mm-53 === [snip my question & some followup]> Now let's consider sets which contain x_0 and are closed under g.> One such set is X itself.> Another such set is the diagonal: {(a,a) : a in N}.> Another such set is the diagonal minus (0,0): {(a,a): a in N, a>0}> Another such set is {(a,b): a,b in N, a>=b}> Another such set is {(a,b): a,b in N, a<=b}> and many more such sets. === >Background:>A triple , where X is a set, g is a unary>operation in X (that is, a function on X into X), and>x_0 is an element of X is a unary system.>Let be a unary system. The set of descendants>of x_0 under g (in symbols D_gx_0) is the intersection of>all subsets A of X, such that x_0 in A and xg in A whenever>x in A (this later requirement will often be phrased as>A is closed under g).> Eech, I hate that notation - note that I'm going to write> g(x) for what you call xg here.>[Set Theory & Logic -Stoll]>My question:>I can't imagine any more than one subset of X satisfying>the above conditions (x_0 in A and A closed under g) so>taking the intersection of all such sets (when I can only>conceive of one) seems odd. Am I missing something?>In the case that I am could you please show a simple>example.> Let X = R, the real numbers. De?e g by g(x) = x for all x.> Let x_0 = 0. Then {x_0} is closed under g, isn't it?Yes.> David C. Ullrich === =Bart Goddardbart god ~ child minder ?my name matches work one in three with skeptics,one in 5 with maths posters,one in 7 on average,but almost unnoticable with atheists.notice the ?st 5 on Pennies list being inventors could? be matchedmaking a 120 (6!) to one proof of numerology, since someoneelse made the list not me I can't be accused of rigging it.Have a go if you want to prove God exists here and now.1st post to uk.org.mensa in year 2000>peoples names determine what scienti? discovery or art>they will uncover____________> Major contribution to the ?st practical nuclear reactor:> Inventor of Nuclear medical tagging:> Discover of DNA structure:> Inventor of high level computer language, ?st computer bug (moth):> Fundamental work in Genetics and member of the national academy of sciences> Jumping genes ( Non-nuclear genetics) and a Nobel Prize.woman from Cold Spring Harbor labs.Lynne Margulis.Madam Wu ( Stonybrook)Emmy NoetherGrace HopperRoberta WoodsRosyln YalowRoslynn FranklinThinking this morning if I could program a rudimentary AI I would have a proof of God.All it has to do is rate the match between every reply to me in googlewith the name, basically it gets a reply with the name stripped, clearsits own memory for a trial, has 2 other names to guess from and if itgets it right then there is a weak association between the reply to meand the authors name, as is my paranormal claim.then it clears its memory of who the authors are again, and tries repeatedlywith different options from the other 2 choices and gets a statistic like :92% of the time this post was recognised to its correct author from a list of 3 total authors.Then it increases the number of options to select from, say this post:---James Randi will test you when you properly apply here.---and the list of names:TomJerryRich ShewmakerSilviaBartReally long alias---The actual author was Rich Shewmaker, people have picked it out from4 options, because James Randi literally makes magic shows and offers riches,So that author would probably stand out against 1000 names ! no joke!check the post : http://tinyurl.com/dhojTwo 1000 to ones consecutively and I win the 1,000,000 dollars!But if the list had 2000 names, then MagicTester, Competitor, and othersuch aliases would eventually cap the ranking, say with 2000 namesin the list to select from, the AI might only guess 50% of the time that theauthor is Rich Shewmaker.So we do a chart of all the correlation values of replies to me in google, anddo the same for a sample of 10 other posters with similar posting history.and get the stats :Herc : 3to1, 2to1, 1to1, 10to1, 100to1, 4to1, 1to1, 1000to1, 2to1, 500to1....control1 : 2to1, 1to1, 1to1, 1to1, 1to1, 2to1, 5to1, 1to1, 1to1.control2:and looking at the graphs, seeing I've already defeated 10,000,000,000,000,000,000,000to one odds evident in google, seeing the evaluation stats were not biased toany one poster, AND the ?al check , testing the posts making sure I'm notenticing people to rig their posts type about your name!would just about coerce James Randi to set up a booth and test if I can guesspeoples names.Ofcoarse you could replace AI here with anyone who gives a dam!Herc === > With constant proper acceleration g the speed as seen by> the initial rest frame is> v(t) = g*t / sqrt[1+(g*t/c)^2]> Doesn't sound right. I worked it out once and it had hyperbolic> functions in it. You have a derivation or reference you care> to post?Suppose a constant force F is applied along the line of motion,and the rest mass m is constant, then you get F = dp/dt = d(m*gamma(v)*v)/dt where gamma(v) = 1/sqrt[1-v^2/c^2] = m * d(gamma(v)*v)/dtthen F/m = d(gamma(v)*v)/dtSince this is supposed to be constant we can call F/m = gand we get gamma(v)*v = g*tor v/sqrt[1-v^2/c^2] = g*tsolving for v gives v(t) = g*t / sqrt[1+(g*t/c)^2]which is a plain vanilla hyperbola. Draw it to convinceyourself :-)You probably worked out v(t') where t' is the proper timeusually called tau.Using the connection t' = c/g * argsinh( g/c*t )which is equivalent with t = c/g * sinh( g/c*t' )and combining this with v(t) = g*t / sqrt[1+(g*t/c)^2]you easily get v(t') = c*tanh( g/c*t' )Dirk Vdm === =[snip]> You probably worked out v(t') where t' is the proper time> usually called tau.> Using the connection> t' = c/g * argsinh( g/c*t )By the way, in case you wonder where this one wascoming from: dt/dt' = gamma(v) = 1/sqrt[1-v^2/c^2]so dt' = sqrt[ 1-v^2/c^2 ] dtwhich combined with our previous result v(t) = g*t / sqrt[1+(g*t/c)^2]gives dt' = sqrt[ 1- ( g*t / sqrt[1+(g*t/c)^2] )^2 / c^2 ] dtwhich simpli?s to dt' = 1/sqrt( 1 + (g/c*t)^2 ] dtwhich gives after integration with t'(0) = 0 t'(t) = c/g * argsinh( g/c*t )Dirk Vdm === > >Since faster-than-light travel has no meaning (in our current >understanding of physics), we can't really answer this question, we can >only speculate.> I beg to differ. I just recently bought some underwear which will> travel faster than light! It's emblazoned right right where the brand name> should be: F T L. > I think I like this better than the singing bunch of grapes.> daveSo have you worn this underwear? Are you aware of the possibleconsequences? A premature ejaculation at this level of prematuritycould result in your siring your own dad. Where can I buy thisunderwear?Max Maximal === Start with a square with side-length 1.> I am wondering what is the best way to divide this square into 3> equal-area, but not necessarily equally shaped, pieces.> By 'best' way, I mean that the sum of the 3 maximum diameters of> the pieces is minimized. By diameter I mean, as is what I think is> the usual de?ition of the general sense of this word, the maximum> (maximum, in the case of this question) distance between 2 parallel> lines that the piece can ? between, the perimeter of the piece> touching both lines (but crossing neither line).> The legal ?e-print:> The pieces are to be simply-connected (as I understand the term), but> not necessarily convex.> And each piece has an area of 1/3, and has no area intersecting with> any other piece.> The problem can most-likely be described in much more friendly and> understandable terms. I hope that, not only is the above clear, but> that I did not confuse the matter with an incomplete discription of> the problem or with ambiguous language.> This must be a well-known problem. What about generalizing it to> n-pieces?> As an example, for 4=n, if we simply divide the square, with two> perpendicular lines that each bisect the square, into four identical> squares, the smaller squares each have a maximum-diameter of> squareroot(1/2). So the sum of these maximum-diameters is sqrt(8).> Intuitively to me, this seems like a minimization of the sum of> diameters. If we had instead divided the square into 4 equal triangles> by connecting diametrically-opposed vertexes, each diameter would be 1> (measured along the triangles' hypotenuses=the square's edge). So this> division would not be the solution for n=4, since> 4 > sqrt(8).> How about if we restrict the pieces to being formed by taking thesquare and making only 3 straight cuts, each from one of three pointson the square's perimeter to a single point in the square's interior?It seems that, under this restriction, the optimal solution might bemore easily ?dable, either analytically or numerically (using acomputer).Am I correct in assuming that if we take 3 points on the square'sperimeter, then there is at-most (if any such point exists) oneinterior point which would subdivide the square into 3 equal-areapieces?Leroy Quet === Message-id: Phil Carmody scribbled the following:> I just purchased a copy of the classic book The Lore of Large Numbers by> Philip J. Davis.> In it, he noted that Webster's Dictionary only listed number names up to> the vigintillion. The Latin word for twenty-one didn't lend itself easily> to creating a number name; that didn't stop a web site I saw which gave a> large Mersenne prime spelled out in words (using the British, rather than> the American, convention, since the -iard suf? appeared).> But one can have, after the vigintillion, a trigintillion, a> quadragintillion, a quinquagintillion, a sexagintillion, a> septuagintillion, an octogintillion, a nonagintillion, and a centillion.> Just as an American billion can be called a thousand million (instead of a> milliard), the lack of an unvigintillion or whatever could be dealt with> by following the vigintillion with a thousand vigintillion, a million> vigintillion, a billion vigintillion, up to a nonillion vigintillion.> Doing it that way would avoid having to coin some rather clumsy> Latin-based terms, and would still allow conventional English-language> names for much higher numbers.> The numerical ?clue' in nonillion vigintillion is 29. How does 29 map to> the exponent? If you use the US scheme, you're in a mire. If, however, you> use the English scheme, you'll be ?e. Which of the two schemes did you> wish it to apply to? It looks like the US scheme. Yeugh. > Somehow, though, I doubt that I've thought of this *?st*.> Yeah, and some people thought it would be cool for X-illion to represent > 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> hands I blame. Therefore latinised(29)-illion being 10^(3*29+6) I sincerely> hope would never catch on, doubling as it does the US error. However,>within> the European scheme it has no problems with it at all.>This is exactly the reply I wished to make but you beat me to it.>Surely a simpler system would be better than a more complicated one, if>there are no other drawbacks to it? If the Americans want to be>needlessly complicated, why not use an even more complicated formula?>Something like 10^(X^2 - 3X + 3) for example. That would make a>million equal 10^1, a billion equal 10^1, a trillion equal>10^3, a quadrillion equal 10^7, a quintillion equal 10^13 and so>on.>What is the most infuriating thing here is that the USA is like>Microsoft - whatever they decide to use, no matter how brain-dead it>is, everyone else copies it from them, willingly or otherwise.If you're going to use the SI pre?es kilo-, mega-, giga- etc., doesn't itmake sense to name the numbers the same way?>-- >/-- Joona Palaste (palaste@cc.helsinki.? -->| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|>| http://www.helsinki.?~palaste W++ B OP+ |>- Finland rules! --/>'It can be easily shown that' means ?I saw a proof of this once (which I>didn't>understand) which I can no longer remember'.> - A maths teacher>--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === =: The numerical ?clue' in nonillion vigintillion is 29. How does 29 map to: the exponent? If you use the US scheme, you're in a mire. If, however, you: use the English scheme, you'll be ?e. Which of the two schemes did you: wish it to apply to? It looks like the US scheme. Yeugh. You do have a valid point. Actually, I thought that the Americans wouldapply it to the American scheme, and the British would apply it to theBritish scheme.: Yeah, and some people thought it would be cool for X-illion to represent : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling: hands I blame.I think I can guess what the cause of this was. Billion was the nextthing after million, and really big numbers were, back in the days thedifferentiated American scheme became established, very rarely used.So, instead of being careful, and remembering that after the million,there is the thousand million, and then the billion, billion was justcarelessly used for the thousand million.Sadly, it's too late to change; and with the U.S. leadership in economicsand technology, the U.S. is the chief user of big numbers these days.A milliard is also a thousand million. Perhaps we need to coin newunambiguous words...so a billiad would be an American billion, and a billiord would be aBritish billion, for example.Or we could start over, with jillion, zillion, bazillion and gazillion...John Savard === =: If you're going to use the SI pre?es kilo-, mega-, giga- etc., doesn't it: make sense to name the numbers the same way?That's an idea; switch to Greek!We don't need the megillion, since we already have the unabiguous wordmillion, but a thousand million could be a gigillion, and a millionmillion a terillion!John Savard === : If you're going to use the SI pre?es kilo-, mega-, giga- etc., doesn'tit> : make sense to name the numbers the same way?> That's an idea; switch to Greek!> We don't need the megillion, since we already have the unabiguous word> million, but a thousand million could be a gigillion, and a million> million a terillion!> John SavardI actually do this or equivalent.If I'm speaking to anyone who has a use fro numbers bigger that a (US)trillion I usually refer to it by the Exponent in Sci Not.ExampleE12 snorks or a Terasnorkfor pure numbers I would just say something like 1 E 21If you like Oldspeak, I guess you cloud say ten to the twenty-?st power'Rj Pease === =: : The numerical ?clue' in nonillion vigintillion is 29. How does 29 map to: : the exponent? If you use the US scheme, you're in a mire. If, however, you: : use the English scheme, you'll be ?e. Which of the two schemes did you: : wish it to apply to? It looks like the US scheme. Yeugh. : You do have a valid point. Actually, I thought that the Americans would: apply it to the American scheme, and the British would apply it to the: British scheme.: : Yeah, and some people thought it would be cool for X-illion to represent : : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling: : hands I blame.: I think I can guess what the cause of this was. Billion was the next: thing after million, and really big numbers were, back in the days the: differentiated American scheme became established, very rarely used.: So, instead of being careful, and remembering that after the million,: there is the thousand million, and then the billion, billion was just: carelessly used for the thousand million.: Sadly, it's too late to change; and with the U.S. leadership in economics: and technology, the U.S. is the chief user of big numbers these days.: A milliard is also a thousand million. Perhaps we need to coin new: unambiguous words...: so a billiad would be an American billion, and a billiord would be a: British billion, for example.: Or we could start over, with jillion, zillion, bazillion and gazillion...It might also be noted that instead of the unambiguous thousand million,or the term milliard, in the context of the British system, an Americanbillion could be called...a sesquillion.Unfortunately, we don't have good Latin pre?es for two and a half, threeand a half, four and a half, and so on...John Savard === =: Somehow, though, I doubt that I've thought of this *?st*.Indeed, I hadn't. The idea was advanced by one Dimitri A. Borgmann, in hisbook Language on Vacation (1965), Charles Scribner's Sons, p. 225.Also, the term sesquillion has already been used as well; a Googlesearch turned up three hits, including one that almost quali?s as partof the scienti? literature.John Savard === =: If the system's broke, it's their meddling: hands I blame. Therefore latinised(29)-illion being 10^(3*29+6) I sincerely: hope would never catch on, doubling as it does the US error.Actually, the Americans were but following the French...John Savard === : Yeah, and some people thought it would be cool for X-illion to represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.Actually, the US got it from the French. (Since then the Frenchchanged their usage...) === > : Yeah, and some people thought it would be cool for X-illion to represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)The French also changed from their own imperial units to the metricsystem. But I guess that if the US would be following them *now* thepeople would riot, accusing anything that comes from Europe of beingcommunistic.