement(s) > depends on the intelligence and awareness of the student. If the student > never pondered the logic, he would feel that any day is a possible quiz > day. That is of course one way to settle the problem. However, if there _are_ intelligent students in the class, you end up there anyway. (As in most puzzles or games, stupid individuals disturb the solution. I guess many of my students would be amazed the following Thursday even if I say IÕll going to have small test on Thursday.) -- Jon Haugsand Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24 92 === Subject: Re: SmullyanÕs Quiz Problem > * Randy Yates > Raymond Smullyan presents the following (paraphrased) riddle in his > book, Forever Undecided: > > On Monday a professor says to his class, I will give you a surprise > examination some day this week. You will not know that there is an > examination when the class begins. A student then reasoned, I > canÕt get the quiz on Friday because if I havenÕt gotten it by > Friday I will know the quiz must be that day. Similarly he reasoned > for Thursday all the way to Monday. Where is the error in his logic? > > My question is this: Why is there any inconsistency in the professor > giving the quiz to the class on Monday? > > Now if the professor had stated to his class on Friday, I will give > you a surprise examiniation some day next week, then there would be > inconsistency problems for all days of the week, but thatÕs not the > way the problem is stated. > Cannot see that there are any differences if the statement is made on > Friday from giving it on the Monday. The difference is that in the first problem statement, the students would have no knowledge of any quiz (whether they had pondered it or not) on Monday when they came to class, so having the quiz on Monday would not contradict his statement. However, in the second problem statement the students would already know that there was going to be a quiz, possibly on Monday, that week. -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124 === Subject: Gamma function/Mills ratio/Ineq posting-account=OyMMlAwAAADyhoVhXYX4Bw0T-1IatpYa Let G be the Gamma function, and f(x)= G(x+0.5)/G(x+ 1) , (x>= 0). Question: to prove or disprove that for each pair (x,y), 0 =< x < y , there exists u(x,y) in ( 0, 1/2 ), such that f(y) ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . f(x) Note:According to G.N.Watson [1] it is knownn hat for every x >= 0 there exists v(x) in ( 1/4, 1/pi ), such that f(x) =1/( sqrt(x+v(x)) ) . Reference: [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh Math.Soc., (2) 11 (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. === Subject: Re: Gamma function/Mills ratio/Ineq > Let G be the Gamma function, and f(x)= G(x+0.5)/G(x+ 1) , (x>= 0). > Question: to prove or disprove that for each pair (x,y), 0 =< x < y , > there exists u(x,y) in ( 0, 1/2 ), such that > f(y) > ----- = sqrt{(x+u(x,y))/(y+u(x,y))} . > f(x) > Note:According to G.N.Watson [1] it is knownn hat for every x >= 0 > there exists v(x) in ( 1/4, 1/pi ), such that > f(x) =1/( sqrt(x+v(x)) ) . > Reference: > [1] G.N. Watson , ,,A note on gamma functions, Proc.Edinburgh > Math.Soc., (2) 11 > (1958/59), Edinburgh Math.Notes 42, (1959) 7-9. I havenÕt seen the reference. But certainly if we intend to allow x to be 0, then the interval for v should actually be ( 1/4, 1/pi ] instead. And it seems that that interval is as tight as possible. Now to your question: I wonÕt be giving a proof here, but surely what you suspect is correct. However, the interval youÕve given for u, ( 0, 1/2 ), is much looser than necessary, it seems to me. I propose that the tightest interval for u is of the form ( 1/4, c ) where c is a constant which is approximately 0.360674 . David Cantrell === Subject: Re: The ßux theory of gravitation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Va31306; >Actually what you said was: >> Now, let us put our money where our mouth is and make this forum >> interesting. I make this challenge: DEFINE A REAL NUMBER The definition >> must be original and different from mine. If you post a correct >> definition, no ßaw, etc. Ill send you a check for $1000, indicate your >> address. However, if your definition is wrong or not original, you >> should send me a check for the same amount. Ill post my address when needed. >IÕm not interested in this challenge, for the following reasons: >1) This is not interesting. Rigourous definitions of the real numbers >have been available for well over 100 years. Look it up in a book. >2) Any correct definition I might make would likely be the same as that in >some book, and therefore not original. >3) I donÕt anticipate IÕd have much chance of collecting your $1000 in >any case, especially if you are the judge of what is correct, no ßaw, >etc. >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~ israel >University of British Columbia Vancouver, BC, Canada It is not challenging because you donÕt know what a number is. If you well-define a number WE can discuss the merit of your definition. My sense is, you probably donÕt know what being well-defined is. It is stated in some of the threads. Present development of mathematics, hence, also definition of number is nonsense. Hence you really need an original one that is different from my definition in the new real number system E. E. Escultura University of the Philippines. === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Vx31296; >I wonder if Mr. Escultura is teaching elementary mathematics to >students in the Phillipines... I do and IÕm telling them whatÕs wrong with mathematics and teaching the right one. === Subject: Re: The ßux theory of gravitation by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3Vw31315; >> Hi Folks, >> Since some of you have made references to the Flux Theory of >> Gravitation that I developed, Ill offer some details. Present >> mathematical physics uses conventional modeling to DESCRIBE nature by >> mathematical spaces, functions, equations and inequalities. Its main >> tool is computation. This is inadequate, the reason there are >> unsolved problems in mathematical physics such as finding the basic >> constituent of matter, the gravitational n-body problem and the >> structure of the electron. To overcome this difficulty I have >> introduced dynamic modeling that EXPLAINS nature, physical systems >> and natural phenomena in terms of the laws of nature. >There is only one catch. Most of what Nature does or is is literally out >of our sight. We can only know it indirectly. Therefore we must rely on >hypotheses and inference, as opposed to direct knowledge. There only way >we can do that is by means of models grounded on physical hypotheses and >mathematical principles. Thus to EXPLAIN means to PREDICT and that is >what our theories do. The soundness of our theories rests on two >requirements. >1. The theories must be internally consistent (in the mathematical and >logical sense). >2. The predcitons must be empirically corroberated. Agreement with >experiment is the sine qua non of a sound theory. >If you think Nature can be deduced a priori from logically necessary >principles, you are in for dissapointment. Science is empirical, right >down to the basement level. >Bob Kolker I fully agree with you. Therefore, I model or explain nature in terms of its laws. I spend my time looking for those laws instead of solving equations. Those laws, and there are now 42 of them, comprise the ßux theory of gravitation. Its domain includes dark matter. E. E. Escultura University of the Philippines === Subject: Re: The ßux theory of gravitation gravitation your mother === Subject: Lyapunov Function by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3UD31245; x= y - 2*x*(y^2) y= -(x^3) + (x^4)*y I have tried V(x,y) of the form: a(x^2) + b(y^2) a(x^2) + b(y^2) + c(x^4) a(x^2) + b(y^2) + c(y^4) a(x^2) + c*x*y + b(y^2) -1x= y - 2*x*(y^2) >y= -(x^3) + (x^4)*y >I have tried V(x,y) of the form: >a(x^2) + b(y^2) >a(x^2) + b(y^2) + c(x^4) >a(x^2) + b(y^2) + c(y^4) >a(x^2) + c*x*y + b(y^2) -1with no luck. I even tried reverse engineering it, but still no luck. A hint or two would be better than an explicit answer, as IÕm trying to learn here ;-) but IÕm glad of any help you can give me. >Mike Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0. Thomas === Subject: Re: Lyapunov Function >>x= y - 2*x*(y^2) >>y= -(x^3) + (x^4)*y >>I have tried V(x,y) of the form: >>a(x^2) + b(y^2) >>a(x^2) + b(y^2) + c(x^4) >>a(x^2) + b(y^2) + c(y^4) >>a(x^2) + c*x*y + b(y^2) -1>with no luck. I even tried reverse engineering it, but still no luck. A hint or two would be better than an explicit answer, as IÕm trying to learn here ;-) but IÕm glad of any help you can give me. >>Mike >Try V(x,y) = x^(2m) + a*y^(2n) with m,n integers >=1 and a>0. To be a little more specific: Use this Ansatz and compute d/dt (V(x(t),y(t)). You will find that there is only one choice for m and n such that the bad terms with odd exponents balance. Next you can find an a which make the bad terms vanish *identically* so that you are only left with even powers. Fortunately this a is positive (otherwise we wouldnÕt have a L.-function and d/dt (V(x(t),y(t)) <= 0 for all t ! That might give you an idea. BTW: You have the solution in your list (you donÕt need that parameters though) - somehow you skipped over it or miscalculated. Thomas >Thomas === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3VH31302; >> I wonder if Mr. Escultura is teaching elementary mathematics to >> students in the Phillipines... >He could be the janitor... You see, in the Philippines, a janitor is smarter than most mathematicians. EEE === Subject: Re: Some questions about the Cauchy distribution by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3SZ31151; >>The mean of n random variables picked from the Cauchy distribution has >>itself the Cauchy distribution. Does this mean that there is no use >>having a larger sample in order to estimate the expected value? (Or wait >>a minute, does the expected value exist? No, right?) >I think what youÕre trying to do is the following: given a random sample >of size n from the Cauchy distribution with density 1/(Pi (1+(x-a)^2)), >estimate the parameter a. This parameter is not the expected value, >because that doesnÕt exist [...] IsnÕt it only a matter of definition of the expectation ? If one integrates x/(Pi (1+(x-a)^2)) from a-t to a+t in the sense of CauchyÕs principal value (instead of Riemann) one gets (2*a*ArcTan[t])/Pi which has Ōaas limit when t goes to infinity. V.Astanoff === Subject: Re: JSH: Simple proof by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3SA31146; >> Why donÕt you simply and brießy state the >>problematic<< property >> of the ring of algebraic integers and/or the statement that you >> want to prove. >The ring of algebraic integers is determined by roots of *monic* >polynomials with integer coefficients. >It is possible to show with basic algebra that there are numbers which >are properly units but because their multiplicative inverse is not the >root of some monic polynomial with integer coefficients they are not >units in the ring of algebraic integers. These numbers are >>properly units<< in which ring? The notion of a unit is defined only with respect to some ring. Your statement makes no sense as it stands. >To see how it works consider that in rationals you can have >(3x + 1)(x + 1) = 3x^2 + 4x + 1 >where, of course, one of the roots is a unit in the ring of algebraic >integers. >But now consider >(3x + u_1)(x + u_2) = 3x^2 + kx + 1, where u_1 u_2 =1, and k is an >integer. >You find that if the uÕs are irrational, then u_1, while an algebraic >integer is not a unit in the ring, while u_2 cannot then even be an >algebraic integer. >My research shows though that both u_1 and u_2 can be units in a ring >where -1 and 1 are the only rational units, and no non-unit member of >the ring is a factor of any two integers that are coprime in the ring >of integers. >You see, I abstracted out two key properties of rings like the ring of >integers and the ring of algebraic integers. I read your contributions about the >>object ring<< as well as the ones of many others. Together we reached a rather satisfying answer concerning the existence of such a ring. >> Using standard mathematical terms and notions (for example from >> commutative algebra) this should be possible in a few lines instead >> of making a long story. >ItÕs not complicated. Basically you canÕt just rely on whether or not >some number is in the ring of algebraic integers when considering >factors of roots of a polynomial. >The mathematics is mostly REALLY simple. This is no clear mathematical statement at all! What is the problem? What did you prove? >> Why do we have to discuss things like >>what is a polynomial?<< >IÕm not discussing that, other posters made a big deal out of it. ...but if such simple things like >>what is the constant term of a polynomial<< are not clear, we have to clarify it before we can proceed. >> here? The notion of a polynomial is defined since a long time and >> can be found in every introductory book on algebra. >So? >> If we consistently use the common definitions of mathematical >> objects like polynomials we should rather quickly be able to >> clarify the situation and avoid all the frustration that frequently >> seems to culminate in personal attacks. >> H >IÕve seen posters come and go, and every once in a while thereÕs a >poster like you who claims to care about working things out. >When it turns out that IÕm right, you go over to the other side, and >either run away, or turn to bizarre behavior. I follow the discussions you initiated since almost one year now and tried to contribute with the aim to clarify things ring, because I found the idea interesting. However I remember clearly that you left the discussion suddenly claiming that you are not interested in the stuff anymore. Strangely you did this when with the help of other posters together we really reached a good overview over the candidates for >>object rings<< within the field of algebraic numbers (and even within larger fields). No >>sides<< in that discussion just mathematics! >Psychologists call it cognitive dissonance. >Basically, deep down you believe that I must be wrong, so your post is >not really in good faith. But simply *saying* certain things that >indicate objectivity or willingness to be objective sets you up >psychologically. >That is, you feel a need to be consistent with what you said. >But later, when you run into the rigid mathematics, which goes against >what you wish to believe, you basically kind of break. Your mind >breaks, and you run away or behave weird. >IÕve seen it lots of times. Do yourself a favor, and just walk away >now. >James Harris What sometimes really makes me angry in the discussions you initiate is the lack of mathematical rigor! Each time one of the participants clearly worked out a point the discussion stops, or turns into personal attacks etc. but it never ends with a statement like: now we have seen that this and this is true. Lets keep this result and proceed until we reach the final conclusion. Why donÕt you answer to Nora BaronÕs recent posts? They are mathematically very clear. The critical points are obvious to everyone who understands basic mathematics - as you also say. Forget about >>sides<< and all that personal stuff. There is the mathematics - answer to it in mathematical terms and with the same clarity. That would lead to a real discussion and eventually to a clear result. Why for example didnÕt we finish the discussion about object rings. We obtained a result. We could have stated it clearly. We would not need to start all over again discussing statements like >>no non-unit member of the ring is a factor of any two integers that are coprime in the ring of integers<<. It has been stated over and over again that if the ring R you are considering here contains the integers, then this statement is satisfied for every ring! It is nothing particular! Check this out: coprime integers generate the whole of Z (integers) as an ideal. Lifting this ideal to the ring R leads to the whole ring too. Thus the two numbers cannot be divided by a common non-unit of R. This is exhausting - one feels like a hamster running in his wheel without ever moving one centimeter ahead. H === Subject: Re: JSH: Simple proof What sometimes really makes me angry in the discussions you > initiate is the lack of mathematical rigor! Each time one of > the participants clearly worked out a point the discussion > stops, or turns into personal attacks etc. but it never ends > with a statement like: now we have seen that this and this > is true. Lets keep this result and proceed until we reach > the final conclusion. YouÕre silly then. IÕve talked to some very elite mathematicians, in discussions that arenÕt on Usenet. I donÕt hear the same crap from them that you people whine about constantly. I just donÕt get that kind of crap from leading mathematicians, and donÕt dream that I just get blown off all the time. IÕve had some involved discussons over these ideas, and there just isnÕt ANY of the crap that you people keep tossing up, year after year, after year. So why should I think anything of you? You say youÕre angry. So what? I think youÕre some weirdly emotional person who gets excited for nothing. > Why donÕt you answer to Nora BaronÕs recent posts? They are > mathematically very clear. The critical points are obvious The poster Nora Baron is not even a girl. That poster is a guy, playing a girl. IÕve explained to that poster everything repeatedly over a period of years. IÕve shot down each and every objection, only to see it raised again, and again, and again. And I have to tell you people, leading mathematicians donÕt make those objections. Barry Mazur didnÕt. Ralph McKenzie didnÕt. Andrew Granville didnÕt. You people are in a world by yourselves living in a fantasy that you are actually good at mathematics and discussing it. The proÕs donÕt talk like you do. > to everyone who understands basic mathematics - as you also > say. Forget about >>sides<< and all that personal stuff. > There is the mathematics - answer to it in mathematical terms > and with the same clarity. That would lead to a real discussion > and eventually to a clear result. > Why for example didnÕt we finish the discussion about object > rings. We obtained a result. We could have stated it clearly. > We would not need to start all over again discussing statements > like >>no non-unit member of the ring is a factor of any two > integers that are coprime in the ring of integers<<. > It has been stated over and over again that if the ring R you > are considering here contains the integers, then this statement > is satisfied for every ring! It is nothing particular! Really? > Check this out: coprime integers generate the whole of Z (integers) > as an ideal. Lifting this ideal to the ring R leads to > the whole ring too. Thus the two numbers cannot be divided by > a common non-unit of R. No. Maybe later I need to come and check KummerÕs work as well for errors. > This is exhausting - one feels like a hamster running in his > wheel without ever moving one centimeter ahead. Why bother? You are kind of annoying, slightly. Why canÕt you people get it through your heads that I have a paper at a journal? You donÕt have to do anything, but wait. Quit with the drama. IÕm not interested in histrionics from posters like you. I can create all the drama I wonÕt by myself. James Harris === Subject: Re: JSH: Simple proof > The poster Nora Baron is not even a girl. That poster is a guy, > playing a girl. A perfect match for yourself -- as I discovered when I found the anagram of your name which reveals your true identity: James S. Harris = Ms. J. Harrie Ass (note the gender) -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: Simple proof > The poster Nora Baron is not even a girl. That poster is a guy, > playing a girl. Why does that bother you so much, James? Is it too reminiscent of your dating experiences? -- Wayne Brown (HPCC #1104) | When your tailÕs in a crack, you improvise fwbrown@bellsouth.net | if youÕre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: Simple proof Discussion, linux) > Quit with the drama. IÕm not interested in histrionics from posters > like you. > I can create all the drama I wonÕt by myself. ^^^^^ Dammit, James, you just ruined a perfect .sig. -- There was an accident in the air. There was a sign saying, ŌPlanes donÕt go here. The clouds have to be fixed.Õ -- Quincy P. Hughes applies the lessons of Dutch train travel to pretending about airplanes. === Subject: Re: JSH: Simple proof > Quit with the drama. IÕm not interested in histrionics from posters > like you. > I can create all the drama I wonÕt by myself. > ^^^^^ > Dammit, James, you just ruined a perfect .sig. HeÕs not adopted an explicitly stream-of-consciousness style before, but thereÕs a first time for everything: perhaps this is to be read as I can create all the drama [if I want to]; [but] I wonÕt by myself [so letÕs keep playing] -- Larry Lard Replies to group please === Subject: Re: JSH: Simple proof <87fz2nbuq4.fsf@phiwumbda.org> !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw >> Quit with the drama. IÕm not interested in histrionics from posters >> like you. > I can create all the drama I wonÕt by myself. >> ^^^^^ >> Dammit, James, you just ruined a perfect .sig. > HeÕs not adopted an explicitly stream-of-consciousness style before, > but thereÕs a first time for everything: perhaps this is to be read > as > I can create all the drama [if I want to]; [but] I wonÕt by myself > [so letÕs keep playing] Uh, whatÕs this stream-of-conscious nonsense? His sentence makes perfect sense without any such tags. It is equivalent to I have the ability to also create by myself all the drama which I actually am not going to create. So that simply means that his potential for drama is not restricted at all, even though he might refrain from some drama deliberately. -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof !3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~( 5eZ41to5f%E@ÕELIi $t^ VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw > I can create all the drama I wonÕt by myself. Wow, he smashes theology and philosophy in one fell swoop together with mathematics. make a noise? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum === Subject: Re: JSH: Simple proof > Why donÕt you answer to Nora BaronÕs recent posts? They are > mathematically very clear. The critical points are obvious > to everyone who understands basic mathematics - as you also > say. That is the reason right there. JSH isnÕt interested in a mathematical discussion that doesnÕt lead to a recognition of his genius. When a poster points out clearly the ßaws in his thinking, JSH must either ignore the post or respond with personal attacks. The very clarity of NBÕs posts is what makes them unacceptable to JSH. The reason JSH will never be a mathematician is that he can tolerate error. I believe his emotional need to see himself as a genius is stronger than his rejection of contradictions, so emotions trump logic. Gib === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3UY31284; >>1271503.iARF35e16701@proapp.mathforum.org... YouÕve *proven* that youÕre not using the same language that other professional mathematicians use. Anything else is simply a matter of [probably correct] deduction in your axiom set which simply doesnÕt agree with everyone elseÕs. Norm Unfortunately, mathematics is not a popularity contest; it is a struggle for precision. One of the requirements of mathematics is that every concept must well-defined, that is, its existence, properties and relations with other concepts must be specified by the axioms. The axioms are the basis of proof. Proofs are nonsense unless they follow from the axioms. That is why a true mathematician looks at the foundations of a field before doing anything there. This is where Wiles failed miserably. Critique-rectification naturally involves new language. E. E. Escultura University of the Philppines === Subject: Re: Escultura affair: publication scandal >>1271503.iARF35e16701@proapp.mathforum.org... YouÕve *proven* that youÕre not using the same language that other professional mathematicians use. Anything else is simply a matter of [probably correct] deduction in your axiom set which simply doesnÕt agree with everyone elseÕs. > Norm > Unfortunately, mathematics is not a popularity contest; it is a struggle for precision. One of the requirements of mathematics is that every concept must well-defined, that is, its existence, properties and relations with other concepts must be specified by the axioms. The axioms are the basis of proof. Proofs are nonsense unless they follow from the axioms. That is why a true mathematician looks at the foundations of a field before doing anything there. This is where Wiles failed miserably. Critique-rectification naturally involves new language. > E. E. Escultura > University of the Philppines No, itÕs not a popularity contest -- itÕs a game. Well, at least there was a game called IIRC Wiff-N-Proof that taught the creation of and deduction of correct logical propositions from axioms. If youÕre really a mathematician then you already know that there are several different logical systems based on differing foundational axiom sets and their semantics. Assuming your system is consistent, youÕve simply created another one thatÕs neither more nor less correct than the others. What you see as errors in their system results from applying your axiom scheme to someone elseÕs system -- which isnÕt valid. Norm === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3TM31190; Zentralblatt f.9fr Mathematik< 9 publications of E. Escultura are listed. For 5 of them the Zentralblatt just mentions >not reviewed<, whatever that means. This is an indexing service like so many others. So it simply notes some contributions unless someone wants to try uncharted course. Zentralblatt f?hematik< 9 publications > of E. Escultura are listed. For 5 of them the > Zentralblatt just mentions >not reviewed<, whatever > that means. > This is an indexing service like so many others. No. It commissions reviews: not reviewed means not reviewed. There are two reasons why a paper might not be reviewed: (i) the editord did not commision a review since they considered it without merit or otherwise unsuitable, (ii) a review was commissioned but never written. There is another reviewing service: Mathematical Reviews. It lists ten items by Escultura. None of them have been reviewed. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Escultura affair: publication scandal > Zentralblatt f.9fr Mathematik< 9 publications > of E. Escultura are listed. For 5 of them the > Zentralblatt just mentions >not reviewed<, whatever > that means. > > This is an indexing service like so many others. > No. It commissions reviews: It was formerly, but havenÕt they changed lately? They publish authorssummaries? Or did I just dream it? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Escultura affair: publication scandal >> Zentralblatt f.9fr Mathematik< 9 publications >> of E. Escultura are listed. For 5 of them the >> Zentralblatt just mentions >not reviewed<, whatever >> that means. >> >> This is an indexing service like so many others. >> No. It commissions reviews: > It was formerly, but havenÕt they changed lately? They publish > authorssummaries? Or did I just dream it? Quote Zentralblatt MATH is the worldÕs most complete and longest running abstracting and reviewing service in pure and applied mathematics. They still publish reviews. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Escultura affair: publication scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3T431194; >>1271503.iARF35e16701@proapp.mathforum.org... === Subject: Re: Escultura affair: publication scandal Author: Norm Dresner 1271503.iARF35e16701@proapp.mathforum.org... YouÕve *proven* that youÕre not using the same language that other professional mathematicians use. Anything else is simply a matter of [probably correct] deduction in your axiom set which simply doesnÕt agree with everyone elseÕs. Norm Unfortunately, mathematics is not a popularity contest; it is a struggle for precision. One of the requirements of mathematics is that every concept must well-defined, that is, its existence, properties and relations with other concepts must be specified by the axioms. The axioms are the basis of proof. Proofs are nonsense unless they follow from the axioms. That is why a true mathematician looks at the foundations of a field before doing anything there. This is where Wiles failed miserably. Critique-rectification naturally involves new language. E. E. Escultura University of the Philppines === Subject: Re: Re Re Publication Scandal by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3UZ31250; >> Now, let us put where our money where our mouth is and make >>this forum interesting. I make this challenge: DEFINE A REAL NUMBER >>The definition must be original and different from mine. I you post a >>correct definition, no ßaw, etc. IÕll send you a check for $1000, >What the heck is a correct definition? For example, why couldnÕt >we define real number to mean the big key on the keyboard that >moves the cursor all the way to the left? >Please use your real number more often. >You can donate my $1000 to charity. === Subject: Re: Re Re Publication Scandal Author: Dave Rusin > Now, let us put where our money where our mouth is and make this forum interesting. I make this challenge: DEFINE A REAL NUMBER > The definition must be original and different from mine. If you post a >correct definition, no ßaw, etc. IÕll send you a check for $1000, < What the heck is a correct definition? For example, why couldnÕt we define real number to mean the big key on the keyboard that moves the cursor all the way to the left? dave As a topologist you know what well-defined means and that a concept does not exist if it isnt well-defined. I require more: The existence, properties, relationship of a concept with others, etc. must be specified by the axioms. In the case of a real number every digit must be known or computable, that is, there is an algorithm for computing it. That is why a nonperiodic number or irrational is nonsense because no one knows its, say, ten millionth digit. Your suggestion does not sound mathematical at all; a robot can do what you are suggesting. You ainÕt got nothin for charity. E. E. Escultura University of the Philippines >dave === Subject: Re: The-State-of-the-Art in Mathematics by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id iB2E3UM31258; >Hey, Escultura, y. Brouwer constructed a counterexample to it which, incidentally, implies that the real numbers have no natural ordering. Moreover, Banach Tarski constructed a counterexample to the completeness axiom of the real number system. Ross Finlayson To sum up, the present axiomatization of the real numbers is terribly wrong. Most of the concepts are ill-defined, two axioms are false and the topics on fractions are redundant because fractions are decimals. All to the best Eddie Escultura University of the Philippines === Subject: Re: The-State-of-the-Art in Mathematics >Hey, Escultura, > Dichotomy axiom says: Given real numbers x, y, one and only one holds: x < y or x = y or x > y. I think thatÕs called trichotomy around here, with dichotomy, for cutting into two parts instead of cutting into three parts, dichotomy is x=y. Anyways, trichotomy holds well within the real numbers. When you get outside of the real numbers, it might not. > Brouwer constructed a counterexample to it which, incidentally, implies that the real numbers have no natural ordering. Please state it here. > Moreover, Banach Tarski constructed a counterexample to the completeness axiom of the real number system. I have not heard of that. Please state it here, on this thread or some other, on this newsgroup where you have brought forth these claims, so that others may validate your claims prima facie, that means from reading only the content on this newsgroup. > number? Do you consider the nonstandard constructions of the surreal > numbers, or hyperreals or colonies of germs and germinal functions? > Nothing to do with rurreals, etc. I have a new construction without these false axioms using only three simple axioms. This relieves the real numbers of all contradictions while at the same time enriching them. > No, I wont define a real number because Im giving a price of $1000 who can define it correctly but differently from mine. The offer expires this weekend. > Yes, its healthy to disagree, thats called critical thinking, the beginning of great things. > Ross Finlayson > To sum up, the present axiomatization of the real numbers is terribly wrong. Most of the concepts are ill-defined, two axioms are false and the topics on fractions are redundant because fractions are decimals. > All to the best > Eddie Escultura > University of the Philippines ThatÕs a fine claim, but itÕs worthless only because you have not presented your reasoning to the point where someone has to be able to explain it to another without saying only Escultura knows. I think youÕre wrong, not in spirit, because there are some few limits of the standard real number constructions, but in basically all details. So, present your actual arguments to sci.math, there are competent reviewers who will read it and put forth their comments. Ross === Subject: The Parageodetic Distance Formula (UBasic) posting-account=o2aKmAwAAABN3qhbZeFPxY770sPh3HDq 
After some reßection--prompted by an e-mail inquiry--this
writer/
programmer has decided to first release the parageodetic
(inverse
problem) formulary core, before the more dressed, integrated
quadratic-mean-spherical, parageodetic and geodetic inverse
and
direct GODx program core.
But, rather than just spit out the program listing in
numerical
address order, however, it is being presented
subroutine-with-execution,
thereby providing another opportunity to disseminate this
unique
formulary approach (including all trigonometric quadrantic
adjustments
and division by zero bulletproofing) for mathematical study 
and
discussion.
As the concept of the parageodesic appears little
acknowledged--if at
all even recognized--it would seem prudent to comparatively
define the
geodesic and parageodesic from the conceptual approach used
here (to
the non-technical layman, the term graticular can be
considered
synonymous with spherical):
Geodesic: The conformally delineated great-arc segment
between two points;
Parageodesic: The graticularly delineated great-arc segment
between two points.
Thus, with an ellipsoid, the traditional geodesic is the
pulled
string path and distance on an elliptically correct model,
whereas
the parageodesic is the pulled string path and distance, first
DEFINED on a spherical model of the ellipsoid, then the
sphere is
reformed to its proper ellipsoidal shape, at which time the
elliptically correct arcradius and angle/azimuth AT ANY GIVEN
POINT
ALONG THE DEFINED SEGMENT can be found (i.e., it is the local,
elliptically correct Great-Circle valuation!).
UBasicÕs system program can be found here:
ftp://rkmath.rikkyo.ac.jp/pub/ubibm
(this web page is translated through Google, thus some of
the wording is off)
UBasicÕs command dictionary can be found here:
http://ed-thelen.org/bab/bab-ubhelp.html
Of course, while it is presented here out of numerical
address order,
once the program code is cut and pasted into UBasic, it will
revert to
its natural arrangement.
-------------------------------------------------------------
---------------
------------------
3000 GoSub 58000
58000 Ō*** Radial Parameters ***
58010 a=6378.135:b=6356.75:b=b::GoTo 58200
58020 a=10000:b=5000
58200 Oz=2*Atan((a-b)^.5/(a+b)^.5)
10000 BLat=+43.66111111111111111111:BLong=
-70.25583333333333333333
10010
DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
1Parageodetic = 4092.6024524649626 Km
291.6521536/73.6370729
1Geodetic = 4092.6016550196721 Km
291.7142510/73.5743344
1Q-M-Spherical = 4082.4708695785296 Km
291.7214263/73.5858490
2200 Pi=#Pi:PH=.5*Pi:PD=2*Pi
2300 RF=Pi/180
55110
BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>Pi
Then
MD=MD-Sgn(MD)*PD
11000 GoSub 55000
1Õ###########################################################
#
55000 Ō*** Spherical, ParaGeodetic Valuations ***
55100 ID%=0
55120 WBLr=DLr:WDLr=BLr:GoSub
55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
55130 WBLr=BLr:WDLr=DLr:GoSub
55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
1Õ##########
55400 Ō*** Spherical Elements ***
55410 GoSub 55420:GoSub 55440:Return
55420 SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)0 Then
SA=#Eps
55422 If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
55425
SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
2150 IO=10^100
55430 WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
55440
TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
55445 If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
55439 Return
55449 Return
1Õ----------
55125 GoSub 55190:DAz(1)=WAz(1)
55135 GoSub 55190:BAz(1)=WAz(1)
1Õ##########
55190 ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub
55430:GoSub
55445:ID%=0:Return
1Õ----------
2100 Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
55140 If Cos(BAz(0))<0 Then TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
55142 If Cos(DAz(0))>0 Then TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
55145 If TvL(5)0 Then DAz(ID%)=PD-DAz(ID%)
55489 BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
1Õ----------
55189 Return
1Õ-----------------------------------------------------------
-
11100 GoSub 59100:KmDx(1)=a*EGDD*AD(0)
1Õ###########################################################
#
59100 Ō*** PEOx ***
59000 Ō*** Integration ***
65100
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*
Cos(TvL)*Sin(Oz))
^2)^.5)
1= Elliptic Integrand Of The Second Kind
65110 .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
1= Elliptic Integrand Of The First Kind
1x a = N (Normal) = Transverse Equatorial Arcradius
65120 .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
1x a = M (Meridian) = Vertical Meridional Arcradius
65130
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,
TvL)^2-.E0p(AP,TvL
)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
1x a = O (Omniversal) = Transverse Meridional Arcradius
1Õ#########################
11020 XF=1:::For UL=2 To 20 Step 2::GoSub 60000
1Integration Convergency for EGDD to .1^20 @ cos{Oz}, for
XFxUL: cos{Oz}
= 1xUL/2xUL/3xUL
1.9999 = 6/8/6; .995 = 8/12/15; .99 = 8/12/15; .9 =
11/16/18;
1.75 = 13/18/24; .5 = 17/24/30; .25 = 27/36/45; .1 =
42/60/75;
1.01 = --/--/231; (.001 ~=~ 20x94 = 1880; .0001 ~=~ 200x97 
=
19400)
60000 Ō*** (Gaussian) Amplitudal Kernal Expansion ***
65000
.AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5
*AEg(NS))^2))^.5
):If AE>.5*Pi Then AE=Pi-AE
65009 Return(AE)
60100 DU=2*UL:For NS=1 To
UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
60200 LP(1)=1:LP=1:For NP=1 To DU
60210
LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2))
)/NP:Next NP
60220 DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
60230 CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
60235 If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
60300 AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
60310 OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next
NS:Return:::::?NS;
;Using(,30),AEg(NS)/RF; ;OE(NS)/RF:Next NS:List
60010:End:Return
1Õ-------------------------
59110 EGDD=0:For NS=1 To UL:For NX=1 To XF
59120 AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
59130 TW=XF*UL:ABq=Cos(AE)*ADq 0:OW=OE(NS)/TW
59140
UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
59150
EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),
LMP)+.EGp(AP(0),LL
P))*OW
59160 Next NX:Next NS:Return
1Õ-----------------------------------------------------------
-
10 ClS
1000 Dim
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1
),BAz(1),DAz(1)
2000 Ō*** Program Constants/Standard Values ***
2400 Deg$=Chr(248):Õ(Degree Sign)
58300 ?Tab(20);************************************:?Tab(19);/
Parageodetic Distance Formula Core :? *----------------* (
cos{Oz}
=;Using(,20),Cos(Oz); ) *-------------------*
58310 ?Using(,13), | a =;a;
Km;;:?Tab(55-Len(Str(Int(b))));Using(,13),b =;b; Km; |:?
*------------------------------------------------------------
---------------*
:Return
11003 ?Tab(7);BTvL, DTvL:
;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
11005 ?Tab(25);ADg: ;Using(,20),ADg;Deg$
11007 ?Tab(25);APg: ;Using(,20),APg;Deg$
11010 ?Tab(5);BAz g, DAz g:
;Using(,20),BAz(0);Deg$,DAz(0);Deg$
11012 ?Tab(5);BAz e, DAz e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
11015 ?Tab(5);APe b, APe d:
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL(5))/RF;
Deg$
65010
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*
RF)*Sin(Oz))^2)^
.5+1)))
1= Reduced Oz: cos{ROz{0}} = cos{Oz},
cos{ROz{90}} = 1
65020
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)
^2-Sin(AP)^2)^(-
.5)))
11017 GoSub 11990:?Tab(16);Cos{ROz{APg}}
=;Using(,20),Cos(.Roz(APg)):?
11900 ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km
(;Using(,25),EGDD; ):Next:?
11910 GoSub 11990
11990
?Tab(4);*----------------------------------------------------
--------------
---*:Return
11980 ?:List 11020:List 58010-58030:List 10000-10010:End
64000 Ō *** Functions ***
1 Point 7
Ō ~=~ .1^33
-------------------------------------------------------------
---------------
------------------
This program is presented as a workbench specimen, intended
for
dissection and analysis.
For just a no-frills output, the following modification can be
made.
-------------------------------------------------------
Delete 11003-11900
4000 XF=1:UL=10:GoSub 60000
11100 GoSub 59100:KmDx(1)=A*EGDD*AD(0)
11200 ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
11300 ?Tab(5);BAz e, DAz e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
11980 ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
58010 a=6378.135:b=6356.75
58020
-------------------------------------------------------
~Kaimbridge M. GoldChild~
=== Subject: Re: The Parageodetic Distance Formula (UBasic) 
Gee, arenÕt you a clever boy.

 After some reßection--prompted by an e-mail
inquiry--this writer/
> programmer has decided to first release the parageodetic
(inverse
> problem) formulary core, before the more dressed, integrated
> quadratic-mean-spherical, parageodetic and geodetic inverse
and
> direct GODx program core.
> But, rather than just spit out the program listing in
numerical
> address order, however, it is being presented
subroutine-with-execution,
> thereby providing another opportunity to disseminate this
unique
> formulary approach (including all trigonometric quadrantic
adjustments
> and division by zero bulletproofing) for mathematical study
and
> discussion.
> As the concept of the parageodesic appears little
acknowledged--if at
> all even recognized--it would seem prudent to comparatively
define the
> geodesic and parageodesic from the conceptual approach used
here (to
> the non-technical layman, the term graticular can be
considered
> synonymous with spherical):
> Geodesic: The conformally delineated great-arc segment
> between two points;
> Parageodesic: The graticularly delineated great-arc segment
> between two points.
> Thus, with an ellipsoid, the traditional geodesic is the
pulled
> string path and distance on an elliptically correct model,
whereas
> the parageodesic is the pulled string path and distance,
first
> DEFINED on a spherical model of the ellipsoid, then the
sphere is
> reformed to its proper ellipsoidal shape, at which time the
> elliptically correct arcradius and angle/azimuth AT ANY
GIVEN POINT
> ALONG THE DEFINED SEGMENT can be found (i.e., it is the
local,
> elliptically correct Great-Circle valuation!).
> UBasicÕs system program can be found here:
> ftp://rkmath.rikkyo.ac.jp/pub/ubibm
> rkmath.rikkyo.ac.jp/~kida/ubasic.htm
> (this web page is translated through Google, thus some of
> the wording is off)
> UBasicÕs command dictionary can be found here:
> http://ed-thelen.org/bab/bab-ubhelp.html
> Of course, while it is presented here out of numerical
address order,
> once the program code is cut and pasted into UBasic, it
will revert to
> its natural arrangement.
>
-------------------------------------------------------------
--------------
> -------------------
> 3000 GoSub 58000
> 58000 Ō*** Radial Parameters ***
> 58010 a=6378.135:b=6356.75:b=b::GoTo 58200
> 58020 a=10000:b=5000
> 58200 Oz=2*Atan((a-b)^.5/(a+b)^.5)
> 10000 BLat=+43.66111111111111111111:BLong=
-70.25583333333333333333
> 10010
DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
> 1Parageodetic = 4092.6024524649626 Km 291.6521536
/73.6370729
> 1Geodetic = 4092.6016550196721 Km 291.7142510 
/73.5743344
> 1Q-M-Spherical = 4082.4708695785296 Km 291.7214263
/73.5858490
> 2200 Pi=#Pi:PH=.5*Pi:PD=2*Pi
> 2300 RF=Pi/180
> 55110
BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>P
> i Then MD=MD-Sgn(MD)*PD
> 11000 GoSub 55000
>
1Õ###########################################################
#
> 55000 Ō*** Spherical, ParaGeodetic Valuations ***
> 55100 ID%=0
> 55120 WBLr=DLr:WDLr=BLr:GoSub
55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
> 55130 WBLr=BLr:WDLr=DLr:GoSub
55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
> 1Õ##########
> 55400 Ō*** Spherical Elements ***
> 55410 GoSub 55420:GoSub 55440:Return
> 55420 SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)0 Then S
> A=#Eps
> 55422 If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
> 55425
SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
> 2150 IO=10^100
> 55430 WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
> 55440
TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
> 55445 If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
> 55439 Return
> 55449 Return
> 1Õ----------
> 55125 GoSub 55190:DAz(1)=WAz(1)
> 55135 GoSub 55190:BAz(1)=WAz(1)
> 1Õ##########
> 55190 ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub
55430:GoSub
> 55445:ID%=0:Return
> 1Õ----------
> 2100 Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
> 55140 If Cos(BAz(0))<0 Then
TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
> 55142 If Cos(DAz(0))>0 Then
TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
> 55145 If TvL(5) 55150 AD(0)=TvL(5)-TvL(1):ADq
0=0.25*AD(0):BTvL=TvL(1)/RF:DTvL=TvL
> (5)/RF:ADg=DTvL-BTvL
> 55155 TvL(2)=TvL(1)+ADq 0:TvL(3)=TvL(1)+2*ADq
0:TvL(4)=TvL(1)+3*ADq 0
> 55147 GoSub 55450:APg=AP(0)/RF:GoSub 55485:ID%=1:GoSub 55485
> 1Õ##########
> 55450
AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Cos(WAz(ID%
))^2+
> (Sin(WBLr)*Sin(WAz(ID%)))^2)^.5+1)):Return
> 55485 If MD<0 Then BAz(ID%)=PD-BAz(ID%)
> 55487 If MD>0 Then DAz(ID%)=PD-DAz(ID%)
> 55489 BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
> 1Õ----------
> 55189 Return
>
1Õ-----------------------------------------------------------
-
> 11100 GoSub 59100:KmDx(1)=a*EGDD*AD(0)
>
1Õ###########################################################
#
> 59100 Ō*** PEOx ***
> 59000 Ō*** Integration ***
> 65100
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*
Cos(TvL)
> *Sin(Oz))^2)^.5)
> 1= Elliptic Integrand Of The Second Kind
> 65110 .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
> 1= Elliptic Integrand Of The First Kind
> 1x a = N (Normal) = Transverse Equatorial Arcradius
> 65120 .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
> 1x a = M (Meridian) = Vertical Meridional Arcradius
> 65130
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,
TvL)^2-.E
> 0p(AP,TvL)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
> 1x a = O (Omniversal) = Transverse Meridional Arcradius
> 1Õ#########################
> 11020 XF=1:::For UL=2 To 20 Step 2::GoSub 60000
> 1Integration Convergency for EGDD to .1^20 @ cos{Oz}, 
for
XFxUL:
cos{Oz}
> = 1xUL/2xUL/3xUL
> 1.9999 = 6/8/6; .995 = 8/12/15; .99 = 8/12/15; .9 = 11/1
> 6/18;
> 1.75 = 13/18/24; .5 = 17/24/30; .25 = 27/36/45; .1 = 
42/6
> 0/75;
> 1.01 = --/--/231; (.001 ~=~ 20x94 = 1880; .0001 ~=~ 
200x9
> 7 = 19400)
> 60000 Ō*** (Gaussian) Amplitudal Kernal Expansion ***
> 65000
.AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5
*AEg(
> NS))^2))^.5):If AE>.5*Pi Then AE=Pi-AE
> 65009 Return(AE)
> 60100 DU=2*UL:For NS=1 To
UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
> 60200 LP(1)=1:LP=1:For NP=1 To DU
> 60210
LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2)))
> /NP:Next NP
> 60220 DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
> 60230 CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
> 60235 If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
> 60300 AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
> 60310 OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next
NS:Return:::::?NS;
;Usin
> g(,30),AEg(NS)/RF; ;OE(NS)/RF:Next NS:List 60010:End:Return
> 1Õ-------------------------
> 59110 EGDD=0:For NS=1 To UL:For NX=1 To XF
> 59120 AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
> 59130 TW=XF*UL:ABq=Cos(AE)*ADq 0:OW=OE(NS)/TW
> 59140
UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
> 59150
EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),
LMP)+.E
> Gp(AP(0),LLP))*OW
> 59160 Next NX:Next NS:Return
>
1Õ-----------------------------------------------------------
-
> 10 ClS
> 1000 Dim
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1
),B
> Az(1),DAz(1)
> 2000 Ō*** Program Constants/Standard Values ***
> 2400 Deg$=Chr(248):Õ(Degree Sign)
> 58300
?Tab(20);************************************:?Tab(19);/
Parageod
> etic Distance Formula Core :? *----------------* ( cos{Oz}
=;Using(,2
> 0),Cos(Oz); ) *-------------------*
> 58310 ?Using(,13), | a =;a;
Km;;:?Tab(55-Len(Str(Int(b))));Using(,1
> 3),b =;b; Km; |:?
*-------------------------------------------------
> --------------------------*:Return
> 11003 ?Tab(7);BTvL, DTvL:
;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
> 11005 ?Tab(25);ADg: ;Using(,20),ADg;Deg$
> 11007 ?Tab(25);APg: ;Using(,20),APg;Deg$
> 11010 ?Tab(5);BAz g, DAz g:
;Using(,20),BAz(0);Deg$,DAz(0);Deg$
> 11012 ?Tab(5);BAz e, DAz e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11015 ?Tab(5);APe b, APe d:
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(
> AP(0),TvL(5))/RF;Deg$
> 65010
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*
RF)*Sin
> (Oz))^2)^.5+1)))
> 1= Reduced Oz: cos{ROz{0 }} = cos{Oz}, cos{ROz{90
> }} = 1
> 65020
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)
^2-Sin(
> AP)^2)^(-.5)))
> 11017 GoSub 11990:?Tab(16);Cos{ROz{APg}}
=;Using(,20),Cos(.Roz(APg)):?
> 11900 ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km
(;Using(,
> 25),EGDD; ):Next:?
> 11910 GoSub 11990
> 11990
?Tab(4);*----------------------------------------------------
------
> -----------*:Return
> 11980 ?:List 11020:List 58010-58030:List 10000-10010:End
> 64000 Ō *** Functions ***
> 1 Point 7
> Ō ~=~ .1^33
>
-------------------------------------------------------------
--------------
> -------------------
> This program is presented as a workbench specimen, intended
for
> dissection and analysis.
> For just a no-frills output, the following modification can
be
> made.
> -------------------------------------------------------
> Delete 11003-11900
> 4000 XF=1:UL=10:GoSub 60000
> 11100 GoSub 59100:KmDx(1)=A*EGDD*AD(0)
> 11200 ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
> 11300 ?Tab(5);BAz e, DAz e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11980 ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
> 58010 a=6378.135:b=6356.75
> 58020
> -------------------------------------------------------
> ~Kaimbridge M. GoldChild~
>
[FN42nn] === Subject: The Parageodetic Distance Formula (UBasic) 
Aftersome[NonBreakingSpace]reßection--prompted[
NonBreakingSpace]by[NonBr
eakingSpace]an[NonBreakingSpace]e-mail[NonBreakingSpace]
inquiry--this[NonB
reakingSpace]writer/
programmerhas[NonBreakingSpace]decided[NonBreakingSpace]to[
NonBreakingSpa
ce]first[NonBreakingSpace]release[NonBreakingSpace]the[
NonBreakingSpace]pa
rageodetic[NonBreakingSpace](inverse
problem)formulary[NonBreakingSpace]core,[NonBreakingSpace]
before[NonBreak
ingSpace]the[NonBreakingSpace]more[NonBreakingSpace]dressed,[
NonBreakingSp
ace]integrated
quadratic-mean-spherical,parageodetic[NonBreakingSpace]and[
NonBreakingSpac
e]geodetic[NonBreakingSpace]inverse[NonBreakingSpace]and
directGODx[NonBreakingSpace]program[NonBreakingSpace]core.
But,rather[NonBreakingSpace]than[NonBreakingSpace]just[
NonBreakingSpace]s
pit[NonBreakingSpace]out[NonBreakingSpace]the[
NonBreakingSpace]program[No
nBreakingSpace]listing[NonBreakingSpace]in[NonBreakingSpace]
numerical
addressorder,[NonBreakingSpace]however,[NonBreakingSpace]it[
NonBreakingSp
ace]is[NonBreakingSpace]being[NonBreakingSpace]presented[
NonBreakingSpace]
subroutine-with-execution,
therebyproviding[NonBreakingSpace]another[NonBreakingSpace]
opportunity[No
nBreakingSpace]to[NonBreakingSpace]disseminate[
NonBreakingSpace]this[NonBr
eakingSpace]unique
formularyapproach[NonBreakingSpace](including[
NonBreakingSpace]all[NonBre
akingSpace]trigonometric[NonBreakingSpace]quadrantic[
NonBreakingSpace]adjus
tments
anddivision[NonBreakingSpace]by[NonBreakingSpace]zero[
NonBreakingSpac
e]bulletproofing)[NonBreakingSpace]for[NonBreakingSpace]
mathematical[NonBr
eakingSpace]study[NonBreakingSpace]and
discussion.
Asthe[NonBreakingSpace]concept[NonBreakingSpace]of[
NonBreakingSpace]the[
NonBreakingSpace]parageodesic[NonBreakingSpace]appears[
NonBreakingSpace]lit
tle[NonBreakingSpace]acknowledged--if[NonBreakingSpace]at
alleven[NonBreakingSpace]recognized--it[NonBreakingSpace]
would[NonBreakin
gSpace]seem[NonBreakingSpace]prudent[NonBreakingSpace]to[
NonBreakingSpace]
comparatively[NonBreakingSpace]define[NonBreakingSpace]the
geodesicand[NonBreakingSpace]parageodesic[NonBreakingSpace]
from[NonBreaki
ngSpace]the[NonBreakingSpace]conceptual[NonBreakingSpace]
approach[NonBreak
ingSpace]used[NonBreakingSpace]here[NonBreakingSpace](to
thenon-technical[NonBreakingSpace]layman,[NonBreakingSpace]
the[NonBreakin
gSpace]term[NonBreakingSpace]graticular[NonBreakingSpace]can[
NonBreaki
ngSpace]be[NonBreakingSpace]considered
synonymouswith[NonBreakingSpace]spherical):
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace]Geodesic:[
NonBreakingSpace][NonBreaki
ngSpace]The[NonBreakingSpace]conformally[NonBreakingSpace]
delineated[NonBr
eakingSpace]great-arc[NonBreakingSpace]segment
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace]between[NonBreakingSpace]two[
NonBreakingSpace]points;
[NonBreakingSpace][NonBreakingSpace]Parageodesic:[
NonBreakingSpace][NonB
reakingSpace]The[NonBreakingSpace]graticularly[
NonBreakingSpace]delineated
[NonBreakingSpace]great-arc[NonBreakingSpace]segment
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace]between[NonBreakingSpace]two[
NonBreakingSpace]points.
Thus,with[NonBreakingSpace]an[NonBreakingSpace]ellipsoid,[
NonBreakingSpac
e]the[NonBreakingSpace]traditional[NonBreakingSpace]geodesic[
NonBreakingSp
ace]is[NonBreakingSpace]the[NonBreakingSpace]pulled
stringpath[NonBreakingSpace]and[NonBreakingSpace]distance[
NonBreakingSp
ace]on[NonBreakingSpace]an[NonBreakingSpace]elliptically[
NonBreakingSpace]
correct[NonBreakingSpace]model,[NonBreakingSpace]whereas
theparageodesic[NonBreakingSpace]is[NonBreakingSpace]the[
NonBreakingSpace
]pulled[NonBreakingSpace]string[NonBreakingSpace]path[
NonBreakingSpace
]and[NonBreakingSpace]distance,[NonBreakingSpace]first
DEFINEDon[NonBreakingSpace]a[NonBreakingSpace]spherical[
NonBreakingSpace]
model[NonBreakingSpace]of[NonBreakingSpace]the[
NonBreakingSpace]ellipsoid,
[NonBreakingSpace]then[NonBreakingSpace]the[NonBreakingSpace]
sphere[NonBr
eakingSpace]is
reformedto[NonBreakingSpace]its[NonBreakingSpace]proper[
NonBreakingSpace]
ellipsoidal[NonBreakingSpace]shape,[NonBreakingSpace]at[
NonBreakingSpace]w
hich[NonBreakingSpace]time[NonBreakingSpace]the
ellipticallycorrect[NonBreakingSpace]arcradius[
NonBreakingSpace]and[NonBr
eakingSpace]angle/azimuth[NonBreakingSpace]AT[
NonBreakingSpace]ANY[NonBrea
kingSpace]GIVEN[NonBreakingSpace]POINT
ALONGTHE[NonBreakingSpace]DEFINED[NonBreakingSpace]SEGMENT[
NonBreakingSpa
ce]can[NonBreakingSpace]be[NonBreakingSpace]found[
NonBreakingSpace](i.e.,
[NonBreakingSpace]it[NonBreakingSpace]is[NonBreakingSpace]the
[NonBreaking
Space]local,
ellipticallycorrect[NonBreakingSpace]Great-Circle[
NonBreakingSpace]val
uation!).
UBasicÕssystem[NonBreakingSpace]program[NonBreakingSpace]can[

NonBreakingS
pace]be[NonBreakingSpace]found[NonBreakingSpace]here:
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace]ftp://rkmath.rikkyo.ac.jp
/pub/ubibm
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace](
this[NonBreakingSpace]web[NonBreakingSpace]page[
NonBreakingSpace]is[NonBr
eakingSpace]translated[NonBreakingSpace]through[
NonBreakingSpace]Google,[N
onBreakingSpace]thus[NonBreakingSpace]some[NonBreakingSpace]
of[NonBreaking
Space]the[NonBreakingSpace]wording[NonBreakingSpace]is[
NonBreakingSpace]of
f)
UBasicÕscommand[NonBreakingSpace]dictionary[NonBreakingSpace]

can[NonBreak
ingSpace]be[NonBreakingSpace]found[NonBreakingSpace]here:
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace]http://ed-thelen.org/bab/bab-ubhelp.html
Ofcourse,[NonBreakingSpace]while[NonBreakingSpace]it[
NonBreakingSpace]is
[NonBreakingSpace]presented[NonBreakingSpace]here[
NonBreakingSpace]out[No
nBreakingSpace]of[NonBreakingSpace]numerical[NonBreakingSpace
]address[NonB
reakingSpace]order,
oncethe[NonBreakingSpace]program[NonBreakingSpace]code[
NonBreakingSpace]i
s[NonBreakingSpace]cut[NonBreakingSpace]and[NonBreakingSpace]
pasted[NonBr
eakingSpace]into[NonBreakingSpace]UBasic,[NonBreakingSpace]it
[NonBreakingS
pace]will[NonBreakingSpace]revert[NonBreakingSpace]to
itsnatural[NonBreakingSpace]arrangement.
-------------------------------------------------------------
---------------
------------------
3000[NonBreakingSpace][NonBreakingSpace]GoSub[
NonBreakingSpace]58000
58000[NonBreakingSpace]Õ***[NonBreakingSpace]Radial[
NonBreakingSpace]Para
meters[NonBreakingSpace]***
58010[NonBreakingSpace]a=6378.135:b=6356.75:b=b::GoTo[
NonBreakingSpace]582
00
58020[NonBreakingSpace]a=10000:b=5000
58200[NonBreakingSpace]Oz=2*Atan((a-b)^.5/(a+b)^.5)
10000[NonBreakingSpace]BLat=+43.66111111111111111111:BLong=[
NonBreakingSpa
ce]-70.25583333333333333333
10010[NonBreakingSpace]DLat=+45.52305555555555555556:DLong=-
122.67583333333
333333333
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace]Parageodetic[
NonBreakingSpace]=[Non
BreakingSpace]4092.6024524649626[NonBreakingSpace]Km[
NonBreakingSpace][Non
BreakingSpace][NonBreakingSpace]291.6521536/73.6370729
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace]Geodetic[
NonBreakingSpace]=[NonBreak
ingSpace]4092.6016550196721[NonBreakingSpace]Km[
NonBreakingSpace][NonBreak
ingSpace][NonBreakingSpace]291.7142510/73.5743344
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace]Q-M-Spherical[NonBreakingSpace]=[
NonBreakingSpace]4082
.4708695785296[NonBreakingSpace]Km[NonBreakingSpace][
NonBreakingSpace][No
nBreakingSpace]291.7214263/73.5858490
2200[NonBreakingSpace][NonBreakingSpace]Pi=#Pi:PH=.5*Pi:PD=2*
Pi
2300[NonBreakingSpace][NonBreakingSpace]RF=Pi/180
55110[NonBreakingSpace]BLr=BLat*RF:DLr=DLat*RF:MD=(
DLong-BLong)*RF:M=Abs(MD
):If[NonBreakingSpace]M>Pi[NonBreakingSpace]Then[
NonBreakingSpace]MD=MD-Sg
n(MD)*PD
11000[NonBreakingSpace]GoSub[NonBreakingSpace]55000
1Õ###########################################################
#
55000[NonBreakingSpace]Õ***[NonBreakingSpace]Spherical,[
NonBreakingSpace]
ParaGeodetic[NonBreakingSpace]Valuations[NonBreakingSpace]***
55100[NonBreakingSpace]ID%=0
55120[NonBreakingSpace]WBLr=DLr:WDLr=BLr:GoSub[
NonBreakingSpace]55400:DAz(
0)=WAz(0):TvL(5)=TvL(0)
55130[NonBreakingSpace]WBLr=BLr:WDLr=DLr:GoSub[
NonBreakingSpace]55400:BAz(
0)=WAz(0):TvL(1)=TvL(0)
1Õ##########
55400[NonBreakingSpace]Õ***[NonBreakingSpace]Spherical[
NonBreakingSpace]E
lements[NonBreakingSpace]***
55410[NonBreakingSpace]GoSub[NonBreakingSpace]55420:GoSub[
NonBreakingSpac
e]55440:Return
55420[NonBreakingSpace]SA=Cos(WDLr)*Sin(M):If[
NonBreakingSpace]SA=0[NonBr
eakingSpace]And[NonBreakingSpace]Abs(WDLr)0[NonBreakingSpace]Then[NonBreakingSpace]SA=
#Eps
55422[NonBreakingSpace]If[NonBreakingSpace]Abs(WBLr/RF)>=90[
NonBreakingSp
ace]Or[NonBreakingSpace]Abs(WDLr/RF)>=90[NonBreakingSpace]
Then[NonBreaking
Space]SA=0
55425[NonBreakingSpace]SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(
WBLr-WDLr)*Cos(
.5*M)^2)
2150[NonBreakingSpace][NonBreakingSpace]IO=10^100
55430[NonBreakingSpace]WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
55440[NonBreakingSpace]TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(
WBLr)*Cos(WAz(I
D%)))+1))
55445[NonBreakingSpace]If[NonBreakingSpace]SB<0[
NonBreakingSpace]Then[No
nBreakingSpace]WAz(ID%)=Pi-WAz(ID%)
55439[NonBreakingSpace]Return
55449[NonBreakingSpace]Return
1Õ----------
55125[NonBreakingSpace]GoSub[NonBreakingSpace]55190:DAz(1)=
WAz(1)
55135[NonBreakingSpace]GoSub[NonBreakingSpace]55190:BAz(1)=
WAz(1)
1Õ##########
55190[NonBreakingSpace]ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,
WBLr):GoSub[N
onBreakingSpace]55430:GoSub[NonBreakingSpace]55445:ID%=0:
Return
1Õ----------
2100[NonBreakingSpace][NonBreakingSpace]Dcm=Int(Log(#Eps)/Log
(0.1)-5):Syot
a=0.1^Dcm
55140[NonBreakingSpace]If[NonBreakingSpace]Cos(BAz(0))<0[
NonBreakingSpace
]Then[NonBreakingSpace]TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
55142[NonBreakingSpace]If[NonBreakingSpace]Cos(DAz(0))>0[
NonBreakingSpace
]Then[NonBreakingSpace]TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
55145[NonBreakingSpace]If[NonBreakingSpace]TvL(5)0[
NonBreakingSpace]Then[No
nBreakingSpace]DAz(ID%)=PD-DAz(ID%)
55489[NonBreakingSpace]BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)
/RF:Return
1Õ----------
55189[NonBreakingSpace]Return
1Õ-----------------------------------------------------------
-
11100[NonBreakingSpace]GoSub[NonBreakingSpace]59100:KmDx(1)=a
*EGDD*AD(0)
1Õ###########################################################
#
59100[NonBreakingSpace]Õ***[NonBreakingSpace]PEOx[
NonBreakingSpace]***
59000[NonBreakingSpace]Õ***[NonBreakingSpace]Integration[
NonBreakingSpace
]***
65100[NonBreakingSpace].E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP
)*Sin(Oz))^2+(
Cos(AP)*Cos(TvL)*Sin(Oz))^2)^.5)
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]=
[NonBreakingSpace]Elliptic[NonBreakingSpace]Integrand[
NonBreakingSpace]Of
[NonBreakingSpace]The[NonBreakingSpace]Second[
NonBreakingSpace]Kind
65110[NonBreakingSpace].E1p(AP,TvL):Return(1/.E2p(AP,TvL))
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]=
[NonBreakingSpace]Elliptic[NonBreakingSpace]Integrand[
NonBreakingSpace]Of
[NonBreakingSpace]The[NonBreakingSpace]First[NonBreakingSpace
]Kind
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]N[
NonBreakingSpa
ce](Normal)[NonBreakingSpace]=[NonBreakingSpace]Transverse[
NonBreaking
Space]Equatorial[NonBreakingSpace]Arcradius
65120[NonBreakingSpace].E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,
TvL)^3)
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]M[
NonBreakingSpa
ce](Meridian)[NonBreakingSpace]=[NonBreakingSpace]Vertical[
NonBreaking
Space]Meridional[NonBreakingSpace]Arcradius
65130[NonBreakingSpace].EGp(AP,TvL):Return((.E0p(AP,TvL)^2+
Sin(AP)^2*(IO*(.
E1p(AP,TvL)^2-.E0p(AP,TvL)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(
TvL))^2)+1)))^.5)
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
x[NonBreakingSpace]a[NonBreakingSpace]=[NonBreakingSpace]O[
NonBreakingSpa
ce](Omniversal)[NonBreakingSpace]=[NonBreakingSpace]
Transverse[NonBrea
kingSpace]Meridional[NonBreakingSpace]Arcradius
1Õ#########################
11020[NonBreakingSpace]XF=1:::For[NonBreakingSpace]UL=2[
NonBreakingSpace]
To[NonBreakingSpace]20[NonBreakingSpace]Step[NonBreakingSpace
]2::GoSub[No
nBreakingSpace]60000
1ÕIntegration[NonBreakingSpace]Convergency[NonBreakingSpace]
for[NonBreaki
ngSpace]EGDD[NonBreakingSpace]to[NonBreakingSpace].1^20[
NonBreakingSpace]@
[NonBreakingSpace]cos{Oz},[NonBreakingSpace]for[
NonBreakingSpace]XFxUL:[N
onBreakingSpace][NonBreakingSpace]cos{Oz}[NonBreakingSpace]=[
NonBreakingSp
ace]1xUL/2xUL/3xUL
1Õ[NonBreakingSpace][NonBreakingSpace].9999[NonBreakingSpace]

=[NonBreaki
ngSpace]6/8/6;[NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace][NonB
reakingSpace].995[NonBreakingSpace]=[NonBreakingSpace]8/12/15
;[NonBreaking
Space][NonBreakingSpace][NonBreakingSpace].99[
NonBreakingSpace]=[NonBreak
ingSpace]8/12/15;[NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace].9
[NonBreakingSpace]=[NonBreakingSpace]11/16/18;
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpac
e].75[NonBreakingSpace]=[NonBreakingSpace]13/18/24;[
NonBreakingSpace][Non
BreakingSpace][NonBreakingSpace].5[NonBreakingSpace]=[
NonBreakingSpace]17/
24/30;[NonBreakingSpace][NonBreakingSpace].25[
NonBreakingSpace]=[NonBreak
ingSpace]27/36/45;[NonBreakingSpace][NonBreakingSpace].1[
NonBreakingSpace]
=[NonBreakingSpace]42/60/75;
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpac
e].01[NonBreakingSpace]=[NonBreakingSpace]--/--/231;[
NonBreakingSpace][No
nBreakingSpace][NonBreakingSpace](.001[NonBreakingSpace]~=~[
NonBreakingSpa
ce]20x94[NonBreakingSpace]=[NonBreakingSpace]1880;[
NonBreakingSpace][NonB
reakingSpace][NonBreakingSpace].0001[NonBreakingSpace]~=~[
NonBreakingSpace
]200x97[NonBreakingSpace]=[NonBreakingSpace]19400)
60000[NonBreakingSpace]Õ***[NonBreakingSpace](Gaussian)[
NonBreakingSpace]
Amplitudal[NonBreakingSpace]Kernal[NonBreakingSpace]Expansion
[NonBreakingS
pace]***
65000[NonBreakingSpace].AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5
*AEg(NS))^2)/(
NX-Sin(.5*AEg(NS))^2))^.5):If[NonBreakingSpace]AE>.5*Pi[
NonBreakingSpace]Th
en[NonBreakingSpace]AE=Pi-AE
65009[NonBreakingSpace]Return(AE)
60100[NonBreakingSpace]DU=2*UL:For[NonBreakingSpace]NS=1[
NonBreakingSpace
]To[NonBreakingSpace]UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
60200[NonBreakingSpace]LP(1)=1:LP=1:For[NonBreakingSpace]NP=1
[NonBreaking
Space]To[NonBreakingSpace]DU
60210[NonBreakingSpace]LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*
LP(1)*CK)-((NP-
1)*LP(2)))/NP:Next[NonBreakingSpace]NP
60220[NonBreakingSpace]DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
60230[NonBreakingSpace]CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
60235[NonBreakingSpace]If[NonBreakingSpace]Abs(CK-CK(1))>(10*
Syota)[NonBr
eakingSpace]And[NonBreakingSpace]CK<>CK(2)[NonBreakingSpace]
Then[NonBreaki
ngSpace]60200
60300[NonBreakingSpace]AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
60310[NonBreakingSpace]OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):
Next[NonBreaking
Space]NS:Return:::::?NS;[NonBreakingSpace];Using(,30),AEg(NS)
/RF;[Non
BreakingSpace][NonBreakingSpace][NonBreakingSpace];OE(NS)/RF:
Next[NonBre
akingSpace]NS:List[NonBreakingSpace]60010:End:Return
1Õ-------------------------
59110[NonBreakingSpace]EGDD=0:For[NonBreakingSpace]NS=1[
NonBreakingSpace]
To[NonBreakingSpace]UL:For[NonBreakingSpace]NX=1[
NonBreakingSpace]To[NonB
reakingSpace]XF
59120[NonBreakingSpace]AE=AEg(NS):If[NonBreakingSpace]XF>1[
NonBreakingSpa
ce]Then[NonBreakingSpace]AE=.AE(NS,NX,XF)
59130[NonBreakingSpace]TW=XF*UL:ABq=Cos(AE)*ADq_0:OW=OE(NS)/TW
59140[NonBreakingSpace]UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(
2)+ABq:LLP=TvL
(2)-ABq
59150[NonBreakingSpace]EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(
AP(0),UMP)+.EGp
(AP(0),LMP)+.EGp(AP(0),LLP))*OW
59160[NonBreakingSpace]Next[NonBreakingSpace]NX:Next[
NonBreakingSpace]NS:
Return
1Õ-----------------------------------------------------------
-
[NonBreakingSpace][NonBreakingSpace]10[NonBreakingSpace][
NonBreakingSpac
e]ClS
1000[NonBreakingSpace][NonBreakingSpace]Dim[NonBreakingSpace]
AEg(101),OE(
101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1),BAz(1),DAz
(1)
2000[NonBreakingSpace][NonBreakingSpace]Õ***[NonBreakingSpace

]Program[No
nBreakingSpace]Constants/Standard[NonBreakingSpace]Values[
NonBreakingSpace]
***
2400[NonBreakingSpace][NonBreakingSpace]Deg$=Chr(248):Õ(
Degree[NonBreakin
gSpace]Sign)
58300[NonBreakingSpace]?Tab(20);*****************************
*******:?T
ab(19);/[NonBreakingSpace]Parageodetic[NonBreakingSpace]
Distance[NonBrea
kingSpace]Formula[NonBreakingSpace]Core[NonBreakingSpace]:?[
NonBreak
ingSpace]*----------------*[NonBreakingSpace]([
NonBreakingSpace]cos{Oz}[No
nBreakingSpace]=;Using(,20),Cos(Oz);[NonBreakingSpace])[
NonBreakingSpac
e]*-------------------*
58310[NonBreakingSpace]?Using(,13),[NonBreakingSpace]|[
NonBreakingSpace
][NonBreakingSpace]a[NonBreakingSpace]=;a;[NonBreakingSpace]
Km;;:?Ta
b(55-Len(Str(Int(b))));Using(,13),b[NonBreakingSpace]=;b;[
NonBreaking
Space]Km;[NonBreakingSpace]|:?[NonBreakingSpace]*------------
----------
-----------------------------------------------------*:Return
11003[NonBreakingSpace]?Tab(7);BTvL,[NonBreakingSpace]DTvL:[
NonBreaking
Space];Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
11005[NonBreakingSpace]?Tab(25);ADg:[NonBreakingSpace];Using(
,20),ADg;
Deg$
11007[NonBreakingSpace]?Tab(25);APg:[NonBreakingSpace];Using(
,20),APg;
Deg$
11010[NonBreakingSpace]?Tab(5);BAz_g,[NonBreakingSpace]DAz_g:
[NonBreaki
ngSpace];Using(,20),BAz(0);Deg$,DAz(0);Deg$
11012[NonBreakingSpace]?Tab(5);BAz_e,[NonBreakingSpace]DAz_e:
[NonBreaki
ngSpace];Using(,20),BAz(1);Deg$,DAz(1);Deg$
11015[NonBreakingSpace]?Tab(5);APe_b,[NonBreakingSpace]APe_d:
[NonBreaki
ngSpace];Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL
(5))/RF;Deg$
65010[NonBreakingSpace].ROz(APg):Return(Atan(IO*Cos(APg*RF)*
Sin(Oz)/(IO*(1-
(Cos(APg*RF)*Sin(Oz))^2)^.5+1)))
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
1Õ[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace
][NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
[NonBreakingSpace]=[NonBreakingSpace]Reduced[NonBreakingSpace
]Oz:[Non
BreakingSpace][NonBreakingSpace]cos{ROz{0}}[NonBreakingSpace]
=[No
nBreakingSpace]cos{Oz},[NonBreakingSpace]cos{ROz{90}}[
NonBreakingS
pace]=[NonBreakingSpace]1[NonBreakingSpace]
65020[NonBreakingSpace].APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p
(AP,TvL)^4*.EG
p(AP,TvL)^2-Sin(AP)^2)^(-.5)))
11017[NonBreakingSpace]GoSub[NonBreakingSpace]11990:?Tab(16);
Cos{ROz{APg
}}[NonBreakingSpace]=;Using(,20),Cos(.Roz(APg)):?
11900[NonBreakingSpace]?TW[NonBreakingSpace]=;UL*XF;Tab(10);
PEODx:[
NonBreakingSpace];Using(,20),KmDx(1);[NonBreakingSpace]Km[
NonBreakingSp
ace](;Using(,25),EGDD;[NonBreakingSpace]):Next:?
11910[NonBreakingSpace]GoSub[NonBreakingSpace]11990
11990[NonBreakingSpace]?Tab(4);*-----------------------------
------------
----------------------------*:Return
11980[NonBreakingSpace]?:List[NonBreakingSpace]11020:List[
NonBreakingSpac
e]58010-58030:List[NonBreakingSpace]10000-10010:End
64000[NonBreakingSpace]Õ[NonBreakingSpace]***[
NonBreakingSpace][NonBreak
ingSpace]Functions[NonBreakingSpace][NonBreakingSpace]***
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace]1[
NonBreakingSpace
][NonBreakingSpace]Point[NonBreakingSpace]7
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
Ō[NonBreakingSpace]~=~[NonBreakingSpace].1^33
-------------------------------------------------------------
---------------
------------------
Thisprogram[NonBreakingSpace]is[NonBreakingSpace]presented[
NonBreakingSpa
ce]as[NonBreakingSpace]a[NonBreakingSpace]workbench[
NonBreakingSpace]s
pecimen,[NonBreakingSpace]intended[NonBreakingSpace]for
dissectionand[NonBreakingSpace]analysis.
Forjust[NonBreakingSpace]a[NonBreakingSpace]no-frills[
NonBreakingSpac
e]output,[NonBreakingSpace]the[NonBreakingSpace]following[
NonBreakingSpace
]modification[NonBreakingSpace]can[NonBreakingSpace]be
made.
-------------------------------------------------------
Delete11003-11900
4000[NonBreakingSpace][NonBreakingSpace]XF=1:UL=10:GoSub[
NonBreakingSpace
]60000
11100[NonBreakingSpace]GoSub[NonBreakingSpace]59100:KmDx(1)=A
*EGDD*AD(0)
11200[NonBreakingSpace]?:?Tab(12);PEODx:[NonBreakingSpace];
Using(,20),
KmDx(1);[NonBreakingSpace]Km:?
11300[NonBreakingSpace]?Tab(5);BAz_e,[NonBreakingSpace]DAz_e:
[NonBreaki
ngSpace];Using(,20),BAz(1);Deg$,DAz(1);Deg$
11980[NonBreakingSpace]?:?:List[NonBreakingSpace]58010:?:?:?:
List[NonBrea
kingSpace]10000:?:?:List[NonBreakingSpace]10010:End
58010[NonBreakingSpace]a=6378.135:b=6356.75
58020
-------------------------------------------------------
[NonBreakingSpace][NonBreakingSpace][NonBreakingSpace][
NonBreakingSpace]
~Kaimbridge[NonBreakingSpace]M.[NonBreakingSpace]GoldChild~
=== Subject: Re: The Parageodetic Distance Formula (UBasic) Whoa!! Do you like to hear yourself or what? > 
 After some reßection--prompted by an e-mail
inquiry--this writer/
> programmer has decided to first release the parageodetic
(inverse
> problem) formulary core, before the more dressed, integrated
> quadratic-mean-spherical, parageodetic and geodetic inverse
and
> direct GODx program core.
> But, rather than just spit out the program listing in
numerical
> address order, however, it is being presented
subroutine-with-execution,
> thereby providing another opportunity to disseminate this
unique
> formulary approach (including all trigonometric quadrantic
adjustments
> and division by zero bulletproofing) for mathematical study
and
> discussion.
> As the concept of the parageodesic appears little
acknowledged--if at
> all even recognized--it would seem prudent to comparatively
define the
> geodesic and parageodesic from the conceptual approach used
here (to
> the non-technical layman, the term graticular can be
considered
> synonymous with spherical):
> Geodesic: The conformally delineated great-arc segment
> between two points;
> Parageodesic: The graticularly delineated great-arc segment
> between two points.
> Thus, with an ellipsoid, the traditional geodesic is the
pulled
> string path and distance on an elliptically correct model,
whereas
> the parageodesic is the pulled string path and distance,
first
> DEFINED on a spherical model of the ellipsoid, then the
sphere is
> reformed to its proper ellipsoidal shape, at which time the
> elliptically correct arcradius and angle/azimuth AT ANY
GIVEN POINT
> ALONG THE DEFINED SEGMENT can be found (i.e., it is the
local,
> elliptically correct Great-Circle valuation!).
> UBasicÕs system program can be found here:
> ftp://rkmath.rikkyo.ac.jp/pub/ubibm
jp/~kida/ubasic.htm
> (this web page is translated through Google, thus some of
the wording is
off)
> UBasicÕs command dictionary can be found here:
> http://ed-thelen.org/bab/bab-ubhelp.html
> Of course, while it is presented here out of numerical
address order,
> once the program code is cut and pasted into UBasic, it
will revert to
> its natural arrangement.
>
-------------------------------------------------------------
-------------
--------------------
> 3000 GoSub 58000
> 58000 Ō*** Radial Parameters ***
> 58010 a=6378.135:b=6356.75:b=b::GoTo 58200
> 58020 a=10000:b=5000
> 58200 Oz=2*Atan((a-b)^.5/(a+b)^.5)
> 10000 BLat=+43.66111111111111111111:BLong=
-70.25583333333333333333
> 10010
DLat=+45.52305555555555555556:DLong=-122.67583333333333333333
> 1Parageodetic = 4092.6024524649626 Km
291.6521536/73.6370729
> 1Geodetic = 4092.6016550196721 Km
291.7142510/73.5743344
> 1Q-M-Spherical = 4082.4708695785296 Km
291.7214263/73.5858490
> 2200 Pi=#Pi:PH=.5*Pi:PD=2*Pi
> 2300 RF=Pi/180
> 55110
BLr=BLat*RF:DLr=DLat*RF:MD=(DLong-BLong)*RF:M=Abs(MD):If M>Pi
Then
MD=MD-Sgn(MD)*PD
> 11000 GoSub 55000
>
1Õ###########################################################
#
> 55000 Ō*** Spherical, ParaGeodetic Valuations ***
> 55100 ID%=0
> 55120 WBLr=DLr:WDLr=BLr:GoSub
55400:DAz(0)=WAz(0):TvL(5)=TvL(0)
> 55130 WBLr=BLr:WDLr=DLr:GoSub
55400:BAz(0)=WAz(0):TvL(1)=TvL(0)
> 1Õ##########
> 55400 Ō*** Spherical Elements ***
> 55410 GoSub 55420:GoSub 55440:Return
> 55420 SA=Cos(WDLr)*Sin(M):If SA=0 And Abs(WDLr)0 Then
SA=#Eps
> 55422 If Abs(WBLr/RF)>=90 Or Abs(WDLr/RF)>=90 Then SA=0
> 55425
SB=(Sin(WBLr+WDLr)*Sin(.5*M)^2)-(Sin(WBLr-WDLr)*Cos(.5*M)^2)
> 2150 IO=10^100
> 55430 WAz(ID%)=Abs(Atan(IO*SA/(IO*SB+1)))
> 55440
TvL(0)=Atan(IO*Sin(WBLr)/(Abs(IO*Cos(WBLr)*Cos(WAz(ID%)))+1))
> 55445 If SB<0 Then WAz(ID%)=Pi-WAz(ID%)
> 55439 Return
> 55449 Return
> 1Õ----------
> 55125 GoSub 55190:DAz(1)=WAz(1)
> 55135 GoSub 55190:BAz(1)=WAz(1)
> 1Õ##########
> 55190 ID%=1:SA=SA*.E1p(0,WBLr):SB=SB*.E0p(0,WBLr):GoSub
55430:GoSub
55445:ID%=0:Return
> 1Õ----------
> 2100 Dcm=Int(Log(#Eps)/Log(0.1)-5):Syota=0.1^Dcm
> 55140 If Cos(BAz(0))<0 Then
TvL(1)=Sgn(IO*BLr+Syota)*Pi-TvL(1)
> 55142 If Cos(DAz(0))>0 Then
TvL(5)=Sgn(IO*DLr+Syota)*Pi-TvL(5)
> 55145 If TvL(5) 55150
AD(0)=TvL(5)-TvL(1):ADq_0=0.25*AD(0):BTvL=TvL(1)/RF:DTvL=TvL(
5)/RF:ADg=DTvL-
BTvL
> 55155
TvL(2)=TvL(1)+ADq_0:TvL(3)=TvL(1)+2*ADq_0:TvL(4)=TvL(1)+3*ADq
_0
> 55147 GoSub 55450:APg=AP(0)/RF:GoSub 55485:ID%=1:GoSub 55485
> 1Õ##########
> 55450
AP(ID%)=Atan(Abs(IO*Cos(WBLr)*Sin(WAz(ID%)))/(IO*(Cos(WAz(ID%
))^2+(Sin(WBLr)
*Sin(WAz(ID%)))^2)^.5+1)):Return
> 55485 If MD<0 Then BAz(ID%)=PD-BAz(ID%)
> 55487 If MD>0 Then DAz(ID%)=PD-DAz(ID%)
> 55489 BAz(ID%)=BAz(ID%)/RF:DAz(ID%)=DAz(ID%)/RF:Return
> 1Õ----------
> 55189 Return
>
1Õ-----------------------------------------------------------
-
> 11100 GoSub 59100:KmDx(1)=a*EGDD*AD(0)
>
1Õ###########################################################
#
> 59100 Ō*** PEOx ***
> 59000 Ō*** Integration ***
> 65100
.E2p(AP,TvL):Return((Cos(Oz)^2+(Sin(AP)*Sin(Oz))^2+(Cos(AP)*
Cos(TvL)*Sin(Oz)
)^2)^.5)
> 1= Elliptic Integrand Of The Second Kind
> 65110 .E1p(AP,TvL):Return(1/.E2p(AP,TvL))
> 1= Elliptic Integrand Of The First Kind
> 1x a = N (Normal) = Transverse Equatorial Arcradius
> 65120 .E0p(AP,TvL):Return(Cos(Oz)^2*.E1p(AP,TvL)^3)
> 1x a = M (Meridian) = Vertical Meridional Arcradius
> 65130
.EGp(AP,TvL):Return((.E0p(AP,TvL)^2+Sin(AP)^2*(IO*(.E1p(AP,
TvL)^2-.E0p(AP,Tv
L)^2)/(IO*(Cos(TvL)^2+(Sin(AP)*Sin(TvL))^2)+1)))^.5)
> 1x a = O (Omniversal) = Transverse Meridional Arcradius
> 1Õ#########################
> 11020 XF=1:::For UL=2 To 20 Step 2::GoSub 60000
> 1Integration Convergency for EGDD to .1^20 @ cos{Oz}, 
for
XFxUL:
cos{Oz}
= 1xUL/2xUL/3xUL
> 1.9999 = 6/8/6; .995 = 8/12/15; .99 = 8/12/15; .9 =
11/16/18;
> 1.75 = 13/18/24; .5 = 17/24/30; .25 = 27/36/45; .1 =
42/60/75;
> 1.01 = --/--/231; (.001 ~=~ 20x94 = 1880; .0001 ~=~
200x97 = 19400)
> 60000 Ō*** (Gaussian) Amplitudal Kernal Expansion ***
> 65000
.AE(NS,NX,XF):AE=2*Atan(((XF-NX+Sin(.5*AEg(NS))^2)/(NX-Sin(.5
*AEg(NS))^2))^.
5):If AE>.5*Pi Then AE=Pi-AE
> 65009 Return(AE)
> 60100 DU=2*UL:For NS=1 To
UL:CK=Cos(.5*Pi*(((4*NS)-1)/((4*UL)+1)))
> 60200 LP(1)=1:LP=1:For NP=1 To DU
> 60210
LP(2)=LP(1):LP(1)=LP:LP=((((2*NP)-1)*LP(1)*CK)-((NP-1)*LP(2))
)/NP:Next NP
> 60220 DP=DU*(LP(1)-(LP*CK))/(1-CK^2)
> 60230 CK(2)=CK(1):CK(1)=CK:CK=CK(1)-(LP/DP)
> 60235 If Abs(CK-CK(1))>(10*Syota) And CK<>CK(2) Then 60200
> 60300 AEg(NS)=2*Atan((1-CK)^.5/(1+CK)^.5)
> 60310 OE(NS)=Sin(AEg(NS))^2/(DU*LP(1)^2):Next
NS:Return:::::?NS;
;Using(,30),AEg(NS)/RF; ;OE(NS)/RF:Next NS:List
60010:End:Return
> 1Õ-------------------------
> 59110 EGDD=0:For NS=1 To UL:For NX=1 To XF
> 59120 AE=AEg(NS):If XF>1 Then AE=.AE(NS,NX,XF)
> 59130 TW=XF*UL:ABq=Cos(AE)*ADq_0:OW=OE(NS)/TW
> 59140
UUP=TvL(4)+ABq:UMP=TvL(4)-ABq:LMP=TvL(2)+ABq:LLP=TvL(2)-ABq
> 59150
EGDD=EGDD+0.25*(.EGp(AP(0),UUP)+.EGp(AP(0),UMP)+.EGp(AP(0),
LMP)+.EGp(AP(0),L
LP))*OW
> 59160 Next NX:Next NS:Return
>
1Õ-----------------------------------------------------------
-
> 10 ClS
> 1000 Dim
AEg(101),OE(101),LP(2),CK(2),AD(1),AP(1),TvL(5),KmDx(1),WAz(1
),BAz(1),DAz(1)
> 2000 Ō*** Program Constants/Standard Values ***
> 2400 Deg$=Chr(248):Õ(Degree Sign)
> 58300
?Tab(20);************************************:?Tab(19);/
Parageodetic Distance Formula Core :? *----------------* (
cos{Oz}
=;Using(,20),Cos(Oz); ) *-------------------*
> 58310 ?Using(,13), | a =;a;
Km;;:?Tab(55-Len(Str(Int(b))));Using(,13),b =;b; Km; |:?
*------------------------------------------------------------
---------------
*:Return
> 11003 ?Tab(7);BTvL, DTvL:
;Using(,20),TvL(1)/RF;Deg$,TvL(5)/RF;Deg$
> 11005 ?Tab(25);ADg: ;Using(,20),ADg;Deg$
> 11007 ?Tab(25);APg: ;Using(,20),APg;Deg$
> 11010 ?Tab(5);BAz_g, DAz_g:
;Using(,20),BAz(0);Deg$,DAz(0);Deg$
> 11012 ?Tab(5);BAz_e, DAz_e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11015 ?Tab(5);APe_b, APe_d:
;Using(,20),.APe(AP(0),TvL(1))/RF;Deg$,.APe(AP(0),TvL(5))/RF;
Deg$
> 65010
.ROz(APg):Return(Atan(IO*Cos(APg*RF)*Sin(Oz)/(IO*(1-(Cos(APg*
RF)*Sin(Oz))^2)
^.5+1)))
> 1= Reduced Oz: cos{ROz{0}} = cos{Oz},
cos{ROz{90}} = 1
> 65020
.APe(AP,TvL):Return(Atan(Sin(AP)*(.E2p(AP,TvL)^4*.EGp(AP,TvL)
^2-Sin(AP)^2)^(
-.5)))
> 11017 GoSub 11990:?Tab(16);Cos{ROz{APg}}
=;Using(,20),Cos(.Roz(APg)):?
> 11900 ?TW =;UL*XF;Tab(10);PEODx: ;Using(,20),KmDx(1); Km
(;Using(,25),EGDD; ):Next:?
> 11910 GoSub 11990
> 11990
?Tab(4);*----------------------------------------------------
-------------
-
---*:Return
> 11980 ?:List 11020:List 58010-58030:List 10000-10010:End
> 64000 Ō *** Functions ***
> 1 Point 7
> Ō ~=~ .1^33
>
-------------------------------------------------------------
-------------
--------------------
> This program is presented as a workbench specimen, intended
for
> dissection and analysis.
> For just a no-frills output, the following modification can
be
> made.
> -------------------------------------------------------
> Delete 11003-11900
> 4000 XF=1:UL=10:GoSub 60000
> 11100 GoSub 59100:KmDx(1)=A*EGDD*AD(0)
> 11200 ?:?Tab(12);PEODx: ;Using(,20),KmDx(1); Km:?
> 11300 ?Tab(5);BAz_e, DAz_e:
;Using(,20),BAz(1);Deg$,DAz(1);Deg$
> 11980 ?:?:List 58010:?:?:?:List 10000:?:?:List 10010:End
> 58010 a=6378.135:b=6356.75
> 58020
> -------------------------------------------------------
> ~Kaimbridge M. GoldChild~
>
=== Subject: Re: The Parageodetic Distance Formula (UBasic) > After some reßection--prompted by an e-mail inquiry--this writer/ > programmer has decided to first release the parageodetic (inverse > problem) formulary core, before the more dressed, integrated > quadratic-mean-spherical, parageodetic and geodetic inverse and > direct GODx program core. As the concept of the parageodesic appears little acknowledged--if > at all even recognized--it would seem prudent to comparatively > define the geodesic and parageodesic from the conceptual approach > used here (to the non-technical layman, the term graticular can > be considered synonymous with spherical): > Geodesic: The conformally delineated great-arc segment > between two points; > Parageodesic: The graticularly delineated great-arc segment > between two points. If the formatting came out garbled (i.e., non-fixed-width or altered characters--such as happened with the original Google posting), or cutting and pasting the UBasic code didnÕt work (e.g., broken lines), try one of these: http://mathforum.org/discuss/sci.math/t/658788 ~Kaimbridge~ ----- WantedKaimbridge (w/mugshot!): http://www.angelfire.com/ma2/digitology/Wanted_KMGC.html ---------- DigitologyThe Grand Theory Of The Universe: http://www.angelfire.com/ma2/digitology/index.html ***** Void Where Permitted; Limit 0 Per Customer. ***** === Subject: Jesus christ WhatÕs going on? There is a work around GoogleÕs awful new usenet interface, at least temporarily. Use Google.fr or Google.ru or probably any other foreign Google domain. They still have the old interface up. Of course the interface will mostly not be in Englsih. === Subject: Re: Jesus christ WhatÕs going on? Vanya infrared: >There is a work around GoogleÕs awful new usenet interface, at least >temporarily. Use Google.fr or Google.ru or probably any other foreign >Google domain. They still have the old interface up. Of course the >interface will mostly not be in Englsih. For searching - which is all I ever use Google Groups for - the following still works: http://www.exit109.com/%7Ejeremy/news/deja.html ItÕs just the Google search with the advertising and graphics stripped off, which speeds things up a bit. Originally designed as an interface to DejaNews, but he kept it working after Google bought the DejaNews database. -- Peter Moylan peter at ee dot newcastle dot edu dot au http://eepjm.newcastle.edu.au (OS/2 and eCS information and software) === Subject: Re: Jesus christ WhatÕs going on? > Vanya infrared: >>There is a work around GoogleÕs awful new usenet interface, at least >>temporarily. Use Google.fr or Google.ru or probably any other foreign >>Google domain. They still have the old interface up. Of course the >>interface will mostly not be in Englsih. > For searching - which is all I ever use Google Groups for - the > following still works: > http://www.exit109.com/%7Ejeremy/news/deja.html I still get there with just: http://www.deja.com -- Theodore (Ted) Heise Bloomington, IN, USA === Subject: Google [was: Jesus christ WhatÕs going on?] >> [...] >> For searching - which is all I ever use Google Groups for - the >> following still works: >> http://www.exit109.com/%7Ejeremy/news/deja.html >I still get there with just: > http://www.deja.com Why are we looking for tricks to get to what worked for us a day ago? This is an utterly irresponsible move by Google. All Usenet authors and readers should strongly object to their messing with the content that was in a sense entrusted to them. The big lesson here: never assume that a corporation (Netscape, Qualcomm, Google, countless others) will forever be a benevolent steward of something that was placed in their care. My exchanges with Google can be summarized as screw you, we are going ahead. Is there an open source parallel of Google? I doubt that, and now it may be too late to create one. For those who love the new interface: I donÕt care how fast are the posts showing up, for posting and reading I use a newsreader. But Google is simply invaluable as an archive, search tool, etc. All that depends on the interface. The new interface doesnÕt let me see the postings verbatim. It cuts out things I want to see, and adds stuff to them that I donÕt care to see. It presents them in proportional font, which ruins the scores of posts which contain ASCII art or carefully aligned math formulas (Usenet has always been and should be ASCII only for maximum compatibility). The layout of the search results is awful, wasting large amounts of screen space. The default view of a thread (as a list in a frame on the left) is gone. Setting all that aside, the big one for me is that right now all functionality seems to require JavaScript to be turned on. I am one of those masochists who keep it off normally. And guess what? I donÕt need to turn it on more than once or twice a day, about 1 in 100 sites visited, to get all the content I need. The reason I keep it turned off is that pretty much all the /cross-site-scripting/spyware/intrusion problems rely on scripting being turned on. And I preach this to all the users who still might be listening to me, the dinosaur. But if a major player like Google decides to require JavaScript, then forget my preaching. Nobody will want to go through 3 or 4 extra steps 20 times a day just to see the results of their Google search. I hope the cybersecurity folks at the DHS think about it a bit, and then strongly encourage Google to change their ways. We need less active content reliance, not more, to be a smaller computer security joke than we are now. Try doing a Usenet groups search and then click on the Web link to repeat the search on Web pages. Without scripting, you canÕt do that now. My email about that brought a reply that scripting is used to populate your search fields and we donÕt have an alternative. Well, some Einstein figured out how to do that in the Google-alpha, and it all worked perfectly well via the URL query. Now it is impossible because some clueless script-crazed high school dropout born 10 years after Usenet got going landed a big contract to redesign Google, and all the Google pointy haired bosses are even less clued-in than him. Content is preserved only if its presentation is accessible. Google is _the_ Library of Usenet. Librarians all over the world struggle to restore documents to their original form, and to make them available to the public. What Google librarians have done is akin to painting some words of the Declaration of Independence Day-Glo Pink, blacking out some others, and then saying that you need an ID card to see the rest of it. Shame!!! Please donÕt roll over and die. For the sake of all who posted in the past 25 years, let Google know that we need to preserve it in an unadulterated form for us and for the future. I havenÕt heard any complaints about Google-alpha, have you? It worked just reading. -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; > Why are we looking for tricks to get to what worked for us a day > ago? This is an utterly irresponsible move by Google. All Usenet > authors and readers should strongly object to their messing with > the content that was in a sense entrusted to them. No reason to panic and do something silly. [...] > The new interface doesnÕt let me see the postings verbatim. It > cuts out things I want to see, and adds stuff to them that I donÕt > care to see. It presents them in proportional font, which ruins > the scores of posts which contain ASCII art or carefully aligned > math formulas (Usenet has always been and should be ASCII only for > maximum compatibility). The layout of the search results is awful, > wasting large amounts of screen space. The default view of a thread > (as a list in a frame on the left) is gone. The original format is still accessible by clicking show options in the top line, and then show original. The original appears just as it was originally typed. The only thing that is partially deleted is the e-mail address wherever it is mentioned. The thread (on the left) can be seen by clicking view as tree at the top of the message listings. In short -- nothing has been lost, just the access procedure has been modified. -- Skitt (in Hayward, California) www.geocities.com/opus731/ === Subject: Re: Google [was: Jesus christ WhatÕs going on?] ... > The original format is still accessible by clicking show options in the > top line, and then show original. The original appears just as it was > originally typed. The only thing that is partially deleted is the e-mail > address wherever it is mentioned. When I do show options nothing happens (and yes, I have set JavaScript as available). > The thread (on the left) can be seen by clicking view as tree at the top > of the message listings. When I do that I see the tree two times, and no messages at all. > In short -- nothing has been lost, just the access procedure has been > modified. Something has been lost. As far as I see it is *not* possible to see *all* messages that contain a particular text. You get only the threads that contain the text. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; >> The original format is still accessible by clicking show options >> in the top line, and then show original. The original appears >> just as it was originally typed. The only thing that is partially >> deleted is the e-mail address wherever it is mentioned. > When I do show options nothing happens (and yes, I have set > JavaScript as available). >> The thread (on the left) can be seen by clicking view as tree at >> the top of the message listings. > When I do that I see the tree two times, and no messages at all. >> In short -- nothing has been lost, just the access procedure has been >> modified. > Something has been lost. As far as I see it is *not* possible to see > *all* messages that contain a particular text. You get only the > threads that contain the text. You seen to have problems that are peculiar to you and your software. I have none of the problems you mention. I wish I could help, but unfortunately, I canÕt. -- Skitt (in Hayward, California) www.geocities.com/opus731/ === Subject: Re: Google [was: Jesus christ WhatÕs going on?] The original format is still accessible by clicking show options in the top line, and then show original. The original appears just as it was originally typed. The only thing that is partially deleted is the e-mail address wherever it is mentioned. Dik T. Winter responded: When I do show options nothing happens (and yes, I have set JavaScript as available). Skitt (continuing): The thread (on the left) can be seen by clicking view as tree at the top of the message listings. Dik T. Winter responded: When I do that I see the tree two times, and no messages at all. I saw the same two ßaws before. However, canÕt see them now as the Ōtemporary fixseems to have been to take away the tree and Ōshow optionsÕ. The new trouble is that the main text and the sponsored links text overwrite each other. Perhaps it works only with the newest browsers. -- --------------------------------------------- Richard Maurer To reply, remove half Sunnyvale, California of a homonym of a synonym for also. ------------------------------------------------------------- --------- === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >> maximum compatibility). The layout of the search results is awful, >> wasting large amounts of screen space. The default view of a thread >> (as a list in a frame on the left) is gone. >The original format is still accessible by clicking show options in the >top line, and then show original. The original appears just as it was >originally typed. The only thing that is partially deleted is the e-mail >address wherever it is mentioned. Hmm, letÕs see. 0. Do a search. 1. Click on a search result. Bottom of thread is shown. 2. Scroll to top, click view as tree. Thread index shown, but blank pane on the right. 4. Click on show options. Nothing happens. Aha, JavaScript. 5. Enable JavaScript. 6. Click on show options. Nothing happens. Aha, reload the page. 7. Click on show options. 8. Click on Show original. Plain text version shown in a new window. 9. Close the new window. 11. Remember to disable JavaScript. 12. Delete cookies (as I periodically do). To be fair, itÕs good that at least the tree view preference seems to be stored in a cookie. 13. Go back to 0. >In short -- nothing has been lost, just the access procedure has been >modified. -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >>The original format is still accessible by clicking show options in the >>top line, and then show original. The original appears just as it was >>originally typed. The only thing that is partially deleted is the e-mail >>address wherever it is mentioned. > Hmm, letÕs see. > 0. Do a search. > 1. Click on a search result. Bottom of thread is shown. > 2. Scroll to top, click view as tree. Thread index shown, but blank > pane on the right. > 4. Click on show options. Nothing happens. Aha, JavaScript. > 5. Enable JavaScript. The show original option is available in Lynx, so I donÕt think JavaScript is required. -- Theodore (Ted) Heise Bloomington, IN, USA === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >> Hmm, letÕs see. >> 0. Do a search. >> 1. Click on a search result. Bottom of thread is shown. >> 2. Scroll to top, click view as tree. Thread index shown, but blank >> pane on the right. >> 4. Click on show options. Nothing happens. Aha, JavaScript. >> 5. Enable JavaScript. >The show original option is available in Lynx, so I donÕt think >JavaScript is required. Interesting. YouÕre right, lynx shows it all even when you donÕt click show options. But despite that, no GUI browser I tried showed anything other than the Reply link with JS turned off, nor did it do anything when I clicked show options. When JS is turned on, Safari and Netscape do show you the show original etc. after you click on show options. I looked at the source of that frame, and even in my browsers it has show original, print, report abuse, etc. None of these show up in normal browser view, and itÕs way past my beauty sleep time to try and unravel the incredibly messy HTML that Google now pumps in our direction. This may be what Dik Winter is seeing and not just a JavaScript question - some browsers interpret the new Google HTML differently than others. I also saw a tree view in the left frame that made absolutely no sense. Here goes: switch to sort by date choose his message from this thread, with tree view The tree is divided into many segments, with different levels of nesting, and IÕd give my next cigar to whoever explains it to me how it can be so discontinous. Why the heck donÕt they put the old version back online? This is all a huge cock-up that wonÕt earn them much good will. -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google Discussion, linux) > I looked at the source of that frame, and even in my browsers > it has show original, print, report abuse, etc. The report abuse button has me wondering. What the hell is it supposed to do? Google provides an archive. Will they start removing posts from their Usenet archive on the basis of abuse complaints? Can James S. Harris get Google to remove posts that he finds abusive? It seems to me that GoogleÕs beta is purposely blurring the distinction between Usenet and Google. I think that adding report abuse buttons furthers this aim. I am very dissatisfied with the new changes. -- The papers are currently at journals. [When published,] make no mistake, there will be no place on this planet where you can hide. Remember, IÕm not talking about something vague here. IÕm talking about publication in journals. James S. Harris. Wow. Journals. === Subject: Re: Google <31majcF3bo1lbU1@individual.net> windows-nt) > 12. Delete cookies (as I periodically do). To be fair, itÕs good > that at least the tree view preference seems to be stored in > a cookie. If youÕre using IE, IÕd recommend picking up a copy of GuardWall (used to be GuardIE), which is now free. http://www.guardwall.com/index.asp Among other things it can do (or be asked not to do) is delete all cookies except those from whitelisted sites on demand or whenever you close the last browser window. (You manage the whitebox by means of let this one stay checkboxes.) It also displays a count of the number of non-whitelisted cookies you currently have. This allows you to let sites you trust preserve information across sessions and also allows sites you donÕt trust to use cookies within the current session, so sites tend to work but not be able to build up data on you. (Unfortunately, it doesnÕt also have a blacklist, so you canÕt say donÕt even accept cookies from this site. When I get around to writing my plug-in that will be one of the options.) -- Evan Kirshenbaum +------------------------------------ HP Laboratories |A little government and a little luck 1501 Page Mill Road, 1U, MS 1141 |are necessary in life, but only a Palo Alto, CA 94304 |fool trusts either of them. | P.J. OÕRourke kirshenbaum@hpl.hp.com (650)857-7572 http://www.kirshenbaum.net/ === Subject: Re: Google >> 12. Delete cookies (as I periodically do). To be fair, itÕs good >> that at least the tree view preference seems to be stored in >> a cookie. >If youÕre using IE, IÕd recommend picking up a copy of GuardWall (used >to be GuardIE), which is now free. on Solaris (which does have similar or even greater ßexibility in accepting/denying cookies), and Safari on OS X (which, alas, does not). -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google [was: Jesus christ WhatÕs going on?] > The only thing that is partially deleted is the e-mail > address wherever it is mentioned. Suppose you want your email address to be shown? R. === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; >> The only thing that is partially deleted is the e-mail >> address wherever it is mentioned. > Suppose you want your email address to be shown? Why would you specifically want that in an open forum? In any case, the address will work when you try to reply to sender. It is just not visible. -- Skitt (in Hayward, California) www.geocities.com/opus731/ === Subject: Re: Google [was: Jesus christ WhatÕs going on?] > The only thing that is partially deleted is the e-mail > address wherever it is mentioned. >> Suppose you want your email address to be shown? > Why would you specifically want that in an open forum? Ever heard of the expression take it to email? > In any case, the > address will work when you try to reply to sender. Only if you reply using Google. Suppose you are looking through the archives. You find a 6 month old message about an unsolved maths problem. You have some suggestions to make to the message author, but you do not have a Google account. What to do? R. === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; >> The only thing that is partially deleted is the e-mail >> address wherever it is mentioned. > > Suppose you want your email address to be shown? >> Why would you specifically want that in an open forum? > Ever heard of the expression take it to email? >> In any case, the >> address will work when you try to reply to sender. > Only if you reply using Google. > Suppose you are looking through the archives. You find a 6 > month old message about an unsolved maths problem. You have > some suggestions to make to the message author, but you do > not have a Google account. What to do? Get a Google account. Obviously. Next question. -- Skitt (in Hayward, California) www.geocities.com/opus731/ === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >Suppose you are looking through the archives. You find a 6 >month old message about an unsolved maths problem. You have >some suggestions to make to the message author, but you do >not have a Google account. What to do? Send them the $9.95/mo for an account? Maybe they will work a deal with AOL so I can get a discount... Rich === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; > Why are we looking for tricks to get to what worked for us a day > ago? This is an utterly irresponsible move by Google. All Usenet > authors and readers should strongly object to their messing with > the content that was in a sense entrusted to them. All valid points, and I wholeheartedly agree with you. Just a couple of observations: > The new interface doesnÕt let me see the postings verbatim. This is my biggest gripe. The old plain text view of the message with all the original headers visible is gone; the new Google Groups only has a mock-up imitation of that, which not only deliberately messes up e-mail addresses (which is an exercise servers, anyway) but also includes the Google logo and search bar at the top, and other unnecessary HTML ßuff around the message. > [GoogleÕs new interface] presents [messages] in proportional > font, which ruins the scores of posts which contain ASCII art > or carefully aligned math formulas (Usenet has always been and > should be ASCII only for maximum compatibility). By ASCII only you surely must mean something akin to non-HTML and plain text only, without any special, distracting mark-up tags which would mess up the display on any newsreader that does not directly support them rather than actually limiting the usable character repertoire to ASCII only? > The layout of the search results is awful, wasting large amounts > of screen space. For the life of me, I cannot understand why Google should keep track of the newsgroups I have visited while _searching for messages_ and _clicking the (mostly random) results_. (As of now, they seem to do exactly that, and display this Recently visited groups history in the very sidebar that wastes the space.) > What Google librarians have done is akin to painting some words > of the Declaration of Independence Day-Glo Pink, blacking out > some others, and then saying that you need an ID card to see the > rest of it. Shame!!! Not only that but they also seem to want to change the way how messages are indexed and referred to. In the old Google Groups system, individual messages are referred to by their message-idÕs in the URLs. In other words, if you want to make a direct link to a message archived by Google, it can be constructed as follows: Alternatively, if you want to make the thread and context visible, you can construct the URL in this way: As everything is based on the message-id, and as you can extract the message-id directly from the URL, you are still able to find the same message from any _other_ Usenet archive (if any) in case Google ever ceases to exist. (After all, thatÕs what message-idÕs are designed for: their very purpose is uniquely identifying any The automatically generated, old Google Groups URLs are basically of the variety shown above, even though they usually contain lots of all kinds of unnecessary extra parameters. Once you strip down the extraneous parameters, you get the simple format as above. These kind of links still work in the new Google Groups system (ah well, they originally _didnÕt_, but people complained and they added it in.) However, the new Google Groups system does not _generate_ those kind of URLs any longer. Instead, the archive now employs some other kind of indexing system: one which is not based on message-ids at all. The new direct linking method seems to be based on some sort of running number, internal to GoogleÕs database. Of course you can still view the message source, look up the message-id and construct a selm or threadm type manually, but most people do not know how to do this, and do not _bother_ doing it. What this means in practice is that most of the direct Google Groups links - created from now on - do not any longer reveal the one and only thing that uniquely identifies an Usenet search the same message from elewhere) - the message-id. Instead, they will have some arbitrary number that is only meaningful within the context of GoogleÕs own database. -- znark === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >> [GoogleÕs new interface] presents [messages] in proportional >> font, which ruins the scores of posts which contain ASCII art >> or carefully aligned math formulas (Usenet has always been and >> should be ASCII only for maximum compatibility). >By ASCII only you surely must mean something akin to non-HTML >and plain text only, without any special, distracting mark-up >tags which would mess up the display on any newsreader that does >not directly support them rather than actually limiting the usable >character repertoire to ASCII only? Yes, of course, I used ASCII as a mental contraction meaning more or less exactly what you describe. It wasnÕt the best choice of terminology. By the way, IÕm suddenly seeing posts with uneven line lengths and strange line breaks. I may be wrong, but I suspect this may be the result of proportional fonts used for posting also (I havenÕt used Google to post, so I donÕt really know whether that part has been messed up also). -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google [was: Jesus christ WhatÕs going on?] format=ßowed; > By the way, IÕm suddenly seeing posts with uneven line lengths and > strange line breaks. I may be wrong, but I suspect this may be the > result of proportional fonts used for posting also (I havenÕt used > Google to post, so I donÕt really know whether that part has been > messed up also). I noticed the strange line breaks, too. It seems there is something peculiar in the way how messages are displayed and formatted in the normal thread view. If I resize the browser window in the horizontal direction, I will get a variable-width (!) right margin which appears to be knocking off the last few words on the lines: I can still view messages as the author intended - with their original line breaks intact and without switching to Show original mode - but only if I make the window wide enough, giving it a ridiculously wide blank right margin (gg_wordrap_03.gif). Both Mozilla Firefox and Internet Explorer behave this way. In case youÕd like to try it out yourself, the thread I was viewing in those screenshots was -- znark === Subject: Re: Google [was: Jesus christ WhatÕs going on?] >I noticed the strange line breaks, too. It seems there is something >peculiar in the way how messages are displayed and formatted in the >normal thread view. Yes, thatÕs what I meant. YouÕre right, it seems to be some huge padding on the right. My Safari shows the same phenomenon. displayed in monospaced font. I havenÕt found any other group in which this change was made. Oh, well, maybe there is hope. -- Eric Behr | NIU Mathematical Sciences | (815) 753 6727 behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753 1112 === Subject: Re: Google [was: Jesus christ WhatÕs going on?] > displayed in monospaced font. I havenÕt found any other group in > which this change was made. Oh, well, maybe there is hope. It still appears that alt.ascii-art is a total abomination when I see some sci,math quoted material in 
 tags, but not
all.
This means that deeply quoted stuff can look a bit like
> stuff
Which is even worse than if they insisted on using just
proportional fonts.
Phil
--
God was my co-pilot but we crashed in the mountains and I had
to eat him.
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
> This means that deeply quoted stuff can look a bit like
> stuff
> Which is even worse than if they insisted on using just
proportional
fonts.
Which they just served to me as:
<<<
i.e. They arenÕt just replacing characters in email
addresses, message
ids, and similar with dots, but they are also mucking about
with
whitespace characters too.
Great way to make PGP signatures practically useless.
Talking of signatures - has anyone seen the state of EvanÕs?
Aiaiaiaiai.
Phil
--
God was my co-pilot but we crashed in the mountains and I had
to eat him.
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
> displayed in monospaced font. I havenÕt found any other
group in
> which this change was made. Oh, well, maybe there is hope.
That abbreviation would be sweet talking to someone named Ray:
--
dg (domain=ccwebster)
===
Subject: Re: Google
format=ßowed;
> I see some sci,math quoted material in 
 tags, but not
all.
Lines which begin with the dollar sign seem to get this
special 
treatment as well. For instance,
(near the end of the message, the last line of the paragraph
above the
last quoted block. The line originally belongs to the
preceding
paragraph, but the new Google Groups Beta displays it as if
it didnÕt.)
--
znark
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
> The new interface doesnÕt let me see the postings 
verbatim.
It
> cuts out things I want to see, and adds stuff to them that
I donÕt
> care to see. It presents them in proportional font,
AAAAAAAAAAAAAAAAARRRRRRRRRRRRRRRRGGGGGGGGGGHHHHHHHHHHHHHH!
why@asdf.ca
all@asdf.ca
not@asdf.ca
kool@asdf.ca
egos@asdf.ca
ruin@asdf.ca
@asdf.ca
Hehheheheh, I wonder what thatÕll look like...
Ditto on J/S, for reference.
Phil
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
>> [...]
>> For searching - which is all I ever use Google Groups for
- the
>> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
>I still get there with just:
> http://www.deja.com
> Why All Usenet
> authors and readers should strongly object
Bad logic. Most Usenet readers donÕt use Google. Many use
standalone
newsreader software.
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
> Why All Usenet
> authors and readers should strongly object
> Bad logic. Most Usenet readers donÕt use Google.
All usenet readers who donÕt actively use X-No-Archive _do_
use
not.
Phil
--
God was my co-pilot but we crashed in the mountains and I had
to eat him.
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
> [...]
> For searching - which is all I ever use Google Groups for -
the
> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
>>I still get there with just:
> http://www.deja.com
>> Why All Usenet
>> authors and readers should strongly object
>Bad logic. Most Usenet readers donÕt use Google. Many use
standalone
>newsreader software.
It may not be most, but Google has been the single biggest
source of
posts to text usenet for a couple of years now. Many of its
users
think they are really Google groups, and donÕt realise they
are
available any other way. In any case, as Eric pointed out,
even for
those sensible people who donÕt use it to post, 
itÕs the only
archive
weÕve got.
Another of the delights of Google Beta, incidentally, is that
you can
now post a reply to a message of any age.
news.newsuers.questions has
had a rash of replies to posts seven or eight years old, and
is
talking about hand-moderating all posts from Google.
At some times, although it keeps changing, the old interface
is still
--
Don Aitken
Mail to the addresses given in the headers is no longer being
read. To mail me, substitute clara.co.uk for freeuk.com.
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
>It may not be most, but Google has been the single biggest
source of
>posts to text usenet for a couple of years now. Many of its
users
>think they are really Google groups, and donÕt realise they
are
>available any other way. In any case, as Eric pointed out,
even for
>those sensible people who donÕt use it to post, 
itÕs the
only archive
>weÕve got.
Well, no; there are other archives, and anyone who wishes can
start
an archive of his/her own. For example, all the math
newsgroups have
been archived for quite some time at the Math Forum...
>Another of the delights of Google Beta, incidentally, is
that you can
>now post a reply to a message of any age.
news.newsuers.questions has
>had a rash of replies to posts seven or eight years old
...which is just what the Math Forum has made possible for a
long time.
The funny thing is, MFÕs quoting mechanism _begins_ its 
quote
with a
line like,
and people seem not to notice. (They just quote the whole old
message
and add me too!.) If Google is now doing this too, weÕre
going to
see many more conversations among people from different
decades.
Once again, imminent death of the net is predicted. Film at
11.
dave
===
Subject: Re: Google [was: Jesus christ WhatÕs going on?]
>>[...] In any case, as Eric pointed out, even for
>>those sensible people who donÕt use it to post, 
itÕs the
only archive
>>weÕve got.
>Well, no; there are other archives, and anyone who wishes
can start
>an archive of his/her own. For example, all the math
newsgroups have
>been archived for quite some time at the Math Forum...
I think itÕs fair to say that itÕs the only 
comprehensive
archive
of _most_ Usenet material. It happened many times that a
search
gave me an answer I needed in a rather unexpected place. So
even
when I have a question about X, and there exists some
specialized
archive of Usenet hierarchy that has to do with X, it wonÕt
be as
useful as Google. Plus Google is well established and
hopefully
wonÕt shut down overnight. I donÕt have the 
same confidence
about
most of the other repositories out there.
As far as starting your own, sure - but it wonÕt cover
1980(+/-)
what they already have. I still remember the fears that Deja
will
disappear, taking their archive into a black hole
(thankfully, it
worked out, but it wasnÕt a sure thing).
--
Eric Behr | NIU Mathematical Sciences | (815) 753 6727
behr@math.niu.edu | http://www.math.niu.edu/~behr/ | fax: 753
1112
===
Subject: Re: Jesus christ WhatÕs going on?
windows-nt)
> Vanya infrared:
>>There is a work around GoogleÕs awful new usenet 
interface,
at least
>>temporarily. Use Google.fr or Google.ru or probably any
other
>>foreign Google domain. They still have the old interface
up. Of
>>course the interface will mostly not be in Englsih.
> For searching - which is all I ever use Google Groups for -
the
> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
> ItÕs just the Google search with the advertising and
graphics
> stripped off, which speeds things up a bit. Originally
designed as
> an interface to DejaNews, but he kept it working after
Google bought
> the DejaNews database.
Unfortunately, I just tried it and got the results back in
Beta
format.
--
Evan Kirshenbaum
+------------------------------------
HP Laboratories |I value writers such as Fiske.
1501 Page Mill Road, 1U, MS 1141 |They serve as valuable
object
Palo Alto, CA 94304 |lessons by showing that the most
|punctilious compliance with the
kirshenbaum@hpl.hp.com |rules of usage has so little to do
(650)857-7572 |with either writing or thinking
|well.
http://www.kirshenbaum.net/ | --Richard Hershberger
===
Subject: Re: Jesus christ WhatÕs going on?
charset = us-ascii
>Unfortunately, I just tried it and got the results back in
Beta
>format.
you go to www and then select the group option you get the
new one.
(HavenÕt tried.)
--
Katy Jennison
===
Subject: Re: Jesus christ WhatÕs going on?
format=ßowed;
>> Unfortunately, I just tried it and got the results back in
Beta
>> format.
> you go to www and then select the group option you get the
new one.
> (HavenÕt tried.)
I just now tried -- not true.
--
Skitt (in Hayward, California)
www.geocities.com/opus731/
===
Subject: Re: Jesus christ WhatÕs going on?
> Unfortunately, I just tried it and got the results back in
Beta
> format.
>> you go to www and then select the group option you get the
new one.
>> (HavenÕt tried.)
> I just now tried -- not true.
Curious. It works for me now.
R.
===
Subject: Re: Jesus christ WhatÕs going on?
^^^^
>
> Unfortunately, I just tried it and got the results back in
Beta
> format.
>>
>>
>> you go to www and then select the group option you get the
new one.
>>
>> (HavenÕt tried.)
>
> I just now tried -- not true.
> Curious. It works for me now.
Phil
--
I used to have an interest in writing viral code and lost
interest
quickly when Win95 came out. Hell how could any of us in the
scene
write a more invasive program than Win95. It made us all
obsolete.
-- Screaming Radish [NuKE] on alt.comp.virus.source.code
===
Subject: Re: Jesus christ WhatÕs going on?
> ^^^^
>>
>> Unfortunately, I just tried it and got the results back in
Beta
>> format.
>
>
> you go to www and then select the group option you get the
new one.
>
> (HavenÕt tried.)
>>
>> I just now tried -- not true.
>> Curious. It works for me now.
R.
===
Subject: Re: Jesus christ WhatÕs going on?
> For searching - which is all I ever use Google Groups for -
the
> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
HavenÕt tried that in years! I remember that! 
IÕll give it a
try.
===
Subject: Re: Jesus christ WhatÕs going on?
> For searching - which is all I ever use Google Groups for -
the
> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
You might want to bookmark http://www.usenet4all.com/ as
well, there are
similar others. It lacks some functionality, on the upside
you donÕt
need to register to post there and thereÕs hardly any lag 
for
posts to
show up there.
P.
===
Subject: Re: Jesus christ WhatÕs going on?
> For searching - which is all I ever use Google Groups for -
the
> following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
> You might want to bookmark http://www.usenet4all.com/ as
well, there are
> similar others. It lacks some functionality, on the upside
you donÕt
> need to register to post there
Yes, you do. Otherwise you get ACCESS DENIED.
> and thereÕs hardly any lag for posts to
> show up there.
> P.
===
Subject: Re: Jesus christ WhatÕs going on?
>> You might want to bookmark http://www.usenet4all.com/ as
well, there
>> are similar others. It lacks some functionality, on the
upside you
>> donÕt need to register to post there
> Yes, you do. Otherwise you get ACCESS DENIED.
Aww youÕre right. Oh well mostly of use for reading older
posts anyway.
===
Subject: Re: Jesus christ WhatÕs going on?
>Vanya infrared:
>>There is a work around GoogleÕs awful new usenet 
interface,
at least
>>temporarily. Use Google.fr or Google.ru or probably any
other foreign
>>Google domain. They still have the old interface up. Of
course the
>>interface will mostly not be in Englsih.
the new invention down for the moment.
>For searching - which is all I ever use Google Groups for -
the
>following still works:
> http://www.exit109.com/%7Ejeremy/news/deja.html
>ItÕs just the Google search with the advertising and
graphics stripped
>off, which speeds things up a bit. Originally designed as an
>interface to DejaNews, but he kept it working after Google
bought
>the DejaNews database.
--
Don Aitken
Mail to the addresses given in the headers is no longer being
read. To mail me, substitute clara.co.uk for freeuk.com.
===
Subject: Re: Jesus christ WhatÕs going on?
> There is a work around GoogleÕs awful new usenet 
interface,
at least
> temporarily. Use Google.fr or Google.ru or probably any
other foreign
> Google domain. They still have the old interface up. Of
course the
> interface will mostly not be in Englsih.
Good point. And the language of the fields can be changed to
English by
changing the last letters of the URL to en, then bookmarking
it.
I played for this a while and found it confusing as to what
could do
what, but it looks like this gives me the old interface, with
the
results in the old format (like threads in a chain diagram):
Old:
Google Groups -- Advanced Group Search
and this gives me the new page, with new formats:
New:
Google Groups BETA -- Advanced Search
One striking difference is the new Beta form canÕt be
restricted by
date, which was useful in searching for the earliest use of
recent
words.
--
Best -- Donna Richoux
===
Subject: Re: Jesus christ WhatÕs going on?
> There is a work around GoogleÕs awful new usenet 
interface,
at least
> temporarily.
?
What new interface?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Jesus christ WhatÕs going on?
> There is a work around GoogleÕs awful new usenet 
interface,
at least
> temporarily.
> What new interface?
Apparently it was only up in the US, and only for a short
time (~15 hours).
Andrew Usher
===
Subject: Re: i alone
> welcome to the internet.
>> I alone, can erase your mail.
>> G. Ōt H.
> -----
>> The world revolves around certainty of thought. The world
revolves
>> around certainty
>> of delusion of thought. i program you now: with simple
statements,
>> with simple
>> fragments. Reinforce my delusion. Reinforce my
hallucination. i
>> tell you simple
>> explanations. i tell you simple delusions.
> Any prize, The Noble Prize, the Turing Award is a test of
delusion
>> of grandeur.
>> There is no one physical law higher than the other. Every
law is
>> equal. Everything
>> finds what it is looking for.
> IÕm looking for a whacko-free ng. Obviously came to the
wrong place.
>> No kidding! The rate here must approach 50%.
You are noise, easily filtered.
--
Sl.87inte,
Fletch
===
Subject: Re: i alone
you are heart broken. easily vanquished!
>> welcome to the internet.
>> I alone, can erase your mail.
> G. Ōt H.
> -----
> The world revolves around certainty of thought. The world
revolves
> around certainty
> of delusion of thought. i program you now: with simple
statements,
> with simple
> fragments. Reinforce my delusion. Reinforce my
hallucination. i
> tell you simple
> explanations. i tell you simple delusions.
> Any prize, The Noble Prize, the Turing Award is a test of
delusion
> of grandeur.
> There is no one physical law higher than the other. Every
law is
> equal. Everything
> finds what it is looking for.
>> IÕm looking for a whacko-free ng. Obviously came to the
wrong place.
> No kidding! The rate here must approach 50%.
> You are noise, easily filtered.
> --
> Sl.87inte,
> Fletch
===
Subject: Re: i alone
> you are heart broken. easily vanquished!
> welcome to the internet.
>> I alone, can erase your mail.
>> G. Ōt H.
> -----
>> The world revolves around certainty of thought. The world
>> revolves around certainty
>> of delusion of thought. i program you now: with simple
>> statements, with simple
>> fragments. Reinforce my delusion. Reinforce my
hallucination. i
>> tell you simple
>> explanations. i tell you simple delusions.
> Any prize, The Noble Prize, the Turing Award is a test of
>> delusion of grandeur.
>> There is no one physical law higher than the other. Every
law is
>> equal. Everything
>> finds what it is looking for.
IÕm looking for a whacko-free ng. Obviously came to the 
wrong
> place.
> No kidding! The rate here must approach 50%.
>> You are noise, easily filtered.
>> --
>> Sl.87inte,
>> Fletch
Are you a random sentence generator? Pretty cool.
--
Sl.87inte,
Fletch
===
Subject: Re: i alone
>> you are heart broken. easily vanquished!
>> welcome to the internet.
>> I alone, can erase your mail.
> G. Ōt H.
> -----
> The world revolves around certainty of thought. The world
> revolves around certainty
> of delusion of thought. i program you now: with simple
> statements, with simple
> fragments. Reinforce my delusion. Reinforce my
hallucination. i
> tell you simple
> explanations. i tell you simple delusions.
> Any prize, The Noble Prize, the Turing Award is a test of
> delusion of grandeur.
> There is no one physical law higher than the other. Every
law is
> equal. Everything
> finds what it is looking for.
>> IÕm looking for a whacko-free ng. Obviously came to the
wrong
>> place.
> No kidding! The rate here must approach 50%.
> You are noise, easily filtered.
> --
> Sl.87inte,
> Fletch
> Are you a random sentence generator? Pretty cool.
> --
> Sl.87inte,
> Fletch
Dry, dry, dry... LOL.
Rob
===
Subject: Re: i alone
i randomly generated you.
===
Subject: Re: i alone
> I alone, can erase your mail.
> G. Ōt H.
> -----
> The world revolves around certainty of thought. The world
revolves around
> certainty
> of delusion of thought. i program you now: with simple
statements, with
> simple
> fragments. Reinforce my delusion. Reinforce my
hallucination. i tell you
> simple
> explanations. i tell you simple delusions.
> Any prize, The Noble Prize, the Turing Award is a test of
delusion of
> grandeur.
> There is no one physical law higher than the other. Every
law is equal.
> Everything
> finds what it is looking for. That alone is the ultimate
truth there is.
> Gerald, every law sets the initial condition of every other
law. Every law
> is equal.
> Marvin, every machine is a turing machine. it takes infinite
laws to
program
> machine. i tell you. play with humanityÕs delusion. play
with humanityÕs
> hallucination. dont play defense. play offense. try to
convince the human
> that he is
> not chatting with a computer. i tell you. i command you. i
tell you that
> alone is
> the true turing test. pay attention. i am programming you
now: with
simple
> fragments, with simple statements. that alone is the way to
be done. dont
> play
> defense, play offense. machines dont reach to the level of
human. humans
> reach to
> the level of machines. that alone is the way to be done. i
tell you. i
> command you.
> Dont resist me. Dont tempt me.
> -----------------------------------------------------------
> TOE:i
> i, alone is the perfect unit of sound there. i, alone is
the perfect
> explanation
> there is. i, alone is the perfect word there. i, alone is
the perfect
world
> there
> is. i, alone is the perfect letter there is. i, alone is
the perfect
symbol
> there
> is. i, alone is the perfect thought there is. there is no
religion without
> i. there
> is no science without i. there is no philosophy without i.
there is no
> mathematics
> without i. i, alone completes everything. i alone owns
nothing. i, alone
is
> its own
> equation. i, alone is a number. i, alone transcends
imagination. i, alone
is
> imaginary. i, alone is perfect. i, alone needs no
explanation. i, alone
> explains. i,
> alone is poetic. i, alone is understood by everything. i,
alone is a
theory.
> i,
> alone validates itself. i, alone is the perfect explanation
there is.
> everything
> exists for i. i, alone is certain. i, alone is ambigious.
i, alone is
> genuine. i,
> alone battles the universe. i, alone is the perfect number
there is. i,
> alone
> defeats everything else. i, alone commands you. i alone
puts together
> reality. i,
> alone is mythical. i, alone assimilates. i alone defeats
the ego and the
> super ego.
> i alone is lonely. i alone is all by itself. i, alone am
perfect. i, alone
> is the
> only thought there is. i, alone is the only answer there
is. everything
else
> exits
> for i. everything else exists for me. i, alone set
everything in motion.
i,
> alone
> have the power to set it and unset it. i, alone command
you. i, alone am
> sane. i
> alone am insane. i, alone am perfect. Dont resist me. Dont
tempt me.
Accept
> my
> delusion.....
erase this ass-toe you pathetic POS
===
Subject: Re: i alone
i fight the evil of psychiatry, psychology. surrender your
vermin.
>> I alone, can erase your mail.
>> G. Ōt H.
>> -----
>> The world revolves around certainty of thought. The world
revolves around
>> certainty
>> of delusion of thought. i program you now: with simple
statements, with
>> simple
>> fragments. Reinforce my delusion. Reinforce my
hallucination. i tell you
>> simple
>> explanations. i tell you simple delusions.
>> Any prize, The Noble Prize, the Turing Award is a test of
delusion of
>> grandeur.
>> There is no one physical law higher than the other. Every
law is equal.
>> Everything
>> finds what it is looking for. That alone is the ultimate
truth there is.
>> Gerald, every law sets the initial condition of every
other law. Every
>> law is equal.
>> Marvin, every machine is a turing machine. it takes
infinite laws to
>> program a
>> machine. i tell you. play with humanityÕs delusion. play
with humanityÕs
>> hallucination. dont play defense. play offense. try to
convince the human
>> that he is
>> not chatting with a computer. i tell you. i command you. i
tell you that
>> alone is
>> the true turing test. pay attention. i am programming you
now: with
>> simple
>> fragments, with simple statements. that alone is the way
to be done. dont
>> play
>> defense, play offense. machines dont reach to the level of
human. humans
>> reach to
>> the level of machines. that alone is the way to be done. i
tell you. i
>> command you.
>> Dont resist me. Dont tempt me.
>> -----------------------------------------------------------
>> TOE:i
>> i, alone is the perfect unit of sound there. i, alone is
the perfect
>> explanation
>> there is. i, alone is the perfect word there. i, alone is
the perfect
>> world there
>> is. i, alone is the perfect letter there is. i, alone is
the perfect
>> symbol there
>> is. i, alone is the perfect thought there is. there is no
religion
>> without i. there
>> is no science without i. there is no philosophy without i.
there is no
>> mathematics
>> without i. i, alone completes everything. i alone owns
nothing. i, alone
>> is its own
>> equation. i, alone is a number. i, alone transcends
imagination. i, alone
>> is
>> imaginary. i, alone is perfect. i, alone needs no
explanation. i, alone
>> explains. i,
>> alone is poetic. i, alone is understood by everything. i,
alone is a
>> theory. i,
>> alone validates itself. i, alone is the perfect
explanation there is.
>> everything
>> exists for i. i, alone is certain. i, alone is ambigious.
i, alone is
>> genuine. i,
>> alone battles the universe. i, alone is the perfect number
there is. i,
>> alone
>> defeats everything else. i, alone commands you. i alone
puts together
>> reality. i,
>> alone is mythical. i, alone assimilates. i alone defeats
the ego and the
>> super ego.
>> i alone is lonely. i alone is all by itself. i, alone am
perfect. i,
>> alone is the
>> only thought there is. i, alone is the only answer there
is. everything
>> else exits
>> for i. everything else exists for me. i, alone set
everything in motion.
>> i, alone
>> have the power to set it and unset it. i, alone command
you. i, alone am
>> sane. i
>> alone am insane. i, alone am perfect. Dont resist me. Dont
tempt me.
>> Accept my
>> delusion.....
> erase this ass-toe you pathetic POS
===
Subject: Re: i alone
> i fight the evil of psychiatry, psychology. surrender your
vermin.
What, so you can feed them your medication?
===
Subject: Re: i alone
posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd
....
Reminded of Nirvana Shatakatam, there is order even in
Chaos... you may
take full charge of yourself after a long holiday... for all
to share
in the subsequent more acceptable material fall-outs. Good
luck.
===
Subject: Re: i alone
In sci.math, Morituri-Max