-- /-- Joona Palaste (palaste@cc.helsinki.? --| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.?~palaste W++ B OP+ |- Finland rules! --/How come even in my fantasies everyone is a jerk? - Daria Morgendorfer === : If you're going to use the SI pre?es kilo-, mega-, giga- etc., doesn't it> : make sense to name the numbers the same way?> That's an idea; switch to Greek!> We don't need the megillion, since we already have the unabiguous word> million, but a thousand million could be a gigillion, and a million> million a terillion!> John SavardAnd we don't need these either. We already have:3 thousand = 3K7 million = 7M12 billion (American), 12 thousand million (French) = 12Gand so on. === > : Yeah, and some people thought it would be cool for X-illion to represent > > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)> The French also changed from their own imperial units to the metric> system. But I guess that if the US would be following them *now* the> people would riot, accusing anything that comes from Europe of being> communistic.Communism has nothing to do with it. When Europeans promoteInternationalism solely as a platform to bash the USA and then turnaround and promote Provincialism solely as a platform to bash the USA,we see through that hypocrisy. === I just purchased a copy of the classic book The Lore of Large Numbers by> Philip J. Davis.> In it, he noted that Webster's Dictionary only listed number names up to> the vigintillion. The Latin word for twenty-one didn't lend itself easily> to creating a number name; that didn't stop a web site I saw which gave a> large Mersenne prime spelled out in words (using the British, rather than> the American, convention, since the -iard suf? appeared).> But one can have, after the vigintillion, a trigintillion, a> quadragintillion, a quinquagintillion, a sexagintillion, a> septuagintillion, an octogintillion, a nonagintillion, and a centillion.> Just as an American billion can be called a thousand million (instead of a> milliard), the lack of an unvigintillion or whatever could be dealt with> by following the vigintillion with a thousand vigintillion, a million> vigintillion, a billion vigintillion, up to a nonillion vigintillion.> Doing it that way would avoid having to coin some rather clumsy> Latin-based terms, and would still allow conventional English-language> names for much higher numbers.> Somehow, though, I doubt that I've thought of this *?st*.> John SavardA LITTLE (snicker) off-topic:What do I call the number 000000000000000000000 (whatever the numberof zeros)?nillionSeriously, I have been reading this thread, and I must say that I doagree that the British/American differences in the names for largenumbers can be a possible (and possibly dangerous, in some situations)problem.But there HAS to be a (political/international) joke/riddle in theresomewhere...Leroy Quet === Message-id: I just purchased a copy of the classic book The Lore of Large Numbers by> Philip J. Davis.> In it, he noted that Webster's Dictionary only listed number names up to> the vigintillion. The Latin word for twenty-one didn't lend itself easily> to creating a number name; that didn't stop a web site I saw which gave a> large Mersenne prime spelled out in words (using the British, rather than> the American, convention, since the -iard suf? appeared).> But one can have, after the vigintillion, a trigintillion, a> quadragintillion, a quinquagintillion, a sexagintillion, a> septuagintillion, an octogintillion, a nonagintillion, and a centillion.> Just as an American billion can be called a thousand million (instead of a> milliard), the lack of an unvigintillion or whatever could be dealt with> by following the vigintillion with a thousand vigintillion, a million> vigintillion, a billion vigintillion, up to a nonillion vigintillion.> Doing it that way would avoid having to coin some rather clumsy> Latin-based terms, and would still allow conventional English-language> names for much higher numbers.> Somehow, though, I doubt that I've thought of this *?st*.> John Savard>A LITTLE (snicker) off-topic:>What do I call the number 000000000000000000000 (whatever the number>of zeros)?>nillion>Seriously, I have been reading this thread, and I must say that I do>agree that the British/American differences in the names for large>numbers can be a possible (and possibly dangerous, in some situations)>problem.>But there HAS to be a (political/international) joke/riddle in there>somewhere...>Leroy Quet>Something like:Q: If the French call a thousand billion a billiard, what do the English callit?A: snooker--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === =mensanator scribbled the following:> > : Yeah, and some people thought it would be cool for X-illion to represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)> The French also changed from their own imperial units to the metric> system. But I guess that if the US would be following them *now* the> people would riot, accusing anything that comes from Europe of being> communistic.> Communism has nothing to do with it. When Europeans promote> Internationalism solely as a platform to bash the USA and then turn> around and promote Provincialism solely as a platform to bash the USA,> we see through that hypocrisy.You're right about that, but what I am proposing is neitherinternationalism or provincialism, it's consistency and simplicity.-- /-- Joona Palaste (palaste@cc.helsinki.? --| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.?~palaste W++ B OP+ |- Finland rules! --/He said: ?I'm not Elvis'. Who else but Elvis could have said that? - ALF === It might also be noted that instead of the unambiguous thousand million,> or the term milliard, in the context of the British system, an American> billion could be called...> a sesquillion.Gets my vote. Or my veto. I can't decide.> Unfortunately, we don't have good Latin pre?es for two and a half, three> and a half, four and a half, and so on...I think ?H' was used to represent two and a half by the Romans, so'hillion' might ? the bill(ion, not).Phillion === mensanator scribbled the following:> : Yeah, and some people thought it would be cool for X-illion to represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)> The French also changed from their own imperial units to the metric> system. But I guess that if the US would be following them *now* the> people would riot, accusing anything that comes from Europe of being> communistic.> Communism has nothing to do with it. When Europeans promote> Internationalism solely as a platform to bash the USA and then turn> around and promote Provincialism solely as a platform to bash the USA,> we see through that hypocrisy.> You're right about that, but what I am proposing is neither> internationalism or provincialism, Ok, but you should leave the out the political agenda, it is off topicin sci.math.> it's consistency and simplicity.My original comment was about consistency also. Since numbers areusually grouped in threes by commas and since SI pre?es use groupsof three digits, the _consistent_ way to name numbers is the Americanway, not the British way. 1,000 kilo- thousand 1,000,000 mega- million1,000,000,000 giga- billion === =mensanator scribbled the following:> mensanator scribbled the following:> : Yeah, and some people thought it would be cool for X-illion to represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)> The French also changed from their own imperial units to the metric> system. But I guess that if the US would be following them *now* the> people would riot, accusing anything that comes from Europe of being> communistic.> Communism has nothing to do with it. When Europeans promote> Internationalism solely as a platform to bash the USA and then turn> around and promote Provincialism solely as a platform to bash the USA,> > we see through that hypocrisy.> You're right about that, but what I am proposing is neither> internationalism or provincialism, > Ok, but you should leave the out the political agenda, it is off topic> in sci.math.> it's consistency and simplicity.> My original comment was about consistency also. Since numbers are> usually grouped in threes by commas and since SI pre?es use groups> of three digits, the _consistent_ way to name numbers is the American> way, not the British way.> 1,000 kilo- thousand> 1,000,000 mega- million> 1,000,000,000 giga- billionNo, I think that's wrong. That's not quite as consistent as theEuropean system. Have you forgotten that billion comes from theLatin word for two, trillion for three, quadrillion for four,etc.? This would make 1000 to the second power a one-illion, 1000to the third power a two-illion, 1000 to the fourth power athree-illion etc.While the European system would simply have 1000000 to the ?stpower a one-illion, 1000000 to the second power a two-illion,1000000 to the third power a three-illion and so on.I don't know about you, but I think matching a number with itselfon both lists is more consistent than matching a number with thesame number plus one.-- /-- Joona Palaste (palaste@cc.helsinki.? --| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.?~palaste W++ B OP+ |- Finland rules! --/A friend of mine is into Voodoo Acupuncture. You don't have to go into herof?e. You'll just be walking down the street and... ohh, that's much better! - Stephen Wright === No, I think that's wrong. That's not quite as consistent as the> European system. Have you forgotten that billion comes from the> Latin word for two, trillion for three, quadrillion for four,> etc.? And million is from Italian, meaning a Big Thousand.So maybe a billion is a big big thousand, and a trillion is a big bigbig thousand. Now if big means 1000 times, we are there! === Message-id: mensanator scribbled the following:> mensanator scribbled the following:> : Yeah, and some people thought it would be cool for X-illion to>represent > : 10^(3X+3) rather than 10^6X. If the system's broke, it's their>meddling> : hands I blame.> I think I can guess what the cause of this was.> Actually, the US got it from the French. (Since then the French> changed their usage...)> The French also changed from their own imperial units to the metric> system. But I guess that if the US would be following them *now* the> people would riot, accusing anything that comes from Europe of being> communistic.> Communism has nothing to do with it. When Europeans promote> Internationalism solely as a platform to bash the USA and then turn> around and promote Provincialism solely as a platform to bash the USA,> we see through that hypocrisy.> You're right about that, but what I am proposing is neither> internationalism or provincialism, > Ok, but you should leave the out the political agenda, it is off topic> in sci.math.> it's consistency and simplicity.> My original comment was about consistency also. Since numbers are> usually grouped in threes by commas and since SI pre?es use groups> of three digits, the _consistent_ way to name numbers is the American> way, not the British way.> 1,000 kilo- thousand> 1,000,000 mega- million> 1,000,000,000 giga- billion>No, I think that's wrong. That's not quite as consistent as the>European system. Have you forgotten that billion comes from the>Latin word for two, trillion for three, quadrillion for four,>etc.? This would make 1000 to the second power a one-illion, 1000>to the third power a two-illion, 1000 to the fourth power a>three-illion etc.No, I haven't forgotten. I also haven't forgotten that cardinals (as opposed toordinals) start with 0, not 1.>While the European system would simply have 1000000 to the ?st>power a one-illion, 1000000 to the second power a two-illion,>1000000 to the third power a three-illion and so on.>I don't know about you, but I think matching a number with itself>on both lists is more consistent than matching a number with the>same number plus one.I have no trouble with the concept that 0 is the ?st number, 1 is the secondnumber, 2 is the third number, etc.>-- >/-- Joona Palaste (palaste@cc.helsinki.? -->| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|>| http://www.helsinki.?~palaste W++ B OP+ |>- Finland rules! --/>A friend of mine is into Voodoo Acupuncture. You don't have to go into her>of?e. You'll just be walking down the street and... ohh, that's much>better!> - Stephen Wright--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === =Mensanator scribbled the following:>Message-id: mensanator scribbled the following:> My original comment was about consistency also. Since numbers are> usually grouped in threes by commas and since SI pre?es use groups> of three digits, the _consistent_ way to name numbers is the American> way, not the British way.> 1,000 kilo- thousand> 1,000,000 mega- million> 1,000,000,000 giga- billion>No, I think that's wrong. That's not quite as consistent as the>European system. Have you forgotten that billion comes from the>Latin word for two, trillion for three, quadrillion for four,>etc.? This would make 1000 to the second power a one-illion, 1000>to the third power a two-illion, 1000 to the fourth power a>three-illion etc.> No, I haven't forgotten. I also haven't forgotten that cardinals (as opposed to> ordinals) start with 0, not 1.If 1000 to the 0th power is 1000 where you live, I'm glad I don't livethere.It's not the ordinals in your 1000 series I'm talking about. It's theexponentials.Look:Thousand = 1000^1 = 1000000^(1/2)Million = 1000^2 = 1000000^1Milliard = 1000^3 = 1000000^(3/2)Billion = 1000^4 = 1000000^2and so on.>While the European system would simply have 1000000 to the ?st>power a one-illion, 1000000 to the second power a two-illion,>1000000 to the third power a three-illion and so on.>I don't know about you, but I think matching a number with itself>on both lists is more consistent than matching a number with the>same number plus one.> I have no trouble with the concept that 0 is the ?st number, 1 is the second> number, 2 is the third number, etc.But you seem to have trouble understanding the concept of naturalexponents.-- /-- Joona Palaste (palaste@cc.helsinki.? --| Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++|| http://www.helsinki.?~palaste W++ B OP+ |- Finland rules! --/Keep shooting, sooner or later you're bound to hit something. - Mis?e === Mensanator scribbled the following:>Message-id: mensanator scribbled the following:> My original comment was about consistency also. Since numbers are> usually grouped in threes by commas and since SI pre?es use groups> of three digits, the _consistent_ way to name numbers is the American> way, not the British way.> 1,000 kilo- thousand> 1,000,000 mega- million> 1,000,000,000 giga- billion>No, I think that's wrong. That's not quite as consistent as the> >European system. Have you forgotten that billion comes from the>Latin word for two, trillion for three, quadrillion for four,>etc.? This would make 1000 to the second power a one-illion, 1000>to the third power a two-illion, 1000 to the fourth power a>three-illion etc.> No, I haven't forgotten. I also haven't forgotten that cardinals (as opposed to> ordinals) start with 0, not 1.> If 1000 to the 0th power is 1000 where you live, I'm glad I don't live> there.