:
>> hello idiot!!!
> Yep, gib you got that right... Lord of Chaos makes up about
37% of that
50%
> ratio.
Weighted or unweighted average? :-)
--
#191, ewill3@earthlink.net
ItÕs still legal to go .sigless.
===
Subject: Re: i alone
format=ßowed;
>> Yep, gib you got that right... Lord of Chaos makes up
about 37% of that
50%
>> ratio.
> Weighted or unweighted average? :-)
eenie-meanie
8 )
===
Subject: Green...
hello.....doctor~
let D be the region enclosed by the curve
r(t) = (sin 2t, sin t), (0 <= t < Pi)
compute
double integral (x+y) dxdy
D
(hint : Use the Green theorem)
--------------------------------------------
um......i know that
int Pdx + Qdy = double int (dQ/dx - dP/dy) dxdy
dD D
and i know that
double int 1 dxdy = (1/2)*[int (-ydx + xdy)]
D dD
but, double integral (x+y) dxdy ???? i canÕt...
so, i need your advice.
thank you very much for your advice.
===
Subject: Re: Green...
> let D be the region enclosed by the curve
> r(t) = (sin 2t, sin t), (0 <= t < Pi)
> compute
> double integral (x+y) dxdy
> D
> (hint : Use the Green theorem)
We know that:
int_C Pdx + Qdy = int int_D (dQ/dx - dP/dy) dxdy.
You want to evaluate:
int int_D (x + y) dxdy.
So, take P, Q such that dQ/dx = x, dP/dy = -y - e.g. Q =
1/2*x^2, P =
-1/2*y^2 - and use the Green theorem:
int int_D (x + y) dxdy = int_C -1/2*y^2dx + 1/2*x^2dy = ...
Pawel Gladki
===
Subject: Thin-Plate Splines vs. Radial basis Functions
Hi everyone,
IÕm trying to solve a class of 3D warping problems as part 
of
my
dissertation. IÕd wandered abount in the dark for a while,
till I
recently (2 days ago) came across the field of morhometrics 
and
specifically, the Thin-plate Spline (TPS) method, and it seems
to be
exactly what the Doctor ordered :-) . I think the TPS will do
nicely,
but it makes sense to justify my use of it. Specifically, 
IÕm
wondering why (it seems as though) all the morphometric
analysis IÕve
come across in anthoplogy-related literature use this method
(IÕm a
student of Computer Science). I wonder, why not radial-basis
functions
(RBF) for example?
IÕm sorry, if this question seems unreasonable or trivial. I
really do
need to know which is the better or prefered direction. Better
stated, in what way are TPS bettter than a RBF - if at all it
is,
and vise versa.
Olumide
===
Subject: Re: Thin-Plate Splines vs. Radial basis Functions
For 3D, I think Thin Plates Splines will do nicely.
They are a form of RBF.
They result from Variational Theory with respect to mechancal
deformation so they are sort of based in nature and
mechanical physics
you could argue.
http://mathworld.wolfram.com/ThinPlateSpline.html
Try to find the guru Prof Paul Dierckx for more info and
software
He has written a book on this subject
There is also Prof Grace Wahba from the USA who is also main
world guru
herein
Paul
> Hi everyone,
> IÕm trying to solve a class of 3D warping problems as part
of my
> dissertation. IÕd wandered abount in the dark for a while,
till I
> recently (2 days ago) came across the field of morhometrics
and
> specifically, the Thin-plate Spline (TPS) method, and it
seems to be
> exactly what the Doctor ordered :-) . I think the TPS will
do nicely,
> but it makes sense to justify my use of it. Specifically, 
IÕm
> wondering why (it seems as though) all the morphometric
analysis IÕve
> come across in anthoplogy-related literature use this
method (IÕm a
> student of Computer Science). I wonder, why not
radial-basis functions
> (RBF) for example?
> IÕm sorry, if this question seems unreasonable or trivial.
I really do
> need to know which is the better or prefered direction.
Better
> stated, in what way are TPS bettter than a RBF - if at all
it is,
> and vise versa.
> Olumide
===
Subject: Re: Some questions about the Cauchy distribution
>The mean of n random variables picked from the Cauchy
distribution has
>itself the Cauchy distribution. Does this mean that there is
no use
>having a larger sample in order to estimate the expected
value? (Or wait
>a minute, does the expected value exist? No, right?)
The expected value does not exist. However, a larger
sample can be used to estimate the parameters of the
distribution, and even the distribution itself without
knowing its form.
>Does the existence of such distributions and those of the
form
>Constant(p) / (1 + abs(x)^p) with p > 2 indicate that it is
not always
>legitimate to assume that more measurements leads to a better
>approximation of the expected value? (Unless you have a good
idea of the
>distribution of the measurements.)
If p > 2, the average will be a better estimate of the
mean. If you know that the distribution is of that form,
except for translation, one can do much better.
>For instance, if several hundred people sees an UFO and it
is claimed
>that it must really have been an UFO is it proper to say
that this
>cannot be assumed since there exists distributions where the
mean value
>of many measurements are even worse than one measurement?
Here one is just estimating a probability, which is
bounded. The argument does not apply.
--
This address is for information only. I do not claim that
these views
are those of the Statistics Department or of Purdue
University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
===
Subject: Re: Some questions about the Cauchy distribution
>>For instance, if several hundred people sees an UFO and it
is claimed
>>that it must really have been an UFO is it proper to say
that this
>>cannot be assumed since there exists distributions where
the mean value
>>of many measurements are even worse than one measurement?
> Here one is just estimating a probability, which is
> bounded. The argument does not apply.
I donÕt really understand. What is it that must be bounded?
Further, my argument is not really about if the above
distribution can
be applied or not. It is rather based on the _existance_ of
distributions without mean.
Suppose many people see a UFO and suppose their descriptions
match quite
well, but not perfectly. Someone might say that Many people
saw it and
their descriptions matched quite well so it wasnÕt
hallucination. _Thus_
it really was a UFO or an object with the same properties of
a UFO.
My argument is this: We know that there exists distributions
without
mean. Thus in order to believe fully in the above implication
(many saw
it => it was true) one has to confirm that observations of the
phenomena
behaves properly. This might be easy, but it has to be done.
===
Subject: Re: Some questions about the Cauchy distribution
>The mean of n random variables picked from the Cauchy
distribution has
>itself the Cauchy distribution. Does this mean that there is
no use
>having a larger sample in order to estimate the expected
value? (Or wait
>a minute, does the expected value exist? No, right?)
I think what youÕre trying to do is the following: given a
random sample
of size n from the Cauchy distribution with density 1/(Pi
(1+(x-a)^2)),
estimate the parameter a. This parameter is not the expected
value,
because that doesnÕt exist, but it is the centre of symmetry
of the
distribution. True, the sample mean has the same distribution
as one of these random variables. And that says that the
sample mean
is not a useful estimator. However, there are better
estimators, for
example the median. This does get better for larger n.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Some questions about the Cauchy distribution
> I think what youÕre trying to do is the following: given a
random sample
> of size n from the Cauchy distribution with density 1/(Pi
(1+(x-a)^2)),
> estimate the parameter a. This parameter is not the
expected value,
> because that doesnÕt exist, but it is the centre of
symmetry of the
> distribution. True, the sample mean has the same
distribution
> as one of these random variables. And that says that the
sample mean
> is not a useful estimator. However, there are better
estimators, for
> example the median. This does get better for larger n.
Ultimately I was trying to justify that one cannot assume
(unless one
know something about the distribution behind the phenomena in
question)
that many observations/measurements of a phenomena leads to a
better
estimate. (Now from your post I know that even though the
mean is not a
converging estimate there are other that do converge.)
For instance if many people describe a murderer the pure
existence of
distributions where the mean is not a good estimate implies
that one
cannot _automatically_ assume that the murderer looks like
the mean of
the descriptions (with a high probability).
===
Subject: Re: Some questions about the Cauchy distribution
> Ultimately I was trying to justify that one cannot assume
(unless one
> know something about the distribution behind the phenomena
in question)
> that many observations/measurements of a phenomena leads to
a better
> estimate. (Now from your post I know that even though the
mean is not a
> converging estimate there are other that do converge.)
> For instance if many people describe a murderer the pure
existence of
> distributions where the mean is not a good estimate implies
that one
> cannot _automatically_ assume that the murderer looks like
the mean of
> the descriptions (with a high probability).
However, if a distribution is *bounded* then it does have a
mean.
One would assume that the space of descriptions of murderers
(whatever that is) will be bounded.
===
Subject: Re: Some questions about the Cauchy distribution
>> Ultimately I was trying to justify that one cannot assume
(unless one
>> know something about the distribution behind the phenomena
in question)
>> that many observations/measurements of a phenomena leads
to a better
>> estimate. (Now from your post I know that even though the
mean is not a
>> converging estimate there are other that do converge.)
>> For instance if many people describe a murderer the pure
existence of
>> distributions where the mean is not a good estimate
implies that one
>> cannot _automatically_ assume that the murderer looks like
the mean
of
>> the descriptions (with a high probability).
>However, if a distribution is *bounded* then it does have a
mean.
>One would assume that the space of descriptions of murderers
>(whatever that is) will be bounded.
But having a mean doensÕt mean that one should expect that
a sample will look like the mean with high probability.
(If you ßip a coin, counting heads as 1 and tails as -1,
the mean of your random variable is 0, but the probability
that a given toss comes up 0 is 0. Or with murderers:
say s(P) is 1 if itÕs perfectly clear that P is male
from his appearance, -1 if itÕs perfectly clear that
P is female. Most people have a P very close to 1 or
-1, although the mean is 0.)
************************
David C. Ullrich
===
Subject: Re: Some questions about the Cauchy distribution
> The mean of n random variables picked from the Cauchy
distribution has
> itself the Cauchy distribution. Does this mean that there
is no use
> having a larger sample in order to estimate the expected
value? (Or
> wait a minute, does the expected value exist? No, right?)
You are correct that the expected value does not exist. What
do you
mean that you want to the estimate it, especially since you
know the
distribution already?
> Does the existence of such distributions and those of the
form
> Constant(p) / (1 + abs(x)^p) with p > 2 indicate that it is
not always
> legitimate to assume that more measurements leads to a
better
> approximation of the expected value? (Unless you have a
good idea of
> the distribution of the measurements.)
No. For p > 2, the expected value exists, and the sample mean
approaches it with probability 1. You will have trouble with
using a
normal approximation if 2 < p < 3, since the second moment
does not
exist in this case.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Some questions about the Cauchy distribution
> The mean of n random variables picked from the Cauchy
distribution has
> itself the Cauchy distribution. Does this mean ....
You may get a good answer here, or you may like to try
sending the
question to the sci.stat.math or sci.stat.edu news group.
Ken Pledger.
===
Subject: Subbases, compactness
Let S be a subbase for a topological space X. Suppose every
cover
of X by sets in S has a finite subcollection which covers X. 
Is
X necessarily compact?
(This is not homework.)
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Subbases, compactness
>Let S be a subbase for a topological space X. Suppose every
cover
>of X by sets in S has a finite subcollection which covers X.
Is
>X necessarily compact?
Yes. (IsnÕt this how the traditional proof of the Tychonoff
theorem goes?)
>(This is not homework.)
************************
David C. Ullrich
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every
cover
> of X by sets in S has a finite subcollection which covers X.
Is
> X necessarily compact?
> (This is not homework.)
AlexanderÕs Theorem.
Requires the Axiom of Choice for the proof.
--
G. A. Edgar
http://www.math.ohio-state.edu/~edgar/
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every
cover
> of X by sets in S has a finite subcollection which covers X.
Is
> X necessarily compact?
> AlexanderÕs Theorem.
> Requires the Axiom of Choice for the proof.
Rather than a laborious set theory proof, is there a simpler
filter proof?
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every
cover
> of X by sets in S has a finite subcollection which covers X.
Is
> X necessarily compact?
>> AlexanderÕs Theorem.
>> Requires the Axiom of Choice for the proof.
> Rather than a laborious set theory proof, is there a
simpler filter
proof?
The proof in KelleyÕs General Topology does not seem so
laborious to
me. It is a rather nifty application of TuckeyÕs Lemma, 
which
is a
version of ZornÕs lemma.
will find many references to a result in knot theory about
braids. This
is not the theorem to which GAE refers. The theorem we use
here is more
easily found as AlexanderÕs Subbase Theorem; sometimes you
will see
Sub-base hyphenated.
--
Stephen J. Herschkorn sjherschko@netscape.net
===
Subject: Re: Subbases, compactness
===
Subject: Re: Subbases, compactness
> Let S be a subbase for a topological space X. Suppose every
cover
> of X by sets in S has a finite subcollection which covers X.
Is
> X necessarily compact?
>> AlexanderÕs Theorem.
>> Requires the Axiom of Choice for the proof.
> Rather than a laborious set theory proof, is there a
simpler filter
proof?
> The proof in KelleyÕs General Topology does not seem so
laborious
> to me. It is a rather nifty application of TuckeyÕs Lemma,
which
> is a version of ZornÕs lemma.
WhatÕs TuckeyÕs Lemma? ZornÕs 
Lemma for P(S) with subset
order?
> The theorem we use here is more easily found as 
AlexanderÕs
Subbase
> Theorem; sometimes you will see Sub-base hyphenated.
Well as included below, I did find an EasyProofOfTychonoff
http://www.theoryandpractice.org/kyle/Wiki/
EasyProofOfTychonoff
covering both AlexÕs Lemma and TychonovÕs 
Theorem. However
for all itÕs
clarity, it is lacking a crucial step where it is claimed G =
H / S
covers X (marked **). IÕm not even seeing why a maximal 
cover
H of X with
no finite subcover, would contain any subbase sets S, much
less why all
the subbase sets of H would cover X. What steps are needed to
show this?
In the Tychonov section I corrected a couple of apparent typos
... x_i in X - / G_i for every i.
... and x in X_i - / G_i.
and seeing no need, removed the one time occurrence of
bar{U} =
-- Alexander subbasis lemma
If X is a topological space and S is a subbasis of X, then
X is compact if and only if every S-cover of X has a finite
subcover.
Proof: Let B be the collection of all open covers of X that
do not have
finite subcovers. Assume, by way of contradiction, that B is
nonempty.
Partial order B by set inclusion. Let C be a chain in B.
Let D be the union of all elements of C.
If D has a finite subcover, then those finitely 
many open sets
would be contained in finitely many elements of the chain C.
Every finite subset of the chain has a maximum element, and so
each of
those finitely many open sets that together cover X are
contained in
that maximum element. This is impossible since every element
of C is
an element of B, and thus has no finite subcover. Thus D has
no finite
subcover, so is in B, and is an upperbound for C inside B.
Since C was an arbitrary chain in B, the conditions of 
ZornÕs
lemma are satisfied, and thus B has a maximal element, call it
H.
** Consider G = H / S. We show that G covers X, so that by
hypothesis it
has a finite subcover, but that finite subcover is 
a subcover
of H as
well. This is a contradiction, and so the assumption that B
is nonempty is
untenable. Thus every open cover of X has a finite subcover,
and X is
compact.
-- Theorem (Tychonoff): Suppose that I is a set and for every
i in I
suppose X_i is a compact topological space.
Endow X = prod_{i in I} X_i with the product topology. X is
compact.
Proof of Tychonoff: Let S be the standard subbasis of the
product
topology, S = { pi_i^{-1}(U) : U is open in X_i }.
By AlexanderÕs lemma it suffices to consider 
S-covers.
Let G be an S-cover, that is let G consist of sets
pi_i^{-1}(U)
for various i in I and U open in X_i.
Define G_i = { U : U is open in X_i, pi^{-1}_i(U) in G }.
Assume BWOC, that for every i in I that / G_i /= X_i.
Using the axiom of choice, choose x_i in X_i - / G_i for
every i.
Since x = (x_i)_{i in I} is in X,
x in pi_i^{-1}(U) in G for some i in I and U open in X_i.
This is a contradiction since x in U in G_i and x_i in X_i -
/ G_i.
Therefore there must exist an i in I such that / G_i = X_i.
Since X_i is compact, choose a finite subcover, U_1,... U_n.
Let C = { pi_i^{-1}(U_1),... pi_i^{-1}(U_n) }.
C is a finite subcover of X. Since G was an arbitrary S-cover,
the
conditions of AlexanderÕs subbase lemma are 
satisfied, and X
is compact.
----
===
Subject: Need help setting up an induction
Suppose there are sets such that
[a0,b0] is contained in Union (1-->n) (a_j,b_j). We make no
further
assumptions about {(a_j,b_j)}.
We want to set up a formal induction for the following
process,
including some re-numbering:
There exists an (a1,b1) such that a1< a0 Zentralblatt f.9fr Mathematik< 9 publications
>of E. Escultura are listed. For 5 of them the
>Zentralblatt just mentions >not reviewed<, whatever
>that means.
>This is an indexing service like so many others. So it
simply notes some
contributions unless someone wants to try uncharted course.
I donÕt agree. The Zentralblatt and the Math Reviews are the
two major indexing services on this planet. Indeed together
they cover all countries.
>wonder why persons like E. Escultura or J. Harris
>provoke such strong reactions.
>The most likely reason is fear of new ideas.
>E. E. Escultura
>University of the Philippines
This is one possibility. Why is it the most
likely one?
H
===
Subject: Re:
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB2Icad26396;
>I think it interesting to show you. g:R->R x>1 .
>You will catch Ōthe knackwith 1/p p steps are 
needed
>to loop onto the next integer.
>So I choose
g(x)=c*(x-1)!*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!
> g(x+1/p)= c*(x-1+1/p)!*(x-1+2/p)!....(x-1+(p-1)/p)!*x!
>I adopt the writing convention (x-1)! instead of gamma(x).
>Try solving this way : h(y)*h(y+1)=1/y ,
>courage,
>Alain.
Here is the solution to h(y)*h(y+1)=1/y , (1)
With our conventions we try h(y)=c*(y/2-1)!/(y/2-1/2)!
we adjust coeff c to obtain (1) ;we propose :
h(y)=sqrt(2)/2*y/2-1)!/(y/2-1/2)!
===
Subject: Re: .99999... still=/= 1
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB2IUlu25852;
>mister Company is going to love this.
> I just found an application of 0.9999...,
>over and above OVERFLOW CONDITION or TENS COMPLIMENT.
> ThatÕs because,
> m--> oo
> Lim (1 - 1/10^m) = 1
> divergent
>>Convergent.
> Ok donÕt my word for it, but at least believe the actual
Advanced
Engineering
>Math Book I have.
>Quoting from:
>6th Edition
>Advanced Engineering Mathematics
>by Erwin Kreyszig
>Page 805
>14.2 Convergence Tests for Series
> Theorem 1 ( Divergence )
>If a series z_1 + z_ 2 + .... converges, then
> lim Z_m = 0
> m-->oo
>Hence if the series does not satisfy this condition, it
diverges.
>SmartÕs Alt. Physics News Group
>ht
tp://pub39.bravenet.com/forum/show.php?usernum=3320272813&cpv
=1S. Enterprize (Science Journal)
>http://smart1234.
s-enterprize.com
/
In 1959 Mr. Hoyt, my Vocational Ag teacher informed my class
that a person
with an I,Q,
of 79 obtained a degree from Iowa State University( a fairly
top notch
Engineering institution). I didnÕt believe him until reading
your verbose
garbage.
I assume that I will find you name enshrined in crankdom
hjs
no one can understand a math subject until they have been
through the next
highter
level, ie you donÕt understand trig until you have completed
calculus
E.E. Doc Smith.
===
Subject: Re: .99999... still=/= 1
>>mister Company is going to love this.
>> I just found an application of 0.9999...,
>>over and above OVERFLOW CONDITION or TENS COMPLIMENT.
> ThatÕs because,
> m--> oo
>> Lim (1 - 1/10^m) = 1
> divergent
>Convergent.
>> Ok donÕt my word for it, but at least believe the actual
Advanced
>Engineering
>>Math Book I have.
>>Quoting from:
>>6th Edition
>>Advanced Engineering Mathematics
>>by Erwin Kreyszig
>>Page 805
>>14.2 Convergence Tests for Series
>> Theorem 1 ( Divergence )
>>If a series z_1 + z_ 2 + .... converges, then
>> lim Z_m = 0
>> m-->oo
>>Hence if the series does not satisfy this condition, it
diverges.
>>SmartÕs Alt. Physics News Group
>>http://pub39.bravenet.com/forum/show.php?usernum=3320272813
&cpv=1
>>S. Enterprize (Science Journal)
>>http://smart1234.s-enterprize.com/
>In 1959 Mr. Hoyt, my Vocational Ag teacher informed my class
that a
person
>with an I,Q,
>of 79 obtained a degree from Iowa State University( a fairly
top notch
>Engineering institution). I didnÕt believe him until 
reading
your
verbose
>garbage.
>I assume that I will find you name enshrined in crankdom
>hjs
>no one can understand a math subject until they have been
through the
next
>highter
>level, ie you donÕt understand trig until you have 
completed
calculus
E.E.
>Doc Smith.
I did make a note of the mistake and made the corrections. I
used the Lim
Partial Sums to determine that particular convergence test of,
m-->oo
lim z_m = 0
The mistake was that I used,
m-->oo
lim SUM z_m = 0
But, even though .999... converges to 1,
.999... still=/= 1
Convergence doesnÕt mean equal to in any case or test.
It is interesting that the Partial Sums lim would fail that
particular
convergence test. When you SUM the total series at the lim at
infinity, it
fails the convergence test. So, if you SUM .999... totally,
it doesnÕt
converge, but if you take only the mth term it does.
In fact, a harmonic series can diverge and converge at the
same time.
Why?
Because the number 1 is involved.
But I think I still won the debate because the original
question was,
does
.999... = 1 and it doesnÕt, it converges to 1.
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
> It is interesting that the Partial Sums lim would fail that
particular
> convergence test. When you SUM the total series at the lim
at infinity,
it
> fails the convergence test. So, if you SUM .999... totally,
it doesnÕt
> converge, but if you take only the mth term it does.
There is no at infinity.
===
Subject: Re: .99999... still=/= 1
>> It is interesting that the Partial Sums lim would fail that
particular
>> convergence test. When you SUM the total series at the lim
at infinity,
it
>> fails the convergence test. So, if you SUM .999...
totally, it doesnÕt
>> converge, but if you take only the mth term it does.
>There is no at infinity.
Hey why try to change the subject?
.999... =/= 1
It converges to 1.
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Re: .99999... still=/= 1
> But, even though .999... converges to 1,
> .999... still=/= 1
Can you, will you dig the wax out of your ears and the sleepy
seeds from
your eyes?
To say Sum (9/10^n) [n >= 1] = 1 MEANS (by DEFINITION) that:
the limit of the sequence of partial sums is 1. Got it? It
does not say
that 1 = a partial sum for some n. It says 1 is the LIMIT of
the partial
sums. Which it to say that for any teeny tiny number eps > 0
we can
find an integer N(eps) such that for all partial sums for n >
N(eps) the
partial sums for n lie in the interval (1 - eps, 1). Can you
get that
through your thick head, you benighted putz?
Bob Kolker
===
Subject: Re: .99999... still=/= 1
>> But, even though .999... converges to 1,
>> .999... still=/= 1
>Can you, will you dig the wax out of your ears and the
sleepy seeds from
>your eyes?
>To say Sum (9/10^n) [n >= 1] = 1 MEANS (by DEFINITION) that:
>the limit of the sequence of partial sums is 1. Got it? It
does not say
>that 1 = a partial sum for some n. It says 1 is the LIMIT of
the partial
> sums. Which it to say that for any teeny tiny number eps >
0 we can
>find an integer N(eps) such that for all partial sums for n >
N(eps) the
>partial sums for n lie in the interval (1 - eps, 1). Can you
get that
>through your thick head, you benighted putz?
>Bob Kolker
Hey goofball,
.999... =/= 1
it converges to 1 in the series test.
SmartÕs Alt. Physics News Group
http://pub39.bravenet.com/forum/show.php?usernum=3320272813&
cpv=1
S. Enterprize (Science Journal)
http://smart1234.s-enterprize.com/
===
Subject: Equal Area Spherical Triangles Apex Locus on Sphere
AB is a fixed geodesic great circle arc, say a latitude
line,on a
sphere radius R.C moves so that area between arcs AB,BC and CA
{(A+B+C-pi)*R^2} is constant. What is locus of C?
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
ETAtAhUApQL6jYNcW0hWib4EVPgc7y0QsTUCFDITqMFsndm0APzGwrAYok6z6K
eo
ItÕs actually a closed curve, which seems very differentfrom
what we get
on the plane.
Imagine that AB is fixed on the Equator of a spherical Earth,
say in
South America with A in northern Ecuador and B in far nothern
Brazil.
Put C initally in the Carribean Sea and start moving it west,
adjusting
its latitude so that the triangle keeps the same area.
Eventually,
angle A reaches 180 degrees and C accordingly reaches the
Equator -- but
the triangle does not collapse into a straight line.
Conserving the
nonzero area has forced the formation of a lune with B and C
as the
poles (C is now in eastern Indonesia, opposite B). Sides AB
and AC
together form a 180-degree arc on the Equator, while side BC
is an
alternate great circular arc that byapsses the Equator to the
north.
Now you can ßip this lune through line AB, so that BC is now
in the
Southern Hemisphere, and continue to rotate C around the base
AB. C
goes through Bolivia and Brazil, all the way around to western
Indonesia, opposite A, where another lune is formed. Flip the
lune
again to complete tracing out the locus. The net result is a
closed
curve, symmetric about AB, with cusps a the points opposite A
and B.
If you specify a smnall lenght ofr AB and a small are for the
triangle,
the locus in the vicinity of AB looks like the pair of
parallel lines
you get on the plane.
--OL
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
> ItÕs actually a closed curve, which seems very
differentfrom what we get
> on the plane.
> Imagine that AB is fixed on the Equator of a spherical
Earth, say in
> South America with A in northern Ecuador and B in far
nothern Brazil.
> Put C initally in the Carribean Sea and start moving it
west, adjusting
> its latitude so that the triangle keeps the same area.
Eventually,
> angle A reaches 180 degrees and C accordingly reaches the
Equator -- but
> the triangle does not collapse into a straight line.
Conserving the
> nonzero area has forced the formation of a lune with B and
C as the
> poles (C is now in eastern Indonesia, opposite B). Sides AB
and AC
> together form a 180-degree arc on the Equator, while side
BC is an
> alternate great circular arc that byapsses the Equator to
the north.
> Now you can ßip this lune through line AB, so that BC is
now in the
> Southern Hemisphere, and continue to rotate C around the
base AB. C
> goes through Bolivia and Brazil, all the way around to
western
> Indonesia, opposite A, where another lune is formed. Flip
the lune
> again to complete tracing out the locus. The net result is
a closed
> curve, symmetric about AB, with cusps a the points opposite
A and B.
OK, ready on our tracks for a symbolic quantificating
trigonometry...
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
ETAuAhUAl7G/
DoN3ibTTTtFAwwKX0m7AnQcCFQCUOVX9ZFzKEYVRs9Nd4y62DEDKNg==
OK, IÕll start that.
Let LtA be the latitude of A, north positive, LoA be the
longitude of A,
east positive, with similar nomenclature for points B and C.
Set N to
quantities we have:
cos(AB) = sin(LtA)sin(LtB)+cos(LtA)cos(LtB)cos(LoA-LoB)
Likewise draw Triangles NAC and NBC and get analogous
expressions for
cos(AC) and cos(BC). With the sides of Triangle ABC thus
characterized,
obtain the angles in that triangle by using the Law of
Cosines solved
for a vertex angle; for example:
cos C = (cos(AB)-cos(AC)cos(BC))/(sin(AC)sin(BC))
where angle C is taken to be within Triangle ABC (no longer
involving
N). To get the sines in the denominator you must take square
roots (sin
x = (+/-) sqrt(1-cos^2 x)); give the square roots positive
signs because
the arcs measure 180 degrees or less. Take the inverse
cosines of your
angles, add them up and set the sum to the desired value.
The locus may be a simple closed curve, but its algebra is
not very
simple!
--OL
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
> AB is a fixed geodesic great circle arc, say a latitude
line,on a
> sphere radius R.C moves so that area between arcs AB,BC and
CA
> {(A+B+C-pi)*R^2} is constant. What is locus of C?
I think you mean longitude line. Latitude lines are not great
circles,
except for the equator.
--Mark
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
> AB is a fixed geodesic
great circle arc, say a latitude line,on a
>> sphere radius R.C moves so that area between arcs AB,BC
and CA
>> {(A+B+C-pi)*R^2} is constant. What is locus of C?
>I think you mean longitude line. Latitude lines are not great
circles,
>except for the equator.
Either way, I would guess that the answer is 2 circles
centered on the same
axis as AB. For example, if AB is a segment of the equator, C
would be a
pair of latitude lines, 1 north and 1 south.
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Equal Area Spherical Triangles Apex Locus on
Sphere
>> AB is a fixed geodesic great circle arc, say a latitude
line,on a
>> sphere radius R.C moves so that area between arcs AB,BC
and CA
>> {(A+B+C-pi)*R^2} is constant. What is locus of C?
>I think you mean longitude line. Latitude lines are not great
circles,
>except for the equator.
Yes as Mark said, what I meant.
> Either way, I would guess that the answer is 2 circles
centered on the
same
> axis as AB. For example, if AB is a segment of the equator,
C would be a
> pair of latitude lines, 1 north and 1 south.
> --Keith Lewis
If base AB is on equator/parallel circle,we get two parallel
latitudes
as required locus for C, right? If so,that tantamounts to
extending/guessing the area formula when AB is on parallel
circle(ph
latitude,th longitude, points 1,2 for A,B),and we have:
(A+B+C-pi)*R^2 = R^2*(sin(ph1)-sin(ph2)).(th1-th2) Or,
Spherical Excess =(A+B+C-pi) = (sin(ph1)-sin(ph2)).(th1-th2)?
Is this
result mentioned in some spherical trigonometry text-books?
===
Subject: Galois-Theory
I want to solve the following exercise.
Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
normal. Determine Gal E/Q.
I have no idea, what to do. I know two ways to show, that E/Q
is normal.
First: show, that E is the splitting field of a polynomial g
in Q[x].
Second: Q=Inv G, where G is a finite Group of automorphisms of
E.
I think the first method needs too many calculations. But i
dont know how
to
realize the second way. If i assume, that u is not an element
of
Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of
order 8. I think
that G=, where r:sqrt(2)->-sqrt(2),
s:sqrt(3)->-sqrt(3), t:u->-u.
But how to prove it? And how can i show, that u i not an
element of
Q(sqrt(2),sqrt(3))?
Is there another method?
Johannes
===
Subject: Re: Galois-Theory
>I want to solve the following exercise.
>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that
E/Q is normal.
>First: show, that E is the splitting field of a polynomial g
in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>I think the first method needs too many calculations.
Not really. You know the polynomial that will give you
sqrt(2):
x^2-2. You know the one for sqrt(3), namely x^2-3. So if we
let F =
Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
Now, E = F(u), and is the splitting field, over F, of the
polynomial
x^2 - (9-5sqrt(3))(2-sqrt(2)).
So E must be a splitting field over Q as well. To 
find the
polynomial
explicitly, take
x^2 - (9-5sqrt(3))(2-sqrt(2))
x^2 - (9+5sqrt(3))(2-sqrt(2))
x^2 - (9-5sqrt(3))(2+sqrt(2))
x^2 - (9+5sqrt(3))(2+sqrt(2))
and multiply them together to get a polynomial g(x). Then E
is the
spliting field over Q of g(x)(x^2-2)(x^2-3). You 
donÕt have to
calculate g(x) explicitly, just note that it will work.
>But i dont know how to
>realize the second way. If i assume, that u is not an
element of
>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of
order 8. I think
>that G=, where r:sqrt(2)->-sqrt(2),
s:sqrt(3)->-sqrt(3), t:u->-u.
There may be some relations between r, s, and t; you should
figure out
what r and s will do to u (see what they do to u^2). There
arenÕt that
many groups of order 8 (only three abelian and two nonabelian
ones),
so you should be able to figure out which one it is.
>But how to prove it? And how can i show, that u i not an
element of
>Q(sqrt(2),sqrt(3))?
Suppose u is an element of Q(sqrt(2),sqrt(3)). That would
mean that u
is of the form
u = a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6)
for some rational numbers a, b, c, and d. Show that this is
impossible, given what u^2 is supposed to be.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
>>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>>I have no idea, what to do. I know two ways to show, that
E/Q is normal.
>>First: show, that E is the splitting field of a polynomial g
in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>>I think the first method needs too many calculations.
> Not really. You know the polynomial that will give you
sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we
let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
> Now, E = F(u), and is the splitting field, over F, of the
polynomial
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
> So E must be a splitting field over Q as well.
You should be careful; Q(sqrt{2})/Q is normal and
Q(sqrt[4]{2})/Q(sqrt{2}) is normal, but clearly
Q(sqrt[4]{2})/Q is
not normal...
Actually, I think that the second method is better. Since the
degree of
the extension Q(sqrt{2}, sqrt{3}, u)/Q is 8, we have good
candidates
for elements of the Galois group, namely f_1, ..., f_8, which
send
sqrt{2}, sqrt{3} and u to:
sqrt{2} sqrt{3} u
f_1 sqrt{2} sqrt{3} sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
f_2 -sqrt{2} sqrt{3} sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
f_3 sqrt{2} -sqrt{3} sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
f_4 sqrt{2} sqrt{3} -sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
f_5 -sqrt{2} -sqrt{3} sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
f_6 -sqrt{2} sqrt{3} -sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
f_7 sqrt{2} -sqrt{3} -sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
f_8 -sqrt{2} -sqrt{3} -sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
The only thing that we have to worry about is whether f_is are
well-defined, that is whether
(9 +/- 5sqrt{3})(2 +/- sqrt{2})
are squares in Q(sqrt{2}, sqrt{3}, u). Indeed, they are -
take for
example:
(9 - 5sqrt{3})(2 + sqrt{2}) =
= [(9 - 5sqrt{3})( 2 - sqrt{2})(2 + sqrt{2})^2] /
/ [(2 - sqrt{2})(2 + sqrt{2})] =
= u^2*(2 + sqrt{2})^2 / (sqrt{2}^2).
Similar argument works also for other terms, because:
(9 - 5sqrt{3})(9 + 5sqrt{3}) = 81 - 75 = 6 =
(sqrt{2}sqrt{3})^2
(2 - sqrt{2})(2 + sqrt{2}) = 4 - 2 = 2 = (sqrt{2})^2.
Pawel Gladki
===
Subject: Re: Galois-Theory
days. My association with the Department is that of an
alumnus.
>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that
E/Q is normal.
>First: show, that E is the splitting field of a polynomial g
in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>I think the first method needs too many calculations.
>> Not really. You know the polynomial that will give you
sqrt(2):
>> x^2-2. You know the one for sqrt(3), namely x^2-3. So if
we let F =
>> Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
>> Now, E = F(u), and is the splitting field, over F, of the
polynomial
>> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>> So E must be a splitting field over Q as well.
>You should be careful; Q(sqrt{2})/Q is normal and
>Q(sqrt[4]{2})/Q(sqrt{2}) is normal, but clearly
Q(sqrt[4]{2})/Q is
>not normal...
Yes.
>Actually, I think that the second method is better. Since
the degree of
>the extension Q(sqrt{2}, sqrt{3}, u)/Q is 8, we have good
candidates
>for elements of the Galois group, namely f_1, ..., f_8,
which send
>sqrt{2}, sqrt{3} and u to:
> sqrt{2} sqrt{3} u
>f_1 sqrt{2} sqrt{3} sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
>f_2 -sqrt{2} sqrt{3} sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
>f_3 sqrt{2} -sqrt{3} sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
>f_4 sqrt{2} sqrt{3} -sqrt{(9 - 5sqrt{3})( 2 - sqrt{2})}
>f_5 -sqrt{2} -sqrt{3} sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
>f_6 -sqrt{2} sqrt{3} -sqrt{(9 - 5sqrt{3})( 2 + sqrt{2})}
>f_7 sqrt{2} -sqrt{3} -sqrt{(9 + 5sqrt{3})( 2 - sqrt{2})}
>f_8 -sqrt{2} -sqrt{3} -sqrt{(9 + 5sqrt{3})( 2 + sqrt{2})}
>The only thing that we have to worry about is whether f_is
are
>well-defined, that is whether
>(9 +/- 5sqrt{3})(2 +/- sqrt{2})
>are squares in Q(sqrt{2}, sqrt{3}, u).
Which is equivalent to showing that the polynomial g(x) I
described,
which is the product of
x^2 - ri
for r1, r2, r3, r4 equal, successively, to (9 +/-
5*sqrt(3))(2 +/- sqrt(2))
does indeed have all its roots in Q(sqrt(2),sqrt(3),u), and
so that
this field is the splitting field of 
(x^2-2)(x^2-3)g(x).
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
Hello Arturo,
>>I want to solve the following exercise.
>>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>>I have no idea, what to do. I know two ways to show, that
E/Q is normal.
>>First: show, that E is the splitting field of a polynomial g
in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>>I think the first method needs too many calculations.
> Not really. You know the polynomial that will give you
sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we
let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
OK, this is clear to me.
> Now, E = F(u), and is the splitting field, over F, of the
polynomial
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
> So E must be a splitting field over Q as well. To 
find the
polynomial
Is there a theorem like this?: Let E be a splitting field of f
over K, and
K
be a splitting field of g over F, then E is the splitting 
field
of a
polynomial over F.
> explicitly, take
> x^2 - (9-5sqrt(3))(2-sqrt(2))
> x^2 - (9+5sqrt(3))(2-sqrt(2))
> x^2 - (9-5sqrt(3))(2+sqrt(2))
> x^2 - (9+5sqrt(3))(2+sqrt(2))
I dont see, why the roots of these polynomials is in F(u) too?
> and multiply them together to get a polynomial g(x). Then E
is the
> spliting field over Q of g(x)(x^2-2)(x^2-3). You 
donÕt have
to
> calculate g(x) explicitly, just note that it will work.
>>But i dont know how to
>>realize the second way. If i assume, that u is not an
element of
>>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of
order 8. I
think
>>that G=, where r:sqrt(2)->-sqrt(2),
s:sqrt(3)->-sqrt(3), t:u->-u.
> There may be some relations between r, s, and t; you should
figure out
> what r and s will do to u (see what they do to u^2). There
arenÕt that
I have the problem, that i cannot see, why such automorphisms
exists? Well,
the identity on F can be extended to an automorphism t of E
sending u->-u
since these are roots of an minimum polynomial m of u over F.
But why can i
extend the automorphisms 
rÕ,sÕ:sqrt(k)->-sqrt(k) k=2,3 on F to
automorphisms of E? To show this i must know that rÕ(m)
[transformed of
Minimum Polynomial of u under rÕ] has also roots in F(u)?
> many groups of order 8 (only three abelian and two
nonabelian ones),
> so you should be able to figure out which one it is.
>>But how to prove it? And how can i show, that u i not an
element of
>>Q(sqrt(2),sqrt(3))?
> Suppose u is an element of Q(sqrt(2),sqrt(3)). That would
mean that u
> is of the form
> u = a + b*sqrt(2) + c*sqrt(3) + d*sqrt(6)
> for some rational numbers a, b, c, and d. Show that this is
> impossible, given what u^2 is supposed to be.
I thought, this would be too complicate. But i will try it
again.
Johannes
===
Subject: Re: Galois-Theory
days. My association with the Department is that of an
alumnus.
>Hello Arturo,
>I want to solve the following exercise.
>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that
E/Q is normal.
>First: show, that E is the splitting field of a polynomial g
in Q[x].
>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>I think the first method needs too many calculations.
>> Not really. You know the polynomial that will give you
sqrt(2):
>> x^2-2. You know the one for sqrt(3), namely x^2-3. So if
we let F =
>> Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
>OK, this is clear to me.
>> Now, E = F(u), and is the splitting field, over F, of the
polynomial
>> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>> So E must be a splitting field over Q as well. To 
find the
polynomial
>Is there a theorem like this?: Let E be a splitting field of
f over K, and
K
>be a splitting field of g over F, then E is the splitting
field of a
>polynomial over F.
Not as such, no; I may have been too hasty there. In fact,
you notice
the problem below: it is easy enough to construct a
polynomial with
coefficients in Q that has u as a root, but there is no
guarantee in
general that K(u) will contain all those roots. In general,
normal
over normal is not necessarily normal.
>> explicitly, take
>> x^2 - (9-5sqrt(3))(2-sqrt(2))
>> x^2 - (9+5sqrt(3))(2-sqrt(2))
>> x^2 - (9-5sqrt(3))(2+sqrt(2))
>> x^2 - (9+5sqrt(3))(2+sqrt(2))
>I dont see, why the roots of these polynomials is in F(u)
too?
Yes, youÕre right. That is not immediate. Sorry about that.
Let me get
back to you.
>> and multiply them together to get a polynomial g(x). Then
E is the
>> spliting field over Q of g(x)(x^2-2)(x^2-3). You 
donÕt have
to
>> calculate g(x) explicitly, just note that it will work.
>But i dont know how to
>realize the second way. If i assume, that u is not an
element of
>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of
order 8. I
think
>that G=, where r:sqrt(2)->-sqrt(2),
s:sqrt(3)->-sqrt(3), t:u->-u.
>> There may be some relations between r, s, and t; you
should figure out
>> what r and s will do to u (see what they do to u^2). There
arenÕt that
>I have the problem, that i cannot see, why such
automorphisms exists?
Well,
>the identity on F can be extended to an automorphism t of E
sending u->-u
>since these are roots of an minimum polynomial m of u over
F. But why can
i
>extend the automorphisms 
rÕ,sÕ:sqrt(k)->-sqrt(k) k=2,3 on F
to
>automorphisms of E? To show this i must know that rÕ(m)
[transformed of
>Minimum Polynomial of u under rÕ] has also roots in F(u)?
This will follow once you know that E is a splitting field (in
characteristic zero, anyway); it is one of the basic
properties of
splitting fields: if E/k is a splitting field, and 
F is a
subfield of
E containing k, then every automorphism of E that fixes k
pointwise
can be extended to an automorphism of F.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Re: Galois-Theory
Hello Arturo,
>>I want to solve the following exercise.
>>Show that E=Q(sqrt(2),sqrt(3),u) where
u^2=(9-5sqrt(3))(2-sqrt(2)) is
>>normal. Determine Gal E/Q.
>I have no idea, what to do. I know two ways to show, that
E/Q is
normal.
>>First: show, that E is the splitting field of a polynomial g
in Q[x].
>>Second: Q=Inv G, where G is a finite Group of automorphisms
of E.
>I think the first method needs too many calculations.
>
> Not really. You know the polynomial that will give you
sqrt(2):
> x^2-2. You know the one for sqrt(3), namely x^2-3. So if we
let F =
> Q(sqrt(2),sqrt(3)), then F is the splitting field of
(x^2-2)(x^2-3).
>>OK, this is clear to me.
> Now, E = F(u), and is the splitting field, over F, of the
polynomial
>
> x^2 - (9-5sqrt(3))(2-sqrt(2)).
>
> So E must be a splitting field over Q as well. To 
find the
polynomial
>>Is there a theorem like this?: Let E be a splitting field of
f over K,
and
>>K be a splitting field of g over F, then E is the splitting
field of a
>>polynomial over F.
> Not as such, no; I may have been too hasty there. In fact,
you notice
> the problem below: it is easy enough to construct a
polynomial with
> coefficients in Q that has u as a root, but there is no
guarantee in
> general that K(u) will contain all those roots. In general,
normal
> over normal is not necessarily normal.
> explicitly, take
>
> x^2 - (9-5sqrt(3))(2-sqrt(2))
> x^2 - (9+5sqrt(3))(2-sqrt(2))
> x^2 - (9-5sqrt(3))(2+sqrt(2))
> x^2 - (9+5sqrt(3))(2+sqrt(2))
>>I dont see, why the roots of these polynomials is in F(u)
too?
> Yes, youÕre right. That is not immediate. Sorry about 
that.
Let me get
> back to you.
> and multiply them together to get a polynomial g(x). Then E
is the
> spliting field over Q of g(x)(x^2-2)(x^2-3). You 
donÕt have
to
> calculate g(x) explicitly, just note that it will work.
>
>>But i dont know how to
>>realize the second way. If i assume, that u is not an
element of
>>Q(sqrt(2),sqrt(3)), then [E:Q]=8, so G must be a group of
order 8. I
>>think that G=, where r:sqrt(2)->-sqrt(2),
s:sqrt(3)->-sqrt(3),
>>t:u->-u.
>
> There may be some relations between r, s, and t; you should
figure out
> what r and s will do to u (see what they do to u^2). There
arenÕt that
>>I have the problem, that i cannot see, why such
automorphisms exists?
>>Well, the identity on F can be extended to an automorphism
t of E sending
>>u->-u since these are roots of an minimum polynomial m of u
over F. But
>>why can i extend the automorphisms 
rÕ,sÕ:sqrt(k)->-sqrt(k)
k=2,3 on F to
>>automorphisms of E? To show this i must know that rÕ(m)
[transformed of
>>Minimum Polynomial of u under rÕ] has also roots in F(u)?
> This will follow once you know that E is a splitting field
(in
> characteristic zero, anyway); it is one of the basic
properties of
> splitting fields: if E/k is a splitting field, 
and F is a
subfield of
> E containing k, then every automorphism of E that fixes k
pointwise
> can be extended to an automorphism of F.
Did you exchange E an F? Then I know this theorem, but the
thing is, that
we
dont know, whether E is a splitting field over Q, otherwise it
follows
directly, that E is normal over Q, since Q is perfect.
Or do you mean another way to apply the above property?
Johannes
===
Subject: Re: Galois-Theory
days. My association with the Department is that of an
alumnus.
>I have the problem, that i cannot see, why such
automorphisms exists?
>Well, the identity on F can be extended to an automorphism t
of E
sending
>u->-u since these are roots of an minimum polynomial m of u
over F. But
>why can i extend the automorphisms 
rÕ,sÕ:sqrt(k)->-sqrt(k)
k=2,3 on F to
>automorphisms of E? To show this i must know that rÕ(m)
[transformed of
>Minimum Polynomial of u under rÕ] has also roots in F(u)?
>> This will follow once you know that E is a splitting field
(in
>> characteristic zero, anyway); it is one of the basic
properties of
>> splitting fields: if E/k is a splitting field, 
and F is a
subfield of
>> E containing k, then every automorphism of E that fixes k
pointwise
>> can be extended to an automorphism of F.
>Did you exchange E an F?
Oops. Yes. (Not my thread...)
>Then I know this theorem, but the thing is, that we
>dont know, whether E is a splitting field over Q, otherwise
it follows
>directly,
Well, yes. But thatÕs why I said that it will follow ->once
you know
that E is a splitting field<-.
--
ItÕs not denial. IÕm just very selective 
about
what I accept as reality.
--- Calvin (Calvin and Hobbes)
Arturo Magidin
magidin@math.berkeley.edu
===
Subject: Polynomials
i want to show, that if r is a root in a splitting field of
f=x^3+x^2-2x-1
over Q, then rÕ=r^2-2 is another root.
I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but how
can I see,
that r^2-2 is a root of the quadratic factor?
Johannes
===
Subject: Re: Polynomials
> i want to show, that if r is a root in a splitting field of
f=x^3+x^2-2x-1
> over Q, then rÕ=r^2-2 is another root.
> I split f into factors f=(x-r)(x^2+(r+1)x+(r^2+r-2)), but
how can I see,
> that r^2-2 is a root of the quadratic factor?
factorization of f(x) by just dividing it by (x - r).
If thatÕs the case, then you should have _no_ trouble
finishing this -- just use polynomial long division
to compute (x^2 + (r_1)x + (f^2+r-2))/(x - (r^2-2)) ...
itÕll work out nicely _because_ f(r) = 0 -- kinda
neat, actually :-) (As a bonus, youÕll also determine
the third root of f in terms of r: itÕs r^2+r-1 ... )
> Johannes
===
Subject: Re: Polynomials
> i want to show, that if r is a root in a splitting field of
f=x^3+x^2-2x-1
> over Q, then rÕ=r^2-2 is another root.
r being a root of f(x) = x^3 + x^2 - 2x - 1 implies that
r^3 = 1 + 2r - r^2.
Then r^4 = r + 2r^2 - r^3 = r + 2r^2 - (1 + 2r - r^2) = -1 -
r + 3r^2.
Similarly you can express r^5 and r^6 in terms of 1, r, and
r^2.
Now multiply out f(r^2 - 2) = (r^2 - 2)^3 + ...
and reduce all the high powers of r as weÕve just seen is
possible,
and see what happens.
Alternatively, let rÕ= (rÕ)^2 
- 2; this must also be a root.
rÕ= (r^2 - 2)^2 - 2 = r^4 - 4 r^2 + 2 = (-1 
- r + 3r^2) - 4
r^2 + 2
= 1 - r - r^2. Now show that r, rÕ, and 
rÕare zeros of f.
The easy part is r + r+ rÕ= 
-1.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: 2-manifold metric spaces with many symmetries
> [...]
>> Here, I hadnÕt scanned up to see the use of n as the
dimension
>> of the ambient space, and used it incorrectly as the
dimension of
>> the submanifold; IÕve replaced n by k in this paragraph,
and note
>> that it refers to the dimension of the submanifold:
> The result you want requires the manifold to have a
transitive
> action of isometries by some Lie group G; in addition, this
> action must contain the rotation group SO(k) in the isotropy
> subgroup of G at x for each element x. That means that your
> k-manifold has a symmetry group of dimension at least
> k + k(k-1)/2 = k(k+1)/2. I note that this is the dimension
> of the group SO(k+1), the group of isometries of the
k-sphere.
> IÕll note that among the various differential structures 
on
> S^k (for k >= 7, these are non-diffeomorphic smooth
structures
> on the same topological space S^k), the only one that has a
> group of isometries of this dimension is the standard, round
> sphere.
> I suspect that this condition poses a severe constraint on
> the manifold, since itÕs as symmetric as the k-sphere.
> Am I right that you have made a generalization to
> k-manifolds, emmbedded in some euclidean space, with k>=2?
Yes, I suppose thatÕs correct. I had thought you were using
the term surface to refer to a closed manifold, rather than
the strict interpretation as a 2-dimensional object. The
question does make sense in higher dimensions, so itÕs a
reasonable generalization.
> IÕm trying to formulate the problem for a k-manifold M:
> Condition (i): For any x, y in M, there is an isometry of M
sending
> x to y. [ Transitivity ]
> Condition (ii): For any x in M, the subgroup of isometries
that leave
> x fixed is such that ___________ {something}.
> I donÕt know what to put in place of ___________.
This depends on your intent.
If all you want is for any triangle to be mappable to any
other
congruent triangle via an isometry, then the following is
sufficient:
the action is transitive on the
space of orthonormal 2-frames in
TM_x
An n-frame F in the vector space V is an ordered n-tuple of
linearly independent vectors in V. If V has an inner product,
an orthonormal n-frame is an n-frame for which the members
are mutually orthogonal, and each has norm 1.
If you have a k-manifold and want all congruent d-simplices
(that
is, the d-dimensional analogue of triangles [d=2]) to be
mappable
to one another via isometries, then you need this:
the action is transitive on the
space of orthonormal d-frames in
TM_x
For d=k, that is you want all congruent k-simplices to be
mappable
to one another via isometries, this is (obviously)
the action is transitive on the
space of orthonormal k-frames in
TM_x
But the space of orthonormal k-frames is the full orthogonal
group O(k). Since the isotropy subgroup associated to a fixed
point of an isometry must be orthogonal (i.e., a subgroup of
O(k)), you get this:
the group is O(k)
You may restrict to orientation-preserving maps, in which the
group would be SO(k).
IÕll note that having the condition met for d-simplices 
forces
the condition to hold for all dimensions <= d.
So, hereÕs what this all amounts to:
1. Your group of isometries acts transitively on M.
That means that M is whatÕs called a *homogeneous space*, 
that
is, a manifold with a transitive action by a Lie group. Given
any
homogeneous space M, one can construct a metric for which the
action
is via isometries (well, I know how to do this for actions by
discrete
groups, by averaging the metric over orbits; I think this
still works
for actions by compact Lie groups in general, by integrating
the
metric wrt Haar measure).
2. The isotropy subgroup at x acts transitively on the space
of orthonormal d-frames in TM_x.
The clasification of subgroups of O(k) that act transitively 
on
the space of orthonormal d-frames in R^k (alias the Stiefel
manifold V(d,k-d)) may be well understood; it isnÕt
well-understood
by me, so this poses a bit of a problem.
However, if d=k, then you get the isotropy subgroups all being
equal to the orthogonal group O(k), or if you only want
orientation
preserving transformations, then the special orthogonal group
SO(k).
In that case, hereÕs how to construct an example:
Take a compact Lie group L of dimension k(k+1)/2,
that contains SO(k) as a subgroup. Then form the
quotient L/SO(k), and that will be your manifold
M. The action of L on M is via multiplication:
L x L ---> L
except you need to divide out by the subgroup SO(k):
L x (L/SO(k)) ---> L x (L/SO(k))
This is not so hard, since the cosets of SO(k) in L
are of the form
x = a SO(k),
for a being any element of the group L. Multiplication
by any b in L is then well-defined:
b (a SO(k)) = ba SO(k)
and you find that if b fixes a particular element 
x of M,
then
b x = b (a SO(k)) = (ba) SO(k)
so we find that
(ba) SO(k) = a SO(k)
or rather
ab a in SO(k), (where ais a^(-1))
or
b in a SO(k) aÕ
That is, the isotropy subgroup at x = a SO(k) is conjugate
(and thus
isomorphic) to the subgroup SO(k).
For the orientation-preserving case, and requiring oriented
congruent
k-simplices to be mapped isometrically to one another, this
construction
will produce all examples.
> David Bernier
Dale
===
Subject: Re: inverse modulos
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB2JbUg32369;
>If two numbers x and m are relatively prime, so that gcd(x,
m) = 1 is
>true, then x has a unique multiplicative inverse modulo m,
a, so that ax
>= 1 (mod m).
>Knowing only the multiplicative inverse, a modulo m, and m,
is it
>possible to find x?
>Is it true that a*x - 1 = k*m, for some k, possibly
negative? Or
>multiple kÕs? How to find the multiple 
kÕs so that x could be
found?
>Is that possible at all, so that the candidate kÕs can be
found knowing
>only the inverse a modulo m and m?
>If there is no single way to find it, is there a reasonable
trial and
>error process to go through to find x given a and m?
Using the ideas from extended GCD or continued fraction.
Take the continued fraction expansion of the fraction a/m,
and truncate it at various stages to get good rational
approximations to a/m with small numerator and denominator.
One of those will be k/x.
===
Subject: Re: inverse modulos
> Using the ideas from extended GCD or continued fraction.
> Take the continued fraction expansion of the fraction a/m,
> and truncate it at various stages to get good rational
> approximations to a/m with small numerator and denominator.
> One of those will be k/x.
Is there somewhere on the web that explains how I can carry
out these
use it for this... it seems to be only for trig functions).
When should
I truncate? And then how would I separate the k and the x in
k/x?
===
Subject: sci.physics , sci.math is nonsense
shut up when i m talking to you. shut up!!
===
Subject: Re: sci.physics , sci.math is nonsense
> shut up when i m talking to you. shut up!!
Speak louder we canÕt hear you.
===
Subject: Re: sci.physics , sci.math is nonsense
shut up when i m taking to you. shut up, shut up. shut up
when i m taking to
you!!!
===
Subject: Re: sci.physics , sci.math is nonsense
is that what your mama said to you over and over?
> shut up when i m taking to you. shut up, shut up. shut up
when i m taking
to
> you!!!
===
Subject: Re: sci.physics , sci.math is nonsense
hello punk, certified punk
> is that what your mama said to you over and over?
>> shut up when i m taking to you. shut up, shut up. shut up
when i m
taking
> to
>> you!!!
===
Subject: Re: sci.physics , sci.math is nonsense
format=ßowed;
reply-type=response
> hello punk, certified punk
are you boring on purpose?
===
Subject: Re: sci.physics , sci.math is nonsense
<319evkF38gkbvU1@individual.net>
<319v8eF2qj9e4U1@individual.net>
<31bnb4F3agmorU1@individual.net>