> It's not the ordinals in your 1000 series I'm talking about. It's the> exponentials.> Look:> Thousand = 1000^1 = 1000000^(1/2)> Million = 1000^2 = 1000000^1> Milliard = 1000^3 = 1000000^(3/2)> Billion = 1000^4 = 1000000^2> and so on.Fractional exponents? This is getting ridiculous.>While the European system would simply have 1000000 to the ?st>power a one-illion, 1000000 to the second power a two-illion,>1000000 to the third power a three-illion and so on.>I don't know about you, but I think matching a number with itself>on both lists is more consistent than matching a number with the> >same number plus one.> I have no trouble with the concept that 0 is the ?st number, 1 is the second> number, 2 is the third number, etc.> But you seem to have trouble understanding the concept of natural> exponents.So your natural exponents work. So what? The SI units use a base of1000 not 1000000. The fact that you can contrive the exponent of yourbase to match the name (bi = 2) is irrelevant. A base of 1000000 is ofno use to anyone, so why do you care if it is consistent?Sure, it's unfortunate that in the American nomenclaturebi = 3tri = 4quad = 5etc.but you're not going to make the general issue of cardinals notmatching ordinals go away by solving a single problem. You would haveto scrap the entire language and start over. === =Here's a problem I've been puzzling about, andI can't decide if it's elementary or related to somethingknown.Let r be a positive integer. Problem A: Does there exist a positive integer s suchthat the binary expansion of rs is a palindrome?(For instance, the factors of a Mersenne or Fermatnumber ? the bill here.)Problem B1: In case problem A is easy, given r, is itpossible to construct s so that s is squarefree?Problem B2: Even more, can s be made to be the productof squarefree primes of the shape 8k+1?Problem C: Can a prime s with this property always befound? Especially one of the shape 8k+1?Uberproblem: In every case, ?d the minimal s that works.If this is easy, I'd be delighted. If it's been workedon, I'd be delighted. If anyone has any references, orthoughts, I'd be delighted.Bart === Let r be a positive integer. > Problem A: Does there exist a positive integer s such> that the binary expansion of rs is a palindrome?> (For instance, the factors of a Mersenne or Fermat> number ? the bill here.)But every odd r is a factor of a Mersenne number - r divides 2^q - 1, where q = phi(r). And even values of r don't compute. Don't know about the other problems. > Problem B1: In case problem A is easy, given r, is it> possible to construct s so that s is squarefree?> Problem B2: Even more, can s be made to be the product> of squarefree primes of the shape 8k+1?> Problem C: Can a prime s with this property always be> found? Especially one of the shape 8k+1?> Uberproblem: In every case, ?d the minimal s that works.-- === Here's a problem I've been puzzling about, and> I can't decide if it's elementary or related to something> known.> Let r be a positive integer. > Problem A: Does there exist a positive integer s such> that the binary expansion of rs is a palindrome?> Are you going to consider 0110 a binary expansion? If not, then r even will cause a problem.> If this is easy, I'd be delighted. If it's been worked> on, I'd be delighted. If anyone has any references, or> thoughts, I'd be delighted.> Bart-- Will Twentyman === > Let r be a positive integer. > Problem A: Does there exist a positive integer s such> that the binary expansion of rs is a palindrome?> But every odd r is a factor of a Mersenne number - r divides > 2^q - 1, where q = phi(r). And even values of r don't compute. Yes, A is just the warm-up problem. And as far as even numbers go, since I'm not interested in them, either allowleading zeros, or restrict to r odd. The case I'm mostinterested in is B2: > Problem B2: Even more, can s be made to be the product> of squarefree primes of the shape 8k+1?Especially if r is a given prime of the same shape.And I assume that C is intractable.:> Problem C: Can a prime s with this property always be> found? Especially one of the shape 8k+1?Bart === =is always an integer for n>=k (n and k non negative integers).for n=k it's trivial since n choose n = 1so we can assume that n > k and then prove that k!*(n-k)! divides n!when n > k but i have no clue how to show this.thanks in advance === is always an integer for n>=k (n and k non negative integers).> for n=k it's trivial since n choose n = 1> so we can assume that n > k and then prove that k!*(n-k)! divides n!> when n > k but i have no clue how to show this.> thanks in advanceFor instance, you can notice that:For any prime p,sum(?/p^r),r>=1)+sum(?n-k)/p^q),q>=1) <= sum(?/p^r),r>=1)Hence k!*(n-k)! divides n! [I used the fact that the power of the prime p inthe arithmetic decomposition of n! is sum(?/p^r),r>=1)].-- Julien Santini,CMI Technop.99le de Ch.89teau-Gombert,France === is always an integer for n>=k (n and k non negative integers).Use induction with (n+1)_C_r = n_C_r + n_C_(r-1) === =Or you can just use, that you can ?d the binomical coe?ient inPascal's triangle (:-)), which is created by adding two integers, soit is integer.But it is better to prove it according to the formula. <3f09f6d3.19319908@netnews.att.net> <2a0cceff.0307091307.3f2604cb@posting.google.com> <17a4a089.0307092008.241d177f@posting.google.com> <3f0d76df.44848355@netnews.att.net> <585ab5d8.0307101339.41f5a702@posting.google.com> <3f0f2b02.589930@netnews.att.net> <3F0F5CC0.7030908@xympatico.ca> <3f0f64d7.4450528@netnews.att.net> <3F0F8920.3040500@xympatico.ca> <3f107096.6738119@netnews.att.net> <2a0cceff.0307150932.483ccd2a@posting.google.com> === =In message <2a0cceff.0307150932.483ccd2a@posting.google.com>, Edward > Sectioned is le mot juste here. If this discussion [*] continues much> longer, somebody probably will be.> [*] if that's how you describe several hundred posts patiently and> painstakingly reiterating Calculus 101 in the face of one who cannot or> will not understand them. Or has the uk.miscreants trolling competition> started early this year?>[Rolls on ?laughing]Why can't you behave?>[Sings]:> There's _no_ miscreant like a _UK_ miscreant,> There's _no_ miscreant I know!> Everything about them is appealing!> Everything about them is a show!>A troupe of strolling players are we...>[continued when muse strikes].I'd continue it myself, but it's too darn hot. Where is the life that late I led?>Miscreant is also le mot just. U-Kites* just have a way with words.>[*] One of two nineteenth century political parties, the other of>which, the U-cons, ? Alaska to avoid persecution.Yes, there really is an annual trolling contest participated in by some of the denizens of the newsgroup uk.misc.-- Richard Herring === My understanding of in?itesimal integrals is more or less as follows>Please remember that the only evidence we have that there is something>you are understanding (as opposed to something you dreamed up) is>your claim that Edward Green says that in?itesimal integrals>_exist_. Oh I see. I dream things up whereas you just intuit them. Is that theidea?Well, if you're serious I suggest you take the matter up with Mr.Green. It's his shoestring.>Remember that in the end if you got answers that agreed with ordinary>physics, I would happily skim over bits of your argument I couldn't>understand; but you are using your unde?ed and mysterious notions to>prove that (e.g.) velocities can be taken in a direction orthogonal>to that which ordinary physics and my intuitional understanding say>they _are_ in.Oh. Let me see if I've got this right. I do physics by dreaming thingsup. Whereas you do physics by intuitional understanding.If you were doing physics I would gladly defer to your intuition. > Associated with orthogonal centripetal force is an acceleration> dr/dtdt.> > The integral of dr/dtdt from 0 to dt is equal to dr/dt. This velocity> is also orthogonal or radial in direction but is ?ite and equal to> tangential v in magnitude for circular rotation.> The integral of radial dr/dt from 0 to dt is equal to dr. This dr is> also radial in direction but in?itessimal in magnitude. This is why> r does not change in magnitude as the result of the operation of> dr/dt.>Oh dear, no. Scratch previous calculation. This velocity of 4 e/f>towards the centre was, um, somehow virtual, in?itesimal, or only>occurred for an in?itesimal interval, or got retracted by the BBC,>or something. Since this velocity only occurred ?gly, it doesn't>have any effect - is that it? Then why doesn't the same apply to the>circumferential velocity?It does, Gonzo. The in?itesimal integral of tangential v producesin?itesimal tangential dr. Which when combined vectorially withradial dr produces rotation of dr's in combination.Do try to think your questions through a little. This isembarrassingly trivial. It ought to be intuitively obvious to thecasual observer. But apparently not to you. What did you do with allthe money your parents gave you for college?>r is the position vector. Consider the snail at 12 o'clock.I know, the big snail is on the 12 and the little snail in on the 24.>My understanding of elementary calculus is that to consider the slope>of a curve, you take a series of approximations, successively smaller>delta-x and delta-y, where the gradient approximation is>delta-y/delta-x. Now in the limit, as delta-x approaches zero,>delta-y also approaches zero, *but* the ratio delta-y/delta-x> >approaches the true gradient. If you write this gradient as dy/dx (as>we* do of course), you have to be aware that it is not *actually* the>ratio of two quantities, both of which are zero.> I think the two quantities are not zero but are in?itesimally small.> There is a difference because the ratio between them is or can be> ?ite even though the individual quantities considered in isolation> are in?itesimal. But they are de?itely not zero.>Well, at the _very_ bottom of page 98, it says (explaining>differentiating y = x^2):>dy/dx = Lim[delta-x -> 0] (delta-y/delta-x) = 2 x>Note that it says limit as delta-x goes to [->] zero. Not as>delta-x becomes an in?itesimal quantity. I believe that of Newton>and Leibnitz one (can't remember which) had a treatment which involved>talking about in?itesimals, but I understand that doing so is>something you have to be _very_ careful about (or you get answers at>90 degrees to reality ^_^).>Does this mean the big snail goes to zero or the little snail goes tozero or both go to zero?> Let me just add that the idea of the in?itesimal integral here was> never intended to represent any kind of radical mathematical or> mechanical revisionism requiring de?ition. You seem to understand> the concept well enough. And I'm employing it just to facilitate the> interpretation of the consequences of centripetal orthogonal> acceleration.>No, you're exploiting the fact that you're not prepared to de?e>_exactly_ what you mean, when _no-one_ else understands what you do>mean, to facilitate the obfuscation required to be able to claim that>(e.g.) in circular motion the velocity and dr/dt are at right angles>to each other, a prima facie contradiction, since in the only>_de?ed_ mathematics we have available they are only two ways of>saying the same thing.Did you actually go to college to learn all this or did you get it outof the Penguin dictionary too?Maybe it can tell you what the de?ition of is is as well. Frankly Ipreferred it when you were trying to de?e antipenultimate. So muchmore stylish and melli?even if comparably inept. === > > My understanding of in?itesimal integrals is more or less as follows>Please remember that the only evidence we have that there is something>you are understanding (as opposed to something you dreamed up) is> >your claim that Edward Green says that in?itesimal integrals>_exist_. > Oh I see. I dream things up whereas you just intuit them. Is that the> idea?Impressive tirade, but where did you get this intuit stufffrom?I'm afraid you, pardon the expression, just dreamed it up.It's a straw man.> > Well, if you're serious I suggest you take the matter up with Mr.> Green. It's his shoestring.Well, what we have is your word for that. So far some of usare skeptical that Mr. Green ever said anything aboutin?itesimal integrals, and he hasn't yet spoken up inhis own defense.>Remember that in the end if you got answers that agreed with ordinary> >physics, I would happily skim over bits of your argument I couldn't>understand; but you are using your unde?ed and mysterious notions to>prove that (e.g.) velocities can be taken in a direction orthogonal>to that which ordinary physics and my intuitional understanding say>they _are_ in.> Oh. Let me see if I've got this right.OK.> I do physics by dreaming things up. Whereas you do > physics by intuitional understanding.Nope, you didn't get it right. That bears no resemblance tothe ideas expressed in the previous paragraph.> It does, Gonzo. The in?itesimal integral of tangential v produces> in?itesimal tangential dr.You are asserting a result of a calculation which none ofus has ever seen or can really dream of how to do.You can't just keep asserting that this ?titiouscalculation has this supposed result, when neither younor anybody else knows how to do that calculation. Animpossible calculation does not produce anything. - Randy === > My understanding of in?itesimal integrals is more or less as follows> >Please remember that the only evidence we have that there is something>you are understanding (as opposed to something you dreamed up) is>your claim that Edward Green says that in?itesimal integrals>_exist_. > Oh I see. I dream things up whereas you just intuit them. Is that the> idea?>Impressive tirade, but where did you get this intuit stuff>from?Actually I got it from an old episode of McCloud.>I'm afraid you, pardon the expression, just dreamed it up.>It's a straw man.> Well, if you're serious I suggest you take the matter up with Mr.> Green. It's his shoestring.>Well, what we have is your word for that. So far some of us>are skeptical that Mr. Green ever said anything about>in?itesimal integrals, and he hasn't yet spoken up in>his own defense.Ah well, I've got a hundred bucks on the quote. >Remember that in the end if you got answers that agreed with ordinary>physics, I would happily skim over bits of your argument I couldn't>understand; but you are using your unde?ed and mysterious notions to>prove that (e.g.) velocities can be taken in a direction orthogonal> >to that which ordinary physics and my intuitional understanding say>they _are_ in.> Oh. Let me see if I've got this right.>OK.> I do physics by dreaming things up. Whereas you do > physics by intuitional understanding.>Nope, you didn't get it right. That bears no resemblance to>the ideas expressed in the previous paragraph.Are you on drugs? Or just illiterate?> It does, Gonzo. The in?itesimal integral of tangential v produces> in?itesimal tangential dr.>You are asserting a result of a calculation which none of>us has ever seen or can really dream of how to do.>You can't just keep asserting that this ?titious>calculation has this supposed result, when neither you>nor anybody else knows how to do that calculation. An>impossible calculation does not produce anything.>Sure it does. It produces a couple of in?itesimal idiots. === > In message , Paul Curran >Now you are just being elliptical. Which reminds me: all this talk of>circular motion is just a special case. Real celestial orbits are> >ellipses, where r and v are almost never perpendicular.> Uhh, yes they are. What makes you say that they are not?>A simple diagram, maybe? (Depending on how strict one wants to be about> >perpendicular.)> --John Park> >In all my books the the v-vector is perpendicular to the R-vector.>No, that's true only for circular orbits> >Its true for plametary and comet orbits> If v were perpendicular to r, the magnitude of r could never change. We > have a word for an orbit with constant |r|. It's circular.>But orbits are circular. That's because they are sections cut from>the crystal spheres which the planets and other celestial objects are>embedded in.>Don't you know anything? === =[Attribution lost to the quote below]> >Remember that in the end if you got answers that agreed with ordinary>physics, I would happily skim over bits of your argument I couldn't>understand; but you are using your unde?ed and mysterious notions to>prove that (e.g.) velocities can be taken in a direction orthogonal>to that which ordinary physics and my intuitional understanding say>they _are_ in.> Oh. Let me see if I've got this right.>OK.> I do physics by dreaming things up. Whereas you do > physics by intuitional understanding.>Nope, you didn't get it right. That bears no resemblance to>the ideas expressed in the previous paragraph.> Are you on drugs? Or just illiterate?I have frequently wondered both things in connection withyour posts. It is a hallmark of your opus that what youquote bears no material to the quoted text. Butlet's just analyze, shall we?Sentence 1:> Remember that in the end if you got answers that agreed with ordinary> physics, I would happily skim over bits of your argumentHere he says that if you had a standard physics resultyou were proving, he would skim over your argument. Thereis no claim here that you are dreaming things up, nor isthere anything about how he does physics by intuitionalunderstanding. He is merely saying that a proof of astandard textbook physics result would not merit closestandards of rigor or close inspection for validity.I am by the way not claiming telepathy either. I am readingthe words that are written and responding only to them.Sentence 2:>I would happily skim over bits of your argument I couldn't>understand;This is a follow on to the previous. Again he says that i? were a standard result you were claiming, it wouldn'tbother him to ?d parts of your argument that he couldn'tunderstand. He would skim over them.Nothing there about dream up. Nothing there about doingphysics by intuitional understanding.Sentence 3:> but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics and my intuitional understanding say> they _are_ in.This is a rather complex one. It contains several thoughts: a. Your argument is proving that certain vectors can be taken in a direction orthogonal to the direction that standard physics assigns them. b. The arguments you use to establish this use unde?ed and mysterious notions. c. Therefore (this is the meaning of the word but, in contrast with the previous sentence fragment) your stuff can't be just skimmed, but you need to provide enough detail to give a complete logical argument. d. Your results are also counter-intuitive.Now here at last we have the word intuitive, but not inthe sense you used it. This poster very clearly is talkingabout mathematical arguments and ordinary physics. He isvery clearly contrasting your argument with the kind we?d in physics books, which rely on mathematics and noton intuition.He ALSO says that in addition to the derivation of thisstuff, he has an intuition about which direction thesevectors go in. That is very different from your paraphrase,I do physics by intuitional understanding. Instead whatit says is I do physics my mathematical derivation, andthe results as far as the directions of certain vectorsmake intuitive sense to me.Just to reiterate: intuititional understanding is CLEARLYapplied to the directions of certain vectors and isCLEARLY an add-on to the main intent of the sentence,which is about ordinary physics, not intuition. Keytest: If you drop the phrase and my intuitional understandingfrom that sentence, you do not change the meaning of thesentence, or of the paragraph. Let's just try that,shall we?> Remember that in the end if you got answers that agreed with ordinary> physics, I would happily skim over bits of your argument I couldn't> understand; but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics says they _are_ in.Can you see the meaning of this paragraph now? Do younotice that the meaning is no different?The meaning of you just dream things up is also absentfrom this paragraph. What he says, and I agree, is that yourargument keeps falling back on processes that are notspelled out (but you claim are intuitively obvious...need I point out that YOU are the one saying you shoulddo it by intuition, not by mathematics?) and terms thatare not de?ed. When asked to de?e a term, you throwin another unde?ed term and another unde?ed process.It's turtles all the way down.So... are you on drugs? Or just illiterate? - Randy === >Sentence 2:>I would happily skim over bits of your argument I couldn't>understand;Which seems to be all of it.>This is a follow on to the previous. Again he says that if>it were a standard result you were claiming, it wouldn't>bother him to ?d parts of your argument that he couldn't>understand. He would skim over them.If it were a standard result I claim, why bother?I would have a lot more respect for the analysis if it were made bythe addressee.> === >[Attribution lost to the quote below]You've lost a bit more than the attribution, sport.>Remember that in the end if you got answers that agreed with ordinary>physics, I would happily skim over bits of your argument I couldn't>understand; but you are using your unde?ed and mysterious notions to>prove that (e.g.) velocities can be taken in a direction orthogonal> >to that which ordinary physics and my intuitional understanding say>they _are_ in.> Oh. Let me see if I've got this right.>OK.> I do physics by dreaming things up. Whereas you do > physics by intuitional understanding.>Nope, you didn't get it right. That bears no resemblance to>the ideas expressed in the previous paragraph.> Are you on drugs? Or just illiterate? === >Sentence 3:> but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics and my intuitional understanding say> they _are_ in.>This is a rather complex one. It contains several thoughts:Unlike your posts I daresay which reiterate the same thought. > a. Your argument is proving that certain vectors can be> taken in a direction orthogonal to the direction that standard> physics assigns them.> b. The arguments you use to establish this use unde?ed and> mysterious notions.> c. Therefore (this is the meaning of the word but, in contrast> with the previous sentence fragment) your stuff can't be just> skimmed, but you need to provide enough detail to give a> complete logical argument.> d. Your results are also counter-intuitive.>Now here at last we have the word intuitive, but not in>the sense you used it. This poster very clearly is talking>about mathematical arguments and ordinary physics. He is>very clearly contrasting your argument with the kind we>?d in physics books, which rely on mathematics and not>on intuition.Except I'm not so sure why the physics texts are intuitive and notcounter intuitive. It would seem that if something is counterintuitive, it must be counter something intuitive. But thenconsistency is the hobgoblin of small minds and you are if nothingelse consistent.>He ALSO says that in addition to the derivation of this>stuff, he has an intuition about which direction these>vectors go in.And I suggest that he needs plenty of intuition because he has noanalytical basis for that direction.> That is very different from your paraphrase,>I do physics by intuitional understanding.So you and your cohorts are allowed literary license but I am not? Whydidn't you explain this to me in the beginning instead of merely apingthe conventional approach? Do I need you et al to explain the obvious?> Instead what>it says is I do physics my mathematical derivation, and>the results as far as the directions of certain vectors>make intuitive sense to me.What else does he have to rely on besides intuition?>Just to reiterate: intuititional understanding is CLEARLY>applied to the directions of certain vectors and is>CLEARLY an add-on to the main intent of the sentence,>which is about ordinary physics, not intuition. Key>test: If you drop the phrase and my intuitional understanding>from that sentence, you do not change the meaning of the>sentence, or of the paragraph. Let's just try that,>shall we?You really need to consider biblical exegesis as a profession. > Remember that in the end if you got answers that agreed with ordinary> physics, I would happily skim over bits of your argument I couldn't> understand; but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics says they _are_ in.>Can you see the meaning of this paragraph now? Do you>notice that the meaning is no different?>The meaning of you just dream things up is also absent>from this paragraph. What he says, and I agree, is that your>argument keeps falling back on processes that are not>spelled out (but you claim are intuitively obvious...>need I point out that YOU are the one saying you should>do it by intuition, not by mathematics?) and terms that>are not de?ed. When asked to de?e a term, you throw>in another unde?ed term and another unde?ed process.>It's turtles all the way down.>So... are you on drugs? Or just illiterate?Well, actually both. I just wondered if I had found some ignoranthomies. I've always found that classical angular mechanical exegeticsgo better with coke. It all seems to make so much sense when you don'tthink about it.By the way, do let me know when you need a reply. So far I've justassumed you're doing stream of consciousness. Was there something youwanted to say? Oh, well, not important as I imagine one or the otherof you has said it all before. Many times. Please let me know when youget an original thought.> - Randy === >Sentence 3:> > but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics and my intuitional understanding say> they _are_ in.> >This is a rather complex one. It contains several thoughts:> > Unlike your posts I daresay which reiterate the same thought.My, my, are we getting a little temper? You realize that youget extra crank points for changing the subject and switchingto personal insults, right?>Now here at last we have the word intuitive, but not in>the sense you used it. This poster very clearly is talking>about mathematical arguments and ordinary physics. He is>very clearly contrasting your argument with the kind we>?d in physics books, which rely on mathematics and not>on intuition.> Except I'm not so sure why the physics texts are intuitive and not> counter intuitive.They don't need to be. In fact I have always believed thatmuch of physics, especially non-Newtonian physics, is counter-intuitive, and that the process is: ?st you learn themathematics and then after a great deal of practice youabsorb it into your gut. In other words, that our intuitiongets trained.> It would seem that if something is counter> intuitive, it must be counter something intuitive.I'm sure that means something to you, but it doesn'tto me. Counter intuitive means goes against our intuition.Not everybody has the same intuition.> But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.your posting history, I am 50% sure that you imagined the postwhere I said angular momentum is a scalar.The Emerson quote by the way has the keyword foolish in it.Look it up.>He ALSO says that in addition to the derivation of this>stuff, he has an intuition about which direction these>vectors go in.> And I suggest that he needs plenty of intuition because he has no> analytical basis for that direction.Nope, see you're misinterpreting AGAIN, in the same way.You start with the analysis. There is no room fordifferent directions in the analysis. It's done mathematically.Not only is there plenty of analytical basis for thatdirection, you've been shown it. Many time. Are youattention de?it?Here's one: if the difference between two vectors isin a certain direction, we say the vector differenceis in that direction. That difference is CALCULATED.Here's one: We have shown you the equations for motionin a circle. We have shown you the ANALYTICAL derivativeof that motion. We have shown you MATHEMATICALLY thatdr/dt and r are vectors in orthogonal directions, becausetheir dot product is zero. What do you think peoplehave been doing in your silly direction threadsexcept give you ANALYTICAL reasons why the velocitymust be in a certain direction and can't just bearbitrarily taken to be in another?> You really need to consider biblical exegesis as a profession. Short attention span. We were arguing about the meaningof a passage. Now you think you can get away with sayingwhy the hell are you talking about the meaning of thispackage?. I don't really care, but since that was thetopic of discussion in the post previous to this one,I backed up my claim about the meaning. You, on the otherhand, just make a lot of noise. I'm sorry about yourshort-term memory problem. Perhaps you need to consulta neurologist. - Randy === They don't need to be. In fact I have always believed that> much of physics, especially non-Newtonian physics, is counter-> intuitive ...I've long believed all of physics is intuitive, it's just a matter of...<...> ... that our intuition gets trained.I've never like the crutch it's counter-intuitive. This is lazynessor inability. When we ?d something counter-intuitive, it's up tous to retrain our intuition.> The Emerson quote by the way has the keyword foolish in it.> Look it up.You've seen the movie Next Stop, Wonderland?... === I've never like the crutch it's counter-intuitive. This is lazyness> or inability. When we ?d something counter-intuitive, it's up to> us to retrain our intuition.That's a remarkable observation. It suggests that a signi?ant componentof learning mathematics is retraining our intuition. I second the motion. === > >Sentence 3:> but you are using your unde?ed and mysterious notions to> prove that (e.g.) velocities can be taken in a direction orthogonal> to that which ordinary physics and my intuitional understanding say> they _are_ in.>This is a rather complex one. It contains several thoughts:> Unlike your posts I daresay which reiterate the same thought.>My, my, are we getting a little temper? You realize that you>get extra crank points for changing the subject and switching>to personal insults, right?>Now here at last we have the word intuitive, but not in>the sense you used it. This poster very clearly is talking>about mathematical arguments and ordinary physics. He is>very clearly contrasting your argument with the kind we>?d in physics books, which rely on mathematics and not>on intuition.> Except I'm not so sure why the physics texts are intuitive and not> counter intuitive.>They don't need to be. In fact I have always believed that>much of physics, especially non-Newtonian physics, is counter->intuitive, and that the process is: ?st you learn the>mathematics and then after a great deal of practice you>absorb it into your gut. In other words, that our intuition>gets trained.> It would seem that if something is counter> intuitive, it must be counter something intuitive.>I'm sure that means something to you, but it doesn't>to me. Counter intuitive means goes against our intuition.>Not everybody has the same intuition.> But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.>your posting history, I am 50% sure that you imagined the post>where I said angular momentum is a scalar.>The Emerson quote by the way has the keyword foolish in it.>Look it up.>He ALSO says that in addition to the derivation of this>stuff, he has an intuition about which direction these>vectors go in.> And I suggest that he needs plenty of intuition because he has no> analytical basis for that direction.>Nope, see you're misinterpreting AGAIN, in the same way.>You start with the analysis. There is no room for>different directions in the analysis. It's done mathematically.>Not only is there plenty of analytical basis for that>direction, you've been shown it. Many time. Are you>attention de?it?>Here's one: if the difference between two vectors is>in a certain direction, we say the vector difference>is in that direction. That difference is CALCULATED.>Here's one: We have shown you the equations for motion>in a circle. We have shown you the ANALYTICAL derivative>of that motion. We have shown you MATHEMATICALLY that>dr/dt and r are vectors in orthogonal directions, because>their dot product is zero. What do you think people>have been doing in your silly direction threads>except give you ANALYTICAL reasons why the velocity>must be in a certain direction and can't just be>arbitrarily taken to be in another?