posting-account=YlfmdwwAAACC77JOQxzd52GbZnwIMh2h
> hello punk, certified punk
> are you boring on purpose?
You know, for some reason, Lord of Chaos seems...
ŌFrazir-likeÕ.
-Mark Martin
===
Subject: Re: sci.physics , sci.math is nonsense
your mom again
>> hello punk, certified punk
> are you boring on purpose?
===
Subject: Re: Personal conjectures
> Hello sci.math readers,
> . IÕm sure that every mathematician has one or more
> personal conjectures that he/she would give his/her soul to
prove.
> IÕd like to see the conjectures that you have thought up,
and that
> you think about whenever you have a spare moment.
> Asger.
> I love Conjectures about prime numbers. This one I framed
many years
> ago:
> For each integer N >= 1 there is an X>=0 that produces a
prime
> number with the formula:
> 6(N - X^2) + 1
> Also for each integer N>=1 there is an Y>=0 that produces a
prime with
> the formula:
> 6(N - Y^2) - 1
> That is, before X or Y attain sqr(N) it will result a prime
number .
> Ludovicus
Naturally the value of X augments with N, but what about the
relation :
r = X / [log(N)]^2 for the first prime found? In this case my
record
in the formula 6(N - X^2) + 1 is 0.585 for the values N =
71160 , X
= 73.
And r = 0.528 in the formula 6(N - Y^2) - 1 N = 16892, Y = 50.
Ludovicus
===
Subject: i clean up sci.physics, sci.math, rec.mensa.org
gentleman bets
taken!!!
===
Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org
gentleman bets
taken!!!
plonk
===
Subject: Re: i clean up sci.physics, sci.math, rec.mensa.org
gentleman bets
taken!!!
pink
> plonk
===
Subject: talk science and mathematics, you faggots.
i m supreme. i alone command you.
===
Subject: Re: talk science and mathematics, you faggots.
format=ßowed;
reply-type=response
> i m supreme. i alone command you.
still boring...
===
Subject: Re: talk science and mathematics, you faggots.
>> i m supreme. i alone command you.
> still boring...
Lunacy is never boring. BTW CHAOS stands for Could (do to)
Have Another
Olanzapine Soon.
===
Subject: Re: talk science and mathematics, you faggots.
In sci.math, Lord of Chaos(Suresh Devanathan)

<319ß1F36rekiU1@individual.net>:
> i m supreme. i alone command you.
[1] A sultan commands his driver to get from point A to point
B
60 miles distant in 1 hour. The driver, for various reasons,
drives the first 30 minutes at 30 miles per hour. How fast
would he have to drive the rest of the distance in order to
make it, and how far would he have to travel?
[2] A proposed US single-stage rocket craft has a thrust of
36 megaNewtons for 120 seconds, and an exhaust velocity of 4
km/s.
Assuming that the fuel tanks comprise 3% of the rocketÕs
mass (sans payload), the radius of the Earth is 6378 km, and
the
surface g is 9.805 m/s/s, can the rocket make it to a circular
orbit 200 km above the surface, and if so, how much payload
could it carry? Neglect rotation and air friction in the
computation. (Hint: Tsiolkovsky.)
[3] Same as [2], only this time the question is whether it can
escape EarthÕs gravitational pull altogether, and how much
payload it can carry.
[4] A construction company has been contracted to build a dam
out
of cement, and the specification requires a certain thickness
of
cement at the bottom of the dam. This thickness can hold
back 100 megaNewtons of pressure. However, the thickness
also has to contend with its own weight, which is 9.805 m/s/s
= N/kg.
The mass of the concrete is 2400 kg/m^3. Assuming that one
can fashion a triangular dam wedge, what is the maximum depth
of water one can hold back with this concrete, and how thick
is the dam at its base? Assume a water density of 1000 kg/m^3.
[5] A ßat iron bar of indefinite length and square cross
section of 1 cm
is propped up on at least two round rods, in such a fashion
that it can travel freely. If the round rods are each 1 mm in
diameter, how far apart can they be spaced without the iron
bar
touching the ground? Assume a rigidity modulus of 82
gigaPascal
and a density of 7874 kg/m^3; neglect distortion in the
support rods.
Have fun, especially since I donÕt know the answers either!
(Except
to #1. No, itÕs not infinite speed.)
--
#191, ewill3@earthlink.net -- no, I donÕt do this for a 
living
ItÕs still legal to go .sigless.
===
Subject: Re: talk science and mathematics, you faggots.
We ARE discussing science and mathematics!
> i m supreme. i alone command you.
===
Subject: Re: talk science and mathematics, you faggots.
Yes you are... yes you are
> We ARE discussing science and mathematics!
>> i m supreme. i alone command you.
===
Subject: Re: talk science and mathematics, you faggots.
> i m supreme. i alone command you.
Hey stooopid,
1) Turbulence scales as the cube of the distance.
2) Graph your pathetic little brainfart on hyperbolic axes.
3) Your trailer park called - its garbage is missing.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
===
Subject: Re: talk science and mathematics, you faggots.
shut up! shut up when i m talking to you!
>> i m supreme. i alone command you.
> Hey stooopid,
> 1) Turbulence scales as the cube of the distance.
> 2) Graph your pathetic little brainfart on hyperbolic axes.
> 3) Your trailer park called - its garbage is missing.
> --
> Uncle Al
> http://www.mazepath.com/uncleal/
> (Toxic URL! Unsafe for children and most mammals)
> http://www.mazepath.com/uncleal/qz.pdf
===
Subject: Re: talk science and mathematics, you faggots.
> shut up! shut up when i m talking to you!
> i m supreme. i alone command you.
>> Hey stooopid,
>> 1) Turbulence scales as the cube of the distance.
>> 2) Graph your pathetic little brainfart on hyperbolic axes.
>> 3) Your trailer park called - its garbage is missing.
>> --
>> Uncle Al
>> http://www.mazepath.com/uncleal/
>> (Toxic URL! Unsafe for children and most mammals)
>> http://www.mazepath.com/uncleal/qz.pdf
Lord of boring.
--
Sl.87inte,
Fletch
===
Subject: Re: talk science and mathematics, you faggots.
lord of you
===
Subject: Re: talk science and mathematics, you faggots.
you are not talking to him, you are typing at him.
===
Subject: Re: talk science and mathematics, you faggots.
> you are not talking to him, you are typing at him.
Not me, I bozobinned him.
Pepe le Pew
--
PT Barnum was right !
===
Subject: Re: talk science and mathematics, you faggots.
> i m supreme. i alone command you.
Chaos = 1/0
Lord of Chaos = (1/0)^2
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Universal sentences of Z and ZxZ
How to prove that the universal sentences of Z are the one of
ZxZ, both
considered as additive groups ?
A proof using ultraproduct will be much appreciated. And to
avoid
discimination, others also!
===
Subject: how to measure entropy of music?
I have vaguely heard about this method... and I am very
interested in it...
could anybody give me some pointers?
After learning how to measure entropy of music... I can begin
to measure
entropy of texts, etc.. thatÕs going to be fun!
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very
interested in
it...
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can
begin to measure
> entropy of texts, etc.. thatÕs going to be fun!
Well, theoretically speaking, you should just treat the notes
of
the music as symbols for the computation of the entropy. Then,
there is no difference between finding the entropy of text and
finding
the entropy of music.
But if you sample the signal of music, encode it, and compute
the
entropy, I doubt if it makes sense to do that given the fact
that you
already have the notes of the music.
If you just want to know what is about entropy of data, you
might
just start with any textbook for Information theory or some
websites
for some ideas, e.g.,
http://www.ScienceOxygen.com/signal.html
http://www.ScienceOxygen.com/electrical.html
Have fun,
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very
interested in
it...
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can
begin to measure
> entropy of texts, etc.. thatÕs going to be fun!
Compress it using a suitable lossless compressor. Provided
that the
compressor knows how to detect (some amount of) musical
redundancy,
this will give you the Kolmogorov complexity with respect to a
particular non-universal computer, e.g. algorithmic entropy
which is
more fundamental than Shannon entropy.
This has already been exploited using the information
distance of
Vitanyi et al. Precise clustering of music files has been
achieved
with ordinary compressors like bzip2. Find the paper
Algorithmic
Clustering of Music by Rudi Cilibrasi and Paul Vitanyi.
Needless to
say, you have to compress uncompressed wav files...
--
Eray Ozkural
===
Subject: Re: how to measure entropy of music?
(snip, and previously snipped discussion of music entropy)
> Compress it using a suitable lossless compressor. Provided
that the
> compressor knows how to detect (some amount of) musical
redundancy,
> this will give you the Kolmogorov complexity with respect
to a
> particular non-universal computer, e.g. algorithmic entropy
which is
> more fundamental than Shannon entropy.
> This has already been exploited using the information
distance of
> Vitanyi et al. Precise clustering of music files has been
achieved
> with ordinary compressors like bzip2. Find the paper
Algorithmic
> Clustering of Music by Rudi Cilibrasi and Paul Vitanyi.
Needless to
> say, you have to compress uncompressed wav files...
I would think a pure sine wave should be low complexity, so
maybe
you should compress the Fourier transform. Maybe that wonÕt
quite do it, but I can imagine wav files of low complexity
audio signals not compressing very well.
Has anyone ever done an FFT of a whole CD?
-- glen
===
Subject: Re: how to measure entropy of music?
> I would think a pure sine wave should be low complexity, so
maybe
> you should compress the Fourier transform.
I think a vocoder does something similar: linear predictive
coding
tries to fit an AR model to the input, effectively searching
for the
formants (resonant peaks) of the vocal tract transfer
function. If
you initialize an IIR filter with the LPC 
coefficients and the
states
(delays) with the signal, you can let it ring like an
oscillator -
the output will converge to a sum of pure damped sine waves
at the
estimated frequencies of the formants. However, instead of
storing the
sine waves, the vocoder stores the LPC coefficients and the
residue
(prediction error).
To take this back to entropy, the energy of the prediction
error could
well be regarded as the entropy of a vocal signal - it gives
a good
measure on how unexpected the signal is, given its past
history.
Recent work has shown that this method is also suited for
general
audio signals, not just speech. The order of the AR model just
increases drastically (for interpolation, orders of 1000 -
3000 are
used). The same applies here for the prediction error and the
entropy.
> Has anyone ever done an FFT of a whole CD?
If you take a music CD, I would expect a linear trend of the
form
1/f^a. This is a common model for music signals, and it
agrees well
with findings of long-range correlated time series.
For speech, I would guess a rectangular pulse response. The
formants
vary across a certain frequency range, but are limited from
above and
below.
> -- glen
Andor
===
Subject: Re: how to measure entropy of music?
> I have vaguely heard about this method... and I am very
interested in
it...
> could anybody give me some pointers?
> After learning how to measure entropy of music... I can
begin to measure
> entropy of texts, etc.. thatÕs going to be fun!
Learn all you want and more at Brian WhitmanÕs site...

Whitman is also the producer of Eigenradio (their motto:
Statistically
their iTunes and WMP links...

--
Remove _me_ for e-mail address
===
Subject: Re: how to measure entropy of music?
is this some joke ?!
>I have vaguely heard about this method... and I am very
interested in it...
>could anybody give me some pointers?
> After learning how to measure entropy of music... I can
begin to measure
> entropy of texts, etc.. thatÕs going to be fun!
===
Subject: Re: how to measure entropy of music?
>> After learning how to measure entropy of music... I can
begin to measure
>> entropy of texts, etc.. thatÕs going to be fun!
>is this some joke ?!
The model of entropy that can be readily recognized:
In music--how often can an informed listener infer the next
note in a
phrase; ie how many bits are needed to specifiy the next note.
In sound--how many bits are needed to specify the next value
of the
signal.
For written language, the analog is how many bits are needed
to
confirm an informed guess as to the next letter in a text.
Perhaps no
more than three. For some TV shows these days, its only two.
For music, the question gets really challenging as one
considers the
entropy of scores--the parts played by accompanying
instruments and
the choices of these instruments requires a lot of encoding.
In this
case, the number of bits needed to feed a high level
orchestral
synthesizer might establish a lower bound.
Using a predictor such as a Hidden Markov Model using the
Viterbi
algorithm or one of its descendants might go a long way in
evaluating
the predictability/aka entropy of sounds. (This is already
the case in
speech recognition.)
A really interesting question would be to rank famous
composers by the
average entropy of their music--eg Bach, Mozart, the Beatles
at the
top, Andrew Lloyd Webber, Elton John, Cole Porter next--well,
you get
the idea.
The basics are at:
http://www.music-cog.ohio-state.edu/Music829D/Notes/
Infotheory.html
key words: hmm, entropy, and music I have learned the
question is of
significance today for serveral reasons:
1) Recognizing when TV commercials are playing.
2) Separating out background music behind speech in speech
recognition.
3) Locating specific music contained in a large database of
sound
Try:
http://crl.research.compaq.com/publications/techreports/
techreports.html
search the page for Logan or music.
Also:
http://www.idiap.ch/publications/ajmera-rr-01-26.bib.abs.html
Speech/Music Discrimination using Entropy and Dynamism
Features in a
HMM Classification Framework
Speech/Music Discrimination Using Discrete Hidden Markov
Models
http://crl.research.compaq.com/publications/techreports/
reports/2000-1.pdf
MUSIC SUMMARY USING KEY PHRASES
M. Brand, Structure Discovery in Conditional Probability
Models
via
an Entropic Prior and Parameter Extinction, Neural
Computation,
July
1999
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: Re: how to measure entropy of music?
> For written language, the analog is how many bits are
needed to
> confirm an informed guess as to the next letter in a text.
Perhaps no
> more than three. For some TV shows these days, its only two.
> For music, the question gets really challenging as one
considers the
> entropy of scores--the parts played by accompanying
instruments and
> the choices of these instruments requires a lot of
encoding. In this
> case, the number of bits needed to feed a high level
orchestral
> synthesizer might establish a lower bound.
Very much a lower bound, I suspect, because even a fantastic
orchestral
synthesizer would be to an orchestra as a typesetter would be
to the
handwritten word: you donÕt only have the notes, you have 
the
playing of
each of them, and the interplay of the playing with the
acoustics of the
hall, etc.
Still, itÕs a neat thought experiment. And you could indeed
do an
interesting analysis of musical scores, rather than muscial
recordings.
--
Andrew
===
Subject: Re: how to measure entropy of music?
 The model of entropy that can be readily recognized:
> In music--how often can an informed listener infer the next
note in a
> phrase; ie how many bits are needed to specifiy the next
note.
> In sound--how many bits are needed to specify the next
value of the
> signal.