> You really need to consider biblical exegesis as a profession. >Short attention span. We were arguing about the meaning>of a passage. Now you think you can get away with saying>why the hell are you talking about the meaning of this>package?. I don't really care, but since that was the>topic of discussion in the post previous to this one,>I backed up my claim about the meaning. You, on the other>hand, just make a lot of noise. I'm sorry about your>short-term memory problem. Perhaps you need to consult>a neurologist.>See, the problem here is that you're trying to speak for someone else.In other words it's not best evidence. Presumably the originaladdressee is capable of speaking for himself. Your comments areincompetent, irrelevant, and immaterial. They have no actual bearingon what the original poster meant when he said any of these things.All you can speak to is what the post would have meant were it basedon your remarks and interpretations, which it is not.The fact is that you don't know whether Newtonian or non Newtonianphysics is or is not intuitive because you don't know much abouteither one except what you've intuited. And anyone who considersintuition or counter intuition a suf?ient basis for scienti?knowledge ipso facto disquali?s himself as a scienti? thinker.If you had any imagination, intuition, or even any counter intuitionyou could see what I'm driving at and why. The only reason theclassical analysis of macro angular momentum in ?ls and sucheven works at all is because radial dr/dt and tangential v aremaintained in lockstep through tensile forces.If you tried to explain celestial orbital angular mechanics on asimilar basis you'd be laughed out of the house. It's a joke. Youcan't even explain the analytical basis of Planck's constant on such abasis. Talk about counter intuitive. Go back to your relativistic andquantum magic formalisms to describe what you can't explain. === > They don't need to be. In fact I have always believed that> much of physics, especially non-Newtonian physics, is counter-> intuitive ...>I've long believed all of physics is intuitive, it's just a matter of>...The source of all scienti? discoveries is intuitive in origin. Butthe science in scienti? discoveries is never intuitive.><...> ... that our intuition gets trained.>I've never like the crutch it's counter-intuitive. This is lazyness>or inability. When we ?d something counter-intuitive, it's up to>us to retrain our intuition.> The Emerson quote by the way has the keyword foolish in it.> Look it up.I know it does. These are just stupid word games. A foolishconsistency is the hobgoblin of small minds. To which one might addthat accurate quotations represent a foolish consistency when dealingwith small minds and limited imaginations.>You've seen the movie Next Stop, Wonderland?>... === > But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.Yes, that's a classic.> your posting history, I am 50% sure that you imagined the post> where I said angular momentum is a scalar.I don't think he did, Randy. I remember it, and assumed it was a typoor similar. But otherwise, Lester has gone into his Very Silly mode,and doesn't seem worth trying to respond to. In the end I think he'sjust a stray pomo who likes the sound of some strings of words.[This ?he' is me, by the way]>He ALSO says that in addition to the derivation of this>stuff, he has an intuition about which direction these>vectors go in.> And I suggest that he needs plenty of intuition because he has no> analytical basis for that direction.> Nope, see you're misinterpreting AGAIN, in the same way.> You start with the analysis. There is no room for> different directions in the analysis. It's done mathematically.> Not only is there plenty of analytical basis for that> direction, you've been shown it. Many time. Are you> attention de?it?I think that when comments reach a certain level of silliness, it'scounterproductive to try to argue. I was hoping everyone would goquiet so we could all hear the New Truth about Planck's constant. Butno chance...Brian Chandler-geo://Sano.Japan.Planet_3Jigsaw puzzles from Japan at:http://imaginatorium.org/shop/ === If you had any imagination, intuition, or even any counter intuition> you could see what I'm driving at and why. The only reason the> classical analysis of macro angular momentum in ?ls and such> even works at all is because radial dr/dt and tangential v are> maintained in lockstep through tensile forces.Despite what I just said, this is all fascinating. Do I understandcorrectly that you derive your radial dr/dt by this process unknownto most of us called in?itesimal integration? Are you prepared tobelieve that most of us use the symbols dr/dt to represent thederivative of r (the position vector) with respect to time (t)? Andfurther, that we also use ?v' as another name for the _same_ thing(dr/dt). Can you also derive _your_ dr/dt by a process ofdifferentiation? How do you obtain your v? Is it necessarily by adifferent process?> If you tried to explain celestial orbital angular mechanics on a> similar basis you'd be laughed out of the house. It's a joke.Naruhodo. So does this mean that if I calculate the period of orbit ofa satellite (for example), using the principles I learned at school, Iget the wrong answer? (I just tried - being desperately rusty at thissort of thing - and got an approximation of 84 minutes for the time ofan orbit at the radius of the earth. That sounds in line with what Iremember for the early satellites.) Is this wrong? Or have I just gotthe wrong names for things somewhere? Or were the satellites allfaked, or something? I didn't use either of your notions (as far as Iknow); are you saying that gravity as a force isn't up to keepingthings in lockstep?Brian Chandler-geo://Sano.Japan.Planet_3Jigsaw puzzles from Japan at:http://imaginatorium.org/shop/...consistency is the hobgoblin of small minds (Lester Zick) === > But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.>Yes, that's a classic.> your posting history, I am 50% sure that you imagined the post> where I said angular momentum is a scalar.>I don't think he did, Randy. I remember it, and assumed it was a typo>or similar.OK, thanks. I had a nagging suspicion it might be true. That's why Iset the bar at only 50%, because I do make stupid typos. It would havebeen a lot lower than 50% if it was anyone but Lester reporting mywords.> But otherwise, Lester has gone into his Very Silly mode,>and doesn't seem worth trying to respond to. In the end I think he's>just a stray pomo who likes the sound of some strings of words.What's a pomo?>[This ?he' is me, by the way]> >He ALSO says that in addition to the derivation of this> >stuff, he has an intuition about which direction these> >vectors go in.> And I suggest that he needs plenty of intuition because he has no> analytical basis for that direction.> Nope, see you're misinterpreting AGAIN, in the same way.> You start with the analysis. There is no room for> different directions in the analysis. It's done mathematically.> Not only is there plenty of analytical basis for that> direction, you've been shown it. Many time. Are you> attention de?it?>I think that when comments reach a certain level of silliness, it's>counterproductive to try to argue. I was hoping everyone would go>quiet so we could all hear the New Truth about Planck's constant. But>no chance...OK, I'll shut up now. It wasn't important, and Lester and I bothmanaged to exhaust each others' patience on this particular sillyargument. Though now I'm wondering what a counter intuition is.Apparently he feels it's the noun that goes with counter-intuition. - Randy === > But otherwise, Lester has gone into his Very Silly mode,>and doesn't seem worth trying to respond to. In the end I think he's>just a stray pomo who likes the sound of some strings of words.> What's a pomo?postmodernist. Google for the Sokal affair sometime you're tired ofWeinberg -turns out it was a physicist friend of SW's who said that in facingdeath, he drew some consolation from the re?n that he wouldnever again have to look up the word hermeneutic in the dictionary.Brian Chandler-geo://Sano.Japan.Planet_3Qualia explained - http://imaginatorium.org/stuff/qualia.htm === > But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.>Yes, that's a classic.> your posting history, I am 50% sure that you imagined the post> where I said angular momentum is a scalar.>I don't think he did, Randy. I remember it, and assumed it was a typo>or similar. But otherwise, Lester has gone into his Very Silly mode,>and doesn't seem worth trying to respond to. In the end I think he's>just a stray pomo who likes the sound of some strings of words.>[This ?he' is me, by the way]>He ALSO says that in addition to the derivation of this>stuff, he has an intuition about which direction these>vectors go in.> And I suggest that he needs plenty of intuition because he has no> > analytical basis for that direction.> Nope, see you're misinterpreting AGAIN, in the same way.> You start with the analysis. There is no room for> different directions in the analysis. It's done mathematically.> Not only is there plenty of analytical basis for that> direction, you've been shown it. Many time. Are you> attention de?it?>I think that when comments reach a certain level of silliness, it's>counterproductive to try to argue. I was hoping everyone would go>quiet so we could all hear the New Truth about Planck's constant. But>no chance...>Well, well. We ?ally get an intelligent perspective on the presentseries of inane observations. First Randy got back from vacation andwanted to talk about it then you wanted to talk about. I've beentrying unsuccessfully to beg off from the current discussion for thelast two weeks.By the way the analytical essence of what I use to explain Planck'sconstant was posted 7/9/03 about two weeks ago to these same groupsunder the title Analytical Origin of Planck's Constant. It shouldstill be available. You're welcome to check it out. What I intend topost next week is just a further series of historical and analyticalnotes. === > If you had any imagination, intuition, or even any counter intuition> you could see what I'm driving at and why. The only reason the> classical analysis of macro angular momentum in ?ls and such> even works at all is because radial dr/dt and tangential v are> maintained in lockstep through tensile forces.>Despite what I just said, this is all fascinating. Do I understand>correctly that you derive your radial dr/dt by this process unknown>to most of us called in?itesimal integration? Are you prepared to>believe that most of us use the symbols dr/dt to represent the>derivative of r (the position vector) with respect to time (t)? And>further, that we also use ?v' as another name for the _same_ thing>(dr/dt). Can you also derive _your_ dr/dt by a process of>differentiation? How do you obtain your v? Is it necessarily by a>different process?> If you tried to explain celestial orbital angular mechanics on a> similar basis you'd be laughed out of the house. It's a joke.>Naruhodo. So does this mean that if I calculate the period of orbit of>a satellite (for example), using the principles I learned at school, I>get the wrong answer? (I just tried - being desperately rusty at this>sort of thing - and got an approximation of 84 minutes for the time of>an orbit at the radius of the earth. That sounds in line with what I>remember for the early satellites.) Is this wrong? Or have I just got>the wrong names for things somewhere? Or were the satellites all>faked, or something? I didn't use either of your notions (as far as I>know); are you saying that gravity as a force isn't up to keeping>things in lockstep?>Oh gravity is but only because it's an inverse square force. The onetime this is irrelevant is in the case of circular rotation - which iswhy I posed the problem the way I did.In ordinary macro angular mechanical contexts this is all we havebecause internal tensile forces keep radial dr/dt equal to tangentialv. In orbital angular momentum this is not the case. Gravitation hasto maintain an inverse square relationship between radial dr/dt inorder to maintain a constant ratio between radial dr/dt and vthroughout the orbit whatever the values of r and t.And by the way I said explain orbital angular mechanics not just do acalculation. Radial dr/dt is commonly used to explain circular orbits- for example - as manifestations of equal centripetal and tangentialvelocities. As the orbiting body falls toward the attracting body itsorbital velocity moves it by exactly the same amount sideways. Thusthe value of r never changes and the circular orbit is maintainedexactly. === > But then> consistency is the hobgoblin of small minds and you are if nothing> else consistent.>Yes, that's a classic.> your posting history, I am 50% sure that you imagined the post> where I said angular momentum is a scalar.>I don't think he did, Randy. I remember it, and assumed it was a typo>or similar.>OK, thanks. I had a nagging suspicion it might be true. That's why I>set the bar at only 50%, because I do make stupid typos. It would have>been a lot lower than 50% if it was anyone but Lester reporting my>words.I just hope you'll remember this as an offer of good faith in myattitude. It would have been easy enough but completely pointless topost an acerbic reply to what was obviously a mistaken comment.> But otherwise, Lester has gone into his Very Silly mode,>and doesn't seem worth trying to respond to. In the end I think he's>just a stray pomo who likes the sound of some strings of words.>What's a pomo?>[This ?he' is me, by the way]> >He ALSO says that in addition to the derivation of this> >stuff, he has an intuition about which direction these> >vectors go in.> And I suggest that he needs plenty of intuition because he has no> analytical basis for that direction.> Nope, see you're misinterpreting AGAIN, in the same way.> You start with the analysis. There is no room for> different directions in the analysis. It's done mathematically.> Not only is there plenty of analytical basis for that> direction, you've been shown it. Many time. Are you> attention de?it?>I think that when comments reach a certain level of silliness, it's>counterproductive to try to argue. I was hoping everyone would go>quiet so we could all hear the New Truth about Planck's constant. But>no chance...>OK, I'll shut up now. It wasn't important, and Lester and I both>managed to exhaust each others' patience on this particular silly>argument. Though now I'm wondering what a counter intuition is.>Apparently he feels it's the noun that goes with counter-intuition.>I haven't given up trying to explain the error of dr parallel to p.But I would rather wait some while for an opportune opportunity topursue the subject. === Well, well. We ?ally get an intelligent perspective on the present> series of inane observations. First Randy got back from vacation and> wanted to talk about it then you wanted to talk about. I've been> trying unsuccessfully to beg off from the current discussion for the> last two weeks.Every time you make a remark like there's no analyticalbasis for saying dr/dt and v are in the same directionyou are doing the opposite of begging off. You are beingprovocative. Very successfully.> By the way the analytical essenceYou clearly have no idea what analysis is. Yes, thatstatement is deliberately provocative.> of what I use to explain Planck's> constant was posted 7/9/03 about two weeks ago to these same groups> under the title Analytical Origin of Planck's Constant. - Randy === >OK, thanks. I had a nagging suspicion it might be true. That's why I>set the bar at only 50%, because I do make stupid typos. It would have>been a lot lower than 50% if it was anyone but Lester reporting my>words.> I just hope you'll remember this as an offer of good faith in my> attitude. It would have been easy enough but completely pointless to> post an acerbic reply to what was obviously a mistaken comment.Mostly you retain a sense of humor and avoid the personalinsults which is why I keep taking the time to reply.>OK, I'll shut up now. It wasn't important, and Lester and I both>managed to exhaust each others' patience on this particular silly>argument. Though now I'm wondering what a counter intuition is.>Apparently he feels it's the noun that goes with counter-intuition.Here I see one of my stupid typos. The last word wascounter-intuitive. You have left me baf?ce againwith (1) the mental leap from counter-intuitive tocounter intuition, and (2) what it is you might mean bythe statement that if something is counter intuitive itmust be counter something intuitive. Clearly you havetaken yet another common phrase (counter-intuitive) andgiven it a meaning all your own, one which I am unable tofathom. - Randy === > Well, well. We ?ally get an intelligent perspective on the present> series of inane observations. First Randy got back from vacation and> wanted to talk about it then you wanted to talk about. I've been> trying unsuccessfully to beg off from the current discussion for the> last two weeks.>Every time you make a remark like there's no analytical>basis for saying dr/dt and v are in the same direction>you are doing the opposite of begging off. You are being>provocative. Very successfully.> By the way the analytical essence>You clearly have no idea what analysis is. Yes, that>statement is deliberately provocative.> of what I use to explain Planck's> constant was posted 7/9/03 about two weeks ago to these same groups> under the title Analytical Origin of Planck's Constant.>In any event I'll be posting a supplemented version in reply to thethread Analytical Origin of Planck's Constant tomorrow morning. Thereare only some additional comments regarding the historical perspectivefor angular mechanics.By the way I haven't forgotten your analytical approach to thedirection of dr/dt, which I believe is the classical approach. I'vebeen puzzling over the treatment and I think I have the solution. ButI'd rather get on to Planck's constant at the moment. === In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.That could appear in the New Heritage Dictionary as a textualillustration of gobbledegook. You still appear completely confusedover the possibility of two meanings for r, and run them togetherinto some kind of meaning soup.r may either mean a scalar (single number), commonly denoting thelength of a radius, or else r may mean a vector (triple of numbersin three space), typically denoting the position vector of a pointrelative to a particular origin.Now, don't be coy, and don't say both: which concept do you want?In the ?st case dr/dt is the rate of change of the radius, and is... among a myriad of other possibilities ... zero for a circularorbit. If we are talking about orbits.In the second case dr/dt is the (vector) velocity, by de?ition(time rate of change of position), a convention which is inutterablynot open to discussion or equivocation at the hands of Lester Zick. dr/dt = v is true for orbits, choo-choo trains, and the ?n gpast your obstructionist nose. v has no other meaning than dr/dtin this context. To suggest otherwise is to betray a complete lack ofunderstanding of the velocity vector; not to mention vectors,calculus and a myriad of other important topics.Did I neglect to mention that v in this context is a vector, justlike r? > And by the way I said explain orbital angular mechanics not just doa> calculation. Radial dr/dt is commonly used to explain circular orbits> - for example - as manifestations of equal centripetal and tangential> velocities. As the orbiting body falls toward the attracting body its> orbital velocity moves it by exactly the same amount sideways. Thus> the value of r never changes and the circular orbit is maintained> exactly.If you are going to use expressions like radial or tangentialdr/dt, you are implicitly adopting the vector meaning. In this case,for circular orbits, dr/dt is purely tangential ... OTOH, if you saythe value of r never changes, you are clearly using the ?st,scalar, sense of r. Cunningly, you move between the two senses of rin the same paragraph! Now, why don't you behave?I must take a seminar offered by the Learning Annex: the one onself-empowerment (what else?) where one of the deadly errors you aretaught to avoid is arguing with the irrational.That wouldn't be _you_, would it, Mr. Zick? Ah yes ... I'm an oldhypocrite ... as Groucho Marx said, we need the eggs. Sigh.answer, do ... === > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?The last time I checked vectors have both magnitude and direction.Maybe something has changed. Is there a question congealed in all thepreceeding? Analytically I've been dealing with vectors and themagnitude of vectors.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.Ah, well, you know that isn't quite true. Kepler thought it was true.Perhaps some force of angels is the concept you're looking for. Quiteironic that major luminaries in the ?ld of angular mechanics ?d itnecessary to revert to anachronistic pre Newtonian explanations.>In the second case dr/dt is the (vector) velocity, by de?ition>(time rate of change of position), a convention which is inutterably>not open to discussion or equivocation at the hands of Lester Zick.Or Newton apparently. >dr/dt = v is true for orbits, choo-choo trains, and the ?n g>past your obstructionist nose. v has no other meaning than dr/dt>in this context. To suggest otherwise is to betray a complete lack of>understanding of the velocity vector; not to mention vectors,>calculus and a myriad of other important topics.>Did I neglect to mention that v in this context is a vector, just>like r?> And by the way I said explain orbital angular mechanics not just do>a> calculation. Radial dr/dt is commonly used to explain circular orbits> - for example - as manifestations of equal centripetal and tangential> velocities. As the orbiting body falls toward the attracting body its> orbital velocity moves it by exactly the same amount sideways. Thus> the value of r never changes and the circular orbit is maintained> exactly.>If you are going to use expressions like radial or tangential>dr/dt, you are implicitly adopting the vector meaning. In this case,>for circular orbits, dr/dt is purely tangential ... OTOH, if you say>the value of r never changes, you are clearly using the ?st,>scalar, sense of r. Cunningly, you move between the two senses of r>in the same paragraph! Now, why don't you behave?Cunningly? You're suggesting I have the guile to confuse the issue butnot to analyze it correctly? Do you understand what the de?ition o? is? Perhaps I should run for president. I could use the job.>I must take a seminar offered by the Learning Annex: the one on>self-empowerment (what else?) where one of the deadly errors you are>taught to avoid is arguing with the irrational.>That wouldn't be _you_, would it, Mr. Zick? Ah yes ... I'm an old>hypocrite ... as Groucho Marx said, we need the eggs. Sigh.>answer, do ...Yeah, I have a hard time telling whether you're part of the thread ornot. You come and go at various times with next to no special insighton the basic analysis of circular rotation. I could and have found thesame explanations in any textbook or in the posts of any of a hundredother posters. Of course you need the eggs. In your position whowouldn't?If you want to participate in the conversation, I strongly suggest youdo so instead of acting like some deus ex dropped on a confused andbewildered group of analytical amateurs and miscreants.I only have one question to answer yours. If there is no radial dr/dt,what causes tangential v to rotate? Do you tie an ethereal string toit? Does centripetal force somehow interact with v directly withoutthe intercession of radial dr/dt? It's true of course that I haveheard of vector addition of velocities before. I've just never heardof the vector addition of a force and velocity before.And if dL/dt = 0 because dr/dt lies in the direction of v, why does protate? The quality of your prose is high. Not quite up to mystandards but certainly good enough for government work. Yet you don'teven apparently understand what you're saying much less what I'msaying. <3f156fe7.34641502@netnews.att.net> <585ab5d8.0307161343.c0ba484@posting.google.com> <3f15fcdc.37862304@netnews.att.net> <585ab5d8.0307171038.3435db69@posting.google.com> <3f1751e3.41708891@netnews.att.net> <585ab5d8.0307180501.fafaafc@posting.google.com> <3f18a02a.547318@netnews.att.net> <3f19abe4.5169345@netnews.att.net> <2a0cceff.0307220427.13bc8570@posting.google.com> <3f1d5002.16373751@netnews.att.net> === =In message <3f1d5002.16373751@netnews.att.net>, Lester Zick > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.Hardly an answer to the question.>Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.That and is precisely the problem.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true.For what circular orbit can you not de?e an origin such that d|r|/dt is always zero?[...]>answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. I could and have found the>same explanations in any textbook or in the posts of any of a hundred>other posters.And you don't draw the obvious conclusion?-- Richard Herring === In message <3f1d5002.16373751@netnews.att.net>, Lester Zick > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.>Hardly an answer to the question.>Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>That and is precisely the problem.Except of course that either can be dealt with independent of theother.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true.>For what circular orbit can you not de?e an origin such that d|r|/dt >is always zero?For any circular orbit.>[...]>answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. I could and have found the>same explanations in any textbook or in the posts of any of a hundred>other posters.>And you don't draw the obvious conclusion?Yeah. That both you and Edward need the eggs.> === > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.Yes, it depends on what his de?itions of r are (Clintonesque?)>Now, don't be coy, and don't say both: which concept do you want?>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>In the second case dr/dt is the (vector) velocity, by de?ition>(time rate of change of position), a convention which is inutterably>not open to discussion or equivocation at the hands of Lester Zick. >dr/dt = v is true for orbits, choo-choo trains, and the ?n g>past your obstructionist nose. v has no other meaning than dr/dt>in this context. To suggest otherwise is to betray a complete lack of>understanding of the velocity vector; not to mention vectors,>calculus and a myriad of other important topics.>Did I neglect to mention that v in this context is a vector, just>like r?> And by the way I said explain orbital angular mechanics not just do>a> calculation. Radial dr/dt is commonly used to explain circular orbits> - for example - as manifestations of equal centripetal and tangential> velocities. As the orbiting body falls toward the attracting body its> orbital velocity moves it by exactly the same amount sideways. Thus> the value of r never changes and the circular orbit is maintained> exactly.>If you are going to use expressions like radial or tangential>dr/dt, you are implicitly adopting the vector meaning. In this case,>for circular orbits, dr/dt is purely tangential ... OTOH, if you say>the value of r never changes, you are clearly using the ?st,>scalar, sense of r. Cunningly, you move between the two senses of r>in the same paragraph! Now, why don't you behave?>I must take a seminar offered by the Learning Annex: the one on>self-empowerment (what else?) where one of the deadly errors you are>taught to avoid is arguing with the irrational.>That wouldn't be _you_, would it, Mr. Zick? Ah yes ... I'm an old>hypocrite ... as Groucho Marx said, we need the eggs. Sigh.>answer, do ...--John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing === > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.>Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true. Kepler thought it was true.>Perhaps some force of angels is the concept you're looking for. Quite>ironic that major luminaries in the ?ld of angular mechanics ?d it>necessary to revert to anachronistic pre Newtonian explanations.the observed non-circular planet orbits as elliptic. Who are the major luminaries you refer to? And are you telling us thatyou have discovered something new that is unknown in physics? In other words, what is the point of this discussion? >In the second case dr/dt is the (vector) velocity, by de?ition>(time rate of change of position), a convention which is inutterably>not open to discussion or equivocation at the hands of Lester Zick.>Or Newton apparently.>dr/dt = v is true for orbits, choo-choo trains, and the ?n g>past your obstructionist nose. v has no other meaning than dr/dt>in this context. To suggest otherwise is to betray a complete lack of>understanding of the velocity vector; not to mention vectors,>calculus and a myriad of other important topics.>Did I neglect to mention that v in this context is a vector, just>like r?> And by the way I said explain orbital angular mechanics not just do>a> calculation. Radial dr/dt is commonly used to explain circular orbits> - for example - as manifestations of equal centripetal and tangential> velocities. As the orbiting body falls toward the attracting body its> orbital velocity moves it by exactly the same amount sideways. Thus> the value of r never changes and the circular orbit is maintained> exactly.You've garbled up a perfectly good description here. There is no radialcomponent of velocity required for this to happen, because the radialdirection is constantly changing. See below.>If you are going to use expressions like radial or tangential>dr/dt, you are implicitly adopting the vector meaning. In this case,>for circular orbits, dr/dt is purely tangential ... OTOH, if you say>the value of r never changes, you are clearly using the ?st,>scalar, sense of r. Cunningly, you move between the two senses of r>in the same paragraph! Now, why don't you behave?>Cunningly? You're suggesting I have the guile to confuse the issue but>not to analyze it correctly? Do you understand what the de?ition of>is is? Perhaps I should run for president. I could use the job.>I must take a seminar offered by the Learning Annex: the one on>self-empowerment (what else?) where one of the deadly errors you are>taught to avoid is arguing with the irrational.>That wouldn't be _you_, would it, Mr. Zick? Ah yes ... I'm an old>hypocrite ... as Groucho Marx said, we need the eggs. Sigh.>answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. I could and have found the>same explanations in any textbook or in the posts of any of a hundred>other posters. Of course you need the eggs. In your position who>wouldn't?>If you want to participate in the conversation, I strongly suggest you>do so instead of acting like some deus ex dropped on a confused and>bewildered group of analytical amateurs and miscreants.>I only have one question to answer yours. If there is no radial dr/dt,>what causes tangential v to rotate? Do you tie an ethereal string toI think I see the problem through all this haze. The change in thevector v (may we speak of the rate of change of the v vector, i.e.,acceleration?) is due to an acceleration component, not a velocitycomponent. There is indeed a radial acceleration (dv/dt) directed towardthe center of rotation for the circular motion you describe. It's calledcentripetal acceleration.>it? Does centripetal force somehow interact with v directly without>the intercession of radial dr/dt? It's true of course that I haveYes. In a circular orbit the vectors v and r rotate at the same angularrate, so the v vector is always perpendicular to the r vector. The rvector has a constant magnitude and the v vector has no radial componentbecause it is always perpendicular to the r vector. Think of it as arotational carrot-on-a-stick.Any radial component of velocity would be along the instantaneous(rotating) radial direction, not the original one. That's what radialcomponent means. The radial direction is a rotating one.