found music to be interesting to listen to when he had a low
success
rate at predicting what would happen next. I think he was
thinking of
higher level structures than single notes, but couldnÕt the
concept of
entropy apply to these higher structures as well?
===
Subject: Re: how to measure entropy of music?
>As a matter of interest,
>do you consider high entropy good or bad?
>found music to be interesting to listen to when he had a low
success
>rate at predicting what would happen next. I think he was
thinking of
>higher level structures than single notes, but couldnÕt the
concept of
>entropy apply to these higher structures as well?
>Presumably a completely random series of notes would have
very high
entropy,
>while absolute silence has very low entropy.
>I wouldnÕt have thought either was very enjoyable.
Both of your comments are quite stimulating. Of course! The
idea
that good music is neither too rote nor too random is not new
but the
idea that entropy as a concept allows a deeper investigation
of what
people like is exciting. Conjecture: classes of composers
cluster
around various levels of entropy. On further thought--what is
predictable changes with audience familiarity with a style.
Music
that is on the leading edge of predictability is the most
interesting.
Its probably all about what gives our music neurons a good
workout.
John Bailey
http://home.rochester.rr.com/jbxroads/mailto.html
===
Subject: Re: how to measure entropy of music?
> Try:
>
http://crl.research.compaq.com/publications/techreports/
techreports.html
> search the page for Logan or music.
Some of BethÕs papers are available from her now-HPLabs
web-site:
http://www.hpl.hp.com/research/crl/publications/papers.html
Ciao,
Peter K.
===
Subject: Re: how to measure entropy of music?
> A really interesting question would be to rank famous
composers by the
> average entropy of their music--eg Bach, Mozart, the
Beatles at the
> top, Andrew Lloyd Webber, Elton John, Cole Porter
next--well, you get
> the idea.
As a matter of interest,
do you consider high entropy good or bad?
Presumably a completely random series of notes would have
very high
entropy,
while absolute silence has very low entropy.
I wouldnÕt have thought either was very enjoyable.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2,
Ireland
===
Subject: Re: how to measure entropy of music?
> The basics are at:
>
http://www.music-cog.ohio-state.edu/Music829D/Notes/
Infotheory.html
> key words: hmm, entropy, and music I have learned the
question is of
> significance today for serveral reasons:
> 1) Recognizing when TV commercials are playing.
> 2) Separating out background music behind speech in speech
> recognition.
I wonder if they could figure a why to identify background
music and
separate it from the sound reaching my ears :-)
Steve
===
Subject: Re: how to measure entropy of music?
>> The basics are at:
>>
http://www.music-cog.ohio-state.edu/Music829D/Notes/
Infotheory.html
>> key words: hmm, entropy, and music I have learned the
question is of
>> significance today for serveral reasons:
>> 1) Recognizing when TV commercials are playing.
>> 2) Separating out background music behind speech in speech
>> recognition.
>I wonder if they could figure a why to identify background
music and
>separate it from the sound reaching my ears :-)
Good point! If the predictor works then you can make a pretty
good
Or better yet, a transmogriphier; if itÕs punk metal coming
out of the
speaker and you want classical, you could put some filter and
adaptation circuits in that take the punk metal energy and
convert it
based on the predictor rules for classical. Punk metal out of
the
speaker, classical into your ear.
Or you could buy an MP3 player and some headphones, either
way. ;)
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be IntelÕs opinions.
http://www.ericjacobsen.org
===
Subject: Re: how to measure entropy of music?
>> The basics are at:
>>
http://www.music-cog.ohio-state.edu/Music829D/Notes/
Infotheory.html
> key words: hmm, entropy, and music I have learned the
question is of
>> significance today for serveral reasons:
>> 1) Recognizing when TV commercials are playing.
>> 2) Separating out background music behind speech in speech
>> recognition.
>I wonder if they could figure a why to identify background
music and
>separate it from the sound reaching my ears :-)
> Good point! If the predictor works then you can make a
pretty good
Not sure how you can come to that conclusion. If you have one
microphone
and
a perfect algorithm you could determine what is music and
what is
subtract it from the
sum of music + speech and without a second sensor this would
not be
possible - unless you assume the statistics of the music do
not change
rapidly so that the music during the speech was approx the
same as during
non-speech - an assumption which may be true for some music
maybe?
On the other hand if we used multiple microphones then maybe
we would have
something worthwhile.
Tom
===
Subject: Re: Prestigious mathematics journals
> In biology we have publications such as Nature which are
generally
> considered to be very prestigious, which magazine in
mathematics holds
> the same distinction?
> Aside from subjective opinions, there is the impact factor
computed
> by ISI Web of Knowledge (www.isiknowledge.com).
> Who wants to guess the top mathematics journal on their
list?
What is the scale of the impact factor? and what sort of
impact does
it measure? I guess what I am trying to say is that are the
calculating the impact in the scientific community or general
public
(this wouldnÕt make much sense)
===
Subject: Re: Prestigious mathematics journals
>> In biology we have publications such as Nature which are
generally
>> considered to be very prestigious, which magazine in
mathematics holds
>> the same distinction?
>> Aside from subjective opinions, there is the impact factor
computed
>> by ISI Web of Knowledge (www.isiknowledge.com).
>> Who wants to guess the top mathematics journal on their
list?
> What is the scale of the impact factor? and what sort of
impact does
> it measure? I guess what I am trying to say is that are the
> calculating the impact in the scientific community or
general public
> (this wouldnÕt make much sense)
Impact factors are essentially citation rates.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Products of Sines?
Does the following product have a nice closed-form solution?
sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Products of Sines?
Daryl McCullough  escribi.97:
> Does the following product have a nice closed-form solution?
> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
Let
P(z) = (z^n - 1)/(z - 1) = 1 + z + z^2 + ... + z^(n-1) (#1)
But also
P(z) = Prod(z - z_k, k, 1, n-1) (#2)
being
z_k = cos(k*2pi/n) + i*sen(k*2pi/n), k = 1, 2, ..., n -1
the n-1 n-roots of unity distinct of 1.
|P(1)| = |Prod(z - z_k, k, 1, n-1)|
= Prod(|z - z_k|, k, 1, n-1)
= Prod(d_k, k, 1, n-1)
where d_k is the distance from (1, 0) to (cos(k*2pi/n),
sen(k*2pi/n)). But
d_k = sqrt((1 - cos(k*2pi/n))^2 + sen^2(k*2pi/n))
= sqrt(2 - 2cos(k*2pi/n)) = 2sqrt((1 - 1cos(k*2pi/n))/2)
= 2*sen(kpi/n)
Then
|P(1)| = n = Prod(2sen(k*pi/n), k, 1, n-1) =
2^(n-1)*Prod(sen(k*pi/n), k, 1,
Prod(sen(k*pi/n), k, 1, n-1) = n/2^(n-1)
Curiously, it is the same that the probability of n points
choosen at ramdom
in a circle lie in the same semicircle.
--
Ignacio Larrosa Ca.96estro
A Coru.96a (Espa.96a)
ilarrosaQUITARMAYUSCULAS@mundo-r.com
===
Subject: Re: Products of Sines?
> Does the following product have a nice closed-form solution?
> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
This is very well-known. Set w = exp(pi i/n). The product
is then
(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} -
w^{-n+1})
= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 -
w^{-2n+2}).
Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Products of Sines?
Robin Chapman says...
>> Does the following product have a nice closed-form
solution?
>> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>This is very well-known. Set w = exp(pi i/n). The product
>is then
>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} -
w^{-n+1})
>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1 -
w^{-2n+2}).
>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
Okay, IÕm puzzled by the step
(X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X -
1)
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: Products of Sines?
> Robin Chapman says...
> Does the following product have a nice closed-form solution?
>
> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>>This is very well-known. Set w = exp(pi i/n). The product
>>is then
>>(2i)^{-n+1} (w - w^{-1})(w^2 - w^{-2}) ... (w^{n-1} -
w^{-n+1})
>>= (2i)^{-n+1} w^{n(n-1)/2} (1 - w^{-2})(1 - w^{-4}) ... (1
- w^{-2n+2}).
>>Now (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2})
>>= (X^n - 1)/(X - 1) = 1 + X + X^2 + X^3 + ... + X^{n-1}.
> Okay, IÕm puzzled by the step
> (X - w^{-2})(X - w^{-4}) ... (X - w^{-2n+2}) = (X^n - 1)/(X
- 1)
What is the factorization of X^n - 1 into linear factors?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Products of Sines?
> Does the following product have a nice closed-form solution?
> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
Yup. It is n/2^{n-1}.
Pawel Gladki
===
Subject: Re: Products of Sines?
>> Does the following product have a nice closed-form
solution?
>> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>Yup. It is n/2^{n-1}.
How do you derive that?
--
Daryl McCullough
===
Subject: Re: Products of Sines?
>Does the following product have a nice closed-form solution?
> sin(pi/n) * sin(2pi/n) * sin(3pi/n) * ... * sin((n-1)pi/n)
>>Yup. It is n/2^{n-1}.
> How do you derive that?
See Robin ChapmanÕs answer.
Pawel Gladki
===
Subject: Re: So you want to count an infinite power set ?
posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L
> P_1(SET) = {
> {1000000000000.. AND SET},
> {0100000000000.. AND SET},
> {1100000000000.. AND SET}...}
That is my only introduced notation, but I manually calculate
the step
straight after.
P_1(SET) = {
{1 AND <0,3>, 0 AND <1,1>, 0 AND <2,4> ...},
{0 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...},
{1 AND <0,3>, 1 AND <1,1>, 0 AND <2,4> ...},
...}
1 = true, 0 = false. TRUE AND X = X
Look up Ōbit masking
The original SET is { <0,3>, <1,1>, <2,4>, ...}
i.e. pi 314159.. put into <0 > <1 > <2 > <3 > ...
Its the quickest way I could think of to make infinite
distinct members
that werent trivial N.
I use { } for set and < > for sequence, fairly standard for
programmers
atleast.
Herc
===
Subject: Probability Of Ultimate Extinction.
LetÕs say that you have an amoeba which has a probability q
of dying without
divinding,
and probability p of spliting into two. What is the
probability of ultimate
extinction?
I know that the recursion formula for extinction after n
cycles is
P(n)=q+p*P(n-1)^2 where P(0)=q.
of 3 or 4) is that
the probability of ultimate extinction involves using the
quadratic formula
in the following way.
1(1-4*p*q)^0.5
---------------
2q
where the minimum of the two values achieved by the quadratic
equation is
the desired probability.
What IÕd like to know is, why bother? IÕve 
discovered that in
the above
problem, the probability
of ultimate extinction is p/q = 1 if p>=q. Why bother with
the quadratic
formula?
--
-------------------------------
Patrick D. Rockwell
===
Subject: Re: Probability Of Ultimate Extinction.
>LetÕs say that you have an amoeba which has a probability q
of dying
without
>divinding,
>and probability p of spliting into two. What is the
probability of ultimate
>extinction?
>I know that the recursion formula for extinction after n
cycles is
>P(n)=q+p*P(n-1)^2 where P(0)=q.
>of 3 or 4) is that
>the probability of ultimate extinction involves using the
quadratic formula
>in the following way.
>1(1-4*p*q)^0.5
>---------------
>where the minimum of the two values achieved by the
quadratic equation is
>the desired probability.
The general statement is that if p_j is the probability that
the amoeba
has j children, the probability of ultimate extinction is the
least
nonnegative solution of s = G(s), where G(s) =
sum_{j=0}^infinity p_j s^j.
In your case G(s) = q + p s^2 where q + p = 1. Since
G(s) - s = p s^2 - s + q = (s - 1)(p s - q), the answer is
min(1, q/p).
>What IÕd like to know is, why bother? IÕve 
discovered that
in the above
>problem, the probability
>of ultimate extinction is p/q = 1 if p>=q.
Huh?? Since when is p/q = 1 if p >= q? I think you mean
q/p if p >= q,
1 if p <= q.
>Why bother with the quadratic
>formula?
No need to do so in this case since the polynomial factors so
easily
(s=1 is always a solution). Who does bother with it?
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
===
Subject: Re: Probability Of Ultimate Extinction.
>>LetÕs say that you have an amoeba which has a probability 
q
of dying
>>without
>>divinding,
>>and probability p of spliting into two. What is the
probability of
>>ultimate
>>extinction?
>>I know that the recursion formula for extinction after n
cycles is
>>P(n)=q+p*P(n-1)^2 where P(0)=q.
>>of 3 or 4) is that
>>the probability of ultimate extinction involves using the
quadratic
>>formula
>>in the following way.
>>1(1-4*p*q)^0.5
>>---------------
>>2q
>>where the minimum of the two values achieved by the
quadratic equation is
>>the desired probability.
> The general statement is that if p_j is the probability
that the amoeba
> has j children, the probability of ultimate extinction is
the least
> nonnegative solution of s = G(s), where G(s) =
sum_{j=0}^infinity p_j
s^j.
> In your case G(s) = q + p s^2 where q + p = 1. Since
> G(s) - s = p s^2 - s + q = (s - 1)(p s - q), the answer is
> min(1, q/p).
>>What IÕd like to know is, why bother? IÕve 
discovered that
in the above
>>problem, the probability
>>of ultimate extinction is p/q = 1 if p>=q.
> Huh?? Since when is p/q = 1 if p >= q? I think you mean
> q/p if p >= q,
> 1 if p <= q.
YouÕre right. I should have said q/p. Sorry. :-)
>>Why bother with the quadratic
>>formula?
> No need to do so in this case since the polynomial factors
so easily
> (s=1 is always a solution). Who does bother with it?
I did a Google search and found this .PDF file
I found a few other .PDF files which show the quadratic
equation used in
cases where you canÕt just
divide p/q, but in the case of the amoebe which either
divides into two, or
just dies, it seems easier to
just calculate p/q.
--
-------------------------------
Patrick D. Rockwell
===
Subject: Re: Probability Of Ultimate Extinction.
> LetÕs say that you have an amoeba which has a probability 
q
of dying
without
> divinding,
> and probability p of spliting into two. What is the
probability of
ultimate
> extinction?
As a side note, some say that cells have an inherent average
number of
divisions before senisence, or the inability to subdivide.
Some research
seems to indicate that it may be possible to optimize that
average,
reducing the standard deviation.
===
Subject: Re: Probability Of Ultimate Extinction.
>LetÕs say that you have an amoeba which has a probability q
of dying
without
>divinding,
>and probability p of spliting into two. What is the
probability of ultimate
>extinction?
>I know that the recursion formula for extinction after n
cycles is
>P(n)=q+p*P(n-1)^2 where P(0)=q.
You have not said what a cycle is, but assusming p+q=1 then
if you are looking for the probability of ultimate
extinction, u, then
it is the probability of going extinct without dividing plus
the
probability that it divides and both halves go extinct:
u = q +p*u^2
>of 3 or 4) is that
>the probability of ultimate extinction involves using the
quadratic formula
>in the following way.
>1(1-4*p*q)^0.5
>---------------
You should be dividing by 2p not 2q.
>where the minimum of the two values achieved by the
quadratic equation is
>the desired probability.
>What IÕd like to know is, why bother? IÕve 
discovered that
in the above
>problem, the probability
>of ultimate extinction is p/q = 1 if p>=q. Why bother with
the quadratic
>formula?
Perhaps you have swapped p and q somewhere
If p+q=1 then the quadratic gives the lower of (1-p)/p and 1,
depending on whether p>=1/2 or p<=1/2 which makes sense and
in the
latter case confirms what is obvious.
===
Subject: Speaking of Attacking the Conclusions
JSH constantly whines that everyone is attacking his
conclusions and
that this is a fallacy because his suppositions and logic are
intact.
Aside from the fact that this isnÕt true (people have been
attacking
his logic and pointing out gaping holes) itÕs funny to point
out that
he does the same thing. All heÕll do is attack the
conclusions posters
have drawn based on his inacuracies. He wonÕt actually
directly attack
or even acknowledge the points they make. In essence heÕs
attacking
their conclusions.
===
Subject: Re: Infantile authours degrading this NG.
> In my time as a part-time tutor of mathematics to adults
the more of this thread I read the more I pity those poor
souls
===
Subject: Re: Infantile authours degrading this NG.
> You claim to be a mathematician of 20 yearsstanding.
> I suggest that you do not try to be a teacher of
mathematics.
> Mathematics is an abstract subject for which your students
> must be relaxed in mind in order to be able to take on
> abstractions.
What qualifies a failed software engineer (remember
Westinghouse) to make
such a statement?
What are YOUR qualifications?
Did you ever actually get a degree?
===
Subject: Re: Infantile authours degrading this NG.
> You claim to be a mathematician of 20 yearsstanding.
> I suggest that you do not try to be a teacher of
mathematics.
> Mathematics is an abstract subject for which your students
> must be relaxed in mind in order to be able to take on
> abstractions. Your emotive and insulting posts
> will disrupt such a state of mind. In my time as a part-time
> tutor of mathematics to adults
Sounds a bit far fetched to me. Was that before or after the
Westinghouse
affair?
===
Subject: Re: Infantile authours degrading this NG.
Neener, and Neener
that...
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
Because of your over-riding urge to be insulted.
--
Checkmate
all rights reserved
===
Subject: Re: Infantile authours degrading this NG.
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
Well, here is what David C. Ullrich posted.... (lines with >>
are from Airy
R. BeanÕs previous post, those preceded by 
Ō>are from David
C. UllrichÕs
pose, my comments are embedded with Ō<*<Õ)
>>Show where it is wrong, or else resort
>>to infantile sneering.
>Ok. Took me a minute to find the place
>where you showed us the integration
>by parts:
>>Certainly, if you integrate by parts, you would choose
>>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>>I try this, I get 0!......
>>int(UV) = U.int(V) - int[dU.int(V)] giving.....
>> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>>with f(t).d(t-T) as V and e^(-sT) as U .....
>>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>>f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
>>f(T).e^(-sT) - f(T).e^(-sT)....
>>0.
>Your first step,
> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>is already wrong. My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals. The (definite) integral 
from
>0 to infinity of f(t).d(t-T) is indeed f(T). But
>what we need here is an antiderivative.
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
<**>Now when we
put in the limits t = 0 to t = infinity
>the first term vanishes since F(0) = 0 and e^(-st)
>tends to 0 as t -> infinty (at least if s > 0; if s < 0
>this integration by parts is not going to work.)
>So out Laplace transform becomes
> - int_0^infinity (e^(-st)/-s . F(t))
>Since F(t) = 0 for t < T and f(T) for t > T
>this equals
> f(T) int_T^infinity (e^(-st)s ).
<*Again, itÕs easy to evaluate the last integral;
>int_T^infinity (e^(-st)s ) = e^(-sT), so we
>finally get f(T)e^(-sT) for the Laplace transform.
<**>
daestrom
===
Subject: Re: Infantile authours degrading this NG.
> Well, here is what David C. Ullrich posted.... (lines with
>> are from
Airy
> R. BeanÕs previous post, those preceded by 
Ō>are from
David C. UllrichÕs
> pose, my comments are embedded with Ō<*<Õ)
Due to the high noise-to-signal ratio and the many ad-hominem
attacks,
itÕs difficult to grasp for a newcomer what this 
thread is all
about.
As far as I can see, one of the subjects is the
(double-sided) Laplace
Transform of the function f(t).d(t-T), where d is the
delta-function.
Ullrich correctly concludes that:
>> so we finally get f(T)e^(-sT) for the Laplace transform.
> <**>
I wouldnÕt have expected otherwise. But I 
donÕt understand
why there is
so much clumsiness involved in deriving this result. Is it
because of
Airy R. BeanÕs original posters? And what different result
did *he* find
then? Weird thread anyway ...
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>> Well, here is what David C. Ullrich posted.... (lines with
>> are from
>> Airy R. BeanÕs previous post, those preceded by 
Ō>are
from David C.
>> UllrichÕs pose, my comments are embedded with 
Ō<*<Õ)
> Due to the high noise-to-signal ratio and the many
ad-hominem attacks,
> itÕs difficult to grasp for a newcomer what 
this thread is
all about.
> As far as I can see, one of the subjects is the
(double-sided) Laplace
> Transform of the function f(t).d(t-T), where d is the
delta-function.
> Ullrich correctly concludes that:
> so we finally get f(T)e^(-sT) for the Laplace transform.
>> <**> I wouldnÕt have expected otherwise. But I 
donÕt
understand why there is
> so much clumsiness involved in deriving this result. Is it
because of
> Airy R. BeanÕs original posters? And what different result
did *he* find
> then? Weird thread anyway ...
It isnÕt that difficult to derive. But ARB seems 
to have made
a couple of
mistakes when he tried to integrate by parts and came to the
wrong result.
Yes, there seems to be a lot of personal attacks along the
way :-(
His most recent objections seem to revolve around the
particular definition
he is using for the Dirac delta function. I replied with a
couple of links
showing that delta(t) = 0 for all values of t<>0. Hopefully
heÕll take the
time to review the linked information carefully.
daestrom
> Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
> His most recent objections seem to revolve around the
particular
definition
> he is using for the Dirac delta function. I replied with a
couple of
links
> showing that delta(t) = 0 for all values of t<>0. Hopefully
heÕll take
the
> time to review the linked information carefully.
ItÕs not that I want to defend him, but this radio amateur
has some
point in finding a delta function to be the same as a bell
shaped/
Gauss function with a very small value for its spread. As I
pointed
out to Ullrich, also the heaviside function cannot be
distinguished,
physically, from an error function with the same very small
spread.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
DonÕt defend me. Defend the truth.
If you believe something to be true then stand
by your guns.
> ItÕs not that I want to defend him,
===
Subject: Re: Infantile authours degrading this NG.
> DonÕt defend me. Defend the truth.
> If you believe something to be true then stand
> by your guns.
>>ItÕs not that I want to defend him,
Uhm, sorry! But I find this thread such a mess that I cannot
find what
truth I should defend. And what guns to stand by ...
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
The truths that you should defend are those that
you hold to be true.
The opinions of others are irrelevant unless they
serve to change your mind.
> DonÕt defend me. Defend the truth.
> If you believe something to be true then stand
> by your guns.
>>ItÕs not that I want to defend him,
> Uhm, sorry! But I find this thread such a mess that I cannot
find what
> truth I should defend. And what guns to stand by ...
===
Subject: Re: Infantile authours degrading this NG.
>> Well, here is what David C. Ullrich posted.... (lines with
>> are from
Airy
>> R. BeanÕs previous post, those preceded by 
Ō>are from
David C.
UllrichÕs
>> pose, my comments are embedded with Ō<*<Õ)
>Due to the high noise-to-signal ratio and the many
ad-hominem attacks,
>itÕs difficult to grasp for a newcomer what 
this thread is
all about.
>As far as I can see, one of the subjects is the
(double-sided) Laplace
>Transform of the function f(t).d(t-T), where d is the
delta-function.
>Ullrich correctly concludes that:
> so we finally get f(T)e^(-sT) for the Laplace transform.
>> <**>I wouldnÕt have expected otherwise. But I 
donÕt understand
why there is
>so much clumsiness involved in deriving this result.
There isnÕt. See my very first post in this 
thread - deriving
the
result is utterly trivial. Airy objected to the derivation on
totally bogus grounds - at this point weÕre simply humoring
him,
pointing out that the result _also_ follows by the methods
that
he (arbitrarily and incorrectly) declared were the only
allowed
methods.
>Is it because of
>Airy R. BeanÕs original posters? And what different result
did *he* find
>then? Weird thread anyway ...
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
Where do you get that from?
The definite integral int -oo^+oo f(t).d(t-T)
gives us f(T), but there is no information
given to us about the indefinite integral
nor of the anti-derivative.
Indeed, the definite integral gives us f(T), which
is a constant function having no aspect of t in its
evaluation and which graphs as a horizontal
line from -oo to +oo. This comes from the
basic definitions of the Diracian.
Your claimed anti-derivative has a strong
dependence on t, and therefore has not
come from anything formally defined.
Are you making this up as you go along?
Well, here is what David C. Ullrich posted..
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
===
Subject: Re: Infantile authours degrading this NG.
> Where do you get that from?
> The definite integral int -oo^+oo f(t).d(t-T)
> gives us f(T), but there is no information
> given to us about the indefinite integral
> nor of the anti-derivative.
That much is true. BUT, it is only true for *some* definite
integrals
such as from -oo to +oo. It is *not* true for all definite
integrals,
such as from -oo to zero (assuming T > 0).
As X moves from -oo to +oo, the value of int_-oo^X f(t)d(t-T)
dt
starts out as a horizontal line at zero, then at T, it is
discontinous
but with a value now of f(T).
There *is* some information given by the definition of the
delta
function though that can be useful. For example, the following
definite integrals can be found from the 
definition.
Because the Dirac delta function is zero for all values < 0...
For all positive values of m,
lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0
And because the Dirac delta function is zero for all values >
0...
For all positive values of m,
lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0
And finally, since given the two conditions above and the part
of the
definition that says int_-oo^+oo d(t) dt = 1 ...
For all positive values of m,
lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1
In point of fact, these three definite integrals can be
considered the
*definition* of the Dirac delta function.
> Indeed, the definite integral gives us f(T), which
> is a constant function having no aspect of t in its
> evaluation and which graphs as a horizontal
> line from -oo to +oo.
Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current
context, G(t)
= (-s)e^(-st) ). You can not Ōfactorthe 
integral into two
parts
like this.
int_-oo^+oo f(t)d(t-T)G(t) dt
is *NOT* equal to
[int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]
But that is what you are in effect doing when you say...
int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo
(-s)e^(-st)
dt
One thing you *can* do is...
int_-oo^+oo f(t)d(t-T)G(t) dt =
int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt
Where Ōais any arbitrary real number.
Then, if we set things up so that Ōahas a 
value approaching
ŌTfrom
the negative in the first integral, and from the positive in
the
second integral, we can make some progress. So define our 
ŌaÕ
as
(T-m) where Ōmis positive and => 0 as a limit 
in the first
integral
and (T+m) in the second. Substituting this into our problem,
we get...
int_-oo^+oo f(t)d(t-T)G(t) dt =
lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo
f(t)d(t-T)G(t) dt ]
Now, over the definite integral of -oo to (T-m), the value of
int_-oo^(T-m)
f(t)d(t-T) dt is always zero so the first integral is zero.
Over the
definite integral of (T+m) to +oo, the value of int_(T+m)^+oo
f(t)d(t-T) dt is always f(T) so for *THIS PARTICULAR DEFINITE
INTEGRAL*, it can be factored. So over these two particular
definite
integrals, you can factor out the term. But note that it has
different values over these two particular definite integrals.
int_-oo^+oo f(t)d(t-T)G(t) dt =
lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t)
dt ]
And, given G(t) = (-s)e^(-st)...
int_-oo^+oo f(t)d(t-T)G(t) dt =
f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]
The integral of (-s)e^(-st) is easy...
int_-oo^+oo f(t)d(t-T)G(t) dt =
f(T) { e^(-oo) - e^(-sT) } =
f(T) { 0 - e^(-sT) }=
-f(T)e^(-sT)
Substituting into the original integration by parts...
int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))
And since Uint(V) is zero...
int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)
daestrom
P.S. Sorry to all if this post shows up twice. My ISPÕs
news-server
seems to have misplaced the first copy so IÕm 
posting again
directly
===
Subject: Re: Infantile authours degrading this NG.
>> Where do you get that from?
>> The definite integral int -oo^+oo f(t).d(t-T)
>> gives us f(T), but there is no information
>> given to us about the indefinite integral
>> nor of the anti-derivative.
> That much is true. BUT, it is only true for *some* definite
integrals
> such as from -oo to +oo. It is *not* true for all definite
integrals,
> such as from -oo to zero (assuming T > 0).
> As X moves from -oo to +oo, the value of int_-oo^X
f(t)d(t-T) dt
> starts out as a horizontal line at zero, then at T, it is
discontinous
> but with a value now of f(T).
> There *is* some information given by the definition of the
delta
> function though that can be useful. For example, the
following
> definite integrals can be found from the 
definition.
> Because the Dirac delta function is zero for all values <
0...
> For all positive values of m,
> lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0
> And because the Dirac delta function is zero for all values
> 0...
> For all positive values of m,
> lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0
> And finally, since given the two conditions above and the
part of the
> definition that says int_-oo^+oo d(t) dt = 1 ...
> For all positive values of m,
> lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1
> In point of fact, these three definite integrals can be
considered the
> *definition* of the Dirac delta function.
>> Indeed, the definite integral gives us f(T), which
>> is a constant function having no aspect of t in its
>> evaluation and which graphs as a horizontal
>> line from -oo to +oo.
> Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
> solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current
context, G(t)
> = (-s)e^(-st) ). You can not Ōfactorthe 
integral into two
parts
> like this.
> int_-oo^+oo f(t)d(t-T)G(t) dt
> is *NOT* equal to
> [int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]
> But that is what you are in effect doing when you say...
> int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo
(-s)e^(-st)
> dt
> One thing you *can* do is...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt
> Where Ōais any arbitrary real number.
> Then, if we set things up so that Ōahas a 
value
approaching ŌTfrom
> the negative in the first integral, and from the positive in
the
> second integral, we can make some progress. So define our
Ōaas
> (T-m) where Ōmis positive and => 0 as a 
limit in the first
integral
> and (T+m) in the second. Substituting this into our
problem, we get...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo
> f(t)d(t-T)G(t) dt ]
> Now, over the definite integral of -oo to (T-m), the value 
of
> int_-oo^(T-m)
> f(t)d(t-T) dt is always zero so the first integral is zero.
Over the
> definite integral of (T+m) to +oo, the value of 
int_(T+m)^+oo
> f(t)d(t-T) dt is always f(T) so for *THIS PARTICULAR
DEFINITE
> INTEGRAL*, it can be factored. So over these two particular
definite
> integrals, you can factor out the term. But note that it has
> different values over these two particular definite
integrals.
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t)
dt ]
> And, given G(t) = (-s)e^(-st)...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]
> The integral of (-s)e^(-st) is easy...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> f(T) { e^(-oo) - e^(-sT) } =
> f(T) { 0 - e^(-sT) }=
> -f(T)e^(-sT)
> Substituting into the original integration by parts...
> int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))
> And since Uint(V) is zero...
> int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)
> daestrom
I notice that ARB hasnÕt replied to this post, so I can only
assume he
agrees with it. Wonder if he has the courage to acknowledge
his errors???
daestrom
===
Subject: Re: Infantile authours degrading this NG.
Up to that point I had made the mistake of
taking you for a mature contributor.
Shame on you.
> Wonder if he has the courage to acknowledge his errors???
===
Subject: Re: Infantile authours degrading this NG.
I would request that Mr daestrom remove uk.radio.amateur from
any
follow-ups he might make to Bean in this - and indeed any
other -
thread that he might so use.
ItÕs not that we in ukra donÕt like to see 
Bean getting yet
another
drubbing (we have seen it many times on ukra), but that he
chooses to
reply in separate postings to each point that he wishes to
make, which
- like changing the spelling of words in the thread title -
is a very
amateurish practice indeed. TIA
--
from
Aero Spike
===
Subject: Re: Infantile authours degrading this NG.
> I would request that Mr daestrom remove uk.radio.amateur
from any
> follow-ups he might make to Bean in this - and indeed any
other -
> thread that he might so use.
> ItÕs not that we in ukra donÕt like to see 
Bean getting yet
another
> drubbing (we have seen it many times on ukra), but that he
chooses to
> reply in separate postings to each point that he wishes to
make, which
> - like changing the spelling of words in the thread title -
is a very
> amateurish practice indeed. TIA
You poor folks in UKRA have our sincere condolences. How much
would it
cost us to have you keep him there? ;-)
--
Keith
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy.
> You poor folks in UKRA have our sincere condolences. How
much would it
> cost us to have you keep him there? ;-)
===
Subject: Re: Infantile authours degrading this NG.
> Stupid boy.
will you marry me beany?
dr. x
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
IÕll stop you there. to arrive at f(T), you need to 
integrate
over the whole domain of +/- oo.
Once again, yuo are introducing a time-dependency in your
integration evalution that is not present in the definition.
>> Where do you get that from?
> The definite integral int -oo^+oo f(t).d(t-T)
>> gives us f(T), but there is no information
>> given to us about the indefinite integral
>> nor of the anti-derivative.
>> Over the
> definite integral of (T+m) to +oo, the value of 
int_(T+m)^+oo
> f(t)d(t-T) dt is always f(T) so for *THIS PARTICULAR
DEFINITE
> INTEGRAL*, it can be factored.
===
Subject: Re: Infantile authours degrading this NG.
> IÕll stop you there. to arrive at f(T), you need to
integrate
> over the whole domain of +/- oo.
No, you do *not*. It is elementary that if...
int_-oo^+oo delta(t) dt = 1
and...
delta(t) = zero for all values of t <> 0 (a part of the
definition
http://hitoshi.berkeley.edu/221A/delta.pdf ), then
lim m-> 0 [int_-oo^-m delta(t) dt = 0
And since for any definite integral...
int_-oo^+oo g(t) dt = int_-oo^a g(t) dt + int_a^+oo g(t) dt
We can conclude that
int_a^+oo g(t) dt = int_-oo^+oo g(t) dt - int_-oo^a g(t) dt
Thus...
int_+a^+oo f(t)delta(t) dt = int_-oo^+oo f(t) delta(t) dt -
int_-oo^-a
delta(t) dt
when lim ->a = 0
int_a^+oo (ft)delta(t) dt = f(a)
> Once again, yuo are introducing a time-dependency in your
> integration evalution that is not present in the definition.
Once again you are working with an incomplete definition of
the dirac
function. There is more to the definition than just...
int_-oo^+oo delta(t) dt = 1
http://mathworld.wolfram.com/DeltaFunction.html
Look carefully at equation 3 of this page. It shows that in
fact the delta
function is...
int_(a-e)^(a+e) f(x)delta(x-a)dx = f(a)
for all values of e > 0
And equation 4....
delta(x-a) = 0 for all values of x <> a.
These are *exactly* the two characteristics of the delta
function that I
exploited in my solution, yet you maintain they are not part
of the
definition. Quote a credible reference with your unusual
Ōdefinitionof
the Dirac delta function please.
daestrom
http://marr.bsee.swin.edu.au/~dtl/het408/backproj/node12.html
===
Subject: Re: Infantile authours degrading this NG.
WRONG!
> int_a^+oo (ft)delta(t) dt = f(a)
===
Subject: Re: Infantile authours degrading this NG.
If you have something to present in defence of your
argument, then present it yourself.
> http://hitoshi.berkeley.edu/221A/delta.pdf )
> http://mathworld.wolfram.com/DeltaFunction.html
>
http://marr.bsee.swin.edu.au/~dtl/het408/backproj/node12.html
===
Subject: Re: Infantile authours degrading this NG.
.....[Pantomime mode ON].....
Oh, yes! You do!
.....[Pantomime mode OFF].....
Either you are dealing with the function as defined,
or else you are creating your own function.
> IÕll stop you there. to arrive at f(T), you need to
integrate
> over the whole domain of +/- oo.
> No, you do *not*.
===
Subject: Re: Infantile authours degrading this NG.
The discussion has already moved on from that point.
Mr.Ullrich, before he went all infantile and insulting
pointed out the need for an anti-derivative and not
a definite integral, although he then
tried to use the value of the definite integral to determine
the anti-derivative.
>> Indeed, the definite integral gives us f(T), which
>> is a constant function having no aspect of t in its
>> evaluation and which graphs as a horizontal
>> line from -oo to +oo.
> Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
> solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current
context, G(t)
> = (-s)e^(-st) ). You can not Ōfactorthe 
integral into two
parts
> like this.
> int_-oo^+oo f(t)d(t-T)G(t) dt
> is *NOT* equal to
> [int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]
> But that is what you are in effect doing when you say...
> int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo
(-s)e^(-st)
> dt
===
Subject: Re: Infantile authours degrading this NG.
If the definition does not help us in this approach. then
what approach would you suggest for evaluating the
LT of f(t).d(t-T)?
>> Indeed, the definite integral gives us f(T), which
>> is a constant function having no aspect of t in its
>> evaluation and which graphs as a horizontal
>> line from -oo to +oo.
> Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
> solving int_-oo^+oo f(t)d(t-T)G(t) dt.
===
Subject: Re: Infantile authours degrading this NG.
No. Those are your deductions. They are not part of
the definition.
> Because the Dirac delta function is zero for all values <
0...
> For all positive values of m,
> lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0
> And because the Dirac delta function is zero for all values
> 0...
> For all positive values of m,
> lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0
> And finally, since given the two conditions above and the
part of the
> definition that says int_-oo^+oo d(t) dt = 1 ...
> For all positive values of m,
> lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1
> In point of fact, these three definite integrals can be
considered the
> *definition* of the Dirac delta function.
===
Subject: Re: Infantile authours degrading this NG.
> No. Those are your deductions. They are not part of
> the definition.
Ah, but the definition tells us that the Dirac function has a
value of zero
everywhere but at zero. Thus...
lim m-> 0 [int_-oo^(-m) delta dt] =0
If the Dirac function was *not* defined as zero everywhere
else, then we
could not deduce this. But because it is explictly defined to
have a value
of zero everywhere but the input value of zero, this is a
simple deduction.
http://marr.bsee.swin.edu.au/~dtl/het408/backproj/node12.html
daestrom
===
Subject: Re: Infantile authours degrading this NG.
We are agreed then
> No. Those are your deductions. They are not part of
> the definition.
, this is a simple deduction.
===
Subject: Re: Infantile authours degrading this NG.
No. The definition gives f(T) as a result without a
consideration
of how it got there.
If the evaluation was how you describe it below, then it would
include attributes of t as addenda in the definition. It
includes no such
attributes. The only thing we know about it is that the
evaluation
of the integral over the complete domain of the independent
variable yields as f(T).
I will accept what you propose below if you can present it in
such
a way that the first evaluation of the Diracian uses its 
defined
characteristic of int -oo^+oo f(t).d(t-T) being f(T).
It seemed to me that if anyone wishes to argue as you do
below, then
you will be denying the availability of the definite integral
too early
on, but then you go on to use it at the same stage of the
calculation.
>> Where do you get that from?
>> The definite integral int -oo^+oo f(t).d(t-T)
>> gives us f(T), but there is no information
>> given to us about the indefinite integral
>> nor of the anti-derivative.
> As X moves from -oo to +oo, the value of int_-oo^X
f(t)d(t-T) dt
> starts out as a horizontal line at zero, then at T, it is
discontinous
> but with a value now of f(T).
===
Subject: Re: Infantile authours degrading this NG.
The Diracian is so defined, and one therefore uses its
definition in evaluating expressions using it.
>> Where do you get that from?
>> The definite integral int -oo^+oo f(t).d(t-T)
>> gives us f(T), but there is no information
>> given to us about the indefinite integral
>> nor of the anti-derivative.
> That much is true. BUT, it is only true for *some* definite
integrals
> such as from -oo to +oo. It is *not* true for all definite
integrals,
> such as from -oo to zero (assuming T > 0).
===
Subject: Re: Infantile authours degrading this NG.
The Berlin News Server was inaccessible all day Saturday.
I was out on Sunday, and by the time I got in, most of
the NG to which I subscribe (27) were running at over 100
posts - too much too deal with. Result - binned most
of them without reading.
So, no, I neither agree nor disagree. Your post has
remained unconsidered.
So, apologies to those of you who thought that they
had contributed something technically worthwhile - feel
free to repost it and I will consider what you say.
No apologies to those of you who may have contributed
rather silly and childish ad hominem attacks. If there were
any, I say to you.....shame on you, you should know better.
>> Where do you get that from?
> The definite integral int -oo^+oo f(t).d(t-T)
>> gives us f(T), but there is no information
>> given to us about the indefinite integral
>> nor of the anti-derivative.
>> That much is true. BUT, it is only true for *some* definite
integrals
> such as from -oo to +oo. It is *not* true for all definite
integrals,
> such as from -oo to zero (assuming T > 0).
> As X moves from -oo to +oo, the value of int_-oo^X
f(t)d(t-T) dt
> starts out as a horizontal line at zero, then at T, it is
discontinous
> but with a value now of f(T).
> There *is* some information given by the definition of the
delta
> function though that can be useful. For example, the
following
> definite integrals can be found from the 
definition.
> Because the Dirac delta function is zero for all values <
0...
> For all positive values of m,
> lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0
> And because the Dirac delta function is zero for all values
> 0...
> For all positive values of m,
> lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0
> And finally, since given the two conditions above and the
part of the
> definition that says int_-oo^+oo d(t) dt = 1 ...
> For all positive values of m,
> lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1
> In point of fact, these three definite integrals can be
considered the
> *definition* of the Dirac delta function.
>> Indeed, the definite integral gives us f(T), which
>> is a constant function having no aspect of t in its
>> evaluation and which graphs as a horizontal
>> line from -oo to +oo.
> Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
> solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current
context, G(t)
> = (-s)e^(-st) ). You can not Ōfactorthe 
integral into two
parts
> like this.
> int_-oo^+oo f(t)d(t-T)G(t) dt
> is *NOT* equal to
> [int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]
> But that is what you are in effect doing when you say...
> int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo
(-s)e^(-st)
> dt
> One thing you *can* do is...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt
> Where Ōais any arbitrary real number.
> Then, if we set things up so that Ōahas a 
value
approaching ŌTfrom
> the negative in the first integral, and from the positive in
the
> second integral, we can make some progress. So define our
Ōaas
> (T-m) where Ōmis positive and => 0 as a 
limit in the first
integral
> and (T+m) in the second. Substituting this into our
problem, we get...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo
> f(t)d(t-T)G(t) dt ]
> Now, over the definite integral of -oo to (T-m), the value 
of
> int_-oo^(T-m)
> f(t)d(t-T) dt is always zero so the first integral is zero.
Over the
> definite integral of (T+m) to +oo, the value of 
int_(T+m)^+oo
> f(t)d(t-T) dt is always f(T) so for *THIS PARTICULAR
DEFINITE
> INTEGRAL*, it can be factored. So over these two particular
definite
> integrals, you can factor out the term. But note that it has
> different values over these two particular definite
integrals.
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t)
dt ]
> And, given G(t) = (-s)e^(-st)...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]
> The integral of (-s)e^(-st) is easy...
> int_-oo^+oo f(t)d(t-T)G(t) dt =
> f(T) { e^(-oo) - e^(-sT) } =
> f(T) { 0 - e^(-sT) }=
> -f(T)e^(-sT)
> Substituting into the original integration by parts...
> int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))
> And since Uint(V) is zero...
> int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)
> daestrom
> I notice that ARB hasnÕt replied to this post, so I can
only assume he
> agrees with it. Wonder if he has the courage to acknowledge
his
errors???
> daestrom
===
Subject: Re: Infantile authours degrading this NG.
>The Berlin News Server was inaccessible all day Saturday.
LOL! The ISP failure again. Sure sign of a losing stand.
Reference: See your thread about a problem with an
off-the-shelf
amateur transceiver that you bought.
>I was out on Sunday, and by the time I got in, most of
>the NG to which I subscribe (27) were running at over 100
>posts - too much too deal with. Result - binned most
>of them without reading.
Amazing that you binned those posts in a group in which you
have so
many ongoing Ōcontributions. Not.
>So, no, I neither agree nor disagree. Your post has
>remained unconsidered.
Cheap. Very, very cheap.
===
Subject: Re: Infantile authours degrading this NG.
> The Berlin News Server was inaccessible all day Saturday.
> I was out on Sunday, and by the time I got in, most of
> the NG to which I subscribe (27) were running at over 100
> posts - too much too deal with. Result - binned most
> of them without reading.
Stop posting on usenet.
Bob
--
Things should be described as simply as possible, but no
simpler.
A. Einstein
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy.
> Stop posting on usenet.
===
Subject: Re: Infantile authours degrading this NG.
> Stupid boy.
will you marry me beany?
dr. x
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
> The Berlin News Server was inaccessible all day Saturday.
Rubbish. I use the same server and had no problems.
> I was out on Sunday,...
out cold after Saturday?
and by the time I got in, most of
> the NG to which I subscribe (27) were running at over 100
> posts - too much too deal with. Result - binned most
> of them without reading.
You mean you filtered out all the lost arguments.
> So, no, I neither agree nor disagree. Your post has
> remained unconsidered.
Is that because you donÕt have enough of a brain to make 
such
decisions?
> So, apologies to those of you who thought that they
> had contributed something technically worthwhile - feel
> free to repost it and I will consider what you say.
When is your keeper next available to read them to you?
> No apologies to those of you who may have contributed
> rather silly and childish ad hominem attacks. If there were
> any, I say to you.....shame on you, you should know better.
Why? ItÕs all youÕre fit for.
...(_!_)...
===
Subject: Re: Infantile authours degrading this NG.
> Where do you get that from?
> The definite integral int -oo^+oo f(t).d(t-T)
> gives us f(T), but there is no information
> given to us about the indefinite integral
> nor of the anti-derivative.
That much is true. BUT, it is only true for *some* definite
integrals such
as from -oo to +oo. It is *not* true for all definite
integrals, such as
from -oo to zero (assuming T > 0).
As X moves from -oo to +oo, the value of int_-oo^X f(t)d(t-T)
dt starts out
as a horizontal line at zero, then at T, it is discontinous
and Ōjumpsto
of f(T).
There *is* some information given by the definition of the
delta function
though that can be useful. For example, the following definite
integrals
can be found from the definition.
Because the Dirac delta function is zero for all values < 0...
For all positive values of m,
lim_m->0 [ int_-oo^(-m) delta(t) dt ] = 0
And because the Dirac delta function is zero for all values >
0...
For all positive values of m,
lim_m->0 [ int_(+m)^+oo delta(t) dt ] = 0
And finally, since given the two conditions above and the part
of the
definition that says int_-oo^+oo d(t) dt = 1 ...
For all positive values of m,
lim_m->0 [int_(-m)^(+m) d(t) dt ] = 1
In point of fact, these three definite integrals can be
considered the
*definition* of the Dirac delta function.
> Indeed, the definite integral gives us f(T), which
> is a constant function having no aspect of t in its
> evaluation and which graphs as a horizontal
> line from -oo to +oo.
Although the int_-oo^+oo f(t)d(t-T) dt = f(T), that isnÕt
much use in
solving int_-oo^+oo f(t)d(t-T)G(t) dt. (in the current
context, G(t) =
(-s)e^(-st) ). You canÕt Ōfactor
the integral into two parts
like this.
int_-oo^+oo f(t)d(t-T)G(t) dt
is *NOT* equal to
[int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo G(t) dt]
But that is what you are in effect doing when you say...
int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt = f(T) int_-oo^+oo
(-s)e^(-st) dt
One thing you *can* do is...
int_-oo^+oo f(t)d(t-T)G(t) dt =
int_-oo^a f(t)d(t-T)G(t) dt + int_a^+oo f(t)d(t-T)G(t)dt
Where Ōais any arbitrary real number.
Then, if we set things up so that Ōahas a 
value approaching
ŌTfrom the
negative in the first integral, and from the positive in the
second
integral, we can make some progress. So define our 
Ōaas
(T-m) where ŌmÕ
is positive and => 0 as a limit in the first integral and
(T+m) in the
second. Substituting this into our problem, we get...
int_-oo^+oo f(t)d(t-T)G(t) dt =
lim_m->0 [int_-oo^(T-m) f(t)d(t-T)G(t) dt + int_(T+m)^+oo
f(t)d(t-T)G(t)
dt ]
Now, over the definite integral of -oo to (T-m), the value of
int_-oo^(T-m)
f(t)d(t-T) dt is always zero so the first integral is zero.
Over the
definite integral of (T+m) to +oo, the value of int_(T+m)^+oo
f(t)d(t-T) dt
is always f(T) so for *THIS PARTICULAR DEFINITE INTEGRAL*, it
can be
factored. So over these two particular definite integrals, you
can factor
out the term. But note that it has different values over
these two
particular definite integrals.
int_-oo^+oo f(t)d(t-T)G(t) dt =
lim_m->0[ 0 int_-oo^(T-m) G(t) dt + f(T) int_(T+m)^+oo G(t)
dt ]
And, given G(t) = (-s)e^(-st)...
int_-oo^+oo f(t)d(t-T)G(t) dt =
f(T) lim_m->0 [ int_(T+m)^+oo (-s)e^(-st) dt ]
The integral of (-s)e^(-st) is easy...
int_-oo^+oo f(t)d(t-T)G(t) dt =
f(T) { e^(-oo) - e^(-sT) } =
f(T) { 0 - e^(-sT) }=
-f(T)e^(-sT)
Substituting into the original integration by parts...
int_-oo^+oo f(t)d(t-T) e^(-st) dt = Uint(V) - (-f(T)e^(-sT))
And since Uint(V) is zero...
int_-oo^+oo f(t)d(t-T) e^(-st) dt = f(T)e^(-sT)
daestrom
===
Subject: Re: Infantile authours degrading this NG.
>Where do you get that from?
Just explained that in a different post.
>The definite integral int -oo^+oo f(t).d(t-T)
>gives us f(T), but there is no information
>given to us about the indefinite integral
>nor of the anti-derivative.
>Indeed, the definite integral gives us f(T), which
>is a constant function having no aspect of t in its
>evaluation and which graphs as a horizontal
>line from -oo to +oo. This comes from the
>basic definitions of the Diracian.
True, and of no relevance whatever to the question
of what an antiderivative is.
Hint: YouÕre _really_ showing over and over that
you have no understanding of _basic_ calculus.
You keep kindly informing me that IÕm making
a laughingstock of myself, presumably youÕd
want to know how utterly ignorant youÕre
revealing yourself to be.
PS: Look up the definition of ad hominem.
IÕm not saying you must be wrong because 
youÕre
stupid, IÕm saying youÕre looking very stupid
because the things youÕre saying are wrong,
at such a basic level.
>Your claimed anti-derivative has a strong
>dependence on t, and therefore has not
>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
>>Where do you get that from?
>Just explained that in a different post.
>>The definite integral int -oo^+oo f(t).d(t-T)
>>gives us f(T), but there is no information
>>given to us about the indefinite integral
>>nor of the anti-derivative.
>>Indeed, the definite integral gives us f(T), which
>>is a constant function having no aspect of t in its
>>evaluation and which graphs as a horizontal
>>line from -oo to +oo. This comes from the
>>basic definitions of the Diracian.
>True, and of no relevance whatever to the question
>of what an antiderivative is.
>Hint: YouÕre _really_ showing over and over that
>you have no understanding of _basic_ calculus.
>You keep kindly informing me that IÕm making
>a laughingstock of myself, presumably youÕd
>want to know how utterly ignorant youÕre
>revealing yourself to be.
I tangled with him in another newsgroup. He is untrainable
and is proud of that fact.
/BAH

/BAH
Subtract a hundred and four for e-mail.
===
Subject: Re: Infantile authours degrading this NG.
An ad hominem attack from you of no value to the
discussion.
Shame on you.
You should no better.
I do not keep kindly informing you that youÕre making
a laughingstock of myself. I have suggested that once
and once only.
Please get your facts correct.
You say that the things quoted below from me are true
and then you say that IÕm utterly ignorant.
You are inconsistent.
>Where do you get that from?
> Just explained that in a different post.
>The definite integral int -oo^+oo f(t).d(t-T)
>gives us f(T), but there is no information
>given to us about the indefinite integral
>nor of the anti-derivative.
>Indeed, the definite integral gives us f(T), which
>is a constant function having no aspect of t in its
>evaluation and which graphs as a horizontal
>line from -oo to +oo. This comes from the
>basic definitions of the Diracian.
> True, and of no relevance whatever to the question
> of what an antiderivative is.
> Hint: YouÕre _really_ showing over and over that
> you have no understanding of _basic_ calculus.
> You keep kindly informing me that IÕm making
> a laughingstock of myself, presumably youÕd
> want to know how utterly ignorant youÕre
> revealing yourself to be.
> PS: Look up the definition of ad hominem.
> IÕm not saying you must be wrong because 
youÕre
> stupid, IÕm saying youÕre looking very 
stupid
> because the things youÕre saying are wrong,
> at such a basic level.
>Your claimed anti-derivative has a strong
>dependence on t, and therefore has not
>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
===
Subject: Re: Infantile authours degrading this NG.
>An ad hominem attack from you of no value to the
>discussion.
>Shame on you.
>You should no better.
I suppose it should be no surprise that you canÕt
follow the definition of a term like ad hominem
any better than you follow definitions in mathematics.
Look it up. Google ad hominem - the first hit is
http://www.nizkor.org/features/fallacies/ad-hominem.html
ThatÕs a correct definition, and nothing 
IÕve said
here fits. (If you donÕt understand why not, I 
gave
an explanation in the PS below.)
>I do not keep kindly informing you that youÕre making
> a laughingstock of myself. I have suggested that once
>and once only.
Uh, no.
>Please get your facts correct.
>You say that the things quoted below from me are true
>and then you say that IÕm utterly ignorant.
>You are inconsistent.
No. What you say about the value of the definite integral is
correct -
your idea that the integral from -infinity to 
infinity
can be used in place of the antiderivative in an integration
by parts shows amazing ignorance.
>>Where do you get that from?
>> Just explained that in a different post.
>>The definite integral int -oo^+oo f(t).d(t-T)
>>gives us f(T), but there is no information
>>given to us about the indefinite integral
>>nor of the anti-derivative.
>>Indeed, the definite integral gives us f(T), which
>>is a constant function having no aspect of t in its
>>evaluation and which graphs as a horizontal
>>line from -oo to +oo. This comes from the
>>basic definitions of the Diracian.
>> True, and of no relevance whatever to the question
>> of what an antiderivative is.
>> Hint: YouÕre _really_ showing over and over that
>> you have no understanding of _basic_ calculus.
>> You keep kindly informing me that IÕm making
>> a laughingstock of myself, presumably youÕd
>> want to know how utterly ignorant youÕre
>> revealing yourself to be.
>> PS: Look up the definition of ad hominem.
>> IÕm not saying you must be wrong because 
youÕre
>> stupid, IÕm saying youÕre looking very 
stupid
>> because the things youÕre saying are wrong,
>> at such a basic level.
>>Your claimed anti-derivative has a strong
>>dependence on t, and therefore has not
>>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
get lost bean
dr. x
> An ad hominem attack from you of no value to the
> discussion.
> Shame on you.
> You should no better.
> I do not keep kindly informing you that youÕre making
> a laughingstock of myself. I have suggested that once
> and once only.
> Please get your facts correct.
> You say that the things quoted below from me are true
> and then you say that IÕm utterly ignorant.
> You are inconsistent.
>>Where do you get that from?
>> Just explained that in a different post.
>>The definite integral int -oo^+oo f(t).d(t-T)
>>gives us f(T), but there is no information
>>given to us about the indefinite integral
>>nor of the anti-derivative.
>>Indeed, the definite integral gives us f(T), which
>>is a constant function having no aspect of t in its
>>evaluation and which graphs as a horizontal
>>line from -oo to +oo. This comes from the
>>basic definitions of the Diracian.
>> True, and of no relevance whatever to the question
>> of what an antiderivative is.
>> Hint: YouÕre _really_ showing over and over that
>> you have no understanding of _basic_ calculus.
>> You keep kindly informing me that IÕm making
>> a laughingstock of myself, presumably youÕd
>> want to know how utterly ignorant youÕre
>> revealing yourself to be.
>> PS: Look up the definition of ad hominem.
>> IÕm not saying you must be wrong because 
youÕre
>> stupid, IÕm saying youÕre looking very 
stupid
>> because the things youÕre saying are wrong,
>> at such a basic level.
>>Your claimed anti-derivative has a strong
>>dependence on t, and therefore has not
>>come from anything formally defined.
>Are you making this up as you go along?
>Well, here is what David C. Ullrich posted..
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
Hang on a moment... you say that I cannot use the properties
of a definite integral.....and then you go on and do so
yourself!
If I cannot rely on the property of a definite integral that
int -oo^+oo f(t).d(t-T) is f(T) then neither can you!
Well, let us suppose that our f(t) is sin(wt)...how
does your claimed anti-derivative then differentiate
to become sin(wt).d(t-T)?
Or, let us suppose that our f(t) is e^t...how
does your claimed anti-derivative then differentiate
to become e^t.d(t-T)?
So, how does your anti-derivative differentiate to
become two different products of functions?
Remember, until we actually evaluate the definite
integral that weÕre dealing in whole functions over
all t in symbolic maths.
> Well, here is what David C. Ullrich posted
> My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals. The (definite) integral 
from
>0 to infinity of f(t).d(t-T) is indeed f(T). But
>what we need here is an antiderivative.
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
===
Subject: Re: Infantile authours degrading this NG.
>Hang on a moment... you say that I cannot use the properties
>of a definite integral.....and then you go on and do so
yourself!
>If I cannot rely on the property of a definite integral that
>int -oo^+oo f(t).d(t-T) is f(T) then neither can you!
I didnÕt say any such thing. I said that you need an
antiderivative in the integration by parts formula.
That formula you quote is _why_ the antiderivative is
what I say it is.
By definition an antiderivative for f(t).d(t-T)
would be F, where
F(x) = int_0^x f(t).d(t-T).
If x < T then that gives F(x) = 0, while if x > T
that gives F(x) = f(T), precisely because of that
formula above.
>Well, let us suppose that our f(t) is sin(wt)...how
>does your claimed anti-derivative then differentiate
>to become sin(wt).d(t-T)?
Impossible to answer this question since you insist
generalized functions have nothing to do with it -
these antiderivatives are not differentiable in
the classical sense, they have derivatives in the
sense of distributions.
>Or, let us suppose that our f(t) is e^t...how
>does your claimed anti-derivative then differentiate
>to become e^t.d(t-T)?
>So, how does your anti-derivative differentiate to
>become two different products of functions?
>Remember, until we actually evaluate the definite
>integral that weÕre dealing in whole functions over
>all t in symbolic maths.
>> Well, here is what David C. Ullrich posted
>> My guess is because of a confusion
>>over definite integrals versus antiderivatives, ie
>>indefinite integrals. The (definite) integral 
from
>>0 to infinity of f(t).d(t-T) is indeed f(T). But
>>what we need here is an antiderivative.
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
>> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
Apart from throwing in the term generalised functions,
how does the presented mathematics for your
derivatives differ from those of the classical sense?
Remember that any one function in your class of
generalised functions represents the properties of the
whole class.
>Well, let us suppose that our f(t) is sin(wt)...how
>does your claimed anti-derivative then differentiate
>to become sin(wt).d(t-T)?
> Impossible to answer this question since you insist
> generalized functions have nothing to do with it -
> these antiderivatives are not differentiable in
> the classical sense, they have derivatives in the
> sense of distributions.
===
Subject: Re: Infantile authours degrading this NG.
get lost bean
dr. x
> Apart from throwing in the term generalised functions,
> how does the presented mathematics for your
> derivatives differ from those of the classical sense?
> Remember that any one function in your class of
> generalised functions represents the properties of the
> whole class.
>>Well, let us suppose that our f(t) is sin(wt)...how
>>does your claimed anti-derivative then differentiate
>>to become sin(wt).d(t-T)?
>> Impossible to answer this question since you insist
>> generalized functions have nothing to do with it -
>> these antiderivatives are not differentiable in
>> the classical sense, they have derivatives in the
>> sense of distributions.
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Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
>Apart from throwing in the term generalised functions,
>how does the presented mathematics for your
>derivatives differ from those of the classical sense?
The antiderivative is not even continuous, much less
differentiable in the classical sense.
>Remember that any one function in your class of
>generalised functions represents the properties of the
>whole class.
>>Well, let us suppose that our f(t) is sin(wt)...how
>>does your claimed anti-derivative then differentiate
>>to become sin(wt).d(t-T)?
>> Impossible to answer this question since you insist
>> generalized functions have nothing to do with it -
>> these antiderivatives are not differentiable in
>> the classical sense, they have derivatives in the
>> sense of distributions.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
The anti-derivative of
(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
is pretty continuous in the classical sense, and
indeed, in any sense you care to present.
> The antiderivative is not even continuous, much less
> differentiable in the classical sense.
===
Subject: Re: Infantile authours degrading this NG.
>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>indeed, in any sense you care to present.
Uh, yes it is. If that function were a delta function
youÕd have a point.
>> The antiderivative is not even continuous, much less
>> differentiable in the classical sense.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
It is indeed a Delta Function, being the expression
of the Gaussian Normal Distribution.
>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>indeed, in any sense you care to present.
> Uh, yes it is. If that function were a delta function
> youÕd have a point.
===
Subject: Re: Infantile authours degrading this NG.
> It is indeed a Delta Function, being the expression
> of the Gaussian Normal Distribution.
>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>indeed, in any sense you care to present.
>>Uh, yes it is. If that function were a delta function
>>youÕd have a point.
Yep. ThatÕs precisely what I expected.
As a radio amateur, Mr. Bean is in his own right.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>>The anti-derivative of
>>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>>is pretty continuous in the classical sense, and
>>indeed, in any sense you care to present.
> Uh, yes it is. If that function were a delta function
> youÕd have a point.
IÕm at lost. What point? That function isnÕt a 
delta function,
but itÕs quite close to it, indeed. What if you substitute 
it
for the delta, solve your problem (IÕm at lost *what* 
problem,
though) and, in the end, take the limit for eps->0? Would that
work? Think so ...
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>> That function isnÕt a delta function, but 
itÕs quite close
to it,
> No, it isnÕt in any useful meaning of the word. It is 
about
as close
> to the delta function as 1/n is to zero.
IsnÕt 1/n _pretty close_ to zero for n -> oo ?
> Distribution theory is not a whole complicated framework
for nothing.
Depends. I routinely replace delta functions by bell-shapes
(or even
triangular shapes _/_) in my numerical work. And got no
problems at
all.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>>Depends. I routinely replace delta functions by bell-shapes
(or even
>>triangular shapes _/_) in my numerical work. And got no
problems at
>>all.
> Just tell me what kind of bridges are designed with that
kind of
> mathematics so that I can remember to avoid them.
Poor you! Almost every bridge in modern times is designed
with just
that kind of mathematics, as embedded in some Finite Element
Method.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>>Depends. I routinely replace delta functions by bell-shapes
(or even
>>triangular shapes _/_) in my numerical work. And got no
problems at
>>all.
>Just tell me what kind of bridges are designed with that
kind of
>mathematics so that I can remember to avoid them.
>>Poor you! Almost every bridge in modern times is designed
with just
>>that kind of mathematics, as embedded in some Finite
Element Method.
> But by people that have a clue about just what they are
doing.
http://huizen.dto.tudelft.nl/deBruijn/sunall.htm
> Numerics is not weÕll try something and hope it will be
close
> enough. Numerics is about _knowing_ what kind and size of
errors you
> introduce.
Numerical Analysis is about bringing theoretical mathematics
back to
earth in the first place. Most practical problems are too
complicated
to enable a decent error analysis. Your statement is wishful
thinking;
it bears little resemblance to Numerics in practice.
> Anyway, most bridges are calculated completely with a beam
model under
> load where finite elements do not really come into play.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
> That function isnÕt a delta function, but 
itÕs quite close
to it,
>> No, it isnÕt in any useful meaning of the word. It is
about as close
>> to the delta function as 1/n is to zero.
>IsnÕt 1/n _pretty close_ to zero for n -> oo ?
>> Distribution theory is not a whole complicated framework
for nothing.
>Depends. I routinely replace delta functions by bell-shapes
(or even
>triangular shapes _/_) in my numerical work. And got no
problems at
>all.
ThatÕs because those exponential things do indeed converge 
to
a
delta function, _in_ the sense of distributions. Which fact
has no relevance whatever to the question of whether the
antiderivative of a delta function is continuous.
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>indeed, in any sense you care to present.
>> Uh, yes it is. If that function were a delta function
>> youÕd have a point.
>IÕm at lost. What point?
That the antiderivative of a delta function is not
continuous. (That was in reply to AiryÕs question
about why generalized functions were relevant here.)
>That function isnÕt a delta function,
>but itÕs quite close to it, indeed. What if you substitute 
it
>for the delta, solve your problem (IÕm at lost *what*
problem,
>though) and, in the end, take the limit for eps->0? Would
that
>work? Think so ...
Work for _what_? Would it work for showing that the
antiderivative of a delta function is continuous,
which is what Airy seemed to be trying to do? No.
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
>>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>>indeed, in any sense you care to present.
>Uh, yes it is. If that function were a delta function
>youÕd have a point.
>>IÕm at lost. What point?
> That the antiderivative of a delta function is not
> continuous. (That was in reply to AiryÕs question
> about why generalized functions were relevant here.)
The antiderivative of a delta function is a heaviside function
h(x) which is not continuous for x = 0, in the classical
sense:
h(x) = 0 for x < 0
h(x) = 1 for x > 0
When it comes to constructivism, there are problems with
heavisideÕs
definition at x = 0. Constructivists would say that a
heaviside is
not defined at x = 0. Hence it is _not a function_, in the
proper
sense of their words. As physicists, we wouldnÕt go that 
far.
While
mainstream mathematics defines something like h(0) = 1/2, we
would
rather say that h(x) is _multi-valued_ for x = 0. Huh?
>>take the limit for eps->0? Would that work? Think so ...
> Work for _what_? Would it work for showing that the
> antiderivative of a delta function is continuous,
> which is what Airy seemed to be trying to do? No.
A heaviside function may be not continuous, but h(x) is quite
close to
an error function Erf(x/eps) which still _is_ continuous for
x = 0, as
eps approaches zero. Therefore continuous or not for h(0)
seems like
nitpicking for the *real* Applied among us. Like radio
amateurs ...
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>>The anti-derivative of
>(sqrt(2 . PI)/eps) .e^(-x^2/2.eps^2))
>is pretty continuous in the classical sense, and
>indeed, in any sense you care to present.
>Uh, yes it is. If that function were a delta function
>>youÕd have a point.
>IÕm at lost. What point?
>> That the antiderivative of a delta function is not
>> continuous. (That was in reply to AiryÕs question
>> about why generalized functions were relevant here.)
>The antiderivative of a delta function is a heaviside
function
>h(x) which is not continuous for x = 0,
Uh, everyone except Airy understands that.
>in the classical sense:
>h(x) = 0 for x < 0
>h(x) = 1 for x > 0
>When it comes to constructivism, there are problems with
heavisideÕs
>definition at x = 0. Constructivists would say that a
heaviside is
>not defined at x = 0. Hence it is _not a function_, in the
proper
>sense of their words. As physicists, we wouldnÕt go that
far. While
>mainstream mathematics defines something like h(0) = 1/2, we
would
>rather say that h(x) is _multi-valued_ for x = 0. Huh?
ThatÕs nonsense. What the value of this function is at 0 has
nothing whatever to do with constructivism.
In fact if you define h(x) as above for x <> 0 and 
define
h(0) to be anything you like, then the derivative of h,
_in the sense of distributions_, is delta.
Or, if you want a stranger example: define h(x) = 42 if x
is rational, 1 if x is irrational and x > 0, and 0 if
x is irrational and x < 0. Then h is continuous nowhere.
Its distribution derivative is still delta.
(The actual explanation for this mystery is that weÕre
talking about distributions, which are in fact not
functions at all - a distribution does not have a
value at _any_ point, itÕs a certain linear functional.
All those various versions of h define the same distribution.)
>take the limit for eps->0? Would that work? Think so ...
>> Work for _what_? Would it work for showing that the
>> antiderivative of a delta function is continuous,
>> which is what Airy seemed to be trying to do? No.
>A heaviside function may be not continuous, but h(x) is
quite close to
>an error function Erf(x/eps) which still _is_ continuous for
x = 0, as
>eps approaches zero.
So 0 is quite close to 1/2? Fine.
>Therefore continuous or not for h(0) seems like
>nitpicking for the *real* Applied among us. Like radio
amateurs ...
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> ThatÕs nonsense. What the value of this function is at 0 
has
> nothing whatever to do with constructivism.
ItÕs _no nonsense_. ItÕs called 
BrouwerÕs Continuity Theorem.
DonÕt know if it is accepted by all constructivist
mathematicians,
admittedly. That theorem says that h(x) is not even defined
for x = 0,
if h(x) is not continuous there. Look it up!
> In fact if you define h(x) as above for x <> 0 and 
define
> h(0) to be anything you like, then the derivative of h,
> _in the sense of distributions_, is delta.
Ah, thatÕs the good news!
[ .. even better news following, but skipped here .. ]
>>A heaviside function may be not continuous, but h(x) is
quite close to
>>an error function Erf(x/eps) which still _is_ continuous
for x = 0, as
>>eps approaches zero.
> So 0 is quite close to 1/2? Fine.
No, but 0 is quite close to our multivalued 0 < h(0) < 1 . (I
know that
a picture is no proof, but it may be convincing once in a
while ...)
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>> ThatÕs nonsense. What the value of this function is at 0
has
>> nothing whatever to do with constructivism.
>ItÕs _no nonsense_. ItÕs called 
BrouwerÕs Continuity Theorem.
>DonÕt know if it is accepted by all constructivist
mathematicians,
>admittedly. That theorem says that h(x) is not even defined
for x = 0,
>if h(x) is not continuous there. Look it up!
Ok, maybe itÕs not nonsense - I missed your point.
The idea that this has some relevance seems a little
nonsensical,
but never mind that.
>> In fact if you define h(x) as above for x <> 0 and 
define
>> h(0) to be anything you like, then the derivative of h,
>> _in the sense of distributions_, is delta.
>Ah, thatÕs the good news!
>[ .. even better news following, but skipped here .. ]
>A heaviside function may be not continuous, but h(x) is
quite close to
>an error function Erf(x/eps) which still _is_ continuous for
x = 0, as
>eps approaches zero.
>> So 0 is quite close to 1/2? Fine.
>No, but 0 is quite close to our multivalued 0 < h(0) < 1 .
(I know that
>a picture is no proof, but it may be convincing once in a
while ...)
Uh, suppose that x is a very small negative number. Then h(x)
= 0,
and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is
close to
h(x) then it follows that 0 is very close to 1/2. No need to
worry about what happens when x = 0.
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> Uh, suppose that x is a very small negative number. Then
h(x) = 0,
> and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is
close to
> h(x) then it follows that 0 is very close to 1/2.
Not true! Give me such a small number (x) and IÕll give you 
a
small
number (eps) such that Erf(x/eps) is very close to 0 . How
close do
you want? Say | Erf(x/eps) - 0 | < delta. Right? An Erf
function is
monotonously increasing. Thus it can be inverted. Name of
inverse
function let be: Erf(-1). Then we find: | Erf(x/eps) | < delta
==>
| x/eps | > | Erf(-1)(delta) | ==> eps < | x / Erf(-1)(delta)
| .
> No need to worry about what happens when x = 0.
The radio amateur: 0 < h(0) < 1 and Erf(0) = 1/2 is in that
range.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>> Uh, suppose that x is a very small negative number. Then
h(x) = 0,
>> and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is
close to
>> h(x) then it follows that 0 is very close to 1/2.
>Not true! Give me such a small number (x) and IÕll give you
a small
>number (eps) such that Erf(x/eps) is very close to 0 .
ThatÕs true. Otoh given eps > 0 there exists x < 0 such that
the two are far apart.
> How close do
>you want? Say | Erf(x/eps) - 0 | < delta. Right? An Erf
function is
>monotonously increasing. Thus it can be inverted. Name of
inverse
>function let be: Erf(-1). Then we find: | Erf(x/eps) | <
delta ==>| x/eps | > | Erf(-1)(delta) | ==> eps < | x /
Erf(-1)(delta) | .
>> No need to worry about what happens when x = 0.
>The radio amateur: 0 < h(0) < 1 and Erf(0) = 1/2 is in that
range.
>Han de Bruijn
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.