>heard of vector addition of velocities before. I've just never heard>of the vector addition of a force and velocity before.>And if dL/dt = 0 because dr/dt lies in the direction of v, why does p>rotate? The quality of your prose is high. Not quite up to my>standards but certainly good enough for government work. Yet you don't>even apparently understand what you're saying much less what I'm>saying.--John E. PrussingUniversity of Illinois at Urbana-ChampaignDepartment of Aerospace Engineeringhttp://www.uiuc.edu/~prussing === >In message <3f1d5002.16373751@netnews.att.net>, Lester Zick <...> >r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.> >Hardly an answer to the question.>Maybe something has changed. Is there a question congealed in all the> >preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>That and is precisely the problem.> Except of course that either can be dealt with independent of the> other.Are you delibrately obtuse, or only accidentally so?You pretend to understand the difference between a vector and itsmagnitude, then sophistically refuse to pin yourself down to which youintend, and use both concepts interchangeably.It's not enough to assert that you understand, you must demonstrateit.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.> >Ah, well, you know that isn't quite true.>For what circular orbit can you not de?e an origin such that d|r|/dt >is always zero?> For any circular orbit.Well ... I commend you on a concrete assertion. Now we are gettingsomewhere: at least that's wrong! Now, let's pretend for a momentthat you really want to understand, and play a little game: youdevelop the last claim, and I will try to parse your semantics.Note that Richard removed all ambiguity from his question by addingthe vertical bars ... indicating he is refering to the length of avector r.>And you don't draw the obvious conclusion?> Yeah. That both you and Edward need the eggs.Is that an overt admission of trolling? A YHBT? Interesting. === In message <3f1d5002.16373751@netnews.att.net>, Lester Zick >r may either mean a scalar (single number), commonly denoting the> >length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.> Hardly an answer to the question.A politician!I was trembling to mention this, but maybe it would actually help: rin the ?st sense may include the case where it is the _magnitude_ ofr in the second case.Darn, I was going to helpfully note that this concept was alreadyunder control as |r|, but I'm disapointed to see you were the one tointroduce that notation, not the mysterious Mr. Zick. >Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.> That and is precisely the problem.>In the ?st case dr/dt is the rate of change of the radius, and is> >... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true.> For what circular orbit can you not de?e an origin such that d|r|/dt > is always zero?> [...]>answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. I stop by from time to time to add my 2 cents.> I could and have found the>same explanations in any textbook or in the posts of any of a hundred>other posters.And don't pro? from them, evidently, since you use the same symbolinterchangeably for a vector and its magnitude, in the same paragraph. Since you don't seem conscious of this, this may be part of theconfusion ...You are occupying a known Usenet niche, BTW: the person regarded asobviously and obliviously wrong, taking on all comers. I guess thereis some kind of mutual grooming implicit in this behavior: comers andtakers. The parties of the second part, being me, get to feel likethey know something before an audience of peers -- to which theelmentary material ensures their election -- while you, party of the?st part, what do _you_ get out of it? <3f156fe7.34641502@netnews.att.net> <585ab5d8.0307161343.c0ba484@posting.google.com> <3f15fcdc.37862304@netnews.att.net> <585ab5d8.0307171038.3435db69@posting.google.com> <3f1751e3.41708891@netnews.att.net> <585ab5d8.0307180501.fafaafc@posting.google.com> <3f18a02a.547318@netnews.att.net> <3f19abe4.5169345@netnews.att.net> <2a0cceff.0307220427.13bc8570@posting.google.com> <3f1d5002.16373751@netnews.att.net> <3f1db75b.18968483@netnews.att.net> === =In message <3f1db75b.18968483@netnews.att.net>, Lester Zick >In message <3f1d5002.16373751@netnews.att.net>, Lester Zick> In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.>Hardly an answer to the question.>Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>That and is precisely the problem.>Except of course that either can be dealt with independent of the>other.Yet you are consistently (inconsistently?) failing to do so, or at least to distinguish them in what you are saying.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true.>For what circular orbit can you not de?e an origin such that d|r|/dt>is always zero?>For any circular orbit.Well, that's a categorical statement for once. Also false.How do you de?e circular?-- Richard Herring === > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?>The last time I checked vectors have both magnitude and direction.>Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.>Ah, well, you know that isn't quite true. Kepler thought it was true.>Perhaps some force of angels is the concept you're looking for. Quite>ironic that major luminaries in the ?ld of angular mechanics ?d it>necessary to revert to anachronistic pre Newtonian explanations.>the observed non-circular planet orbits as elliptic.Oh do tell. Let me ask you this. If I suggested that in response to QMEinstein thought that god does not play dice with the world, would yourespond that I had misstated what Einstein thought and that in facthis major contribution was relativity?>Who are the major luminaries you refer to?Actually I'm referring to a couple of juvies who have nothing betterto do with their summers than troll the usenet.> And are you telling us that>you have discovered something new that is unknown in physics?Unknown to physics, no. Unknown to angular mechanics, yes. Unknown toyou obviously.>In other words, what is the point of this discussion?What discussion? You've sworn me off twice already. You keep comingback for more. Ask yourself what the point of the discussion is. >In the second case dr/dt is the (vector) velocity, by de?ition>(time rate of change of position), a convention which is inutterably>not open to discussion or equivocation at the hands of Lester Zick.>Or Newton apparently.>dr/dt = v is true for orbits, choo-choo trains, and the ?ing>past your obstructionist nose. v has no other meaning than dr/dt>in this context. To suggest otherwise is to betray a complete lack of>understanding of the velocity vector; not to mention vectors,>calculus and a myriad of other important topics.>Did I neglect to mention that v in this context is a vector, just>like r?> And by the way I said explain orbital angular mechanics not just do>a> calculation. Radial dr/dt is commonly used to explain circular orbits> - for example - as manifestations of equal centripetal and tangential> velocities. As the orbiting body falls toward the attracting body its> orbital velocity moves it by exactly the same amount sideways. Thus> the value of r never changes and the circular orbit is maintained> exactly.>You've garbled up a perfectly good description here. There is no radial>component of velocity required for this to happen, because the radial>direction is constantly changing. See below.>If you are going to use expressions like radial or tangential>dr/dt, you are implicitly adopting the vector meaning. In this case,>for circular orbits, dr/dt is purely tangential ... OTOH, if you say>the value of r never changes, you are clearly using the ?st,>scalar, sense of r. Cunningly, you move between the two senses of r>in the same paragraph! Now, why don't you behave?>Cunningly? You're suggesting I have the guile to confuse the issue but>not to analyze it correctly? Do you understand what the de?ition of>is is? Perhaps I should run for president. I could use the job.>I must take a seminar offered by the Learning Annex: the one on>self-empowerment (what else?) where one of the deadly errors you are>taught to avoid is arguing with the irrational.>That wouldn't be _you_, would it, Mr. Zick? Ah yes ... I'm an old>hypocrite ... as Groucho Marx said, we need the eggs. Sigh.>answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. I could and have found the>same explanations in any textbook or in the posts of any of a hundred>other posters. Of course you need the eggs. In your position who>wouldn't?>If you want to participate in the conversation, I strongly suggest you>do so instead of acting like some deus ex dropped on a confused and>bewildered group of analytical amateurs and miscreants.>I only have one question to answer yours. If there is no radial dr/dt,>what causes tangential v to rotate? Do you tie an ethereal string to>I think I see the problem through all this haze. The change in the>vector v (may we speak of the rate of change of the v vector, i.e.,>acceleration?) is due to an acceleration component, not a velocity>component. There is indeed a radial acceleration (dv/dt) directed toward>the center of rotation for the circular motion you describe. It's called>centripetal acceleration.Do tell. So, what is it we do exactly with this centripetalacceleration? We combine it vectorially directly with tangentialvelocity? My, my. I've heard tell of vector addition. I may even haveused it once or twice. But this is the ?st I've ever heard of thevector addition of a force with a velocity.Are you sure you have academic credentials? You certainly belong in aninstitution.>it? Does centripetal force somehow interact with v directly without>the intercession of radial dr/dt? It's true of course that I have>Yes. In a circular orbit the vectors v and r rotate at the same angular>rate, so the v vector is always perpendicular to the r vector. The r>vector has a constant magnitude and the v vector has no radial component>because it is always perpendicular to the r vector. Think of it as a>rotational carrot-on-a-stick.>Any radial component of velocity would be along the instantaneous>(rotating) radial direction, not the original one. That's what radial>component means. The radial direction is a rotating one.Well, well. We ?ally come to the crux of the problem. Unless you'vebeen lying, any interpretation of the phrase any radial component ofvelocity has to refer to an instantaneous (rotating) radialdirection. And that's what a radial component means.Have I ever suggested otherwise? Have I ever suggested that r remainsin one orientation or that radial dr/dt doesn't lie along it?>heard of vector addition of velocities before. I've just never heard>of the vector addition of a force and velocity before.>And if dL/dt = 0 because dr/dt lies in the direction of v, why does p>rotate? The quality of your prose is high. Not quite up to my>standards but certainly good enough for government work. Yet you don't>even apparently understand what you're saying much less what I'm>saying.> === > In ordinary macro angular mechanical contexts this is all we have> because internal tensile forces keep radial dr/dt equal to tangential> v. In orbital angular momentum this is not the case. Gravitation has> to maintain an inverse square relationship between radial dr/dt in> order to maintain a constant ratio between radial dr/dt and v> throughout the orbit whatever the values of r and t.>That could appear in the New Heritage Dictionary as a textual>illustration of gobbledegook. You still appear completely confused>over the possibility of two meanings for r, and run them together>into some kind of meaning soup.>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point>relative to a particular origin.>Yes, it depends on what his de?itions of r are (Clintonesque?)Or possibly Prussing-Hall-Hogg-Green-Herringesque. Same difference. === > >In message <3f1d5002.16373751@netnews.att.net>, Lester Zick ><...>r may either mean a scalar (single number), commonly denoting the>length of a radius, or else r may mean a vector (triple of numbers>in three space), typically denoting the position vector of a point> >relative to a particular origin.>Now, don't be coy, and don't say both: which concept do you want?> >The last time I checked vectors have both magnitude and direction.>Hardly an answer to the question.> >Maybe something has changed. Is there a question congealed in all the>preceeding? Analytically I've been dealing with vectors and the>magnitude of vectors.>That and is precisely the problem.> Except of course that either can be dealt with independent of the> other.>Are you delibrately obtuse, or only accidentally so?Accidentally so. Or perhaps deliberately so. I haven't decided yet.>You pretend to understand the difference between a vector and its>magnitude, then sophistically refuse to pin yourself down to which you>intend, and use both concepts interchangeably.>It's not enough to assert that you understand, you must demonstrate>it.>In the ?st case dr/dt is the rate of change of the radius, and is>... among a myriad of other possibilities ... zero for a circular>orbit. If we are talking about orbits.> >Ah, well, you know that isn't quite true.>For what circular orbit can you not de?e an origin such that d|r|/dt >is always zero?> For any circular orbit.>Well ... I commend you on a concrete assertion. Now we are getting>somewhere: at least that's wrong! Now, let's pretend for a moment>that you really want to understand, and play a little game: you>develop the last claim, and I will try to parse your semantics.>Note that Richard removed all ambiguity from his question by adding>the vertical bars ... indicating he is refering to the length of a>vector r.Actually Richard needs to be behind vertical bars.>And you don't draw the obvious conclusion?> Yeah. That both you and Edward need the eggs.>Is that an overt admission of trolling? A YHBT? Interesting.If it is I must have bagged my limit of trolls by now. Next step fortrolls the endangered species list. You noticed that I've just pickedup Prussing for a third time. Undoubtedly he needs the eggs too.Now you can salvage the whole situation by explaining what causes v torotate. Prussing allowed that there is a centripetal force and thatany component of radial dr/dt must lie along a rotating radius. === > In message <3f1d5002.16373751@netnews.att.net>, Lester Zick > >answer, do ...>Yeah, I have a hard time telling whether you're part of the thread or>not. You come and go at various times with next to no special insight>on the basic analysis of circular rotation. >I stop by from time to time to add my 2 cents.You can't afford it.> === In message <3f1db75b.18968483@netnews.att.net>, Lester Zick >Ah, well, you know that isn't quite true.>For what circular orbit can you not de?e an origin such that d|r|/dt>is always zero?>For any circular orbit.>Well, that's a categorical statement for once. Also false.>How do you de?e circular?Ah well, methinks this is a trick question.If you cannot show what causes the circular rotation of v there can beno circle.Please, please tell me this is not just a red herring.> === =Can anyone tell me how to calculate the integral ofsin(Pye/2)* e^(-x)?any ideas? === Can anyone tell me how to calculate the integral of> sin(Pye/2)* e^(-x)?> any ideas?Mmm ... Pi not Pye !!!!!!!! === =Julien Santini scribe:> Can anyone tell me how to calculate the integral of> sin(Pye/2)* e^(-x)?> any ideas?> Mmm ... Pi not Pye !!!!!!!!??? ...sin(pi/2)=1=const. And now it is obviously? === Can anyone tell me how to calculate the integral of> sin(Pye/2)* e^(-x)?Hmmm, integrating sin(Pye/2)*e^(-x) with respect to e seems pretty hard. === > Can anyone tell me how to calculate the integral of> sin(Pye/2)* e^(-x)?