<31b2u2F387alsU3@individual.net>
<5vp0r0l9sgivq1g3r2cf91adv1os7a5spq@4ax.com>
<31bonaF380tlpU3@individual.net>
<4iv1r05tl08bo0d4vpq1gipjov7j0r1prl@4ax.com>



<8ft8r052smoav3bug0v8vmcd369hdgolbh@4ax.com>



!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ÕELIi
$t^
VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
>> Uh, suppose that x is a very small negative number. Then
h(x) = 0,
>> and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is
close to
>> h(x) then it follows that 0 is very close to 1/2.
> Not true! Give me such a small number (x) and IÕll give 
you
a small
> number (eps) such that Erf(x/eps) is very close to 0 . How
close do
> you want? Say | Erf(x/eps) - 0 | < delta. Right? An Erf
function is
> monotonously increasing. Thus it can be inverted. Name of
inverse
> function let be: Erf(-1). Then we find: | Erf(x/eps) | <
delta ==> | x/eps | > | Erf(-1)(delta) | ==> eps < | x /
Erf(-1)(delta) | .
Which was pretty much DavidÕs point. Just apply it to your
delta
function depending on eps.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Infantile authours degrading this NG.
>Uh, suppose that x is a very small negative number. Then
h(x) = 0,
>and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is
close to
>h(x) then it follows that 0 is very close to 1/2.
>>Not true! Give me such a small number (x) and IÕll give 
you
a small
>>number (eps) such that Erf(x/eps) is very close to 0 . How
close do
>>you want? Say | Erf(x/eps) - 0 | < delta. Right? An Erf
function is
>>monotonously increasing. Thus it can be inverted. Name of
inverse
>>function let be: Erf(-1). Then we find: | Erf(x/eps) | <
delta ==>>| x/eps | > | Erf(-1)(delta) | ==> eps < | x /
Erf(-1)(delta) | .
> Which was pretty much DavidÕs point. Just apply it to your
delta
> function depending on eps.
If this has been UllrichÕs point, then something must be 
very
wrong
with my understanding of English.
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
>> So 0 is quite close to 1/2? Fine.
> No, but 0 is quite close to our multivalued 0 < h(0) < 1 .
(I know that
> a picture is no proof, but it may be convincing once in a
while ...)
Oops. I meant that Erf(0) = 1/2 is close to our multivalued 0
< h(0) < 1
Han de Bruijn
===
Subject: Re: Infantile authours degrading this NG.
Where does that assertion come from?
Remember - you proscribe the definite
integral int -oo^+oo f(t).d(t-T)
> By definition an antiderivative for f(t).d(t-T)
> would be F, where
> F(x) = int_0^x f(t).d(t-T).
> If x < T then that gives F(x) = 0, while if x > T
> that gives F(x) = f(T), precisely because of that
> formula above.
===
Subject: Re: Infantile authours degrading this NG.
get lost bean
dr. x
> Where does that assertion come from?
> Remember - you proscribe the definite
> integral int -oo^+oo f(t).d(t-T)
>> By definition an antiderivative for f(t).d(t-T)
>> would be F, where
>> F(x) = int_0^x f(t).d(t-T).
>> If x < T then that gives F(x) = 0, while if x > T
>> that gives F(x) = f(T), precisely because of that
>> formula above.
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
By what definition? - You have presented here a
definite integral and not an anti-derivative which
contradicts earlier assertions by you.
Indeed, in the only related definition of which I
am aware, the limits of integration should be +/- oo
and not 0 to x. To which definition in your By 
definition
do you refer?
> By definition an antiderivative for f(t).d(t-T)
> would be F, where
> F(x) = int_0^x f(t).d(t-T).
===
Subject: Re: Infantile authours degrading this NG.
get lost bean
dr. x
> By what definition? - You have presented here a
> definite integral and not an anti-derivative which
> contradicts earlier assertions by you.
> Indeed, in the only related definition of which I
> am aware, the limits of integration should be +/- oo
> and not 0 to x. To which definition in your By 
definition
> do you refer?
>> By definition an antiderivative for f(t).d(t-T)
>> would be F, where
>> F(x) = int_0^x f(t).d(t-T).
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Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
>By what definition? - You have presented here a
>definite integral and not an anti-derivative which
>contradicts earlier assertions by you.
>Indeed, in the only related definition of which I
>am aware, the limits of integration should be +/- oo
>and not 0 to x. To which definition in your By 
definition
>do you refer?
You really need to find a calculus book somewhere.
If F(x) = int_-infinty^x f(t) then F= f; this
is a thing called the Fundamental Theorem of Calculus.
(In the present context int_0^x is the same, since
weÕre assuming that f(t) = 0 for t < 0).
>> By definition an antiderivative for f(t).d(t-T)
>> would be F, where
>> F(x) = int_0^x f(t).d(t-T).
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
You really need to grow up.
This is an international forum and not
the staff room of the infantsÕs school in
which you apparently teach.
YouÕre changing your tune, once again.
The discussion was about int f(t).d(t-T),
now your equivocating about int f(t) only.
Integration and differentiation of products
is not done by simple substitution.
> You really need to find a calculus book somewhere.
> If F(x) = int_-infinty^x f(t) then F= f; 
this
> is a thing called the Fundamental Theorem of Calculus.
> (In the present context int_0^x is the same, since
> weÕre assuming that f(t) = 0 for t < 0).
===
Subject: Re: Infantile authours degrading this NG.
>You really need to grow up.
>This is an international forum and not
>the staff room of the infantsÕs school in
>which you apparently teach.
>YouÕre changing your tune, once again.
>The discussion was about int f(t).d(t-T),
>now your equivocating about int f(t) only.
ItÕs hard to decide. YouÕre just trolling,
or youÕre really unable to follow simple
arguments.
Hint: The f in the statement from me you
quote below is not the same as the f that
weÕve been talking about. If IÕd realized
it was going to cause confusion I would have
used a different letter...
>Integration and differentiation of products
>is not done by simple substitution.
>> You really need to find a calculus book somewhere.
>> If F(x) = int_-infinty^x f(t) then F= f; 
this
>> is a thing called the Fundamental Theorem of Calculus.
>> (In the present context int_0^x is the same, since
>> weÕre assuming that f(t) = 0 for t < 0).
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> YouÕre changing your tune, once again.
Quack? Honk?
===
Subject: Re: Infantile authours degrading this NG.
Yes - I corrected that in a later post. However, the
differentiation and integration was -s/-s giving unity.
> <* is apparent the Ō-was taken out front. But 
since the
beginning, dU
should
> have been Ō-e^(-st)swith the 
Ōsin the numerator, not
denominator. I
> suspect this error came from ARBÕs original attempt at the
integration by
===
Subject: Re: Infantile authours degrading this NG.
You may well have a point here, if my error is in
not evaluating the definite integral over the whole
term.
However, other errors in what you propose nullify your
argument, (see other posts)
> Well, here is what David C. Ullrich posted
>So that line should be
> F(t).e^(-st) ......
===
Subject: Re: Infantile authours degrading this NG.
> You may well have a point here, if my error is in
> not evaluating the definite integral over the whole
> term.
> However, other errors in what you propose nullify your
> argument, (see other posts)
The Ōother errorsare on your part, not 
DavidÕs. (see other
replies)
daestrom
===
Subject: Re: Infantile authours degrading this NG.
Your argument falls down when T = 0,
giving t -T as just t, for then F(0) = f(0),
and not 0.
> Well, here is what David C. Ullrich posted.
>Now when we put in the limits t = 0 to t = infinity
>the first term vanishes since F(0) = 0
===
Subject: Re: Infantile authours degrading this NG.
> Your argument falls down when T = 0,
> giving t -T as just t, for then F(0) = f(0),
> and not 0.
Not true. F(0) = zero, not f(0). This is because F(0) =
delta(0 - T) f(0).
And the Dirac delta function evaluated at (0-T) is zero,
regardless of the
value of f(0). (assuming that T > 0)
daestrom
===
Subject: Re: Infantile authours degrading this NG.
if F(t) = f(t).d(t-T), then with T equal to zero,
this comes down to f(t).d(t).
This might be undefined because it is not expressed
under an integral sign, but it is (probably) not 0.
If it were to be expressed under an integral sign, then
it would be f(0) as I stated.
> Your argument falls down when T = 0,
> giving t -T as just t, for then F(0) = f(0),
> and not 0.
> Not true. F(0) = zero, not f(0). This is because F(0) =
delta(0 - T)
f(0).
> And the Dirac delta function evaluated at (0-T) is zero,
regardless of
the
> value of f(0). (assuming that T > 0)
===
Subject: Re: Infantile authours degrading this NG.
nope, youÕve got it all wrong bean. obviously 
youÕre
incompetent. well we
knew that from the start. need better than that if youÕre
going to stay
here.
dr. x
> if F(t) = f(t).d(t-T), then with T equal to zero,
> this comes down to f(t).d(t).
> This might be undefined because it is not expressed
> under an integral sign, but it is (probably) not 0.
> If it were to be expressed under an integral sign, then
> it would be f(0) as I stated.
>> Your argument falls down when T = 0,
>> giving t -T as just t, for then F(0) = f(0),
>> and not 0.
>> Not true. F(0) = zero, not f(0). This is because F(0) =
delta(0 - T)
> f(0).
>> And the Dirac delta function evaluated at (0-T) is zero,
regardless of
>> the
>> value of f(0). (assuming that T > 0)
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Subject: Re: Infantile authours degrading this NG.
The evaluation yields f(T) which is a constant
and is a graphed as a horizontal straight line
having no expression of time in its evaluation.
This is fundamental to the definition of the Diracian,
where no other properties are specified. This
property is used in other aspects of DSP and so
it must be used consistently here.
> Well, here is what David C. Ullrich posted....
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
===
Subject: Re: Infantile authours degrading this NG.
> The evaluation yields f(T) which is a constant
> and is a graphed as a horizontal straight line
> having no expression of time in its evaluation.
> This is fundamental to the definition of the Diracian,
> where no other properties are specified. This
> property is used in other aspects of DSP and so
> it must be used consistently here.
>> Well, here is what David C. Ullrich posted....
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
No, the antiderivative ( F(t) ) is not a constant. It is a
discontinuous
function of t. Where t T it has a
value of f(T) (note carefully the case of each statement). It
has two
possible values depending on whether you are evaluating it at
tT.
That is not the same thing as what youÕre saying, ... is a
constant...
daestrom
===
Subject: Re: Infantile authours degrading this NG.
That does not agree with the definition with which
I work, which is int -oo^+oo f(t).d(t-T) = f(T).
In that evaluation, all dependence upon t disappears.
What definition are you using to claim discontinuity?
> The evaluation yields f(T) which is a constant
> and is a graphed as a horizontal straight line
> having no expression of time in its evaluation.
> This is fundamental to the definition of the Diracian,
> where no other properties are specified. This
> property is used in other aspects of DSP and so
> it must be used consistently here.
>> Well, here is what David C. Ullrich posted....
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
> No, the antiderivative ( F(t) ) is not a constant. It is a
discontinuous
> function of t.
===
Subject: Re: Infantile authours degrading this NG.
> That does not agree with the definition with which
> I work, which is int -oo^+oo f(t).d(t-T) = f(T).
> In that evaluation, all dependence upon t disappears.
If you integrate it over the entire range from -oo to +oo,
then yes, the
value is f(T). BUT, you canÕt integrate -oo^+oo f(t).d(t-T)
separately from
the Ō(-s)e^(-st)dtterm in the second term of 
your
Ōintegration by partsÕ.
The second term of your Ōintegration by parts
is -int_-oo^+oo
F(t)(-s)e^(-st) dt. That is *not* the same thing as -
(int_-oo^+oo
F(t))
(int_-oo^+oo (-s)e^(-st)dt) You canÕt integrate the two
factors
separately, so you canÕt Ōpull
the f(T) Ōout frontjust yet.
> What definition are you using to claim discontinuity?
The Ōdefinitionis part of the 
Dirac function. The value is
zero for all
values < 0 and all values > 0 and undefined at zero. lim_m->0
[int_-oo^(0-m)delta(t) dt] = 0 and lim_m->0 [int_(0+m)^+oo
delta(t) dt] = 0
and lim_m->0[int_(0-m)^(0+m) delta(t) dt] = 1.0
What you can do is evaluate the integral in steps between -oo
and +oo like
this...
-int_-oo^+oo F(t)(-s)e^(-st) dt = - lim_m->0 [ int_-oo^(T-m)
F(t)(-s)e^(-st)
dt + int_(T+m)^+oo F(t)(-s)e^(-st) dt ]
Since F(t) = 0 for all values of t < T, and F(t) = f(T) for
all values of t
> T ...
-int_-oo^+oo F(t)(-s)e^(-st) dt = - lim_m->0 [ 0 +
int_(T+m)^+oo f(T)
(-s)e^(-st) dt ]
Since this last integral is over the range of T+m to +oo, and
over that
specific range F(t) is a constant, here we can 
Ōpull it out
frontÕ.
= - f(T) lim_m->0[ int_(T+m)^+oo (-s)e^(-st) dt]
= - f(T) [ e^(-s(+oo)) - e^(-sT) ]
As long as s > 0 e^(-s(+oo)) = zero so...
= f(T) [e^(-sT)]
daestrom
===
Subject: Re: Infantile authours degrading this NG.
>That does not agree with the definition with which
>I work, which is int -oo^+oo f(t).d(t-T) = f(T).
>In that evaluation, all dependence upon t disappears.
That evaluation of the _antiderivative_ is simply
_wrong_.
>What definition are you using to claim discontinuity?
>> The evaluation yields f(T) which is a constant
>> and is a graphed as a horizontal straight line
>> having no expression of time in its evaluation.
> This is fundamental to the definition of the Diracian,
>> where no other properties are specified. This
>> property is used in other aspects of DSP and so
>> it must be used consistently here.
>> Well, here is what David C. Ullrich posted....
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
> No, the antiderivative ( F(t) ) is not a constant. It is a
discontinuous
>> function of t.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
That evaluation is the one by which all electronic
engineers are taught.
>That does not agree with the definition with which
>I work, which is int -oo^+oo f(t).d(t-T) = f(T).
>In that evaluation, all dependence upon t disappears.
> That evaluation of the _antiderivative_ is simply
> _wrong_.
===
Subject: Re: Infantile authours degrading this NG.
> That evaluation is the one by which all electronic
> engineers are taught.
But the integral you need for your integration by parts is...
int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt
Not just
int_-oo^+oo f(t)d(t-T) dt
And you canÕt Ōfactorintegrals
int_-oo^+oo f(t)d(t-T)(-s)e^(-st) dt is *not* equal to
[int_-oo^+oo f(t)d(t-T) dt] [int_-oo^+oo (-s)e^(-st) dt]
See my other post for one method of evaluating the term
properly.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
But the gives a different result if the integration
is done in the opposite direction, and therefore
cannot be valid.
Well, here is what David C. Ullrich posted....
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
===
Subject: Re: Infantile authours degrading this NG.
> But the gives a different result if the integration
> is done in the opposite direction, and therefore
> cannot be valid.
> Well, here is what David C. Ullrich posted....
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
Evaluating F(t) where tT you have a
value of f(t).
If evaluated in the reverse order (i.e. over the interval
+inf to zero), you
get -f(t) instead of f(t). This is exactly as it should be.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
If you evaluate the anti-derivative given by Mr.Ullrich
(without accepting it as a correct formula, BTW) then
you would get the properties where F(t) = 0 for t > T
and f(T) for t < T.
> But the gives a different result if the integration
> is done in the opposite direction, and therefore
> cannot be valid.
> Well, here is what David C. Ullrich posted....
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
> Evaluating F(t) where tT you have
a value of
f(t).
> If evaluated in the reverse order (i.e. over the interval
+inf to zero),
you
> get -f(t) instead of f(t). This is exactly as it should be.
> daestrom
===
Subject: Re: Infantile authours degrading this NG.
>> But the gives a different result if the integration
>> is done in the opposite direction, and therefore
>> cannot be valid.
>> Well, here is what David C. Ullrich posted....
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
> Evaluating F(t) where tT you have
a value of
> f(t).
Oops, where t> T you have F(t) = f(T), not f(t). My mistake.
> If evaluated in the reverse order (i.e. over the interval
+inf to zero),
> you get -f(t) instead of f(t). This is exactly as it should
be.
That should be -f(T) instead of f(T), again, my mistake.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
I disagree that it is wrong. The operation being
undertaken is Integration By Parts and
therefore Integration is required and not
Anti-Differentiation.
My f(T).e^(-st) is the result of evaluating
U.int(V), and, as it is part of a sum, evaluting
the integral between the limits of 0^oo to save
further tedious writing.
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
>>Certainly, if you integrate by parts, you would choose
>>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>>I try this, I get 0!......
>>int(UV) = U.int(V) - int[dU.int(V)] giving.....
>> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>>with f(t).d(t-T) as V and e^(-sT) as U .....
>>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>Your first step,
> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>is already wrong. My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals. The (definite) integral 
from
>0 to infinity of f(t).d(t-T) is indeed f(T). But
>what we need here is an antiderivative.
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
===
Subject: Re: Infantile authours degrading this NG.
>I disagree that it is wrong. The operation being
>undertaken is Integration By Parts and
>therefore Integration is required and not
>Anti-Differentiation.
Hilarious. I think you might want to take
Calc 101 again before lecturing us on all
this stuff.
>My f(T).e^(-st) is the result of evaluating
>U.int(V),
If int means int_0^infinty then yes it is, 
thatÕs
exactly whu itÕs wrong.
> and, as it is part of a sum, evaluting
>the integral between the limits of 0^oo to save
>further tedious writing.
Making less and less sense - if youÕd done that
evaluation already then there would be no t
appearing at this point.
ItÕs really good of you to warn me about what a
laughingstock IÕm making of myself, btw.
>> Unfortunately, although he claimed to have posted such, it
was
>> masked by his over-riding urge to be insulting and was
therefore
>> not seen by me.
>Certainly, if you integrate by parts, you would choose
>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>I try this, I get 0!......
>int(UV) = U.int(V) - int[dU.int(V)] giving.....
> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>with f(t).d(t-T) as V and e^(-sT) as U .....
>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>>Your first step,
>> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>>is already wrong. My guess is because of a confusion
>>over definite integrals versus antiderivatives, ie
>>indefinite integrals. The (definite) integral 
from
>>0 to infinity of f(t).d(t-T) is indeed f(T). But
>>what we need here is an antiderivative.
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
>> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
>> I disagree that it is wrong. The operation being
>> undertaken is Integration By Parts and
>> therefore Integration is required and not
>> Anti-Differentiation.
> Hilarious. I think you might want to take
> Calc 101 again before lecturing us on all
> this stuff.
>> My f(T).e^(-st) is the result of evaluating
>> U.int(V),
> If int means int_0^infinty then yes it is, 
thatÕs
> exactly why itÕs wrong.
i just realized now how far this is cross-posted. i suspect
that David is
hanging out on sci.math (but i dunno). David, he really
doesnÕt get it.
and he doesnÕt want to get it. if it were up to Beanie, we
would rewrite
all the calculus textbooks.
letÕs not feed the troll. maybe if we stop feeding it, it
will shrivel up
and blow away.
r b-j
===
Subject: Re: Infantile authours degrading this NG.
> I disagree that it is wrong. The operation being
> undertaken is Integration By Parts and
> therefore Integration is required and not
> Anti-Differentiation.
>> Hilarious. I think you might want to take
>> Calc 101 again before lecturing us on all
>> this stuff.
> My f(T).e^(-st) is the result of evaluating
> U.int(V),
>> If int means int_0^infinty then yes it is, 
thatÕs
>> exactly why itÕs wrong.
>i just realized now how far this is cross-posted. i suspect
that David is
>hanging out on sci.math (but i dunno).
Yes.
> David, he really doesnÕt get it.
>and he doesnÕt want to get it.
up. Was actually trying to help the guy with the math in
my first one or two replies, but since then itÕs 
all been
just for the fun of watching the way the guyÕs technique.
HeÕs pretty good, has things like being insulting and
then whining when people reply insultingly, ignoring
refutations and then claiming they donÕt exist, claiming
that the fact that there have been a lot of replies
proves he must be right, etc down pat. WeÕve got better
here on sci.math but heÕs pretty good. I canÕt 
decide
whether heÕs just trolling or really believes 
heÕs
right...
> if it were up to Beanie, we would rewrite
>all the calculus textbooks.
>letÕs not feed the troll. maybe if we stop feeding it, it
will shrivel up
>and blow away.
What fun would that be?
>r b-j
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
So you are one of the trolls that everybody has been
complaining about.
>since then itÕs all been
> just for the fun of watching the way the guyÕs technique.
===
Subject: Re: Infantile authours degrading this NG.
I see that you cannot resist the temptation to
lower the tone.
> Hilarious. I think you might want to take
> Calc 101 again before lecturing us on all
> this stuff.
> ItÕs really good of you to warn me about what a
> laughingstock IÕm making of myself, btw.
===
Subject: Re: Infantile authours degrading this NG.
>I see that you cannot resist the temptation to
>lower the tone.
If you had any sense youÕd stop making an idiot of
yourself over the math, stop being so utterly
hilarious complaining about the tone of peopleÕs
comments, and just take a calculus class.
>> Hilarious. I think you might want to take
>> Calc 101 again before lecturing us on all
>> this stuff.
>> ItÕs really good of you to warn me about what a
>> laughingstock IÕm making of myself, btw.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
I see that you cannot resist the temptation to
lower the tone.
>I see that you cannot resist the temptation to
>lower the tone.
> If you had any sense youÕd stop making an idiot of
> yourself over the math, stop being so utterly
> hilarious complaining about the tone of peopleÕs
> comments, and just take a calculus class.
>> Hilarious. I think you might want to take
>> Calc 101 again before lecturing us on all
>> this stuff.
>> ItÕs really good of you to warn me about what a
>> laughingstock IÕm making of myself, btw.
===
Subject: Re: Infantile authours degrading this NG.
>I see that you cannot resist the temptation to
> lower the tone.
In an earlier post David pointed out legitimate problems with
the math
behind your argument. Rather than read, and learn from, that
post, you
accidentally deleted it. I suggest you find the post on Google
and read
it. You may learn something.
Charles Perry P.E.
===
Subject: Re: Infantile authours degrading this NG.
I suggest that you read what is written before you open
your mouth and come out with untruths.
You may learn something.
I did not accidentally delete anything. In response to
a barrage of quite unnecessary gratuitous and insulting
remarks by Mr.Ullrich, I _DELIBERATELY_ deleted
things.
If Mr.Ullrich had anything valuable to offer, then it was lost
in the fog of his childish outbursts.
There is no concept of legitimacy. Law does not come into it.
>I see that you cannot resist the temptation to
> lower the tone.
> In an earlier post David pointed out legitimate problems
with the math
> behind your argument. Rather than read, and learn from,
that post, you
> accidentally deleted it. I suggest you find the post on
Google and
read
> it. You may learn something.
===
Subject: Re: Infantile authours degrading this NG.
>I suggest that you read what is written before you open
> your mouth and come out with untruths.
> You may learn something.
> I did not accidentally delete anything. In response to
> a barrage of quite unnecessary gratuitous and insulting
> remarks by Mr.Ullrich, I _DELIBERATELY_ deleted
> things.
One of the posts you _DELIBERATELY_ deleted is the one I
quoted earlier,
where David C. Ullrich, started out quite politely showing
you where you
made your mistakes. And in his more recent post, he states:
If you take f(T) exp(-st), (assume s > 0), evaluate
at t = infinity, t = 0 and subtract you get -f(t).
If F is an antiderivative of f(t) delta(t-T), so
F(t) = 0 for t < T, F(t) = f(T) for t > T, and
you do the same evaluate-subtract with F(t)e^{-st}
you get 0.
This was clear from the difference in your calculation
and my correct version of the same calculuation, btw.
Not much Ōunnecessary gatuitous and insulting 
remarksin
that unless you
consider the last sentence such. And if you do, then you
obviously have a
much thinner Ōskinthan the way you treat 
others.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
letÕs ask crystal, shall we bean?
dr. x
>I suggest that you read what is written before you open
> your mouth and come out with untruths.
> You may learn something.
> I did not accidentally delete anything. In response to
> a barrage of quite unnecessary gratuitous and insulting
> remarks by Mr.Ullrich, I _DELIBERATELY_ deleted
> things.
> If Mr.Ullrich had anything valuable to offer, then it was
lost
> in the fog of his childish outbursts.
> There is no concept of legitimacy. Law does not come into
it.
>>I see that you cannot resist the temptation to
>> lower the tone.
>> In an earlier post David pointed out legitimate problems
with the math
>> behind your argument. Rather than read, and learn from,
that post, you
>> accidentally deleted it. I suggest you find the post on
Google and
>> read
>> it. You may learn something.
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
How does your Indefinite Integral or (Antiderivative)
differ from a Definite Integral when the limits of
integration are imposed?
> Well, here is what David C. Ullrich posted
>is already wrong. My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals.
===
Subject: Re: Infantile authours degrading this NG.
>How does your Indefinite Integral or (Antiderivative)
>differ from a Definite Integral when the limits of
>integration are imposed?
If you take f(T) exp(-st), (assume s > 0), evaluate
at t = infinity, t = 0 and subtract you get -f(t).
If F is an antiderivative of f(t) delta(t-T), so
F(t) = 0 for t < T, F(t) = f(T) for t > T, and
you do the same evaluate-subtract with F(t)e^{-st}
you get 0.
This was clear from the difference in your calculation
and my correct version of the same calculuation, btw.
>> Well, here is what David C. Ullrich posted
>>is already wrong. My guess is because of a confusion
>>over definite integrals versus antiderivatives, ie
>>indefinite integrals.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
No you donÕt.
You get -f(T).
>How does your Indefinite Integral or (Antiderivative)
>differ from a Definite Integral when the limits of
>integration are imposed?
> If you take f(T) exp(-st), (assume s > 0), evaluate
> at t = infinity, t = 0 and subtract you get -f(t).
===
Subject: Re: Infantile authours degrading this NG.
>> Unfortunately, although he claimed to have posted such, it
was
>> masked by his over-riding urge to be insulting and was
therefore
>> not seen by me.
>Well, here is what David C. Ullrich posted.... (lines with
>> are from Airy
>R. BeanÕs previous post, those preceded by 
Ō>are from
David C. UllrichÕs
>pose, my comments are embedded with Ō<*<Õ)
>Show where it is wrong, or else resort
>to infantile sneering.
>>Ok. Took me a minute to find the place
>>where you showed us the integration
>>by parts:
>Certainly, if you integrate by parts, you would choose
>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>I try this, I get 0!......
>int(UV) = U.int(V) - int[dU.int(V)] giving.....
> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>with f(t).d(t-T) as V and e^(-sT) as U .....
>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
>f(T).e^(-sT) - f(T).e^(-sT)....
>0.
>>Your first step,
>> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>>is already wrong. My guess is because of a confusion
>>over definite integrals versus antiderivatives, ie
>>indefinite integrals. The (definite) integral 
from
>>0 to infinity of f(t).d(t-T) is indeed f(T). But
>>what we need here is an antiderivative.
>>An antiderivative of f(t).d(t-T) is given by
>>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>>So that line should be
>> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
><**>>Now when
we put in the limits t = 0 to t = infinity
>>the first term vanishes since F(0) = 0 and e^(-st)
>>tends to 0 as t -> infinty (at least if s > 0; if s < 0
>>this integration by parts is not going to work.)
>>So out Laplace transform becomes
>> - int_0^infinity (e^(-st)/-s . F(t))
>>Since F(t) = 0 for t < T and f(T) for t > T
>>this equals
>> f(T) int_T^infinity (e^(-st)s ).
><*is apparent the Ō-was taken out front. But 
since the
beginning, dU
should
>have been Ō-e^(-st)swith the 
Ōsin the numerator, not
denominator. I
>suspect this error came from ARBÕs original attempt at the
integration by
I pointed out explicitly, in lines you snipped, that Airy had
written e^{-st}/s where it should have been e^{-st} s. I
thought
that IÕd fixed the error at that point, 
itÕs possible that some
erroneous cut&pasted lines were not edited properly.
(Also pointed out at least twice so far that this is somewhat
amusing. Before looking at it I knew that his integration by
parts must be wrong, since it gave the wrong answer, but I
assumed that the problem was with something subtle about
delta functions. Turns out that in addition to not getting
the antiderivative right his amazing conclusion is based
in part on simply differentiating an exponential incorrectly!
ItÕs easy to see why this was the point at which he suddenly
decided he needed to announce that he was not reading any
more replies... Although announcing that heÕd deleted the
replies containing math and then also complaining that
the replies he was reading didnÕt contain any math was
a very bold step.)
>>Again, itÕs easy to evaluate the last integral;
>>int_T^infinity (e^(-st)s ) = e^(-sT), so we
>>finally get f(T)e^(-sT) for the Laplace transform.
><**>daestrom
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> Unfortunately, although he claimed to have posted such, it
was
> masked by his over-riding urge to be insulting and was
therefore
> not seen by me.
>>Well, here is what David C. Ullrich posted.... (lines with
>> are from
>>Airy
>>R. BeanÕs previous post, those preceded by 
Ō>are from
David C.
UllrichÕs
>>pose, my comments are embedded with Ō<*<Õ)
>>Show where it is wrong, or else resort
>>to infantile sneering.
>Ok. Took me a minute to find the place
>where you showed us the integration
>by parts:
>>Certainly, if you integrate by parts, you would choose
>>int(f(t).d(t-T)) as the integrated bit to yield f(T), but
when
>>I try this, I get 0!......
>>int(UV) = U.int(V) - int[dU.int(V)] giving.....
>> int(+/-inf)(f(t).d(t-T).e^(-st)) as .....
>>with f(t).d(t-T) as V and e^(-sT) as U .....
>>f(T).e^(-st) - int(e^(-st)/-s . f(T)).....
>>f(T).e^(-st) - -s.e^(-st)/-s . f(T).....
>>f(T).e^(-sT) - f(T).e^(-sT)....
>>0.
>Your first step,
> f(T).e^(-st) - int(e^(-st)/-s . f(T))
>is already wrong. My guess is because of a confusion
>over definite integrals versus antiderivatives, ie
>indefinite integrals. The (definite) integral 
from
>0 to infinity of f(t).d(t-T) is indeed f(T). But
>what we need here is an antiderivative.
>An antiderivative of f(t).d(t-T) is given by
>F(t), where F(t) = 0 for t < T and f(T) for t > T.
>So that line should be
> F(t).e^(-st) - int(e^(-st)/-s . F(t)).
>><**>Now when
we put in the limits t = 0 to t = infinity
>the first term vanishes since F(0) = 0 and e^(-st)
>tends to 0 as t -> infinty (at least if s > 0; if s < 0
>this integration by parts is not going to work.)
>So out Laplace transform becomes
> - int_0^infinity (e^(-st)/-s . F(t))
>Since F(t) = 0 for t < T and f(T) for t > T
>this equals
> f(T) int_T^infinity (e^(-st)s ).
>><*>It
>>is apparent the Ō-was taken out front. But 
since the
beginning, dU
>>should
>>have been Ō-e^(-st)swith the 
Ōsin the numerator, not
denominator. I
>>suspect this error came from ARBÕs original attempt at the
integration by
> I pointed out explicitly, in lines you snipped, that Airy
had
> written e^{-st}/s where it should have been e^{-st} s. I
thought
> that IÕd fixed the error at that point, 
itÕs possible that
some
> erroneous cut&pasted lines were not edited properly.
> (Also pointed out at least twice so far that this is
somewhat
> amusing. Before looking at it I knew that his integration by
> parts must be wrong, since it gave the wrong answer, but I
> assumed that the problem was with something subtle about
> delta functions. Turns out that in addition to not getting
> the antiderivative right his amazing conclusion is based
> in part on simply differentiating an exponential
incorrectly!
Yes, I was not trying to point the finger at you for the
original error,
just pointing out that at that particular point you corrected
your
explanation (for those that may have wondered how the 
ŌsÕ
term suddenly
moved from denominator to numerator). And that, as I said, I
suspect this
error came from ARBÕs original... posting, not you per se.
Interesting bit of calculus, I may have to go visit sci.math
to see what
other interesting things are bandyÕd about.
daestrom
===
Subject: Re: Infantile authours degrading this NG.
> Interesting bit of calculus, I may have to go visit
sci.math to see what
> other interesting things are bandyÕd about.
> daestrom
Take ARB with you.
===
Subject: Re: Infantile authours degrading this NG.
The bold step that I took was to make a stand
against infantile ad hominem attacks as is your wont.
You continue with your motivation to insult in your
posting below.
If you have anything of value to offer, then offer it
in a mature post.
Shame on you. You should know better.
(If you adopt the same arrogant stance to your
pupils as you do here, and if any of those pupils are
reading this NG, you can be sure that you are the
laughing stock in your school (if not so already))
> ItÕs easy to see why this was the point at which he 
suddenly
> decided he needed to announce that he was not reading any
> more replies... Although announcing that heÕd deleted the
> replies containing math and then also complaining that
> the replies he was reading didnÕt contain any math was
> a very bold step.)
===
Subject: Re: Infantile authours degrading this NG.
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> The bold step that I took was to make a stand
> against infantile ad hominem attacks as is your wont.
> You continue with your motivation to insult in your
> posting below.
> If you have anything of value to offer, then offer it
> in a mature post.
> Shame on you. You should know better.
The irony of it, the irony.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Infantile authours degrading this NG.
>> The bold step that I took was to make a stand
>> against infantile ad hominem attacks as is your wont.
>> You continue with your motivation to insult in your
>> posting below.
>> If you have anything of value to offer, then offer it
>> in a mature post.
>> Shame on you. You should know better.
>The irony of it, the irony.
Indeed. But you left out the best bit, irony-wise:
>(If you adopt the same arrogant stance to your
>pupils as you do here, and if any of those pupils are
>reading this NG, you can be sure that you are the
>laughing stock in your school (if not so already))
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
> The bold step that I took was to make a stand
> against infantile ad hominem attacks as is your wont.
>
> You continue with your motivation to insult in your
> posting below.
>
> If you have anything of value to offer, then offer it
> in a mature post.
>
> Shame on you. You should know better.
>> The irony of it, the irony.
> Indeed. But you left out the best bit, irony-wise:
>> (If you adopt the same arrogant stance to your
>> pupils as you do here, and if any of those pupils are
>> reading this NG, you can be sure that you are the
>> laughing stock in your school (if not so already))
Beanie doesnÕt have pupils (except maybe two beneath his
corneas). Beanie
(real name Gareth) is not competent enough in the subject
matter to be
teaching it to anyone.
despite
http://www.informatics.bangor.ac.uk/public/news/prizes_2002.
shtml
Even kids would be able to figure out quickly that they cannot
learn
(except
by rote, decidedly a bad way to learn math) from Beanie
because what he
says
makes no sense because it is false. even more so with adults.
you cannot
systemically learn underlying concepts from misconceptions.
and you cannot
learn math without learning conceptually.
i imagine that Beanie is unemployed and (due to his character)
unemployable.
r b-j
===
Subject: Re: Infantile authours degrading this NG.
> Beanie doesnÕt have pupils (except maybe two beneath his
corneas).
Beanie
> (real name Gareth) is not competent enough in the subject
matter to be
> teaching it to anyone.
> despite
http://www.informatics.bangor.ac.uk/public/news/prizes_2002.
shtml
Wrong Gareth
The Gareth that is giving you the run around, and that is
exactly what he
is
doing, is in his 50s.
He attended Essex University, although it appears there
little evidence he
graduated, and has an varied employment history. He has
admitted being
dismissed from on job. All the evidence is he screwed up big
time. Other
posts by him suggest that this was not an isolated case.
He trolls varies newsgroups with his wind-ups on a range of
topics, DSP is
one his favorite topics- his theories on sampling first came
to light
several years ago. They have been refuted many times.
Other topics have included a justification for the Soham
murders (for US
readers, two teenage girls were murdered in Soham a few years
ago).
He appears to be unemployable and becomes fixated on those who
have won out
arguements against him.
Do some googling and you will find evidence of all the above.
If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS.
STARVE THE
TROLL.
===
Subject: Re: Infantile authours degrading this NG.
: He appears to be unemployable and becomes fixated on those
who have won
out
: arguements against him.
said NIMROD hiding in the shadows thowing stones from a safe
distance
:
: If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS.
STARVE THE
TROLL.
said NIMROD hiding in the shadows thowing stones from a safe
distance
===
Subject: Re: Infantile authours degrading this NG.
>> Beanie doesnÕt have pupils (except maybe two beneath his
corneas).
Beanie
>> (real name Gareth) is not competent enough in the subject
matter to be
>> teaching it to anyone.
>> despite
http://www.informatics.bangor.ac.uk/public/news/prizes_2002.
shtml
> Wrong Gareth
> The Gareth that is giving you the run around, and that is
exactly what he
is
> doing, is in his 50s.
i found a few old postings where *he* says heÕs class of 
Ō72
(not 2002).
same middle name, same first name, same last name. education 
in
mathematics. lotÕsa coincidences for someone on this side of
the pond.
iÕm
sure you UKers know him better.
> He attended Essex University, although it appears there
little evidence
he
> graduated, and has an varied employment history. He has
admitted
being
> dismissed from on job. All the evidence is he screwed up
big time.
i picked that up.
> Other posts by him suggest that this was not an isolated
case.
> He trolls varies newsgroups with his wind-ups on a range of
topics, DSP
is
> one his favorite topics- his theories on sampling first came
to light
> several years ago. They have been refuted many times.
i know that. another reason i thought it was the young guy in
the photo is
that his arguments are really neophyte (his
integration-by-parts and
his
concepts surrounding bandlimited sampling are indicative of
someone who
hasnÕt really done it) and his means of argument are very
immature. not
reßective of someone in their 50s.
> Other topics have included a justification for the Soham
murders (for US
> readers, two teenage girls were murdered in Soham a few
years ago).
> He appears to be unemployable and becomes fixated on those
who have won
out
> arguements against him.
> Do some googling and you will find evidence of all the 
above.
itÕs exactly what i did.
> If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS.
STARVE THE
TROLL.
i agree.
r b-j
===
Subject: Re: Infantile authours degrading this NG.
>same middle name, same first name, same last name. education
in
>mathematics. lotÕsa coincidences for someone on this side 
of
the pond.
iÕm
>sure you UKers know him better.
There are some historical consistencies with the email
addresses with
which heÕs been associated that put him as an experienced
ringer (a
person who, with other folks, pulls the ropes to toll peals
and carols
from church bell towers), with email traffic to related lists
back to
at least 1998. He was upsetting some of those folks with his
views,
too, and his account of his historical geographic locations is
consistent with where he is now.
So I think the young guy in the website you found is unlikely
to be
Airy. The employment history issues also seem to indicate the
same.
He seems to be interested in and experienced with music,
which is
something he has in common with you and I, though. ;)
Apologies to those of you whoÕve dealt with this for way too
long.
HeÕs still a little new to us at comp.dsp, but 
weÕve had very
similar
experiences with other people.
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be IntelÕs opinions.
http://www.ericjacobsen.org
===
Subject: Re: Infantile authours degrading this NG.
> So I think the young guy in the website you found is
unlikely to be
> Airy. The employment history issues also seem to indicate
the same.
iÕve come around to that conclusion. itÕs just 
that with an
identical
first, middle, and last name and some formal study in
electrical
engineering
and math and being in the UK, that just seemed to be too many
coincidences
for me initially.
> Apologies to those of you whoÕve dealt with this for way
too long.
> HeÕs still a little new to us at comp.dsp, but 
weÕve had
very similar
> experiences with other people.
*very* similar? i think Beanie is in a league of his own. his
trolling
mastery *way* outstrips eBob.
r b-j
===
Subject: Re: Infantile authours degrading this NG.
>*very* similar? i think Beanie is in a league of his own. his
trolling
>mastery *way* outstrips eBob.
Whatever he is I donbelieve he is a troll! He is a very
mixed up
person and can be very vindictive and vicious at times. I
kill-filed
him a long time ago but unfortunately others quote him in
their posts
and reply to the rubbish he puts out,. Only by totally
ignoring him
will he eventually leave the newsgroups that he presently
infests.
Peter
===
Subject: Re: Infantile authours degrading this NG.
>>*very* similar? i think Beanie is in a league of his own.
his
trolling
>>mastery *way* outstrips eBob.
> Whatever he is I donbelieve he is a troll! He is a very
mixed up
> person and can be very vindictive and vicious at times.
If heÕs not a troll, he is one sick pup! HeÕs 
been trolling
under this
pseudonym for some time (which is an anagram of one of his
previous
nemisis). He was trolling comp.arch recently with a wonderful
idea for a
mechanical Turing machine, thinking (I use ther term loosely)
that Alan
had to be working on mechanical beasts.
A half year ago he was berating an alt.folklore.computer
regular for
knowing nothing about computer histor, though the person he
was talking to
was in on the ground-ßoor of the DEC-10, among others. He
couldnÕt even
get the sex right.
> I kill-filed
> him a long time ago but unfortunately others quote him in
their posts
> and reply to the rubbish he puts out,. Only by totally
ignoring him will
> he eventually leave the newsgroups that he presently
infests.
He was sufficiently embarrased that even he decided to leave
AFC and is
likely sucking his tail still. You folks have been doing a
good job of
stuffing the troll too, BTW. ;-)
--
Keith
===
Subject: Re: Infantile authours degrading this NG.
I am not a troll. That you do not agree with me
does not make me one.
I never suggested that Alan Turing had to be thinking on
mechanical lines - I merely said that it was the major
technology extant at the time and that it would be
interesting to make a Turing Machine mechanically.
Insofar as Turing didnÕt make such a machine at the
time of his paper, it is disingenuous to claim that he
would have used other technologies in their infancy.
I have never been on a NG entitled, comp.arch...
are you making this up as you go along in order to
justify your childish burst of tantrum?
It is not possible to determine the sex of a contributor
from the ASCII representation of their Internet Pseudonym.
I have never been embarrassed and felt that I had to leave
a NG. However, when the Mongolian Hordes of infantile
escapees from a school playground gang up, such as yourself
quoted below, then there is no reason to encourage such
behaviour in an international forum. If you say that I had
something
to be embarrassed about - then quote it here, and weÕll 
carry
on
discussing it.....over to you.
This NG is an international public forum
wherein one expects to see technical excellence.
It should be a place for enlightenment
and reasoned technical discussion. It
certainly used to be like that.
For some reason, best known only to
yourself, you have chosen to turn it into
an outpost of the infantsschool
playground using behaviour and emotional
stances left behind years ago by mature contributors.
Such an exhibition by you does nothing for your
reputation, nor for the well-being of this NG.
Shame on you.
You should know better.
Grow up, Keith!
> If heÕs not a troll, he is one sick pup! 
HeÕs been trolling
under this
> pseudonym for some time (which is an anagram of one of his
previous
> nemisis). He was trolling comp.arch recently with a
wonderful idea for a
> mechanical Turing machine, thinking (I use ther term
loosely) that Alan
> had to be working on mechanical beasts.
> A half year ago he was berating an alt.folklore.computer
regular for
> knowing nothing about computer histor, though the person he
was talking
to
> was in on the ground-ßoor of the DEC-10, among others. He
couldnÕt even
> get the sex right.
> He was sufficiently embarrased that even he decided to leave
AFC and is
> likely sucking his tail still. You folks have been doing a
good job of
> stuffing the troll too, BTW. ;-)
===
Subject: Re: Infantile authours degrading this NG.
> I am not a troll. That you do not agree with me
> does not make me one.
You *ARE* a troll! Whether that fact has sunk into that void
between your
ears yet doesnÕt change this fact. You are a know-nothing
that argues
stupid points with those who know better. You argue until
youÕre blue in
the face, or until no one bothers listening to your thines
anymore. ARB
is a well known net loon.
> I never suggested that Alan Turing had to be thinking on
mechanical
> lines - I merely said that it was the major technology
extant at the
> time and that it would be interesting to make a Turing
Machine
> mechanically. Insofar as Turing didnÕt make such a machine
at the time
> of his paper, it is disingenuous to claim that he would
have used other
> technologies in their infancy.
YOu are a loon. YOu have a nearly infinite mechanical tape?
Who cares?
YouÕre still a loon -> troll.
> I have never been on a NG entitled, comp.arch... are you
making this
> up as you go along in order to justify your childish burst
of tantrum?
You canÕt quote-answer either, I guess. I follow a bunch of
groups, but
perhaps it was alt.folklore.computers, *AGAIN*. Sheesh, what
a luser!
> It is not possible to determine the sex of a contributor
from the ASCII
> representation of their Internet Pseudonym.
If youÕd hung around the group a while before you made such
an asshole of
yourself you *should* have been able to figure out who /BAH
was, instead
of making a perfect *ass* of yourself (as usual).
> I have never been embarrassed and felt that I had to leave
a NG.
ThatÕs why you went away in a huff. ...yeah, right!
> However, when the Mongolian Hordes of infantile escapees
from a school
> playground gang up, such as yourself quoted below, then
there is no
> reason to encourage such behaviour in an international
forum. If you say
> that I had something to be embarrassed about - then quote
it here, and
> weÕll carry on discussing it.....over to you.
Yep, I know how to quote and respond, there olBeanie! If
anyone one
cares to search out olbeanie heÕs easy 
enough.
alt.folklore.computers
was only *one* place heÕs been outed.
> This NG is an international public forum wherein one
expects to see
> technical excellence.
Why donÕt you go eleswhere then. You certainly 
havenÕt a clue
here.
> It should be a place for enlightenment and reasoned
technical
> discussion. It certainly used to be like that.
...until you showed up.
> For some reason, best known only to
> yourself, you have chosen to turn it into an outpost of the
infantsÕ
> school
> playground using behaviour and emotional stances left
behind years ago
> by mature contributors.
NOpe Beanie, you do that all by yourself. IÕm just warnign
people who
theyÕre dealing with, as others have done with me.
> Such an exhibition by you does nothing for your reputation,
nor for the
> well-being of this NG.
My reputation isnÕt in question here ol
Beanie. Yours is
shot, however.
...and donÕt blame the messenger. YouÕre 
perfectly capable of
commiting
intellectual suicide.
> Shame on you.
Oh, IÕm *SO* glad you care.
> You should know better.
Ah, ainÕt that sweet!
> Grow up, Keith!
Shut up Gareth! YouÕve proven that you canÕt 
run with the big
dogs. Go
back to your porch.
--
Keith
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy.
> You *ARE* a troll! Whether that fact has sunk into that
void between
your
> ears yet doesnÕt change this fact. You are a know-nothing
that argues
> stupid points with those who know better. You argue until
youÕre blue in
> the face, or until no one bothers listening to your thines
anymore. ARB
> is a well known net loon.
> YOu are a loon. YOu have a nearly infinite mechanical tape?
Who cares?
> YouÕre still a loon -> troll.
> You canÕt quote-answer either, I guess. I follow a bunch 
of
groups, but
> perhaps it was alt.folklore.computers, *AGAIN*. Sheesh,
what a luser!
> If youÕd hung around the group a while before you made 
such
an asshole of
> yourself you *should* have been able to figure out who /BAH
was, instead
> of making a perfect *ass* of yourself (as usual).
> ThatÕs why you went away in a huff. ...yeah, right!
> Yep, I know how to quote and respond, there olBeanie! If
anyone one
> cares to search out olbeanie heÕs easy 
enough.
alt.folklore.computers
> was only *one* place heÕs been outed.
> Why donÕt you go eleswhere then. You certainly 
havenÕt a
clue here.
> ...until you showed up.
> NOpe Beanie, you do that all by yourself. IÕm just warnign
people who
> theyÕre dealing with, as others have done with me.
> My reputation isnÕt in question here ol
Beanie. Yours is
shot, however.
> ...and donÕt blame the messenger. YouÕre 
perfectly capable
of commiting
> intellectual suicide.
> Oh, IÕm *SO* glad you care.
> Ah, ainÕt that sweet!
> Shut up Gareth! YouÕve proven that you canÕt 
run with the
big dogs. Go
> back to your porch.
===
Subject: Re: Infantile authours degrading this NG.
> Stupid boy.
will you marry me beany?
dr. x
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
> *very* similar? i think Beanie is in a league of his own.
his
trolling
> mastery *way* outstrips eBob.
>> Whatever he is I donbelieve he is a troll! He is a very
mixed up
>> person and can be very vindictive and vicious at times.
> If heÕs not a troll, he is one sick pup!
...
> You folks have been doing a good job of stuffing the troll
too, BTW. ;-)
i dunno which of us folks you mean, but us folks on comp.dsp
are tired of
it. iÕd like to see the stuffing beaten out of 
this troll.
this asshole
is
so transparently obnoxious for so long that the humor of the
situation is
long gone. and i havenÕt had to suffer from it as long as 
the
others on
comp.dsp.
--
r b-j rbj@audioimagination.com
Imagination is more important than knowledge.
===
Subject: Re: Infantile authours degrading this NG.
>
>> *very* similar? i think Beanie is in a league of his own.
his
trolling
>> mastery *way* outstrips eBob.
>
> Whatever he is I donbelieve he is a troll! He is a very
mixed up
> person and can be very vindictive and vicious at times.
>> If heÕs not a troll, he is one sick pup!
> ...
>> You folks have been doing a good job of stuffing the troll
too, BTW.
;-)
> i dunno which of us folks you mean, but us folks on
comp.dsp are tired of
> it.
Yes, you good folks on .dsp. ;-)
> iÕd like to see the stuffing beaten out of 
this troll. this
asshole is
> so transparently obnoxious for so long that the humor of
the situation
> is long gone. and i havenÕt had to suffer from it as long
as the others
> on comp.dsp.
You should have heard his nonsense on alt.folklore.computers
(donÕt recall
if the nut posted to comp.arch too)! He has a long history of
infesting
newsgroups. Trust me, itÕs best to deal with vermin as they
show up!
--
Keith
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy.
> You should have heard his nonsense on
alt.folklore.computers (donÕt
recall
> if the nut posted to comp.arch too)! He has a long history
of infesting
> newsgroups. Trust me, itÕs best to deal with vermin as 
they
show up!
===
Subject: Re: Infantile authours degrading this NG.
> Stupid boy.
will you marry me beany?
dr. x
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
he just wants to be loved. im close to him now, i know what
area of the
country he lives in.
wonÕt be long before i can come along and save him, poor old
bean.
dr. x
> i dunno which of us folks you mean, but us folks on
comp.dsp are tired of
> it. iÕd like to see the stuffing beaten out of 
this troll.
this asshole
> is
> so transparently obnoxious for so long that the humor of
the situation is
> long gone. and i havenÕt had to suffer from it as long as
the others on
> comp.dsp.
> --
> r b-j rbj@audioimagination.com
> Imagination is more important than knowledge.
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
I am not a troll, and neither am I mixed up, vindictive or
vicious.
The person speaking below has a particularly nasty
attitude problem which manifests itself in his
behaving like a 5 year old in these NG, an example
of which can be seen below. Notice that is is a
gratuitous contribution unrelated to any other matter.
Peter Day is kill-filing me because he made a fool of himself
by uttering repeated abusive remarks to which I did not rise;
I chastised him for his foolish childishness and now he
is sulking in a kill file.
Stupid boy!
>*very* similar? i think Beanie is in a league of his own. his
trolling
>mastery *way* outstrips eBob.
> Whatever he is I donbelieve he is a troll! He is a very
mixed up
> person and can be very vindictive and vicious at times. I
kill-filed
> him a long time ago but unfortunately others quote him in
their posts
> and reply to the rubbish he puts out,. Only by totally
ignoring him
> will he eventually leave the newsgroups that he presently
infests.
> Peter
===
Subject: Re: Infantile authours degrading this NG.
>I am not a troll, and neither am I mixed up, vindictive or
vicious.
WouldnÕt know about the vindictive or vicious, but if
youÕre not a troll then you are indeed _very_ mixed up.
>The person speaking below has a particularly nasty
>attitude problem which manifests itself in his
>behaving like a 5 year old in these NG, an example
>of which can be seen below. Notice that is is a
>gratuitous contribution unrelated to any other matter.
>Peter Day is kill-filing me because he made a fool of himself
>by uttering repeated abusive remarks to which I did not rise;
>I chastised him for his foolish childishness and now he
>is sulking in a kill file.
>Stupid boy!
>>*very* similar? i think Beanie is in a league of his own.
his
>trolling
>>mastery *way* outstrips eBob.
>> Whatever he is I donbelieve he is a troll! He is a very
mixed up
>> person and can be very vindictive and vicious at times. I
kill-filed
>> him a long time ago but unfortunately others quote him in
their posts
>> and reply to the rubbish he puts out,. Only by totally
ignoring him
>> will he eventually leave the newsgroups that he presently
infests.
>> Peter
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy - once again you resort to the use of gratuitous
personal remarks, merely because you disagree with
an opinion expressed.
Shame on you - you should know better.
> WouldnÕt know about the vindictive or vicious, but if
> youÕre not a troll then you are indeed _very_ mixed up.
===
Subject: Re: Infantile authours degrading this NG.
>Stupid boy - once again you resort to the use of gratuitous
>personal remarks, merely because you disagree with
>an opinion expressed.
Not a personal remark at all. And the idea that these
things are matters of opinion is hilarious. YouÕre
entitled to whatever opinion you wish. But this
is math - the fact that youÕre entitled to your
opinion doesnÕt change the fact that your opinions
are ludicrously wrong.
>Shame on you - you should know better.
>> WouldnÕt know about the vindictive or vicious, but if
>> youÕre not a troll then you are indeed _very_ mixed up.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
Indeed, it was a personal remarks, and followed the
several that you have indulged yourself with; the
several that showed that you were incapable of
partaking in an adult discussion or even
an acadaemic discussion.
All mathematics is a matter of interpretation by
practitioners. What is bizarre about your
quote below is that your opinions are
ludicrously wrong and yet you say that
my opinions are ludicrously wrong.
You say that the Diracian is not continuous,
but the Gaussian form is continuous,
differentiable and integrable using classical
mathematics. You see - your opinions
are ludicrously wrong.
>Stupid boy - once again you resort to the use of gratuitous
>personal remarks, merely because you disagree with
>an opinion expressed.
> Not a personal remark at all. And the idea that these
> things are matters of opinion is hilarious. YouÕre
> entitled to whatever opinion you wish. But this
> is math - the fact that youÕre entitled to your
> opinion doesnÕt change the fact that your opinions
> are ludicrously wrong.
===
Subject: Re: Infantile authours degrading this NG.
>Indeed, it was a personal remarks, and followed the
>several that you have indulged yourself with; the
>several that showed that you were incapable of
>partaking in an adult discussion or even
>an acadaemic discussion.
>All mathematics is a matter of interpretation by
>practitioners. What is bizarre about your
>quote below is that your opinions are
>ludicrously wrong and yet you say that
>my opinions are ludicrously wrong.
>You say that the Diracian is not continuous,
>but the Gaussian form is continuous,
>differentiable and integrable using classical
>mathematics. You see - your opinions
>are ludicrously wrong.
Ok. HereÕs a personal remark, just so you can see
the difference: If youÕre not just trolling then
youÕre either an idiot, a lunatic or both.
HereÕs a proof that 0 > 0, using the same reasoning
as youÕre using to show that a delta function is
continuous: 1/n > 0, and 1/n is a form of 0.
>>Stupid boy - once again you resort to the use of gratuitous
>>personal remarks, merely because you disagree with
>>an opinion expressed.
>> Not a personal remark at all. And the idea that these
>> things are matters of opinion is hilarious. YouÕre
>> entitled to whatever opinion you wish. But this
>> is math - the fact that youÕre entitled to your
>> opinion doesnÕt change the fact that your opinions
>> are ludicrously wrong.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
This NG is an international public forum
wherein one expects to see technical excellence.
It should be a place for enlightenment
and reasoned technical discussion. It
certainly used to be like that.
For some reason, best known only to
yourself, you have chosen to turn it into
an outpost of the infantsschool
playground using behaviour and emotional
stances left behind years ago by mature contributors.
Such an exhibition by you does nothing for your
reputation, nor for the well-being of this NG.
Shame on you.
You should know better.
> Ok. HereÕs a personal remark, just so you can see
> the difference: If youÕre not just trolling then
> youÕre either an idiot, a lunatic or both.
===
Subject: Re: Infantile authours degrading this NG.
>This NG is an international public forum
>wherein one expects to see technical excellence.
>It should be a place for enlightenment
>and reasoned technical discussion. It
>certainly used to be like that.
>For some reason, best known only to
>yourself, you have chosen to turn it into
>an outpost of the infantsschool
>playground using behaviour and emotional
>stances left behind years ago by mature contributors.
>Such an exhibition by you does nothing for your
>reputation, nor for the well-being of this NG.
>Shame on you.
Have you noticed that it never seems to bother anyone
when you say shame on you like this?
>You should know better.
>> Ok. HereÕs a personal remark, just so you can see
>> the difference: If youÕre not just trolling then
>> youÕre either an idiot, a lunatic or both.
************************
David C. Ullrich
===
Subject: Re: Infantile authours degrading this NG.
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> This NG is an international public forum
> wherein one expects to see technical excellence.
Then go away.
> It should be a place for enlightenment
> and reasoned technical discussion.
Then go away.
> It certainly used to be like that.
It will be a step in that direction if you go away.
> For some reason, best known only to
> yourself, you have chosen to turn it into
> an outpost of the infantsschool
> playground using behaviour and emotional
> stances left behind years ago by mature contributors.
You are still confusing first and second person pronouns.
> Such an exhibition by you does nothing for your
> reputation, nor for the well-being of this NG.
If you care for the well-being of this NG, go away.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Infantile authours degrading this NG.
Stupid boy.
> Then go away.
> Then go away.
> It will be a step in that direction if you go away.
> You are still confusing first and second person pronouns.
> If you care for the well-being of this NG, go away.
===
Subject: An infantile author was Re: Infantile authours
degrading this NG.
> You say that the Diracian is not continuous,
> but the Gaussian form is continuous,
> differentiable and integrable using classical
> mathematics. You see - your opinions
> are ludicrously wrong.
Alas, this first sentence is ludicrously wrong :-(
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: Infantile authours degrading this NG.
ho ho bean!
your intelligence astounds me. tell me, where the hell did
you manage to
spell the word up???
good lord, what intelligence you have there, i cannot wait to
hear another
one of your ramblings, maybe i might have some luck as i
scroll down this
newsgroup further, maybe i might start having uncontrollable
orgasms at your
intelligence.
dr. x
>I am not a troll, and neither am I mixed up, vindictive or
vicious.
> The person speaking below has a particularly nasty
> attitude problem which manifests itself in his
> behaving like a 5 year old in these NG, an example
> of which can be seen below. Notice that is is a
> gratuitous contribution unrelated to any other matter.
> Peter Day is kill-filing me because he made a fool of 
himself
> by uttering repeated abusive remarks to which I did not
rise;
> I chastised him for his foolish childishness and now he
> is sulking in a kill file.
> Stupid boy!
>>*very* similar? i think Beanie is in a league of his own.
his
> trolling
>>mastery *way* outstrips eBob.
>> Whatever he is I donbelieve he is a troll! He is a very
mixed up
>> person and can be very vindictive and vicious at times. I
kill-filed
>> him a long time ago but unfortunately others quote him in
their posts
>> and reply to the rubbish he puts out,. Only by totally
ignoring him
>> will he eventually leave the newsgroups that he presently
infests.
>> Peter
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
===
Subject: Re: Infantile authours degrading this NG.
> I am not a troll, and neither am I mixed up, vindictive or
vicious.
> The person speaking below has a particularly nasty
> attitude problem which manifests itself in his
> behaving like a 5 year old in these NG, an example
> of which can be seen below. Notice that is is a
> gratuitous contribution unrelated to any other matter.
> Peter Day is kill-filing me because he made a fool of 
himself
> by uttering repeated abusive remarks to which I did not
rise;
> I chastised him for his foolish childishness and now he
> is sulking in a kill file.
> Stupid boy!
Interesting.
Peter has you in his killfile and he doesnÕt 
respond to you.
Dr Reay has you in his killfile and doesnÕt 
respond to you.
You have claimed to have both in you killfile but respond to
them both.
Seems mixed up to me.
===
Subject: Re: Infantile authours degrading this NG.
Nimrod for a while caused the following to be typed:
> If you want rid of him DO NOT RESPOND TO ANY OF HIS POSTS.
STARVE THE
> TROLL.
Best advice IÕve heard on this subject so far.
--
Radio glossary #21
Packet: An item which causes your XYL to not speak to you for
five days
when it arrives along with a Waters and Stanton invoice.
===
Subject: Re: Infantile authours degrading this NG.
brießy between bottles, to write:
>He appears to be unemployable and becomes fixated on those
who have won
out
>arguements against him.
Which letÕs be honest, is pretty much everybody.
Nick.
===
Subject: Re: Infantile authours degrading this NG.
>Beanie doesnÕt have pupils (except maybe two beneath his
corneas). Beanie
>(real name Gareth) is not competent enough in the subject
matter to be
>teaching it to anyone.
>despite
http://www.informatics.bangor.ac.uk/public/news/prizes_2002.
shtml
...
>i imagine that Beanie is unemployed and (due to his
character)
unemployable.
>r b-j
IÕm donÕt think thatÕs the same 
Gareth Evans (way too young).
Maybe
he can enlighten us. It was evident to me yesterday that
thatÕs
apparently not an uncommon name in the UK (or Australia).
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be IntelÕs opinions.
http://www.ericjacobsen.org
===
Subject: Re: Infantile authours degrading this NG.
>IÕm donÕt think thatÕs the 
same Gareth Evans (way too
young). Maybe
>he can enlighten us. It was evident to me yesterday that
thatÕs
>apparently not an uncommon name in the UK (or Australia).
Gareth is a pretty common Welsh name. Evans is very common
Welsh name.
Put them together and you get a fairly common, and very Welsh
name :-)
Steve
===
Subject: Re: Infantile authours degrading this NG.
> Beanie doesnÕt have pupils (except maybe two beneath his
corneas). Beanie
>(real name Gareth) is not competent enough in the subject
matter to be
>teaching it to anyone.
>despite
http://www.informatics.bangor.ac.uk/public/news/prizes_2002.
shtml
>Even kids would be able to figure out quickly that they
cannot learn
(except
>by rote, decidedly a bad way to learn math) from Beanie
because what he
says
>makes no sense because it is false. even more so with
adults. you cannot
>systemically learn underlying concepts from misconceptions.
and you
cannot
>learn math without learning conceptually.
>i imagine that Beanie is unemployed and (due to his
character)
unemployable.
>r b-j
Not too sure about that logic.
When I was doing A-level maths (A-levels are the UK exams at
18, before
entering university) we had a really bad maths teacher.
Everyone knew
she was really bad, so we largely ignored her. We worked
extra hard on
our own from books, and did lots of exercising with the
published past
papers. The final result was we all achieved really excellent
grades in
which is a bit sad, but the bottom line is her incompetance
actually
achieved good results. :-)
Steve
===
Subject: Re: Infantile authours degrading this NG.
>> Beanie doesnÕt have pupils (except maybe two beneath his
corneas).
Beanie
>> (real name Gareth) is not competent enough in the subject
matter to be
>> teaching it to anyone.
...
>> Even kids would be able to figure out quickly that they
cannot learn
(except
>> by rote, decidedly a bad way to learn math) from Beanie
because what he
says
>> makes no sense because it is false.
...
> Not too sure about that logic.
> When I was doing A-level maths (A-levels are the UK exams
at 18, before
> entering university) we had a really bad maths teacher.
Everyone knew
> she was really bad, so we largely ignored her.
makes sense. i was saying that they cannot learn ... *from*
Beanie....
> We worked extra hard on
> our own from books, and did lots of exercising with the
published past
> papers. The final result was we all achieved really
excellent grades in
> which is a bit sad, but the bottom line is her incompetance
actually
> achieved good results. :-)
just because of two outcomes that might be correlated, it
does not mean
that
one was the cause (ty teacher) and the other was the effect
(good test
results). you do not know how well youÕld have done with a
really good
teacher. perhaps it is that you did so well despite the poor
teacher not
because of the poor teacher.
i am an advocate of good teaching. good, effective, teachers
have to be,
at
the very least, competent in the subject and in related
subjects that they
are teaching. my $0.02 .
r b-j
===
Subject: Question about numbers in arithmetic progression
Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
What
is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
Rich
===
Subject: Re: Question about numbers in arithmetic progression
>Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
What
>is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
Regardless of gcd, the trivial solution applies.
The minimum value is 1/max(|x_1|,|x_n|).
--Keith Lewis klewis {at} mitre.org
The above may not (yet) represent the opinions of my employer.
===
Subject: Re: Question about numbers in arithmetic progression
>>Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
>What
>>is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
>Regardless of gcd, the trivial solution applies.
>The minimum value is 1/max(|x_1|,|x_n|).
I intended that the a_i be integers. My apologies for not
making this
clear.
Rich
===
Subject: Re: Question about numbers in arithmetic progression
> Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
What
> is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
Well, that would depend on x_1, n, and the common difference
d,
so I suppose you want some sort of bound (I donÕt think
youÕll get
an exact expression) in terms of those parameters, yes?
If x_1 & x_2 have any common factor then all the x_i have it,
so itÕs enough to assume x_1 & x_2 are relatively prime.
LetÕs further assume 0 < x_1 < x_2.
Then certainly thereÕs a solution with |a_1| < x_2 and |a_2|
< x_1
and a_i = 0 for i = 3, .... - that gives 2 x_1 + d - 2 as a
bound,
independent of n.
--
Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email)
===
Subject: Re: Question about numbers in arithmetic progression
>> Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
>What
>> is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
>Well, that would depend on x_1, n, and the common difference
d,
>so I suppose you want some sort of bound (I donÕt think
youÕll get
>an exact expression) in terms of those parameters, yes?
Add the condition 0> Suppose x_1,..,x_n are in arithmetic progression and
gcd(x_1,..,x_n)=1.
> What
>> is the minimum value of |a_1|+...+|a_n| where
a_1*x_1+...+a_n*x_n=1?
>Well, that would depend on x_1, n, and the common difference
d,
>so I suppose you want some sort of bound (I donÕt think
youÕll get
>an exact expression) in terms of those parameters, yes?
> Add the condition 0 to above. Here is what I get for m:
> m(2001,2,5)=501 [251*2001-250*2009=1]
> m(20001,2,5)=5001 [2501*20001-2500*20009=1]
> m(200001,2,5)=50001 [25001*200001-25000*200009=1]
> m(1001,3,5)=169
> m(10001,3,5)=1669
> m(100001,3,5)=16669
> m(1001,10,7)=35
> m(10001,10,7)=335
> m(100001,10,7)=3335
> So it seems as though there *may* be an exact expression
for these
parameters.
> Others, like m(1001,31,7), ... are not at all obvious.
> I have a method that seems to compute m(x_1,...,x_n) and am
looking for
inputs
> with known minimums to test against, hence the question.
If you apply the Jacobi method to
a_1*x_1+...+a_n*x_n=1
you get an equation of the form
x_1(a_1 + a_2 + .. a_n) + d(a_2 + 2a_3 + + (n-1)a_m) =1
x_1 A +dB = 1
If P/Q is the penultimate convergent of
x_1/d then x_1 A +dB = 1 has a solution
if you choose appropriate signs for P and Q.
a_2 + 2a_3 + + (n-1)a_n = P =>
|a_n| = INT(p/(n-1))
and some |a_j| (2<= j 
|a_1| = |Q| + |1| +|INT(P/(n-1))
This gives minima which agree with your results.
What values did you get for m(1001,31,7) etc. ?
What is the basis of your method ?
===
Subject: Re: Question about numbers in arithmetic progression
I. M. Davidson wote:
>This gives minima which agree with your results.
But doesnÕt prove these are the actual minima, right?
>What values did you get for m(1001,31,7) etc. ?
I get:
m(1001,31,7)=87
m(10001,31,7)=1417
m(100001,31,7)=6462
m(1000001,31,7)=107548
m(10000001,31,7)=322585
and, curiously,
m(10^12+1,31,7)=64516129040
m(10^22+1,31,7)=322580645161290322585.
>What is the basis of your method ?
Casting in a small pond full of big fish and hoping for the
best?
Seriously, I donÕt have a answer for your question. I will
try to write a
good
description and post it here, though.
Rich
===
Subject: Re: Question about numbers in arithmetic progression
posting-account=YU8ZnAwAAABFplvVaRgMUYVYk_DRcg4s
> I. M. Davidson wote:
>This gives minima which agree with your results.
> But doesnÕt prove these are the actual minima, right?
Precisely. I was thinking that the CF method
gives the smallest (in your sense) solution
to AX + dB =1 and that the smallest solution
to a_2 + d*a_2 + + (n-1)da_n must involve the
a_j wth the largest coefficient and some
other a_k =1. Bit IÕm not sure if this
necessarily follows.
>What values did you get for m(1001,31,7) etc. ?
> I get:
> m(1001,31,7)=87
I get 83 as the minimum here
a_1 = 45, a_5 =-1,a_6 = -37
i.e 45*1001 +1125*(-1) +1187(-37) =1
So your method does not find the minima
in all cases, even though the actual
minimum may be less than 83.
> m(10001,31,7)=1417
Here I get 1954. A considerable
difference.
What are the actual values you
get for the aÕs ?
Whatever they are they must satisfy
the Jacobi reduction.
> m(100001,31,7)=6462
I also have a different value here.
What were the aÕs in this case ?
===
Subject: Re: Question about numbers in arithmetic progression
>> I. M. Davidson wote:
>This gives minima which agree with your results.
>> But doesnÕt prove these are the actual minima, right?
>Precisely. I was thinking that the CF method
>gives the smallest (in your sense) solution
>to AX + dB =1 and that the smallest solution
>to a_2 + d*a_2 + + (n-1)da_n must involve the
>a_j wth the largest coefficient and some
>other a_k =1. Bit IÕm not sure if this
>necessarily follows.
>>What values did you get for m(1001,31,7) etc. ?
>> I get:
>> m(1001,31,7)=87
>I get 83 as the minimum here
>a_1 = 45, a_5 =-1,a_6 = -37
>i.e 45*1001 +1125*(-1) +1187(-37) =1
>So your method does not find the minima
>in all cases, even though the actual
>minimum may be less than 83.
Yes. This is not a real suprise as the problem is quite
difficult. IÕll
have
to work on this one a bit...
>> m(10001,31,7)=1417
>Here I get 1954. A considerable
>difference.
>What are the actual values you
>get for the aÕs ?
1417: -715*10001+689*10187+5*10125+8*10156=1
1433: -723*10001+689*10187+13*10125+8*10032=1
1433: -715*10001+689*10187-8*10094+21*10125=1
1433: -715*10001+689*10187+13*10125+8*10094=1
1443: -728*10001+18*100063+689*100187+8*10094=1
>Whatever they are they must satisfy
>the Jacobi reduction.
>> m(100001,31,7)=6462
>I also have a different value here.
>What were the aÕs in this case ?
6462: 3234*100001-1*100094-3222*100187-5*100125=1
6472: 3223*100001+16*100063-3222*100187-11*100156
Rich
===
Subject: Re: Question about numbers in arithmetic progression
posting-account=YU8ZnAwAAABFplvVaRgMUYVYk_DRcg4s
>> I. M. Davidson wote:
>> m(1001,31,7)=87
>I get 83 as the minimum here
>a_1 = 45, a_5 =-1,a_6 = -37
>i.e 45*1001 +1125*(-1) +1187(-37) =1
>So your method does not find the minima
>in all cases, even though the actual
>minimum may be less than 83.
> Yes. This is not a real suprise as the problem is quite
difficult.
IÕll have
> to work on this one a bit...
realise that I had forgotten about the
ambiguity arising when a CF ends in a 1
- as is the case with 1001/31 and 100001/31.
This gave me values much greater than
they should have been.
>> m(10001,31,7)=1417
>Here I get 1954. A considerable
>difference.
>What are the actual values you
>get for the aÕs ?
> 1417: -715*10001+689*10187+5*10125+8*10156=1
> 1433: -723*10001+689*10187+13*10125+8*10032=1
> 1433: -715*10001+689*10187-8*10094+21*10125=1
> 1433: -715*10001+689*10187+13*10125+8*10094=1
> 1443: -728*10001+18*100063+689*100187+8*10094=1
>Whatever they are they must satisfy
>the Jacobi reduction.
In this case I now get 1411
(-712)*10001 + 699*10187 = 1
>> m(100001,31,7)=6462
>I also have a different value here.
>What were the aÕs in this case ?
> 6462: 3234*100001-1*100094-3222*100187-5*100125=1
> 6472: 3223*100001+16*100063-3222*100187-11*100156
Here too, I get the much lower value than before of 6458
3232*100001 +(-1)*100156) + (-3225)*10187 =1
As the Jacobi reduction is identical
to the original equation, i.e. whenever the
original equation is equal to 1 the Jacobi reduction is
also equal to 1 and vice versa.
As the CF expansion gives the minimum
solution to AX +dY =1,xÕ,ythe minimum
aÕs must be given by the minimum solution
of x= a_1 +a_2 +.. + a_n
y= a_2 + 2*a_3 +.. + (n-1)*a_n
choosing the appropriate signs for xÕ,yÕ
Given the minimum aÕs they must satisfy
the Jacobi reduction and xÕ,yÕmust be minima.
So your initial intuition was correct -
there is a formula (of sorts) for the minimum
solutions.
===
Subject: Re: Question about numbers in arithmetic progression
>So your initial intuition was correct -
>there is a formula (of sorts) for the minimum
> solutions.
Rich
===
Subject: Re: Question about numbers in arithmetic progression
posting-account=YU8ZnAwAAABFplvVaRgMUYVYk_DRcg4s
> I. M. Davidson wote:
>This gives minima which agree with your results.
> But doesnÕt prove these are the actual minima, right?
Precisely. I was thinking that the CF method
gives the smallest (in your sense) solution
to AX + dB =1 and that the smallest solution
to a_2 + d*a_2 + + (n-1)da_n must involve the
a_j wth the largest coefficient and some
other a_k =1. Bit IÕm not sure if this
necessarily follows.
>What values did you get for m(1001,31,7) etc. ?
> I get:
> m(1001,31,7)=87
I get 83 as the minimum here
a_1 = 45, a_5 =-1,a_6 = -37
i.e 45*1001 +1125*(-1) +1187(-37) =1
So your method does not find the minima
in all cases, even though the actual
minimum may be less than 83.
> m(10001,31,7)=1417
Here I get 1954. A considerable
difference.
What are the actual values you
get for the aÕs ?
Whatever they are they must satisfy
the Jacobi reduction.
> m(100001,31,7)=6462
I also have a different value here.
What were the aÕs in this case ?
===
Subject: A first countable separable space that is not second
countable
I need to see an example of a topological space that is first
countable and
separable but not second countable.
Are there any?
===
Subject: Re: A first countable separable space that is not
second countable
> I need to see an example of a topological space that is
first countable
> and separable but not second countable.
DoesnÕt R topologized with basis the sets [a,b) satsify 
this?
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
Lacan, Jacques, 79, 91-92; mistakes his penis for a square
root, 88-9
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
===
Subject: Re: A first countable separable space that is not
second countable
Never mind. I found one.
> I need to see an example of a topological space that is
first countable
and
> separable but not second countable.
> Are there any?
===
Subject: spherical triangle
How is it possible to compute this:
a spherical triangle is given by its three sides (three
angles between
the three vectors). Somewhere in the triangle lies the origin
(North
(angle between vector to point and North pole) to the North
pole. How
can one computes the distance (angle) to the third point?
The coordindates of the points are not given.
S. Nurbe
===
Subject: Re: spherical triangle
ETAsAhRA6YPedgVIROPfRhZHGxprqiLEJgIUI/e7XyhTK3Ni+
O7Ia9MxNHOIlXs=
This problem is easy to solve in spherical trig.
Draw triangle ABC containing the North Pole N inside it. Use
the Law of
Cosines for Arcs to solve for angle ABC:
cos(AC) = cos(AB)cos(BC)+sin(AB)sin(BC)cos(ABC)
where three letters indicate an angle and two letters
indicate an arc.
All arcs in the above are known so the equation can be solved
for the
angle ABC.
Now draw triangle ABN where the arcs AN and BN are also
known. Solve
for angle ABN by the same method as above.
Subtract angle ABN from angle ABC getting angle NBC. Now draw
triangle
NBC and apply the Law of Cosines once more:
cos(CN) = cos(BC)cos(BN)+sin(BC)sin(BN)cos(NBC)
where this time BC, BN, and NBC are known and CN is to be
calculated.
--OL
===
Subject: Re: spherical triangle
<22415-41B3BF7B-38@storefull-3254.bay.webtv.net>
posting-account=umH7TQwAAAAbsM7u2UtD4HnjGexWISl5
Now I have just one more problem. I have to put the
computation of CN
in one equation (which is not that problem) in order to put
this
equation in equation system which I let compute by matlab.
The problem
is that it computes to infinity as long as there are arccos
terms in it
(I donÕt no why). But I canÕt 
find a way to express the
equation of CN
without arccos!
Anybody know?
===
Subject: Re: spherical triangle
> a spherical triangle is given by its three sides (three
angles between
> the three vectors). Somewhere in the triangle lies the
origin (North
> (angle between vector to point and North pole) to the North
pole. How
> can one computes the distance (angle) to the third point?
> The coordindates of the points are not given.
Hint: Stereographic conformal projection from South pole rays
on to a
plane tangent to North pole.
===
Subject: Re: spherical triangle
posting-account=CfSJ5AwAAAD1yt3VP50q913IBHikxMCd
Thought that it would be solvable by the conformally
equivalent
triangle obtained by projecting the geodesic sides on tangent
plane at
one pole from another pole. It is not so, and sorry for this
indication.. The straight forward spherical trigonometry
calculation
outlined by Oscar Lanzi is easier.. often the old problem
(sin(a)/sin(A) equalling to no known property of the
traiangle, that it
is only the Jacobian sin(a)~sin(A) etc.) seems to clog up an
easy
solution (to me), had posted about it earlier also.
===
Subject: Re: spherical triangle
> a spherical triangle is given by its three sides (three
angles between
> the three vectors). Somewhere in the triangle lies the
origin (North
> (angle between vector to point and North pole) to the North
pole. How
> can one computes the distance (angle) to the third point?
> The coordindates of the points are not given.
> Hint: Stereographic conformal projection from South pole
rays on to a
> plane tangent to North pole.
Do you have maybe some more information about this (because I
canÕt
find s.th. proper)?
Or maybe another way?
===
Subject: JSH: Look at it backwards
I usually start with one polynomial and then talk about
dividing 49
off from it, but here IÕll start *after* 49 has been divided
off:
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
and consider the factorization
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
and the assertion that the dÕs must have factors
w_1(x), w_2(x), and w_3(x)
that multiply to give w_1(x) w_2(x) w_3(x) = 7.
But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
functions that are factors of 7?
Ok, maybe that seems unfair, as there could be LOTS of
different
functions you can plug in for the cÕs and dÕs, 
but why should
ANY of
them have functions of x that are factors of 7?
The equation has no memory.
You have a memory, so if I start with the polynomial
multiplied by 49,
then you can say to yourself that thereÕs some dependency on
7. But
itÕs a mirage.
Mathematically a constant multiple is not a big deal. ItÕs
just a
constant multiple that you can divide off, leaving a result
that--you
guessed it--has no memory of the multiple!
One of the weirder things about the discussions here, which I
assume
escapes most of you is some fascinating belief that the
equation has a
memory.
You see something like
P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
and your brain apparently HOLLERS at you that 49 is still
there, but
no, itÕs gone.
When I show something like
P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
you brain insists that 7 is still there.
I mean just LOOK! You can see 7 dividing two of the aÕs and
for GodÕs
sake!!!
ThereÕs a 7 in that last factor, you see it, 
donÕt you?
In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of course 
7
is
still there!!!
So some poster comes at you claiming that 49 divides off from
P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
in a way that varies dependent on the value of x, and your
brain tells
you, OK!!!
You have a memory. To you 7 is still there, even though the
factor is
divided off.
Follwing the sci.mathÕers insistent raving you get
P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
where the wÕs are factors of 7, so that youÕre 
left with
functions of
7.
But they would STILL be there with the factorization of
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
inexplicably still there despite the polynomial not having
any factors
of 7.
Your eyes fool you. Your memory betrays you.
The math doesnÕt bother with such nonsense. A multiple of a
polynomial can be divided off and itÕs gone.
It doesnÕt leave a trace.
In a way whatÕs happening now is an instructive lesson in 
the
limitations of the human brain. Your brains SEE something, and
insistently tell you that 7 is STILL THERE, and so the
arguments go on
for years.
After all, you can SEE the 7Õs. Come on, whoÕs 
fooling who,
right?
Dammit.
You can see the 7Õs in there, canÕt you?
James Harris
===
Subject: Re: JSH: Look at it backwards
> I usually start with one polynomial and then talk about
dividing 49
> off from it, but here IÕll start *after* 49 has been
divided off:
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> and consider the factorization
> S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
> and the assertion that the dÕs must have factors
> w_1(x), w_2(x), and w_3(x)
> that multiply to give w_1(x) w_2(x) w_3(x) = 7.
Let w_1(x)=w_2(x)=w_3(x) = 7^(1/3). Then
the w_i are factors of the d_i in the ring of algebraic
numbers.
(Are they factors in any other ring? Who knows? Who cares?
No one except you has made such a claim.)
> But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
> functions that are factors of 7?
the dÕs have functions that are factors of 7 is gibberish.

-William Hughes
===
Subject: Re: JSH: Look at it backwards
>I usually start with one polynomial and then talk about
dividing 49
>off from it, but here IÕll start *after* 49 has been 
divided
off:
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>and consider the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>and the assertion that the dÕs must have factors
>w_1(x), w_2(x), and w_3(x)
>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
>functions that are factors of 7?
You are getting maximally confused. In your original
approach, you are factoring P(x) as a polynomial in 5,
not as a polynomial in x. The constant term with
respect to 5 and the constant term with respect to
x are different. It was a stupid mistake to think you
were simplifying by treating 5 as if it were the
polynomial variable in the first place. And that is what
is causing your confusion here.
>Ok, maybe that seems unfair, as there could be LOTS of
different
>functions you can plug in for the cÕs and 
dÕs, but why
should ANY of
>them have functions of x that are factors of 7?
>The equation has no memory.
The memory problem here is, you have forgotten what you
started with. To see this, you need to go back to the way you
were doing things originally: for example,
P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2) xu^2
+ u^3 f)
To put this in terms of what you are doing now: replace x in
this
expression by 5, and replace m by x, and replace f by 7. Also
replace u by 1. This gives
P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5
+ 7),
and you can clearly see that, as a polynomial in 5, the
constant
term is 49 * 7 = 7^3, not 49 * 22.
Go back to the expression above for P(m). The constant term
with respect to m is f^2 (3 xu^2 + u^3 f). That happens to
reduce to
49*22 when u = 1, f = 7, and x = 5. But the constant term with
respect to x is u^3 f^3. You have forgotten this and gotten
mixed
up, thatÕs all.
>You have a memory, so if I start with the polynomial
multiplied by 49,
>then you can say to yourself that thereÕs some dependency 
on
7. But
>itÕs a mirage.
>Mathematically a constant multiple is not a big deal. ItÕs
just a
>constant multiple that you can divide off, leaving a result
that--you
>guessed it--has no memory of the multiple!
>One of the weirder things about the discussions here, which
I assume
>escapes most of you is some fascinating belief that the
equation has a
>memory.
>You see something like
>P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
>and your brain apparently HOLLERS at you that 49 is still
there, but
>no, itÕs gone.
No - my brain does no such hollering.
>When I show something like
>P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
Key thing to note here: as I said, the polynomial variable
with respect to which you are factoring is 5, not x. See?
>you brain insists that 7 is still there.
>I mean just LOOK! You can see 7 dividing two of the aÕs and
for GodÕs
>sake!!!
>ThereÕs a 7 in that last factor, you see it, 
donÕt you?
>In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of course
7 is
>still there!!!
>So some poster comes at you claiming that 49 divides off from
>P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
>in a way that varies dependent on the value of x, and your
brain tells
>you, OK!!!
>You have a memory. To you 7 is still there, even though the
factor is
>divided off.
As a polynomial in 5, itÕs still there. It is not a
coincidence
that 22 = 3 * 5 + 7.
>Follwing the sci.mathÕers insistent raving you get
>P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
>7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
>where the wÕs are factors of 7, so that 
youÕre left with
functions of
>But they would STILL be there with the factorization of
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>inexplicably still there despite the polynomial not having
any factors
>of 7.
>Your eyes fool you. Your memory betrays you.
No, itÕs YOUR memory that is the problem. You have forgotten
how you were factoring: not with respect to x, but with
respect
to 5.
>The math doesnÕt bother with such nonsense. A multiple of a
>polynomial can be divided off and itÕs gone.
>It doesnÕt leave a trace.
>In a way whatÕs happening now is an instructive lesson in 
the
>limitations of the human brain. Your brains SEE something,
and
>insistently tell you that 7 is STILL THERE, and so the
arguments go on
>for years.
>After all, you can SEE the 7Õs. Come on, 
whoÕs fooling who,
right?
You are hopelessly tangled in your own dimwitted
oversimplification.
You have forgotten what you were doing.
>Dammit.
>You can see the 7Õs in there, canÕt you?
Sure. On *both* sides. Just look at:
P(x)/49 = (x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2) 5 + 7,
and note that the constant term is ... 7, just like it should
be,
if you regard 5 as the polynomial variable!
Nora B.
>James Harris
===
Subject: Re: JSH: Look at it backwards
>I usually start with one polynomial and then talk about
dividing 49
>off from it, but here IÕll start *after* 49 has been 
divided
off:
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>and consider the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>and the assertion that the dÕs must have factors
>w_1(x), w_2(x), and w_3(x)
>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
>functions that are factors of 7?
> You are getting maximally confused. In your original
> approach, you are factoring P(x) as a polynomial in 5,
> not as a polynomial in x. The constant term with
> respect to 5 and the constant term with respect to
> x are different. It was a stupid mistake to think you
> were simplifying by treating 5 as if it were the
> polynomial variable in the first place. And that is what
> is causing your confusion here.
There is only one variable Nora Baron, so what other variable
would
you have listed with
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
considering that
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22?
So whoÕs confused?
>Ok, maybe that seems unfair, as there could be LOTS of
different
>functions you can plug in for the cÕs and 
dÕs, but why
should ANY of
>them have functions of x that are factors of 7?
>The equation has no memory.
> The memory problem here is, you have forgotten what you
> started with. To see this, you need to go back to the way
you
> were doing things originally: for example,
Now note, the polynomial S(x) doesnÕt have 7 as a factor, as
this time
IÕm starting with the result of dividing off the multiple.
So the poster Nora Baron (actually a guy as revealed in
another post
when he ended with a male name) is trying to show how 7 gets
back into
the expression.
That is, heÕs trying to prove that the factorization DOES
have a
memory of the multiple 49.
> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2)
xu^2 + u^3 f)
> To put this in terms of what you are doing now: replace x
in this
> expression by 5, and replace m by x, and replace f by 7.
Also
> replace u by 1. This gives
> P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2)
5 + 7),
> and you can clearly see that, as a polynomial in 5, the
constant
> term is 49 * 7 = 7^3, not 49 * 22.
No. It can be factored with respect to 5 and 7, but 5 is not a
variable.
The polynomial variable is x.
Now, IÕve explained lots of times to the Nora Baron poster,
and what
I want you all to consider is that the poster isnÕt really
that dense,
but instead knows you better than you know yourselves.
Essentially the basic strategy is just to disagree.
Time after time, and even in answer to surveys that IÕve
done, readers
who normally lurk will admit that they primarily rely on the
fact that
people argue with me, assuming that if I were right, then
others
wouldnÕt disagree!
So, for sci.mathÕers like Nora Baron, the strategy is
clear--just
disagree.
They often donÕt even TRY to actually make mathematical 
senze.
> Go back to the expression above for P(m). The constant term
> with respect to m is f^2 (3 xu^2 + u^3 f). That happens to
reduce to
> 49*22 when u = 1, f = 7, and x = 5. But the constant term
with
> respect to x is u^3 f^3. You have forgotten this and gotten
mixed
> up, thatÕs all.
Now the expressions actually is
S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
where the factorization
S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
is being considered.
Now how do you get 7 into that? Well, thatÕs not the issue
for the
poster!
All he has to do is disagree.
For most readers thatÕs what works.
Well, that enough here.
James Harris
===
Subject: Re: JSH: Look at it backwards
>snip
>There is only one variable Nora Baron . . .
Here I agree with Mr. Harris. I once posted that there are
evidently
a number of real people living in the United States that
really are
named Nora Baron, but those individuals are constant while
our Nora
Baron may indeed be the only one that is a variable --
although that
property has not yet been established for sure.
===
Subject: Re: JSH: Look at it backwards
>snip
>There is only one variable Nora Baron . . .
> Here I agree with Mr. Harris. I once posted that there are
evidently
> a number of real people living in the United States that
really are
> named Nora Baron, but those individuals are constant while
our Nora
> Baron may indeed be the only one that is a variable --
although that
> property has not yet been established for sure.
LOL. IÕve been having a good time today. Lots of fun stuff,
and some
neat accomplishments.
And sci.math posters are actually getting a LOT more
entertaining with
some of their replies. ItÕs a hoot.
The real work though is in working through the details of
*explaining*
my result. Which is an interesting problem, which is also why
I need
so many posts as I just throw a bunch of ideas at the problem
until I
find something that sticks.
ItÕs basically how I found my math results, brainstorming,
but with
expository style.
I think IÕm almost done. IÕve got that 
mathematicians
ignoring my
work are passive-aggressive, and using a passive-aggressive
style. I
have that the work itself resolves down to acceptance of some
VERY
BASIC mathematics known for thousands of years. And I have the
explanation for why some posters obsessively reply to my
posts as
theyÕve learned that simply disagreeing is their most potent
tactic.
The information can be played out in many different
combinations, and
as each one plays out, I use various tools to look for impact.
Eventually IÕll get the right combination.
The key is what IÕm looking for, and IÕm hot 
on the hunt now.
ItÕs
just so, terribly exciting!
James Harris
===
Subject: Re: JSH: Look at it backwards
> The key is what IÕm looking for, and IÕm hot 
on the hunt
now. ItÕs
> just so, terribly exciting!
Be sure and keep your zipper up.
--
There are two things you must never attempt to prove: the
unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Look at it backwards
>>I usually start with one polynomial and then talk about
dividing
49
>off from it, but here IÕll start *after* 49 has been 
divided
off:
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>and consider the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>and the assertion that the dÕs must have factors
>w_1(x), w_2(x), and w_3(x)
>that multiply to give w_1(x) w_2(x) w_3(x) = 7.
>But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
>>functions that are factors of 7?
> You are getting maximally confused. In your original
>> approach, you are factoring P(x) as a polynomial in 5,
>> not as a polynomial in x. The constant term with
>> respect to 5 and the constant term with respect to
>> x are different. It was a stupid mistake to think you
>> were simplifying by treating 5 as if it were the
>> polynomial variable in the first place. And that is what
>> is causing your confusion here.
>There is only one variable Nora Baron,
I didnÕt say there were other variables in your
present polynomial. I was describing its origin,
when there were 4 variables, which explains why
you are now confused about the whole thing.
> so what other variable would
>you have listed with
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>considering that
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22?
>So whoÕs confused?
No question about that. You are.
>>Ok, maybe that seems unfair, as there could be LOTS of
different
>>functions you can plug in for the cÕs and 
dÕs, but why
should ANY of
>>them have functions of x that are factors of 7?
>The equation has no memory.
> The memory problem here is, you have forgotten what you
>> started with. To see this, you need to go back to the way
you
>> were doing things originally: for example,
>Now note, the polynomial S(x) doesnÕt have 7 as a factor,
I DIDNÕT SAY IT DID, ASSHOLE
> as this time
>IÕm starting with the result of dividing off the multiple.
>So the poster Nora Baron (actually a guy as revealed in
another post
>when he ended with a male name) is trying to show how 7 gets
back into
>the expression.
It gets back in in the constant term, because you
have been factoring this function as if it were a polynomial
in 5.
>That is, heÕs trying to prove that the factorization DOES
have a
>memory of the multiple 49.
Not at all. The 49 is obviously gone. But the
constant term at the end, when P(x)/49 is viewed as
a polynomial in 5 [your idea, not mine!], is still
there, and it is 7, not 22.
>> P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1 + mf^2)
xu^2 + u^3 f)
>> To put this in terms of what you are doing now: replace x
in this
>> expression by 5, and replace m by x, and replace f by 7.
Also
>> replace u by 1. This gives
>> P(x) = 49 * ((x^3 7^4 - 3x^2 7^2 + 3x) 5^3 - 3(-1 + x 7^2)
5 + 7),
>> and you can clearly see that, as a polynomial in 5, the
constant
>> term is 49 * 7 = 7^3, not 49 * 22.
>No. It can be factored with respect to 5 and 7, but 5 is not
a
>variable.
You have treated it exactly as such. When you factor P(x)
in the form
P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 a_3(x) + 7),
you are NOT factoring it as a polynomial in x. If you
were, the terms a_1(x), etc., would themselves be
polynomials in x. As you yourself have pointed out MANY
TIMES, this is not a polynomial factorization when
considered as a function of x. (In fact you have claimed
that such Ōnonpolynomial factorizationwas one 
of your
great conceptual breakthroughs. Now you want to deny
it ???). However it CLEARLY has the FORM of a factorization
in 5, where 5 is treated as a polynomial variable.
>The polynomial variable is x.
It most certainly is not [yes, I know you are going
to howl about this, but I stand by what I say]. When you
factor
a b x^2 + (a + b) x + 1
as (a x + 1)(b x + 1),
THEN you are factoring in terms of the polynomial variable
x. The coefficients of x in the factors are functions
of a and b, NOT functions of x, as in your present
factorization of P(x). Have you really no recollection
of how you got started on this track? Originally, x
was m, and 5 was x. At that time you were considering
factors like
a_1(m) x + u f,
etc. This was a factorization in terms of the polynomial
variable x, with coefficients which were algebraic
integer functions of m. Just plug in x = 5, u = 1, f = 7,
and m = x, and you have your present factorization - as
a polynomial in 5, not in x !!
ItÕs easy, actually, to see how you have gotten confused,
especially if (unlike real mathematicians) you have a
poor memory.
>Now, IÕve explained lots of times to the Nora Baron poster,
and what
>I want you all to consider is that the poster isnÕt really
that dense,
>but instead knows you better than you know yourselves.
>Essentially the basic strategy is just to disagree.
I only disagree when you are wrong.
>Time after time, and even in answer to surveys that IÕve
done, readers
>who normally lurk will admit that they primarily rely on the
fact that
>people argue with me, assuming that if I were right, then
others
>wouldnÕt disagree!
I donÕt recall people saying that.
What lurkers do seem to think, when they speak up, is that
YOUR math is vague, hand-waving and illogical. ThatÕs
not my fault. Plus they seem to think that your habit of
making nasty personal attacks damages your credibility.
ThatÕs not my fault either.
>So, for sci.mathÕers like Nora Baron, the strategy is
clear--just
>disagree.
Again, I only disagree when you are wrong. Of course,
that happens to be virtually all the time.
>They often donÕt even TRY to actually make mathematical
senze.
This is gratuitous and false in my case, as you well know.
Also in the cases of Dik Winter, Rick Decker, Arturo Magidin,
ŌRupertÕ, Will Twentyman, Dale Hall, and 
others.
>> Go back to the expression above for P(m). The constant term
>> with respect to m is f^2 (3 xu^2 + u^3 f). That happens to
reduce to
>> 49*22 when u = 1, f = 7, and x = 5. But the constant term
with
>> respect to x is u^3 f^3. You have forgotten this and
gotten mixed
>> up, thatÕs all.
>Now the expressions actually is
>S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
>where the factorization
>S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
>is being considered.
>Now how do you get 7 into that? Well, thatÕs not the issue
for the
>poster!
It was explained in what I posted. You need to consider
how you arrive at S(x), and for that you need to consider
the form of the original function P(x): as a polynomial
in 5, it has constant term 7^3.
I thought you were simply confused on this point, having
forgotten how you got here. I am now inclined to think you
are trying to deny the obvious correct explanation, though
I am not sure why.
Plus throughout your entire reply you seem to be playing
to the grandstand. You whine, exaggerate, and attempt to
discredit. The actual math you have hardly addressed at all.
You seem to think that you can win mathematical arguments
by an appeal to some great silent majority of people out
there who you can cajole into thinking that we critics are
evil, cheating liars who are jealous of your discoveries.
(If that were true, one of us evil cheating liars would have
published it long ago, of course giving no credit to James
S.Harris. Funny, that hasnÕt happened. Wonder why? Why
isnÕt anyone stealing your ideas?)
You seem to have the impression that math is a popularity
contest. ItÕs not. The only way to win is through
rigorous proof. You donÕt have one. ThatÕs the 
main reason
you are not winning.
>All he has to do is disagree.
>For most readers thatÕs what works.
Right. Go ahead and ask your faithful admiring grandstand
if thatÕs what works.
>Well, that enough here.
Yuh - Duh - that enough !
Nora B.
>James Harris
===
Subject: Re: JSH: Look at it backwards
> So the poster Nora Baron (actually a guy as revealed in
another post
> when he ended with a male name) is trying to show how 7
gets back into
> the expression.
What difference does it make what NoraÕs gender is? What 
does
it have to do
with math? Has anybody but you made note of the question of
NoraÕs gender?
No, because it is not relevant. Your obsession with her (or
his) gender just
screams mental illness. It is (probably) just a pseudonym,
nothing more. So
what? My pseudonym is o[CapitalYAcute]in, so do you claim
that I am
pretending to be a
Norse god? Well, nobody cares!!! Idiot!!!
===
Subject: Re: JSH: gametes
So... James Harris got me wondering... which type of gametes
do you actually
produce, Nora Baron?
===
Subject: Re: JSH: gametes
> So... James Harris got me wondering... which type of
gametes do you
actually
> produce, Nora Baron?
See how that works? You start by mocking JSH, and after a
while you
start to think like him. HeÕs addictive. HeÕs 
actually making
some very
good points about the social aspect of proof. Philosophers of
math make
the same arguments ... that a proof is whatever convinces a
majority of
working mathematicians; that whatÕs considered a proof in 
one
era is
regarded as utterly lacking in rigor in another.
And now heÕs got you caring about the gender of a 
palindromic
poster.
===
Subject: Re: JSH: gametes
> See how that works? You start by mocking JSH, and after a
while you
> start to think like him. HeÕs addictive. 
HeÕs actually
making some very
> good points about the social aspect of proof. Philosophers
of math make
> the same arguments ... that a proof is whatever convinces a
majority of
> working mathematicians; that whatÕs considered a proof in
one era is
> regarded as utterly lacking in rigor in another.
> And now heÕs got you caring about the gender of a
palindromic poster.
Yes. You are 100 correct in all that... but now I really do
want to know...
Damn it!!!
===
Subject: Re: JSH: gametes
Nora B. Baron comes from a long line
of palindromic forearmed folks;
if you can make that stick for two generations,
thatÕs farther than IÕve gotten!
> And now heÕs got you caring about the gender of a
palindromic poster.
--Advice, 0.05; free, if wrong!
http://tarpley.net/bush22.htm
===
Subject: Re: JSH: Look at it backwards
> I usually start with one polynomial and then talk about
dividing 49
> off from it, but here IÕll start *after* 49 has been
divided off:
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> and consider the factorization
> S(x) = (c_1(x) + d_1(x))(c_2(x) + d_2(x))(c_3(x) + d_3(x))
Ok, at this point there is no indication as to what these
c_i(x) and
d_i(x) might be, aside from the obvious necessary properties
to generate
the coefficients of S(x) when the factors are multiplied
together. Are
they meant to be continuous functions? Continuous almost
everywhere?
What are their domain/range?
> and the assertion that the dÕs must have factors
> w_1(x), w_2(x), and w_3(x)
> that multiply to give w_1(x) w_2(x) w_3(x) = 7.
If they have such factors, then the range must not be in the
algebraic
integers. Note: I mention that since you usually say
something about
everything is in the algebraic integers.
> But thereÕs no factors of 7 anywhere. So why should the 
dÕs
have
> functions that are factors of 7?
They can quite easily in the appropriate ring. For example,
22 has 7 as
a factor in the rational numbers, algebraic numbers, reals,
etc. What
makes you say thereÕs no factors of 7 anywhere?
> Ok, maybe that seems unfair, as there could be LOTS of
different
> functions you can plug in for the cÕs and 
dÕs, but why
should ANY of
> them have functions of x that are factors of 7?
*IN WHAT RING?* Do you not see that you are making some sort
of
assumption that you have not shared with us?
> The equation has no memory.
Huh? Why would you even speak of such a thing?
> You have a memory, so if I start with the polynomial
multiplied by 49,
> then you can say to yourself that thereÕs some dependency
on 7. But
> itÕs a mirage.
Actually, you are the one who keeps talking about 7 with such
a polynomial.
> Mathematically a constant multiple is not a big deal. ItÕs
just a
> constant multiple that you can divide off, leaving a result
that--you
> guessed it--has no memory of the multiple!
Clue: you are think of things as being processes. Try
thinking of them
as being equivalent statements. You can talk about two
different
equations being equivalent, but they are different equations.
The
notion of a process, or memory of prior steps in the process,
is an
artifact of the work a person does to work towards a goal.
The result
in something like the above is simply a set of equivalent
statements and
some justification for them being equivalent.
> One of the weirder things about the discussions here, which
I assume
> escapes most of you is some fascinating belief that the
equation has a
> memory.
Then why are you the only one who talks about it?
> You see something like
> P(x)/49 = 300125x^3 - 18375 x^2 - 360 x + 22
> and your brain apparently HOLLERS at you that 49 is still
there, but
> no, itÕs gone.
I see it quite clearly. On the left side of the equation.
Where it was
in the previous equation (if at all) is irrelevent.
> When I show something like
> P(x)/49 = (5a_1(x)/7 + 1)(5a_2(x)/7 + 1)(5a_3(x) + 7)
> you brain insists that 7 is still there.
> I mean just LOOK! You can see 7 dividing two of the aÕs 
and
for GodÕs
> sake!!!
I see three 7s, actually.
> ThereÕs a 7 in that last factor, you see it, 
donÕt you?
> In 5a_3(x) + 7. Come on, thereÕs a 7 RIGHT THERE! Of 
course
7 is
> still there!!!
Precisely.
> So some poster comes at you claiming that 49 divides off
from
> P(x) = 49(300125x^3 - 18375 x^2 - 360 x + 22)
> in a way that varies dependent on the value of x, and your
brain tells
> you, OK!!!
Wrong. Simple examples make it clear that, in general, this
is how
things work when viewed from the perspective of working on
*factors* of
a polynomial with a goal of keeping the terms of each factor
in a
particular ring. If you focus on terms of the factors or
donÕt care
about staying within a ring at any particular level, then it
doesnÕt
matter how you write it.
> You have a memory. To you 7 is still there, even though the
factor is
> divided off.
I see only /7, /7, and +7.
> Follwing the sci.mathÕers insistent raving you get
> P(x)/49 = (5a_1(x)/w_1(x) + 7/w_1(x))(5a_2(x)/7/w_2(x) +
> 7/w_2(x))(5a_3(x)/w_3(x) + 7/w_3(x))
> where the wÕs are factors of 7, so that 
youÕre left with
functions of
> 7.
Functions of 7? Huh?
> But they would STILL be there with the factorization of
> S(x) = 300125x^3 - 18375 x^2 - 360 x + 22
> inexplicably still there despite the polynomial not having
any factors
> of 7.
So, it would appear that an appropriate definition for your
functions at
the top of your post, to tie it in with whatÕs down here
would be: given
the a_i(x)s and w_i(x)s, then:
c_i(x) = 5a_i(x)/w_i(x) and d_i(x)=7/w_i(x).
However, that would mean that d_i(x)*w_i(x)=7, which does not
suggest
anything about d_i(x) having a factor of 7 as a factor. So,
assuming
that the d_i(x) was arrived at as above, and that the w_i(x)
are the
same in both cases, where is the notion of w_i(x) being a
factor of
d_i(x) coming from?
> Your eyes fool you. Your memory betrays you.
> The math doesnÕt bother with such nonsense. A multiple of 
a
> polynomial can be divided off and itÕs gone.
> It doesnÕt leave a trace.
> In a way whatÕs happening now is an instructive lesson in
the
> limitations of the human brain. Your brains SEE something,
and
> insistently tell you that 7 is STILL THERE, and so the
arguments go on
> for years.
Apparently you did not understand any of the arguments, or
you wouldnÕt
be saying any of this.
> After all, you can SEE the 7Õs. Come on, 
whoÕs fooling who,
right?
> Dammit.
> You can see the 7Õs in there, canÕt you?
I see two division by 7Õs and a +7. Now, which 7s are
supposed to
still be there?
--
Will Twentyman
email: wtwentyman at copper dot net
===
Subject: Re: Look at it backwards
> The equation has no memory.
This is new! What the are you on about now?
> You have a memory, so if I start with the polynomial
multiplied by 49,
> then you can say to yourself that thereÕs some dependency
on 7. But
> itÕs a mirage.
Mirage? Ummmmm..... Holy moly... I have no words for this...
Ummmm....
OK,
go on....
> Mathematically a constant multiple is not a big deal.
Can you prove that assertion, and can you first 
define
mathematically what a
big deal is. Just joking, I know you are talking
figuratively... but
really James, do you have to add in all this goofy non-math
talk in every
post?
> ItÕs just a
> constant multiple that you can divide off, leaving a result
that--you
> guessed it--has no memory of the multiple!
What does divide off mean? I think I know what you mean, but
can you not
try to use standard terminology?
> Your eyes fool you. Your memory betrays you.
And Wiles was tricked in this way in his FLT proof, right?
===
Subject: Area between these curves..
A friend recently posed this question to me, and IÕd like to
be able to
explain the answer:
integrate[ (sec x)^2 - tan x ] from x=0 to x=1
Simple enough.
the antiderivative of (sec x)^2 = tan x
and
the antiderivative of tan x = -ln(cos x)
So, after calculating F(1) - F(0), the exact answer is:
tan (1) + ln [cos(1)]
Ok, but now say we were to integrate the same function, this
time from x=0
to x=Pi.
There is an asymptote in the given domain shared by both
functions, at Pi /
2.
2 to Pi, the area is also infinite.
Plugging x=Pi into -ln(cos x) results in an imaginary.
Would it be correct to say, that the area between the curves
from 0 to Pi
is
infinite? Or is it really undefined because of the 
imaginary
and/or the
asymptote (non-continuous, therefore non-integrable)?
Or, should the answer be infinity, except at x = Pi / 2?
--Samantha
===
Subject: Re: Area between these curves..
I wasnÕt aware that an absolute value was taken in the
-ln(|cos x|) part,
but this makes sense to me now upon looking closely at it.
And I will be
looking into complex integration.
--Samantha
===
Subject: Re: Area between these curves..
>A friend recently posed this question to me, and IÕd like 
to
be able to
>explain the answer:
>integrate[ (sec x)^2 - tan x ] from x=0 to x=1
>Simple enough.
>the antiderivative of (sec x)^2 = tan x
>and
>the antiderivative of tan x = -ln(cos x)
If you wish to find the antiderivative of 1/u in domains where
u may
be negative, you get ln(|u|) + c. So you would have
-ln(|cos(x)|).
>So, after calculating F(1) - F(0), the exact answer is:
>tan (1) + ln [cos(1)]
>Ok, but now say we were to integrate the same function, this
time from x=0
>to x=Pi.
>There is an asymptote in the given domain shared by both
functions, at Pi
/
>2 to Pi, the area is also infinite.
The integral is called an improper integral because of the
asymptote, and you are correct, it diverges to infinity.
>Plugging x=Pi into -ln(cos x) results in an imaginary.
No, it doesnÕt, see the note above.
>Would it be correct to say, that the area between the curves
from 0 to Pi
is
>infinite?
That is common terminology, or you could say the integral is
divergent.
> Or is it really undefined because of the imaginary
No.
> and/or the asymptote (non-continuous, therefore
non-integrable)?
It is not Riemann integrable, nor does the improper Riemann
integral
converge.
>Or, should the answer be infinity, except at x = Pi / 2?
No.
>--Samantha
--Lynn
===
Subject: Re: Area between these curves..
| Ok, but now say we were to integrate the same function,
this time from
x=0
| to x=Pi.
|
| There is an asymptote in the given domain shared by both
functions, at Pi
/
| 2.
|
| 2 to Pi, the area is also infinite.
Intepreting it physically (area) then itÕs 
infinite (or
undefined!)
|
| Plugging x=Pi into -ln(cos x) results in an imaginary.
Not too sure about this one. Try read up something in the
field of complex
integration.
eg, treat x as z. This is university level maths ok!! :)
===
Subject: JSH: Limitations in thinking
If you follow logic, then thereÕs no room to argue with me.
But if you rely on people relying on their intuitions and
what they
think you see, then you can argue with me for years, as some
sci.mathÕers have done.
For me the realization of what was happening came just a
while back
when I talked with Professor McKenzie at Vanderbilt
University.
I traced out the argument explaining the paper Advanced
Polynomial
Factorization.
his blackboard, and asked pointed questions at key areas.
When he was satisfied, weÕd move on. It took 
about 40 minutes
to go
through it all.
I didnÕt hear *any* of the objections that many of you have
relied on
for years from Professor McKenzie.
I donÕt get those objections anywhere but on Usenet.
Think about it. I thought about it for a while, and I
concluded that
certain posters were *deliberately* putting up objections
that relied
on most of you readers not knowing any better.
They were and are deliberately fooling you, and must be doing
it
consciously as some of them apparently are actually rather
well-versed
in mathematics.
That makes it easier for them to fool you.
It was over a year ago when I realized that and it was
extremely
depressing.
So these people understood what I was talking about,
understood the
mathematics, understood why it was important, but for them,
more
importantly, they understood that they could lie about it
successfully
to people like you, who would either get confused or just
believe
them.
I mean, look at how easily they do it? Not just with my work
on
algebraic number theory, but also with the prime distribution.
ThatÕs with freaking PRIME NUMBERS, and still they confuse
many of
you, and convince many of you with statements that are just
crazy.
Even there, I contacted leading mathematicians, and
corresponded by
email with Lagarias and Odlyzko, mathematicians noted in the
area, and
didnÕt get the objections that sci.mathÕers 
give, until
Odlyzko
replied to me finally, claiming that there were many results
that
while true were not of mathematical interest.
Lagarias never said that or anything like it. HeÕd just
suggest
things like sending it to certain areas, like posting my prime
counting on arxiv.
Well, I told him I couldnÕt get on arxiv. He 
didnÕt reply.
The way this works is that mathematicians I contact by email
or in
person basically just walk away. I explain things to them,
they may
offer a question or two, but thatÕs it. Actually getting to
Professor
McKenzie for an in-depth discussion was a major event for me,
as
mathematicians usually just walk away, which probably
happened because
IÕm an alumnus of Vanderbilt and professors are told to be
nice to us!
So, the difference between Usenet and the rest of the math
world is
that you talk to me, but you make up bizarre objections,
which I know
are really bogus and weird.
Journals represent the area where mathematicians are supposed
to
listen, so I can send a paper to a journal, and if the
editors play by
the rules, it should get a fair hearing.
The Annals at Princeton, will most likely I believe, follow
the rules.
By the rules that math journals follow they are to properly
evaluate
my paper for correctness and relevance.
ThatÕs all I ask.
So why keep posting on Usenet? I think it really is about
boredom and
also about a need just to talk about my research to someone,
even
hostile weird people who make up voodoo mathematics and lie a
lot.
IÕm just short of people to talk to about my research.
On Usenet, I can go on and on and on about it.
James Harris
===
Subject: Re: JSH: Limitations in thinking
> IÕm just short of people to talk to about my research.
> On Usenet, I can go on and on and on about it.
If you really want to communicate, try the following: Write
down
clear, precise definitions for all the terms you use. Put it
on the
web somewhere, and reference it when you post.
For example, you could write
if a is an element of the (unitary) ring R, then we say
a is a unit of (or in) R
iff there exists some c in R so that ca=1.
Then, when you make a post, be careful to always use the
definitions
you have set up and be precise. For example, never write is a
unit,
but is a unit of R.
It really wonÕt take up much time.
> James Harris
Tim Mellor
===
Subject: Re: JSH: Limitations in thinking
> For me the realization of what was happening came just a
while back
> when I talked with Professor McKenzie at Vanderbilt
University.
> I traced out the argument explaining the paper Advanced
Polynomial
> Factorization.
> his blackboard, and asked pointed questions at key areas.
> When he was satisfied, weÕd move on. It took 
about 40
minutes to go
> through it all.
Yes, yes... And what did Professor McKenzie say _at the end_?
If I may be permitted a few words from the peanut gallery,
Harris, old
chap, IÕll try to help you understand why 
youÕre not actually
very
convincing up here. Of course we love to hear you talk the
way you do
- past all the people who post objections, at some imaginary
attentive
audience hanging on your every word, yet dangerously close to
the
slippery slope of being persuaded by the arguments of your
tormentors.
Trouble is, that when one of your detractors says something
that one
of us (if I may presume to be one of ŌusÕ) 
doesnÕt
understand, then we
ask, and typically get a reply we can understand. What is a
unit?
Well, comes the answer, in some ring, dot dot dot.
You, on the other hand, use inscrutable terminology we cannot
find in
books, and no-one else seems to be able to understand either.
When you
say something or other is not a unit in the ring of algebraic
integers, but is properly a unit, what does this mean? May
seem
tough, since others can delegate the responsibility for
answering
elementary questions, but it looks like you are the only one
who can
answer this. So unless you do, we ainÕt believing anything
you say.
Brian Chandler
http://imaginatorium.org
> On Usenet, I can go on and on and on about it.
> James Harris
===
Subject: Re: JSH: Limitations in thinking
>[...]
>Journals represent the area where mathematicians are
supposed to
>listen, so I can send a paper to a journal, and if the
editors play by
>the rules, it should get a fair hearing.
>The Annals at Princeton, will most likely I believe, follow
the rules.
>By the rules that math journals follow they are to properly
evaluate
>my paper for correctness and relevance.
No doubt. And they will _not_ be publishing the paper.
Because theyÕre
going to follow the rules, one of which is that theyÕre not
supposed
to publish obvious nonsense.
WhatÕs not entirely clear is whether youÕll 
get a letter
explaining
that itÕs obvious nonsense or a polite letter saying that 
itÕs
not suitable, maybe some other journal. A letter saying itÕs
obvious
nonsense would really be the best thing for you, in terms of
retaining
some sort of contact with reality, but I doubt theyÕll send
such
a letter, because they wonÕt think it would be polite.
>ThatÕs all I ask.
>So why keep posting on Usenet? I think it really is about
boredom and
>also about a need just to talk about my research to someone,
even
>hostile weird people who make up voodoo mathematics and lie
a lot.
>IÕm just short of people to talk to about my research.
>On Usenet, I can go on and on and on about it.
Uh, weÕre aware that you can go on and on on usenet.
If youÕd actually _learn_ some math then you could make
posts here that people would be interested in.
>James Harris
************************
David C. Ullrich
===
Subject: Re: Limitations in thinking
> If you follow logic, then thereÕs no room to argue with 
me.
Has anybody ever followed your logic? If not, do you think
the problem just
might be in your logic?
> I didnÕt hear *any* of the objections that many of you 
have
relied on
> for years from Professor McKenzie.
Perhaps he was being polite? Perhaps he was scared of you?
Think back to
that day... do you recall him telling you tat you were
correct? If he
thought you were correct, and he understood you logic, why
did he not
recognize the big consequences that you claim about
destroying Galois and
Wiles? Why has he not spread the gospel according to James?
> didnÕt get the objections that sci.mathÕers 
give, until
Odlyzko
> replied to me finally, claiming that there were many results
that
> while true were not of mathematical interest.
Right... if you had what you claim you have, he would have
seen lots of
mathematical interest in your work. What can you infer from
that James?
> Well, I told him I couldnÕt get on arxiv. He 
didnÕt reply.
Oh? He did not reply? Keeping his distance from you? What can
you infer from
that James?
> The way this works is that mathematicians I contact by
email or in
> person basically just walk away.
What can you infer from that James?
> mathematicians usually just walk away, ...
What can you infer from that James?
> The Annals at Princeton, will most likely I believe, follow
the rules.
And they will reject your paper!
> IÕm just short of people to talk to about my research.
What can you infer from that James?
> On Usenet, I can go on and on and on about it.
Yes, for years.... we know James...
===
Subject: JSH: Two sides
IÕve had months to wonder why mathematicians would work to
ignore or
hide my results, and I think there are two sides to it.
For one thing, my discoveries betray much of what theyÕve
been taught
about mathematics and itÕs not just that I was able to 
find a
problem
in core, but also that there were results to find, using 
rather
basic methods.
Somehow, despite what has been often said, all those other
mathematicians working over hundred of years didnÕt cover 
all
the
ground for use of simple arguments and simple ideas.
Like my work with non-polynomial factors relies on something
as simple
as factoring a polynomial with respect to something other
than the
polynomial variable.
Do that and an entire world opens up. Possibilities that
didnÕt exist
with other methods are suddenly easy. Factors of irrational
numbers
can be discerned, even if you canÕt tell *which* root might
have a
number like 7 as a factor, you can know that at least one of
them
must.
And to add insult to injury (from a particular point of view)
I went
on to find a prime counting function that relies on a partial
difference equation.
Hints of it can be seen all over the area, and MeisselÕs
Formula
actually uses a variant of it, while part of it looks (as some
sci.mathÕers have noted) like the recurrence relation that
follows
from LegendreÕs Method, as in fact a relationship can easily
be
proven.
So what was missed?
It looks like no one figured on a multi-variable function, but
instead
kept focusing on the single variable function.
After all IT MAKES SENSE that it be a single variable
function, as you
have one cout of prime numbers for a given value. The count
of primes
up to 10 is 4. The count of primes up to 100 is 25. The count
of
primes up to 1000 is 168.
One variable seems to be all you need.
Yet, here is a simple, compact prime counting function that
captures
the entire prime distribution in a small space, with, a
three-dimensional p(x,y) function.
So yet another simple idea, not far away, but so close.
Yet losing what you had, your beliefs, your communityÕs
shared vision,
is one thing, looking at a world thatÕs not cognizant of it 
is
another.
The second even more brutal side of it for mathematicians may
be in
realizing that the rest of the world doesnÕt have a clue.
After all,
the discoverer was labeled a crank by sci.mathÕers and the
label seems
to have stuck.
Years have gone by, with so much overturned, so many beliefs
shattered, and so much research shown to be invalid, but the
public
doesnÕt notice, or care.
The classes keep coming. The papers are still being published.
Awards and prizes still given out. Day follows day. Month
follows
month, and the world continues on.
That may be the greatest insult of all that maybe none of it
ever
really mattered. That maybe the world never really cared. That
mathematicians were just part of some play, some game, some
empty soap
opera without any real meaning.
After all, some of the most dramatic events in world history
are
passing without a notice. The much vaunted world press 
hasnÕt
a clue.
The supposed intellectuals in all their corners prance, and
preen,
talk as much as ever, and donÕt realize.
The world knows nothing, and that maybe the harshest lesson
of all.
Maybe then, after all, it never really mattered, and what
matters now
is to hold on to what you have.
ItÕs not really possible to completely understand how they
feel, what
they feel, unless youÕre in their shoes. To have so many
beliefs
shattered. To have so much that you thought true, shown to be
false.
To feel betrayed down to the depths of your core...thatÕs
something
you have to live to understand.
James Harris
===
Subject: Re: JSH: Two sides
> IÕve had months to wonder why mathematicians would work to
ignore or
> hide my results,
Ignoring your results takes no effort, hiding them does not
occur.
--
There are two things you must never attempt to prove: the
unprovable --
and the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com
===
Subject: Re: JSH: Two sides
[snip delusion]
What does properly a unit mean?
Can you answer that simple question James Harris?
===
Subject: Re: JSH: Two sides
> [snip delusion]
> What does properly a unit mean?
> Can you answer that simple question James Harris?
JSH does not answer questions.
===
Subject: Re: JSH: Two sides
!3KEIp?*w`|bL5qr,H)LFO6Q=qx~iH4DN;i;/yuIsqbLLCh/!U#X[S~(
5eZ41to5f%E@ÕELIi
$t^
VcLWP@J5p^rst0+(Ō>Er0=^1{]M9!p?&:z]|;&=NP3AhB!B_bi^]Pfkw
> IÕve had months to wonder why mathematicians would work to
ignore or
> hide my results, and I think there are two sides to it.
Ignoring takes no work.
> Like my work with non-polynomial factors relies on
something as
> simple as factoring a polynomial with respect to something
other
> than the polynomial variable.
> Do that and an entire world opens up.
Sure, if you have something more than polynomials in one
variable, you
have lots of opportunities for confusing constants with
regard to one
variable to constants with regard to another.
It is something you have to be aware of as soon as you work
with more
than one variable. Very basic stuff. And where variables are
not
necessarily independent, things get more complicated (partial
differential equations ooze all over with the intricacies).
You donÕt
get it. Not even the basics.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
===
Subject: Re: Two sides
> IÕve had months to wonder why mathematicians would work to
ignore or
> hide my results, and I think there are two sides to it.
Just one side to it.... you are wrong.
===
Subject: Re: Two sides
says...
> IÕve had months to wonder why mathematicians would work to
ignore or
> hide my results, and I think there are two sides to it.
> Just one side to it.... you are wrong.
and INCREDIBLY lonely.
===
Subject: Re: November 25 is Infinite Clause day!!


<9kq772-t9i.ln1@sirius.athghost7038suus.net>

<96c872-j7j.ln1@sirius.athghost7038suus.net>
<87653og9oe.fsf@phiwumbda.org>
posting-account=Qiuj5AwAAACmGnmS12qcvqA9IXzD0s4L
huh! Go figger!
Herc
===
Subject: Example of a formulation in production scheduling
Hello everybody,
I am looking for an example formulation of production
scheduling
including storages with the objective of minimising the
makespan.
I want the MILP formulations of this example to be solved
using
Lp_solve if possible.
===
Subject: FLTMA: S.1 = c
For every triple a,b,c of postive integers there is an
associated series
{S.n)
=
(a^n+b^n) mod c +
(c^n - a^n) mod b +
(c^n - b^n) mod a
and I find that while S.0 is of course 2, that S.1 seems to be
equal to c.
At least in my tests, where I am applying the additional
conditions
aS.1 = a mod c + b mod c + c mod b - a mod b + c mod a - b
mod a
>= a + b + c mod b - a + c mod a - b mod a
>= b - b mod a + c mod b + c mod a
= b - (b-a) + (c-b) + (c-a)
= a + 2c - b - a
= 2c - b
I must have done that wrong.
writing m for mod
S.1=
amc + bmc + cmb - amb + cma - bma=
a + b + (c-b) - a + (c-a) - (b-a)=
a + b + c - b - a + c - a - b + a=
a - a + a - a + b - b - b + c + c=
2c - b.
I will have to show the intermediate sums in Mathcad.
I tolerance everything and tolerate everyone.
I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri,
alphabetically.
I drive: A double-step Thunderbolt with 657% range.
I fight terrorism by: Using less gasoline.
===
Subject: Re: FLTMA: S.1 = c
One more time
c-a < b,
c-b < a
and then making certain assumptions that just sort of come to
mind:
S.1 =
(a mod c + b mod c) mod c +
(c mod b - a mod b) mod b +
(c mod a - b mod a) mod a =
( a + b ) mod c +
(c-b - a) mod b +
(c-a - (b-a)) mod a =
a+b - c +
c - b - a - b +
c - a + b + a - a =
a + b - c + c - b - a - b + c - a + b + a - a =
- c + c + c + b - b + b + a - a - a + a - a =
c + b - a
Which is closer
Still some consideration of the underlying inequalities is
necessary. a
could
be < c/2 or < c/3, etc....
I need to see how the constraint S.n = 0 affects S.1 = c.
I tolerance everything and tolerate everyone.
I love: Dona, Jeff, Kim, Kimmie, Mom, Neelix, Tasha, and Teri,
alphabetically.
I drive: A double-step Thunderbolt with 657% range.
I fight terrorism by: Using less gasoline.
===
Subject: Novice: Pi in other base systems
Hello...
My apologies if this is a silly question. How does pi behave
in
different base systems, for instance binary, base 8 or base
16?
Is it still a mysterious number with endless non-repeating
numbers? Is there any base system where thereÕs some
especially
interesting results?
(I guess the correct way to ask this is how the ration of
circumference to diameter behaves in different number systems,
but still you get the point.)
Just curious.
Thx,
D.
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi
behave in
> different base systems, for instance binary, base 8 or base
16?
> Is it still a mysterious number with endless non-repeating
> numbers? Is there any base system where thereÕs some
especially
> interesting results?
> (I guess the correct way to ask this is how the ration of
> circumference to diameter behaves in different number
systems,
> but still you get the point.)
> Just curious.
> Thx,
> D.
It is possible to compute the n-th digit of pi in hexadecimal
--- and
thus also in binary --- notation without computing all the
previous
digits. (I donÕt have the reference here right now, but 
IÕm
sure you
will not be eventually repeating since pi is not rational.
Hope this helps,
Lasse
---
===
Subject: Re: Novice: Pi in other base systems
In sci.math, piter
<1jRrd.45$wy5.6266@news.uswest.net>:
> Hello...
> My apologies if this is a silly question. How does pi
behave in
> different base systems, for instance binary, base 8 or base
16?
> Is it still a mysterious number with endless non-repeating
> numbers? Is there any base system where thereÕs some
especially
> interesting results?
> (I guess the correct way to ask this is how the ration of
> circumference to diameter behaves in different number
systems,
> but still you get the point.)
> Just curious.
> Thx,
> D.
The digits differ, but the nature of pi never changes. :-)
However, this may interest you; it can be used to implement
progressive digits in base 16. Look at equations (29), (31),
and (32) in
http://mathworld.wolfram.com/PiFormulas.html
--
#191, ewill3@earthlink.net
ItÕs still legal to go .sigless.
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi
behave in
> different base systems, for instance binary, base 8 or base
16?
> Is it still a mysterious number with endless non-repeating
> numbers?
Consider: suppose there were a base b in which the expression
of pi
became repeating. That is, we could write p = pi-3 as:
(0.a_1 a_2 a_3 ... a_n-1 a_n a_n+1 ... a_n+k a_n+1 a_n+k ...)b
where the a_Õs are valid digits in base b, and (x y z)b 
means
Ōrepresentation in base bÕ. Here we have n 
digits then a
repeating
group of k digits. Then we have:
p b^n = (a_1 a_2 ... a_n . a_n+1 ... a_n+k a_n+1 ... a_n+k
...)b
p b^n - (a_1 a_2 ... a_n)b = (0. a_n+1 ... a_n+k a_n+1 ...
a_n+k ...)b
Call this quantity q. Now we have
q b^k = (a_n+1 ... a_n+k . a_n+1 ... a_n+k a_n+1 ... a_n+k
...)b
q b^k - q = (a_n+1 ... a_n+k)b
q = (a_n+1 ... a_n+k)b / (b^k - 1)
Both numerator and denominator here are integers, so this
means q is
rational. We defined q by
q = p b^n - (a_1 a_2 ... a_n)b
so
p = (q + (a_1 a_2 ... a_n)b) / b^n
which will still be rational; but p = pi - 3, and we know pi
to be
irrational. We have a contradiction - so our assumption that
there is
a base b in which the representation of pi becomes repeating
must be
false.
The key point here which makes this not so much a Ōsilly
questionas
a Ōquestion which shows I donÕt know this key 
point:) is
that
rationality <=> a repeating representation in a rational base
You can prove this yourself (both ways) using the same
technique I
just demonstrated. Note that you can regard 
ŌterminatingÕ
expansions
as repeating with a repesting group consisting of 
Ō0- it
makes
proofs shorter. In simple arithmetic terms, it comes down to
noticing
/ knowing / establishing that eg
0.738475 738475 ... (arbitrary digits)
= 738475 / 999999
which is obvious once you know it, or Ōtrivial
as
mathematicians say
:)
> Is there any base system where thereÕs some especially
> interesting results?
One interesting result is the BBP formula
 which can be
used to
obtain the digits of the hexadecimal expansion with relative
ease.
> (I guess the correct way to ask this is how the ration of
> circumference to diameter behaves in different number
systems,
> but still you get the point.)
Remember that bases are not different number systems, they
are just
different *representations* of the same numbers!
--
Larry Lard
Replies to group please
===
Subject: Re: Novice: Pi in other base systems
> Hello...
> My apologies if this is a silly question. How does pi
behave in
> different base systems, for instance binary, base 8 or base
16?
> Is it still a mysterious number with endless non-repeating
> numbers? Is there any base system where thereÕs some
especially
> interesting results?
> (I guess the correct way to ask this is how the ration of
> circumference to diameter behaves in different number
systems,
> but still you get the point.)
Since pi is not a rational number, it is mysterious when
expressed in
any
integer base. There may be some interesting results using a
transcendental
base, such as pi, or e. But then the representations of all
the integers
would be mysterious.
===
Subject: Re: Novice: Pi in other base systems
| Hello...
|
| My apologies if this is a silly question. How does pi
behave in
| different base systems, for instance binary, base 8 or base
16?
| Is it still a mysterious number with endless non-repeating
| numbers? Is there any base system where thereÕs some
especially
| interesting results?
|
| (I guess the correct way to ask this is how the ration of
| circumference to diameter behaves in different number
systems,
| but still you get the point.)
|
| Just curious.
Well, base pi makes it a very succinct number. ______Gerard S.
===
Subject: Re: induction vs Cantor
>> > What I am saying is that if one can consider
the totality of all
>> functions from N to R, without worrying about
constructability, then
>> there is no problem with the diagonal proof of CantorÕs
theorem,
>> since
>> the Cantor anti-diagonal algorithm applied to any member
of that set
of
>> functions is guaranteed to produce a number not in the
image of the
>> function, and that production is what I have called the
Cantor
function
>> from R^N to R.
>> Uses CantorÕs conclusions.
> No. Only his assumptions.
No. His conclusions.
===
Subject: Re: induction vs Cantor
>  What I am saying is that if one can consider the
totality of all
>> functions from N to R, without worrying about
constructability, then
>> there is no problem with the diagonal proof of CantorÕs
theorem,
>> since
>> the Cantor anti-diagonal algorithm applied to any member
of that set
of
>> functions is guaranteed to produce a number not in the
image of the
>> function, and that production is what I have called the
Cantor
function
>> from R^N to R.
> Uses CantorÕs conclusions.
> No. Only his assumptions.
> No. His conclusions.
Which ones? Are CantorÕs conclusions about the value of 1+1
to be
excluded as well as his conclusions about the uncountablity
of reals?
Is anything that tends to lead ultimately to the
uncountablility of the
reals to be excluded?
Uncountability is nowhere a rerequisite of the construction
of what I
have called the Cantor function, it is only a consequence of
it, so it
seems to me that you are outtlawing anything leading to
CantorÕs
conclusion, not merely the conclusion itself.
It is not clear to me that your blanket accusation does not
rule out
even 1+1 = 2, unless you are a bit clearer about just which
conclusions are to be outlawed.
===
Subject: Re: induction vs Cantor
 what I have called the Cantor function from R^N to
R.
Describe this class of functions and show how Cantor used
them in his
proof as you say. IÕm looking for where he 
classified them as
R^N to R
and that they were uncountable.
===
Subject: Re: induction vs Cantor
Poker Joker says...
>> what I have called the Cantor function from R^N to R.
>Describe this class of functions and show how Cantor used
them in his
>proof as you say. IÕm looking for where he 
classified them as
R^N to R
>and that they were uncountable.
I think Virgil means the class of functions f which given
an infinite list of reals returns a new real that is not
on the list.
Cantor came up with only one such function: his
diagonalization
function. R^N just means the set of functions from N to R, or
equivalently, the set of infinite lists of reals, or a set of
reals indexed by the naturals. CantorÕs diagonalization
procedure
defines a function which given an infinite list of 
reals (that
is,
an element of R^N) returns an element of R that is not on the
list.
Cantor didnÕt prove that there were uncountably many such
functions,
he only proved that there was one. ThatÕs all he needs to be
able
to show that no list of reals contains every real.
It is easy enough to show that the cardinality of the set of
such
functions is uncountable, although this fact is not used in
CantorÕs
proof.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor
CantorÕs diagonalization procedure
> defines a function which given an infinite list 
of reals
(that is,
> an element of R^N) returns an element of R that is not on
the list.
> Cantor didnÕt prove that there were uncountably many such
functions,
> he only proved that there was one. ThatÕs all he needs to
be able
> to show that no list of reals contains every real.
Do you think there are any counter-examples to CantorÕs
diagonal
proof?
If so, can you provide a valid modification that keeps the
concept
of providing a real that is proven not in the list and has no
counter-example?
===
Subject: Re: induction vs Cantor
> CantorÕs diagonalization procedure
> defines a function which given an infinite list 
of reals
(that is,
> an element of R^N) returns an element of R that is not on
the list.
> Cantor didnÕt prove that there were uncountably many such
functions,
> he only proved that there was one. ThatÕs all he needs to
be able
> to show that no list of reals contains every real.
> Do you think there are any counter-examples to CantorÕs
diagonal
> proof?
No.
> If so, can you provide a valid modification that keeps the
concept
> of providing a real that is proven not in the list and has
no
> counter-example?
NA.
===
Subject: Re: induction vs Cantor
> ...
>...
> If so, can you provide a valid modification that keeps the
concept
> of providing a real that is proven not in the list and has
no
> counter-example?
> NA.
There exists a countable model of your ZF set.
===
Subject: Re: induction vs Cantor
>> ...
>>...
>> If so, can you provide a valid modification that keeps the
concept
>> of providing a real that is proven not in the list and has
no
>> counter-example?
>> NA.
>There exists a countable model of your ZF set.
Sure, but inside such models it is still the case that |N| <
|R|. Since |N|
< |R| follows from the ZF axioms, it must hold in all models
that satisfy
those axioms, regardless of the size of the model (as viewed
from the
outside).
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: induction vs Cantor
>> ...
>>...
>> If so, can you provide a valid modification that keeps the
concept
>> of providing a real that is proven not in the list and has
no
>> counter-example?
> NA.
>There exists a countable model of your ZF set.
> Sure, but inside such models it is still the case that |N|
< |R|. Since
|N|
> < |R| follows from the ZF axioms, it must hold in all
models that satisfy
> those axioms, regardless of the size of the model (as
viewed from the
> outside).
Hi Barb,
That leads to a problem.
The set N, the natural numbers, in any of those extensions,
contains no
elements not contained in each other copy of the set N, and
it contains each
of
them. The copies are equal and identical.
There is thus identity, a tautology, between any copies of N,
and as well
between any copies of P(N), Then, if there is a bijection
between N and
P(N),
there is a bijection between N and P(N).
theory bogosity is a reaction to Skolem, who otherwise says
infinite sets
are
equivalent.
Look outside.
Do you use transfinite cardinals? Is it for anything besides
transfinite
cardinals?
Some people use them as a definition as part of measure
theory, but thatÕs
because they lack better tools, and the results of measure
theory are
largely
upon continua. That is to say, the useful results of measure
theory are
derivable without the use of transfinite cardinals, and they
should be.
Of your ordinals, is there any that specifically represents
the integer -1?
Is
not the order type of the powerset of the naturals the
successor of the
order
type of the naturals?
If itÕs infinite thereÕs always 
one more. ThatÕs part of the
basis of
induction, that something can be proven to hold for each,
thus that for each
it
is proven, as opposed to proved.
When we get to the contradiction between each infinite set
being equivalent
inductively and the necessity of dual representation via
CantorÕs results,
then
accept that there is dual representation instead of that
induction fails.
That way, youÕre on the path to being inside 
first order
logic, and nobody
can
prove you incomplete, inconsistent, and immaterial.
If you want a consistent theory, start by removing the
inconsistencies, not
reusing them.
Ross F.
===
Subject: Re: induction vs Cantor
>> ...
>>...
> If so, can you provide a valid modification that keeps the
concept
> of providing a real that is proven not in the list and has
no
> counter-example?
> NA.
>There exists a countable model of your ZF set.
>> Sure, but inside such models it is still the case that |N|
< |R|. Since
|N|
>> < |R| follows from the ZF axioms, it must hold in all
models that
satisfy
>> those axioms, regardless of the size of the model (as
viewed from the
>> outside).
>Hi Barb,
>That leads to a problem.
>The set N, the natural numbers, in any of those extensions,
contains no
>elements not contained in each other copy of the set N, and
it contains
each of
>them. The copies are equal and identical.
>There is thus identity, a tautology, between any copies of N,
I agree. N (via von Neumann ordinals) is categorical in ZF,
so the N in
different models would be isomorphic.
>and as well between any copies of P(N),
That looks like a serious misstep. AIUI (which admittedly
isnÕt very
deeply), the downward LST gives a model in which all the
elements correspond
to first-order-definable terms. So, the set of 
such elements of
P(N) in the
LST countable model is very much smaller than P(N) in the
usual
iterative-hierarchy models (which includes the
uncountably-many elements
which are not first-order definable).
(ItÕs rather like the difference between the set of
computable infinite
binary sequences and the set of all infinite binary sequences
computable or
not.)
But |N| < |P(N)| even in the countable model, because the
bijection between
N and the modelÕs P(N) is not itself first-order 
definable and
therefore
does not exist in the model!
ISTM you have strong Constructivist leanings. ThatÕs 
fine, but
itÕs
important then not to mix constructive and non-constructive
entities in the
same argument. If you exclude all non-constructive entities
then you canÕt
argue that |N| = |P(N)| in the countable model, because the
isomorphism
would be non-constructive and therefore not talkable-about.
> Then, if there is a bijection between N and P(N),
>there is a bijection between N and P(N).
>theory bogosity is a reaction to Skolem, who otherwise says
infinite sets
are
>equivalent.
IÕm sure he doesnÕt actually say or imply 
that.
>Look outside.
OK.
>Do you use transfinite cardinals?
I donÕt see any outside, but then again I 
donÕt see ANY
mathematical
entities outside -- there are no natural numbers ßitting
through the
trees.
>Is it for anything besides transfinite cardinals?
Sure. One needs at least P(N) for representing the real
numbers when
putting Analysis on a set-theory foundation.
>Some people use them as a definition as part of measure
theory, but thatÕs
>because they lack better tools, and the results of measure
theory are
largely
>upon continua. That is to say, the useful results of measure
theory are
>derivable without the use of transfinite cardinals, and they
should be.
Giving up continua is asking a lot. Continuum Mechanics is a
useful area
with many practical applications.
--
---------------------------
| BBB b Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
===
Subject: Re: induction vs Cantor
<8JqdnQxnYLcA6zrcRVn-3w@giganews.com>



<41a7dbeb$0$20860$afc38c87@news.optusnet.com.au>

<1fRrd.107121$T02.77638@twister.rdc-kc.rr.com>


<41B40665.2DB637D2@tiki-lounge.com>

<41B4DED8.947CA9BB@tiki-lounge.com>

posting-account=7ryOqgsAAABSV_46k1efyFxO01THH4J8
Hi Barb,
Yeah, it is.
The reals are great. Their simple assumption as a continuum
offers a
necessary tool for the pursuit of many and much mathematics.
Most of the fundamental results about real numbers are in
place without
set theory, eg via Euclid.
The (set of) real numbers is the indefinite contiguous
sequence, the
point set. You use zero, and iota, and integral multiples of
iota, and
those are all the non-negative real numbers.
Mapping the integers to the reals in this way escapes the
consequences
of CantorÕs first proof, which some apply to the 
rational
numbers, and
I to neither.
ZornÕs Lemma, the well-ordering principle, or the Axiom of
Choice allow
methods to guarantee enumerability of the sets. Well-order
the reals
and inductively select elements to inject into the integers,
for the
integers, the existence of integer n guarantees the existence
of
integer n+1.
I agree that IÕm constructivist, but IÕm more
quasi-intuitionist and
not prejudiced.
Also IÕm a platonist because math is real. Math can be quite
irrelevant. The natural numbers are the trees.
Do you use ZF with classes? WhatÕs the class of all classes?
If your
answer is no, can there be more than one proper class?
There are theories with a set of all sets, that set is its own
powerset, there the identity and tautology is between the set
and
itself, its own powerset. ThatÕs for intuitionism to 
resolve,
the
singular excluded excluded middle and the double entendre.
Double entendre: a double entendre.
The proper class is the ur-element, itÕs at once all and
nothing, and
the void and null. Its dualistic nature is its singular
nature, and
vice versa.
Ross F.
===
Subject: Re: induction vs Cantor
<8JqdnQxnYLcA6zrcRVn-3w@giganews.com>



<41a7dbeb$0$20860$afc38c87@news.optusnet.com.au>

<1fRrd.107121$T02.77638@twister.rdc-kc.rr.com>


<41B40665.2DB637D2@tiki-lounge.com>

<41B4DED8.947CA9BB@tiki-lounge.com>

posting-account=WZMvOwwAAAC_B1TaayrVN99BJgDWQkUc
RF> The reals are great.
How the would YOU know THAT?
You donÕt even know a basic definition of what 
they are.
RF> Their simple assumption as a continuum
There is nothing simple about it TO YOU, idiot.
To the rest of us, there is a simple first-order
axiomatization, but it is hardly a simple assumption
as a continuum; it is an assumption that they are
an ordered field that is complete in the sense of containing
all of the things you can produce from it by the operation of
taking limits of COUNTABLY infinite sequences of things
already in it.
RF> offers a necessary tool for the pursuit of many and much
mathematics.
RF> Most of the fundamental results about real numbers are in
place
RF> without set theory,
Yes, there are axioms defining the reals without
mentioning sets.
RF> eg via Euclid.
To prove that Euclid is an example of that,
you would have to give an example FROM Euclid of
some results about reals.
RF> The (set of) real numbers is the indefinite contiguous
sequence,
In modern set theories, you CANÕT just gloss over that in
parentheses:
you have to PICK A SIDE: is the class of all reals proper, or
is it a
set????
RF> the point set. You use zero, and iota, and integral
multiples of
iota,
RF> and those are all the non-negative real numbers.
It is a consequence of the fact that first-order languages are
countable
that you can model any consistent first-order theory
countably, so if
there are denumerably many iota-terms, yes, there is a way
you can
make them serve as a model. BUt that does NOT mean that they
really are the reals.
RF> Mapping the integers to the reals in this way escapes the
RF> consequences of CantorÕs first proof, which 
some apply
RF> to the rational numbers, and I to neither.
NOthing escapes the proof. No matter how many iotaÕs
you use to represent the reals, they will STILL all have
denumerably-long-bit-string representations as well,
and all you have to do to reproduce the proof in the iota-
context is define the function that returns the nth decimal
place (or bit) of a real as output, give the right number of
iotas
as input.
RF> ZornÕs Lemma, the well-ordering principle,
RF> or the Axiom of Choice allow
RF> methods to guarantee enumerability of the sets.
They DO NOT, dumbass. All these things are NON-
constructive. You do get an enumeration or well-ordering
from them but you NEVER get a METHOD!
RF> Well-order the reals and inductively select elements to
RF> inject into the integers,
Obviously, this is not possible; you run out of natnums.
RF> for the integers, the existence of integer n guarantees
RF> the existence of integer n+1.
And the existence of a term with n iotaÕs guarantees the
existence
of a term with n+1 iotaÕs. But nothing guarantees that you 
can
inject the reals into either of those.
RF> Do you use ZF with classes?
NObody uses ZF with classes. ZF DOES NOT HAVE
classes. In ZF, EVERYTHING IN THE ENTIRE UNIVERSE
is a set. Proper classes DO NOT EXIST.
RF> WhatÕs the class of all classes?
A contradiction in terms, THATÕS what it is.
RF> There are theories with a set of all sets, that set is
its own
RF> powerset, there the identity and tautology is between
RF> the set and itself, its own powerset.
You havenÕt actually studied any set theories with a 
universal
set and you DONÕT know WTF you are talking about.
All these theories have some VERY counter-intuitive
consequences
at very simple levels.
===
Subject: Re: induction vs Cantor
<8JqdnQxnYLcA6zrcRVn-3w@giganews.com>



<41a7dbeb$0$20860$afc38c87@news.optusnet.com.au>

<1fRrd.107121$T02.77638@twister.rdc-kc.rr.com>


<41B40665.2DB637D2@tiki-lounge.com>

<41B4DED8.947CA9BB@tiki-lounge.com>

posting-account=XHGPzAwAAACBzDwz8l_9KYOmofDRUylj
> RF> The reals are great.
> How the would YOU know THAT?
> You donÕt even know a basic definition of what 
they are.
> RF> Their simple assumption as a continuum
> There is nothing simple about it TO YOU, idiot.
> To the rest of us, there is a simple first-order
> axiomatization, but it is hardly a simple assumption
> as a continuum; it is an assumption that they are
> an ordered field that is complete in the sense of containing
> all of the things you can produce from it by the operation
of
> taking limits of COUNTABLY infinite sequences of things
> already in it.
Erm, for every cauchy sequence... or for every dedikind
cut... is
second order.
The first order theory of the real field is just 
the theory of
real
closed fields axiomatised via ordered field + 
square roots of
positive
elements + roots of odd degree polynomials.
===
Subject: Re: induction vs Cantor
Poker Joker says...
>>CantorÕs diagonalization procedure
>> defines a function which given an infinite list 
of reals
(that is,
>> an element of R^N) returns an element of R that is not on
the list.
>> Cantor didnÕt prove that there were uncountably many such
functions,
>> he only proved that there was one. ThatÕs all he needs to
be able
>> to show that no list of reals contains every real.
>Do you think there are any counter-examples to CantorÕs
diagonal
>proof?
What do you mean by a counter-example?
>If so, can you provide a valid modification that keeps the
concept
>of providing a real that is proven not in the list and has no
>counter-example?
Sorry, I donÕt know what you mean by counterexample.
--
Daryl McCullough
Ithaca, NY
===
Subject: Re: induction vs Cantor
> Poker Joker says...
>> what I have called the Cantor function from R^N to R.
>Describe this class of functions and show how Cantor used
them in his
>proof as you say. IÕm looking for where he 
classified them as
R^N to R
>and that they were uncountable.
> I think Virgil means the class of functions f which given
> an infinite list of reals returns a new real that is not
> on the list.
Precisely!
> Cantor came up with only one such function: his
diagonalization
> function. R^N just means the set of functions from N to R,
or
> equivalently, the set of infinite lists of reals, or a set 
of
> reals indexed by the naturals. CantorÕs diagonalization
procedure
> defines a function which given an infinite list 
of reals
(that is,
> an element of R^N) returns an element of R that is not on
the list.
> Cantor didnÕt prove that there were uncountably many such
functions,
> he only proved that there was one. ThatÕs all he needs to
be able
> to show that no list of reals contains every real.
> It is easy enough to show that the cardinality of the set
of such
> functions is uncountable, although this fact is not used in
CantorÕs
> proof.
> --
> Daryl McCullough
> Ithaca, NY
===
Subject: Re: induction vs Cantor
>  what I have called the Cantor function from R^N
to R.
> Describe this class of functions and show how Cantor used
them in his
> proof as you say. IÕm looking for where he 
classified them
as R^N to R
> and that they were uncountable.
It is I who call them Cantor functions, Cantor himself never
did, nor,
as far as I am aware did he ever use the notation R^N to
designate the
set of all listings of reals (functions from N to R).
Cantor only needed one such function, which I will denote by
C: R^N -> R: l-> C(l). As described by Cantor, it takes a
list of reals
as argument and produces a real not in that list as its value.
Cantor never classified the cardinality of the set of all
possible such
Cantor functions as uncountable, but it is not difficult to do
so.
All Cantor was interested in was producing one such Cantor
function,
at which he succeeded.
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision:
1.9 primary) id iB34FCx13309;
Particularly dealing with complex curves as Riemann surfaces
& their
topology. Also moduli spaces of curves of genus g.
I own a couple of ShafarevichÕs Ōbasic alg 
geovolumes.
There is a
book by Kirwan on complex algebraic curves that IÕve heard
good
things about. IÕm thinking maybe something like 
ForsterÕs
book on
Riemann surfaces might help some, but IÕm not sure. Hence, 
the
request for references.
===
Subject: Re: GCH vs. Axiom of Choice.
> I got a few questions. IÕve heard that the GCH implies the
axiom of
> choice. Is the GCH true for (the?) (Goedels?) heirarchy of
> constructable sets?
> Yes.
> Is there anything independant of ZF that is false
> of that model that set theorists normally like to be true?
Is every
> formula of FOL with equality and a binary relation either
true or
> false in that model?
> Yes.
> Are there any statements where the truth or
> falseness in that model canÕt be proved?
> Yes, ZF+V=L is incomplete.
Ah, I should have been more clear. L is a model (right?), ZF
is an
axiom system (right?). GCH is true for L (you said so in your
post
right?). You also said that we prove that GCH is true for L?
But
what do we need to prove it? For instance if we had to assume
ZFC+GCH
to prove it, then it wouldnÕt mean much IMO. I assume that 
ZF
can
prove only limited things about L, since ZF is incomplete,
but if I
add GCH to ZF then can ZF+GCH prove something new about L?
How about
ZF+(~GCH), can that also prove things about L? Do the truths
of L
vary depending on what other axioms L is defined with? 
IÕm
trying to
be more clear, but I apologize in advance if I failed again.
===
Subject: Re: GCH vs. Axiom of Choice.
|Ah, I should have been more clear. L is a model (right?),
ItÕs a proper class of sets, the sets satisfying a certain
condition.
(Being constructible.) Another way to describe it is that
itÕs the
minimal model of ZF containing all the ordinals.
| ZF is an
|axiom system (right?). GCH is true for L (you said so in
your post
|right?). You also said that we prove that GCH is true for L?
But
|what do we need to prove it?
ZF is strong enough by itself to prove that GCH holds in L.
ZF is strong enough to prove that V=L holds in L, and that
V=L->GCH.
| For instance if we had to assume ZFC+GCH
|to prove it, then it wouldnÕt mean much IMO.
Heh, itÕs sort of amusing when to prove that some fact holds
true
when relativized to a model one needs to use the same fact not
relativized to the model. I can see how that might seem
underwhelming,
but itÕs not necessarily trivial or uninteresting. For
instance, one
had to prove that the axioms of ZF hold in L, too, and
obviously we
need to assume some of them in order to prove they hold in L.
| I assume that ZF can
|prove only limited things about L, since ZF is incomplete,
but if I
|add GCH to ZF then can ZF+GCH prove something new about L?
IÕm guessing it probably can, but I donÕt know 
of a good
example.
My guess is that some assumption intermediate between ZF and
ZF+V=L
is sufficient to prove (say) that aleph_1 is the same as
aleph_1
relative to L. Certainly V=L implies that!
On the other hand, there arenÕt any first order 
properties of L
that can be proven in ZF+GCH but not in ZF. In fact, for an
arbitrary
sentence X in the language of ZF, the following are
equivalent:
(a) ZF |- (V=L->X)
(b) ZF + V=L |- X
(c) ZF |- L |= X.
(d) ZF + V=L |- L |= X
(a)->(b) is a case of the deduction theorem. (b)->(c) holds
because
itÕs a theorem of ZF that V=L relativized to L is true Given
a model
M of ZF, we can consider constructibility as relativized to
M. The
elements of M satisfying this relativized constructibility
form a
submodel L_M, which is another model of ZF. If we repeat this
process,
to get L_{L_M}, itÕs possible to prove that we 
donÕt shrink
the model
any; L_{L_M} = L_M. So L_M is a model of V=L, the axiom
stating that
every set is in L. So anything thatÕs a consequence of V=L
holds
relative to L. If X is a theorem of ZF+V=L, then that fact
can be
proven in ZF, and then applied to L.
(c)->(d) is obvious.
To prove (d)->(a), assume first that (a) is false. By 
GoedelÕs
completeness theorem, there would exist a model of ZF+V=L+~X.
Suppose M is a model of ZF+V=L+~X. Since the L relativized to
M
is the same as M, we get that M |= ~(L|=X). Hence M is also a
model of ZF+V=L in which L|=X fails.
Since GCH is a consequence of V=L,
(cÕ) ZF + GCH |- L|= X
is also equivalent.
| How about
|ZF+(~GCH), can that also prove things about L?
Since ~GCH is so weak, I donÕt think there are many theorems
about
L in ZF+(~GCH) that arenÕt also theorems in ZF. We can 
deduce
from
~GCH the existence of sets not in L, of course.
| Do the truths of L
|vary depending on what other axioms L is defined with? 
IÕm
trying to
|be more clear, but I apologize in advance if I failed again.
How would whether something is true of L depend on axioms?
Of course, what can be proven to be true of L from a set of
axioms
does depend on the axioms! IÕm not familiar with the 
details,
but I
gather that some large cardinal axioms (I think the existence
of a
measurable cardinal is strong enough) imply a relatively
detailed
account of the structure of L. Note that the existence of a
measurable cardinal is inconsistent with V=L.
This real 0# that is believed to be outside of L encodes some
of the
structure of L (if it exists).
Keith Ramsay
===
Subject: Re: GCH vs. Axiom of Choice.
> I got a few questions. IÕve heard that the GCH implies the
axiom of
> choice. Is the GCH true for (the?) (Goedels?) heirarchy of
> constructable sets?
>
> Yes.
>
> Is there anything independant of ZF that is false
> of that model that set theorists normally like to be true?
Is every
> formula of FOL with equality and a binary relation either
true or
> false in that model?
>
> Yes.
>
> Are there any statements where the truth or
> falseness in that model canÕt be proved?
>
> Yes, ZF+V=L is incomplete.
> Ah, I should have been more clear. L is a model (right?),
ZF is an
> axiom system (right?). GCH is true for L (you said so in
your post
> right?). You also said that we prove that GCH is true for
L? But
> what do we need to prove it? For instance if we had to
assume ZFC+GCH
> to prove it, then it wouldnÕt mean much IMO.
ZF+V=L can prove GCH, and ZF can prove that GCH is true in L.
> I assume that ZF can
> prove only limited things about L, since ZF is incomplete,
but if I
> add GCH to ZF then can ZF+GCH prove something new about L?
How about
> ZF+(~GCH), can that also prove things about L? Do the
truths of L
> vary depending on what other axioms L is defined with? 
IÕm
trying to
> be more clear, but I apologize in advance if I failed again.
If you added other axioms to ZF, they might make a difference
to what
can be proved to be true in L. But ZF can prove that GCH is
true in L.
===
Subject: Re: GCH vs. Axiom of Choice.
> I got a few questions. IÕve heard that the GCH implies the
axiom
of
> choice. Is the GCH true for (the?) (Goedels?) heirarchy of
> constructable sets?
>
> Yes.
>
> Is there anything independant of ZF that is false
> of that model that set theorists normally like to be true?
Is
every
> formula of FOL with equality and a binary relation either
true or
> false in that model?
>
> Yes.
>
> Are there any statements where the truth or
> falseness in that model canÕt be proved?
>
> Yes, ZF+V=L is incomplete.
>
> Ah, I should have been more clear. L is a model (right?),
ZF is an
> axiom system (right?). GCH is true for L (you said so in
your post
> right?). You also said that we prove that GCH is true for
L? But
> what do we need to prove it? For instance if we had to
assume ZFC+GCH
> to prove it, then it wouldnÕt mean much IMO.
> ZF+V=L can prove GCH, and ZF can prove that GCH is true in
L.
> I assume that ZF can
> prove only limited things about L, since ZF is incomplete,
but if I
> add GCH to ZF then can ZF+GCH prove something new about L?
How about
> ZF+(~GCH), can that also prove things about L? Do the
truths of L
> vary depending on what other axioms L is defined with? 
IÕm
trying to
> be more clear, but I apologize in advance if I failed again.
> If you added other axioms to ZF, they might make a
difference to what
> can be proved to be true in L. But ZF can prove that GCH is
true in L.
If you have two axioms A and B such that
Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves about
L consistent with the things that ZF+B proves about L? Even if
Con(ZF) does not imply Con(ZF+A+B)?
ItÕs sort of like using Goedels completeness of the entire 
set
theoretical universe to prove things about L, if ZF+A proves
something
about L that is consistent with what ZF+B proves about L,
then one
might expect that from arbitrary relatively consistent axioms
you
outside of L, that one could prove everything about L using
GoedelÕs
completeness result. But I donÕt know if 
thatÕs correct or
why it
wouldnÕt work if it didnÕt.
A possibly related question is why is V=L considered
restrictive
while foundation or regularity is not considered restrictive?
===
Subject: Re: GCH vs. Axiom of Choice.
> I got a few questions. IÕve heard that the GCH implies the
axiom
of
> choice. Is the GCH true for (the?) (Goedels?) heirarchy of
> constructable sets?
>
> Yes.
>
> Is there anything independant of ZF that is false
> of that model that set theorists normally like to be true?
Is
every
> formula of FOL with equality and a binary relation either
true or
> false in that model?
>
> Yes.
>
> Are there any statements where the truth or
> falseness in that model canÕt be proved?
>
> Yes, ZF+V=L is incomplete.
>
> Ah, I should have been more clear. L is a model (right?),
ZF is an
> axiom system (right?). GCH is true for L (you said so in
your post
> right?). You also said that we prove that GCH is true for
L? But
> what do we need to prove it? For instance if we had to
assume
ZFC+GCH
> to prove it, then it wouldnÕt mean much IMO.
>
> ZF+V=L can prove GCH, and ZF can prove that GCH is true in
L.
>
> I assume that ZF can
> prove only limited things about L, since ZF is incomplete,
but if I
> add GCH to ZF then can ZF+GCH prove something new about L?
How about
> ZF+(~GCH), can that also prove things about L? Do the
truths of L
> vary depending on what other axioms L is defined with? 
IÕm
trying to
> be more clear, but I apologize in advance if I failed again.
>
> If you added other axioms to ZF, they might make a
difference to what
> can be proved to be true in L. But ZF can prove that GCH is
true in L.
> If you have two axioms A and B such that
> Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves about
> L consistent with the things that ZF+B proves about L? Even
if
> Con(ZF) does not imply Con(ZF+A+B)?
No, not necessarily.
> ItÕs sort of like using Goedels completeness of the entire
set
> theoretical universe to prove things about L, if ZF+A
proves something
> about L that is consistent with what ZF+B proves about L,
then one
> might expect that from arbitrary relatively consistent
axioms you
> outside of L, that one could prove everything about L using
GoedelÕs
> completeness result. But I donÕt know if 
thatÕs correct or
why it
> wouldnÕt work if it didnÕt.
> A possibly related question is why is V=L considered
restrictive
> while foundation or regularity is not considered
restrictive?
===
Subject: Re: GCH vs. Axiom of Choice.
> If you have two axioms A and B such that
> Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves about
> L consistent with the things that ZF+B proves about L? Even
if
> Con(ZF) does not imply Con(ZF+A+B)?
> No, not necessarily.
So what is true in L depends on whether L is made in ZF or
ZFC or some
other axioms system that is relatively consistent with ZF,
right?
===
Subject: Re: GCH vs. Axiom of Choice.
> If you have two axioms A and B such that
> Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves
about
> L consistent with the things that ZF+B proves about L? Even
if
> Con(ZF) does not imply Con(ZF+A+B)?
>
> No, not necessarily.
> So what is true in L depends on whether L is made in ZF or
ZFC or some
> other axioms system that is relatively consistent with ZF,
right?
Not what is *true* in L, but what is *provably true* in L,
depends on
which axiom system youÕre working with, yes.
===
Subject: Re: GCH vs. Axiom of Choice.
>
>
> If you have two axioms A and B such that
> Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves
about
> L consistent with the things that ZF+B proves about L? Even
if
> Con(ZF) does not imply Con(ZF+A+B)?
>
> No, not necessarily.
>
> So what is true in L depends on whether L is made in ZF or
ZFC or some
> other axioms system that is relatively consistent with ZF,
right?
> Not what is *true* in L, but what is *provably true* in L,
depends on
> which axiom system youÕre working with, yes.
I asked if the things PROVABLY true in L by ZF+A are
CONSISTENT with
the things PROVABLY ture in L by ZF+B, and you said No, not
necessarily, but without an example. GoedelÕs completeness
says that
you can add axioms to any system (like ZF) to prove any
statement such
that both it and itÕs negation are unprovable without 
causing
a
contradiction as long as the axiom you add is a statement of
the
unprovable statement itself, right? Well, thatÕs a bit silly
for
proving things about the system itself because youÕll 
clearly
get
different results depending on whether you add the statement
or itÕs
negation. But IÕm, thinking that no matter what you add to 
ZF
to
maintain itÕs relative consistency, that you either prove X
about L or
you donÕt, where X is true of L, but that you can never 
prove
~X about
L as long as X is true of L. But since Goedel mentioned that
EVERY
consistent statement can be proved with the right axioms,
that seems
like weÕd be able to prove every truth of L actually proved
by using
every relatively consistent axiom system.
A related example is that someone said that you can make a
model of
Union, Pairing, Empty Set, Foundation, Infinity, Separation,
and
Powers given Union, Pairing, Empty Set, Foundation, Infinity,
Separation, and Powers + Replacement, clearly thatÕs devoid 
of
interest if you had to assume consistentcy of the second
system, so I
assumed it meant you just assumed the truth of the second
system, but
if the first system is large enough to make a model of itself
(which
since it contains omega seems reasonable given
Skolem-Lowenheim) then
maybe the truth of the model existing from the second system
shows
that the model exists and is consistent without begging the
question.
===
Subject: Re: GCH vs. Axiom of Choice.
>
>
> If you have two axioms A and B such that
> Con(ZF)=>[Con(ZF+A)&Con(ZF+B)], then are things that ZF+A
proves
about
> L consistent with the things that ZF+B proves about L? Even
if
> Con(ZF) does not imply Con(ZF+A+B)?
>
> No, not necessarily.
>
> So what is true in L depends on whether L is made in ZF or
ZFC or
some
> other axioms system that is relatively consistent with ZF,
right?
>
> Not what is *true* in L, but what is *provably true* in L,
depends on
> which axiom system youÕre working with, yes.
> I asked if the things PROVABLY true in L by ZF+A are
CONSISTENT with
> the things PROVABLY ture in L by ZF+B, and you said No, not
> necessarily, but without an example. GoedelÕs completeness
says that
> you can add axioms to any system (like ZF) to prove any
statement such
> that both it and itÕs negation are unprovable without
causing a
> contradiction as long as the axiom you add is a statement
of the
> unprovable statement itself, right? Well, thatÕs a bit
silly for
> proving things about the system itself because youÕll
clearly get
> different results depending on whether you add the
statement or itÕs
> negation. But IÕm, thinking that no matter what you add to
ZF to
> maintain itÕs relative consistency, that you either prove 
X
about L or
> you donÕt, where X is true of L, but that you can never
prove ~X about
> L as long as X is true of L. But since Goedel mentioned
that EVERY
> consistent statement can be proved with the right axioms,
that seems
> like weÕd be able to prove every truth of L actually 
proved
by using
> every relatively consistent axiom system.
False axioms can be consistent with ZF. You could add a false
axiom to
ZF and the results that you could prove to be true in L with
this
axiom system might not in fact be true in L.
> A related example is that someone said that you can make a
model of
> Union, Pairing, Empty Set, Foundation, Infinity, Separation,
and
> Powers given Union, Pairing, Empty Set, Foundation, 
Infinity,
> Separation, and Powers + Replacement, clearly thatÕs 
devoid
of
> interest if you had to assume consistentcy of the second
system, so I
> assumed it meant you just assumed the truth of the second
system, but
> if the first system is large enough to make a model of
itself (which
> since it contains omega seems reasonable given
Skolem-Lowenheim) then
> maybe the truth of the model existing from the second
system shows
> that the model exists and is consistent without begging the
question.
You can prove in ZF that Con(Z), without assuming Con(ZF),
yes.
===
Subject: Compute plane with angles and vectors
I have two known vectors with the same origin and a unknown
plain
which goes also through this origin. The two angles between
the two
vectors and the normal of the plane are known, but not the
normal
itself. How can I compute the normal?
Is there another way than this:
a1*n1 + a2*n2 + a3*n3 = cos(alpha1)
b1*n1 + b2*n2 + b3*n3 = cos(alpha2)
n1^2 + n2^2 + n3^2 = 1
Because the result is really huge (over pages)...
There can be two results, right?
Can I fix this with a third vector (and a third angle), such
that I
have only one result?
How would I do this?
S. Nurbe
===
Subject: Re: Compute plane with angles and vectors
>I have two known vectors with the same origin and a unknown
plain
>which goes also through this origin. The two angles between
the two
>vectors and the normal of the plane are known, but not the
normal
>itself. How can I compute the normal?
>Is there another way than this:
>a1*n1 + a2*n2 + a3*n3 = cos(alpha1)
>b1*n1 + b2*n2 + b3*n3 = cos(alpha2)
>n1^2 + n2^2 + n3^2 = 1
>Because the result is really huge (over pages)...
>There can be two results, right?
>Can I fix this with a third vector (and a third angle), such
that I
>have only one result?
>How would I do this?
Yes, with a third vector C it would be easy. You could
compute both
N cross (A cross B) and N cross (A cross C) from your
information,
each of which would be perpendicular to N, giving two vectors
parallel
to the plane. Cross them again to get a normal.
--Lynn
===
Subject: Re: Compute plane with angles and vectors
>I have two known vectors with the same origin and a unknown
plain
>which goes also through this origin. The two angles between
the two
>vectors and the normal of the plane are known, but not the
normal
>itself. How can I compute the normal?
>Is there another way than this:
>a1*n1 + a2*n2 + a3*n3 = cos(alpha1)
>b1*n1 + b2*n2 + b3*n3 = cos(alpha2)
>n1^2 + n2^2 + n3^2 = 1
>Because the result is really huge (over pages)...
>There can be two results, right?
>Can I fix this with a third vector (and a third angle), such
that I
>have only one result?
>How would I do this?
> Yes, with a third vector C it would be easy. You could
compute both
> N cross (A cross B) and N cross (A cross C) from your
information,
> each of which would be perpendicular to N, giving two
vectors parallel
> to the plane. Cross them again to get a normal.
> --Lynn
IÕm sorry, but how would it look like?
equationsystem? Because N is unknown.
===
Subject: Re: Compute plane with angles and vectors
>> Yes, with a third vector C it would be easy. You could
compute both
>> N cross (A cross B) and N cross (A cross C) from your
information,
>> each of which would be perpendicular to N, giving two
vectors parallel
>> to the plane. Cross them again to get a normal.
>> --Lynn
>IÕm sorry, but how would it look like?
>equationsystem? Because N is unknown.
Use the formula:
N cross (A cross B) = (N dot B) A - (N dot A) B.
Think of N as a unit vector, then N dot B = |B| cos(angle)
which you
know, and same for N dot A. You donÕt need a system of
equations.
--Lynn
===
Subject: Re: Compute plane with angles and vectors
>>
>> Yes, with a third vector C it would be easy. You could
compute both
>>
>> N cross (A cross B) and N cross (A cross C) from your
information,
>> each of which would be perpendicular to N, giving two
vectors parallel
>> to the plane. Cross them again to get a normal.
>>
>> --Lynn
>IÕm sorry, but how would it look like?
>equationsystem? Because N is unknown.
> Use the formula:
> N cross (A cross B) = (N dot B) A - (N dot A) B.
> Think of N as a unit vector, then N dot B = |B| cos(angle)
which you
> know, and same for N dot A. You donÕt need a system of
equations.
> --Lynn
Just to prove it:
If all vectors are unit vectors I have the equations:
cos(alpha2)*A - cos(alpha1)*B = R1
cos(alpha3)*A - cos(alpha1)*C = R2
Now cross product of R1 and R2 and the result is my normal.
Right?
P.S. Sorry for the 2nd thread - it was too late...
===
Subject: Re: Compute plane with angles and vectors
>Just to prove it:
>If all vectors are unit vectors I have the equations:
>cos(alpha2)*A - cos(alpha1)*B = R1
>cos(alpha3)*A - cos(alpha1)*C = R2
>Now cross product of R1 and R2 and the result is my normal.
>Right?
Yes, that should do it. Much easier than solving your system
of
equations.
--Lynn
===
Subject: (Finite) Generating Functions
Surely there must be a nice way to solve this problem. I will
post the
problem and my feeble method of solution.
Write an ordinary generating function for the number of ways
of
distributing n pieces of candy to 10 kids if the two youngest
kids get
between 1 and 5 pieces each and each of the other 8 children
gets either
1 or two pieces. Use this generating function to find the
number of
ways to distribute 15 candies.
ogf for youngest: (x + x^2 + ... + x^5) [1]
ogf for others: (x + x^2) [2]
Since two youngest square [1] and since eight others, raise
[2] to the
8th power. Thus, the ogf is:
(x + x^2 + x^3 + x^4 + x^5)^2 * (x + x^2)^8
Using Maple, I get 520 x^15 so there are 520 ways to do this.
My
attempt without Maple, and where things get ugly...
(x + x^2 + x^3 + x^4 + x^5)^2 * (x + x^2)^8
= x^2 * (1 + x^2 + x^3 + x^4)^2 * x^8 * (1 + x)^8
= x^10 * (1 + x^2 + x^3 + x^4)^2 * (1 + x)^8 [3]
(1 - x^5)^2 (1 - x^2)^8
= x^10 * ------------- * -------------
(1 - x)^2 (1 - x)^8
= x^10 * (1 - x)^-10 * (1 - x^5)^2 * (1 - x^2)^8
Now use logic. Since there is an x^10 in front, ignore it
momentarily
and find the coefficient of x^5 in the other three 
terms.
The second term can only contribute for powers of x = {0, 5,
10}.
However, 10 is not valid since we only need x^5.
The third term can only contribute even powers of x = {0, 2,
4, ...,
16}. The only Ōvalidpowers are {0, 2, 4}.
The first term can contribute powers of {0, 1, 2, ..., 10}.
But using
the knowledge of the what the other terms can contribute, we
end up with
only 4 valid combinations
(0, 5, 0)
(1, 0, 4)
(3, 0, 2)
(5, 0, 0)
Now find out the coefficients for each of the 
valid
combinations.
(0, 5, 0) => 1 for x^0 from the first term
=> -2 for x^5 from the second
=> 1 for the x^0 from the last term
=> 1 * -2 * 1 = -2
(1, 0, 4) => C(10,1) = 10 from the first
=> 1 from the second
=> C(8,2) = 28 from the third
=> 10 * 1 * 28 = 280
(3, 0, 2) => C(12,3) = 220 from the first
=> 1 from the second
=> -C(8,1) = -8 from the third
=> 220 * 1 * -8 = -1760
(5, 0, 0) => C(14,5) = 2002 from the first
=> 1 from the second
=> 1 from the third
=> 2002 * 1 * 1 = 2002
Thus there are -2 + 280 - 1760 + 2002 = 520 ways.
Surely there is an easier way to do this, other than the
Maple way. I
thought about using [3] above, but I donÕt think it makes it
any easier.
- Tim
--
Timothy M. Brauch
NSF Fellow
Department of Mathematics
University of Louisville
email is:
news (dot) post (at) tbrauch (dot) com
===
Subject: The Size of GrahamÕs Number
Trying to explain the sheer hugeness of GrahamÕs Number in a
popular
way to someone, I kind of came up short. I have read on the
net
somewhere that even if all matter in the universe were
converted to
pen and paper, it wouldnÕt be enough to write the number
down. But
thatÕs a pretty difficult concept to grasp as 
well. Is there
an easier
way to a) estimate the number of digits in GrahamÕs Number
and b)
express this as something slighty less unfathomable (i.e. it
wouldrequier a hard drive 40 times the size of our galaxy to
store it
or something)
===
Subject: Re: The Size of GrahamÕs Number
> Trying to explain the sheer hugeness of GrahamÕs Number in
a popular
> way to someone, I kind of came up short. I have read on the
net
> somewhere that even if all matter in the universe were
converted to
> pen and paper, it wouldnÕt be enough to write the number
down. But
> thatÕs a pretty difficult concept to grasp as 
well. Is there
an easier
> way to a) estimate the number of digits in GrahamÕs Number
and b)
> express this as something slighty less unfathomable (i.e. it
> wouldrequier a hard drive 40 times the size of our galaxy
to store it
> or something)
The number of digits it would take to write down GrahamÕs
number is just
as unfathomable as the number itself.
===
Subject: Re: The Size of GrahamÕs Number
>The number of digits it would take to write down GrahamÕs
number is just
>as unfathomable as the number itself.
phil had a good point though - couldnÕt 
GrahamÕs number
concievably be
certain precision in a certain volume V? Or is Grahams number
so huge
the precision of the Planck length is less than GrahamÕs
number?
===
Subject: Re: The Size of GrahamÕs Number
>>The number of digits it would take to write down GrahamÕs
number is just
>>as unfathomable as the number itself.
> phil had a good