> Hmmm, integrating sin(Pye/2)*e^(-x) with respect to e seems pretty hard.Seems like it would result in gamma functions of sorts. === =This problem is rather long, so please bear with me as I try toexplain it. I would really appreciate any help on this. It's not fora class, I'm just trying to work all of the problems in a book. For two sets X and Y, assume that there exist two injective functionsf and g such that f:X->Y and g:Y->X. Note that the inverse functionsf^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary xin X, de?e the chain C_x as below: C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... Note that the number of elements to the left of x in the chain may bezero, ?ite, or in?ite, depending on x. Also, note that any twochains are either completely disjoint or identical. De?e the sets Aand B as: A = {C_x : (C_x intersection Y) is a subset of f(X)} B = {C_x : C_x is not a member of A}Exercise (2 parts):---i) Prove that for all C_x in B, there exists y in Y, such that y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),...ii) Let X_1 = A intersection X, X_2 = B intersection X, Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that f:X_1->Y_1 is surjective and g:Y_2->X_2 is surjectiveUse this to prove that there exists a bijective function h:X->Y, andhence X~Y. === = >functions f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective.This is called the Cantor-Bernstein's theorem which is quite a bit to biteoff for as you see, it's famous enuf to be named by those who ?st provedit. You could ?d where this was discussed at sci.math by doing anadvanced google group search. You may also ?d more about CB's theoremgoogling the web. >For arbitrary x in X, de?e the chain C_x as below: > C_x = >...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),.. .Huh? Don't understand, not liking this approach. >Note that the number of elements to the left of x in the chain may >be zero, ?ite, or in?ite, depending on x. Also, note that any >two chains are either completely disjoint or identical. De?e the >sets A and B as: A = {C_x : (C_x intersection Y) is a subset of f(X)} B = {C_x : C_x is not a member of A}Here's from my notes upon the topic.Have you see the notation: g(X) = { g(x) | x in X } ?injections f:A -> B, g:B -> A = A,B equinumerous let p:P(A) -> P(A), X -> X - g(B - f(X)) p ascending function complete lattice; some Z with p(Z) = ZIndeed p(X) = X - g(B - f(X)) has a ?ed point or set.?st prove p is ascending, ie X subset Y = p(X) subset p(Y)Now use theorem, an ascending function over a complete lattice has a ?ed point.To prove that for the complete lattice of sets, show Z = /{ X | X subset p(X) }is ?ed point of p, ie p(Z) = Z where / is great union of all the sodescribed X's. Notice the collection of sets isn't empty as A subset p(A)and the exercise isn't dependent upon p other than being ascending overP(A).As the rest is just notes, you may want some clari?ation or details. Z = Z - g(B - f(Z)); A-Z = g(B - f(Z)) let h(x) = f(x) if x in Z, = g^-1(x) if x in A-Z h surjection. h(Z) = f(Z); h(A-Z) = g^-1(A-Z) = B - f(Z) h injection. If h(x) = h(y): x in Z, y in A-Z not possible if x,y in Z; f(x) = f(y); x = y if x,y in A-Z: g^-1(x) = g^-1(y); x = yCorollarysurjections f:A->B, g:B->A = A,B equinumerous g':B -> A, x -> choose f^-1(x) is injectionChoose is axiom of choice function that takes one element out off^-1(x) = { y | y = f(x) }. g' is just a hint, to use CB thm.You'll want a f' also.Cantor's diagonal theoremsno surjection f:S -> P(S). Otherwise let A = { x in S | x not in f(x) } some a in S with f(a) = A; a in A iff a not in f(a) iff a not in Asurjection f:S -> A^S = |A| <= 1. Otherwise: let g:S -> A, s -> choose { a in A | a /= f(s)(s) } some s in S with f(s) = g; f(s)(s) = g(s) /= f(s)(s) >Exercise (2 parts): >i) Prove that for all C_x in B, there exists y in Y, such that > y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),... >ii) Let X_1 = A intersection X, X_2 = B intersection X, >Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that > f:X_1->Y_1 is surjective > g:Y_2->X_2 is surjective >Use this to prove that there exists a bijective function h:X->Y, and >hence X~Y.Ah er, no thanks. Besides I've already done that.Perhaps you'll ?d the ?ed point method more pleasant.---- === This problem is rather long, so please bear with me as I try to>explain it. I would really appreciate any help on this. It's not for>a class, I'm just trying to work all of the problems in a book.>For two sets X and Y, assume that there exist two injective functions>f and g such that f:X->Y and g:Y->X. Note that the inverse functions>f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary x>in X, de?e the chain C_x as below:>[*] C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... >Note that the number of elements to the left of x in the chain may be>zero, ?ite, or in?ite, depending on x. Also, note that any two>chains are either completely disjoint or identical. De?e the sets A>and B as:> A = {C_x : (C_x intersection Y) is a subset of f(X)}> B = {C_x : C_x is not a member of A}>Exercise (2 parts):>--->i) Prove that for all C_x in B, there exists y in Y, such that> y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),...Start with a C_x in B. For convenience let Y_x be theintersection of C_x with Y; by de?ition of B, Y_x is not asubset of f(X), so there must be some y in Y_x f(X). Thismeans that y is not f(x) for any x in X, i.e., that f^(-1)(y) isempty. On the other hand, y is in C_x. If you look at thede?ition of C_x, you'll see that this is possible only if y isthe ??st' element of C_x in the order listed at [*] above,i.e., that C_x is precisely {y,g(y),f(g(y)),g(f(g(y))),...}.>ii) Let X_1 = A intersection X, X_2 = B intersection X, Technically this is nonsense: elements of A are sets of points,while elements of X are points, so that A intersect X is empty.What's meant is that X_1 should be {z in X : z in C_x for someC_x in A}, i.e., the intersection of X with the union of thecollection A. The de?itions of X_2, Y_1, and Y_2 are marred bythe same sort of sloppiness.>Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that> f:X_1->Y_1 is surjectiveThis is pretty straightforward. Start with an x in X_1 and showthat f(x) must be in Y_1; this is trivial, since if x is in someC_z, then f(x) is in the same C_z. Then start with a y in Y_1and show that it's f(x) for some x in X_1; this is prettyimmediate from the de?ition of A.> and> g:Y_2->X_2 is surjectiveStart with y in Y_2 and show that g(y) is in X_2; then start withx in X_2 and show that it's g(y) for some y in Y_2, for whichyou'll probably want part (i).>Use this to prove that there exists a bijective function h:X->Y, and>hence X~Y.Paste together f | X_1 and g^(-1) | X_2.Brian === This problem is rather long, so please bear with me as I try to> explain it. I would really appreciate any help on this. It's not for> a class, I'm just trying to work all of the problems in a book.> For two sets X and Y, assume that there exist two injective functions> f and g such that f:X->Y and g:Y->X. Note that the inverse functions> f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective. For arbitrary x> in X, de?e the chain C_x as below:> C_x = ...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),... > Note that the number of elements to the left of x in the chain may be> zero, ?ite, or in?ite, depending on x. Also, note that any two> chains are either completely disjoint or identical. De?e the sets A> and B as:> A = {C_x : (C_x intersection Y) is a subset of f(X)}> B = {C_x : C_x is not a member of A}> Exercise (2 parts):> ---> i) Prove that for all C_x in B, there exists y in Y, such that> y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),...> Is that question entirely correct? Let f(X) not be empty, and let C_xbe any chain in B; then there exists y in Y such that y in f(X); andso the statement y not in f(X) --> C_x = y, g(y),... is triviallytrue for that y, since F-->T is always a true statement.Maybe instead of --> they meant and? === =Corrections ensue.>This problem is rather long, so please bear with me as I try to>explain it. I would really appreciate any help on this. It's not>for a class, I'm just trying to work all of the problems in a book.> Ha, as a self learner I tried the same.> However you've the web while I didn't.>For two sets X and Y, assume that there exist two injective>functions f and g such that f:X->Y and g:Y->X. Note that the inverse>functions f^{-1}:f(X)->X and g^{-1}:g(Y)->Y are both bijective.> This is called the Cantor-Bernstein's theorem which is quite a bit to bite> off for as you see, it's famous enuf to be named by those who ?st proved> it. You could ?d where this was discussed at sci.math by doing an> advanced google group search. You may also ?d more about CB's theorem> googling the web.>For arbitrary x in X, de?e the chain C_x as below:> C_x >...,f^{-1}(g^{-1}(x)),g^{-1}(x),x,f(x),g(f(x)),f(g(f(x))),.. .> Huh? Don't understand, not liking this approach.>Note that the number of elements to the left of x in the chain may>be zero, ?ite, or in?ite, depending on x. Also, note that any>two chains are either completely disjoint or identical. De?e the>sets A and B as:> A = {C_x : (C_x intersection Y) is a subset of f(X)}> B = {C_x : C_x is not a member of A}> Here's from my notes upon the topic.> Have you see the notation: g(X) = { g(x) | x in X } ?> injections f:A -> B, g:B -> A = A,B equinumerous> let p:P(A) -> P(A), X -> X - g(B - f(X))> p ascending function complete lattice; some Z with p(Z) = Z>P is the power set of A and p is a function from P(A) to P(A).> Indeed p(X) = X - g(B - f(X)) has a ?ed point or set.> ?st prove p is ascending, ie> X subset Y = p(X) subset p(Y)> Now use theorem,> an ascending function over a complete lattice has a ?ed point.> To prove that for the complete lattice of sets, show> Z = /{ X | X subset p(X) }> is ?ed point of p, ie p(Z) = Z where / is great union of all the so> described X's. Notice the collection of sets isn't empty as A subset p(A)> and the exercise isn't dependent upon p other than being ascending over> P(A).> As the rest is just notes, you may want some clari?ation or details.> Z = Z - g(B - f(Z)); A-Z = g(B - f(Z))> let h(x) = f(x) if x in Z, = g^-1(x) if x in A-Z> h surjection. h(Z) = f(Z); h(A-Z) = g^-1(A-Z) = B - f(Z)> h injection. If h(x) = h(y): x in Z, y in A-Z not possible> if x,y in Z; f(x) = f(y); x = y> if x,y in A-Z: g^-1(x) = g^-1(y); x = y> Corollary> surjections f:A->B, g:B->A = A,B equinumerous> g':B -> A, x -> choose f^-1(x) is injection> Choose is axiom of choice function that takes one element out of> f^-1(x) = { y | y = f(x) }. g' is just a hint, to use CB thm.> You'll want a f' also.>f^-1(x) = { y | x = f(y) } perhaps better statedf^-1(y) = { x | y = f(x) }> Cantor's diagonal theorems> no surjection f:S -> P(S). Otherwise let A = { x in S | x not in f(x) }> some a in S with f(a) = A; a in A iff a not in f(a) iff a not in A> surjection f:S -> A^S = |A| <= 1. Otherwise:> let g:S -> A, s -> choose { a in A | a /= f(s)(s) }> some s in S with f(s) = g; f(s)(s) = g(s) /= f(s)(s)> >Exercise (2 parts):>i) Prove that for all C_x in B, there exists y in Y, such that> y not in f(X) --> C_x = y,g(y),f(g(y)),g(f(g(y))),...>ii) Let X_1 = A intersection X, X_2 = B intersection X,>Y_1 = A intersection Y, and Y_2 = B intersection Y. Show that> f:X_1->Y_1 is surjective> g:Y_2->X_2 is surjective>Use this to prove that there exists a bijective function h:X->Y, and>hence X~Y.> Ah er, no thanks. Besides I've already done that.> Perhaps you'll ?d the ?ed point method more pleasant.> ---- === =http://tinyurl.com/fuf82000 quote : she looks exactly like Laurie Holden maybe I could see her againhttp://tinyurl.com/fuf22000 quote : treat my loungeroom like a hollywood basement government has spiedon me so longI.E. The truman show is based on me1998 Truman released starring Jim Carrey, man who is on camera all his life seeking a lady2002 Majestic released starring Jim Carrey and Laurie Holden, a romance. TRUMAN 2no comment ?its a simple question in propositional logic, circumstantial evidence I'm the Truman or not!Its not PROOF I know, but answer this question:If you were informed there is a real Truman, then do these 2 URLS identify him accurately?Herc-- __ / / / / / / / / / / / /__/__ /________ ___________/ www.A1Sites.com === no comment ?Yeah. You need professional help.Oh and -- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion___________________________________________________ _____________...peace increases our peril by making discipline less urgent, encouraging some of our worst instincts, in depriving us of some of our best leaders.- Michael Ledeen, neoconservative leader & full time foreign policy adviser to Karl Rove === =http://tinyurl.com/fuf8 she looks exactly like Laurie Holdenhttp://tinyurl.com/fuf2 government has spied on me so longYou're in a closet or your not that bright.Why use the actress I pinpointed as the trumans affection? AND call it TRUEman?dozen people yell out to me every day ?I saw the truman, I played my part in the truman show'media is a big f*cking lier to all youHerc === =pushed hard, and farted out the following message in> http://tinyurl.com/fuf8> 2000 quote : she looks exactly like Laurie Holden maybe I could see> her again > http://tinyurl.com/fuf2> 2000 quote : treat my loungeroom like a hollywood basement> government has spied on me so long> I.E. The truman show is based on me> 1998 Truman released starring Jim Carrey, man who is on camera all his> life seeking a lady 2002 Majestic released starring Jim Carrey and> Laurie Holden, a romance. TRUMAN 2 > no comment ?> its a simple question in propositional logic, circumstantial evidence> I'm the Truman or not! > Its not PROOF I know, but answer this question:> If you were informed there is a real Truman, then do these 2 URLS> identify him accurately? Has anyone ever told you you're a bit delusional?-- Mekkala, Atheist #2148When did I realize I was God? Well, I was praying and I suddenly realized I was talking to myself!--Peter O'Toole. === =|-|erc warmed at our ?e and told this tale:>http://tinyurl.com/fuf8 she looks exactly like Laurie Holden>http://tinyurl.com/fuf2 government has spied on me so long>You're in a closet or your not that bright.>Why use the actress I pinpointed as the trumans affection? AND call it TRUEman?>dozen people yell out to me every day ?I saw the truman, I played my part in the truman show'>media is a big f*cking lier to all youYou exhibit classical signs of schizophrenia. Seek professional help.--Douglas Berry gridlore@mindspring.comhttp:// gridlore.home.mindspring.comAtheist #2147, Atheist Vet #5Ezekiel 13:20 Wherefore thus saith the Lord GOD; Behold, I am against your pillows === pushed hard, and farted out the following message in> http://tinyurl.com/fuf8> 2000 quote : she looks exactly like Laurie Holden maybe I could see> her again> http://tinyurl.com/fuf2> 2000 quote : treat my loungeroom like a hollywood basement> government has spied on me so long> I.E. The truman show is based on me> 1998 Truman released starring Jim Carrey, man who is on camera all his> life seeking a lady 2002 Majestic released starring Jim Carrey and> Laurie Holden, a romance. TRUMAN 2> no comment ?> its a simple question in propositional logic, circumstantial evidence> I'm the Truman or not!> Its not PROOF I know, but answer this question:> If you were informed there is a real Truman, then do these 2 URLS> identify him accurately?> Has anyone ever told you you're a bit delusional?Yes, but they were all secret government agents hired by the global mediaconglomerate as part of the plot to a new movie about to be released,starring James Dean and Katherine Hepburn. I bet you can't ?d anyone on*any* of the newsgroups Herc posts to, who is from Gilboa, NY, and who willcategorically deny that I am the one, true, Luke Skywalker. They all knowit. Proof by absence of denial.With apologies to The Simpsons/Leonard Nemoy:Sure he's delusional. But they're amusing delusions, and isn't that the*real* reality? The answer is no. === =http://tinyurl.com/fuf8 she looks exactly like Laurie Holdenhttp://tinyurl.com/fuf2 government has spied on me so longjust check the urls before you all post up your ignorance.Nic Cage - nic prison cage prison face off prison move conair prison moviekylie minogue - mini god techno midgit divathe truman show, True Showi dont' think any of you read the urls, any further comments have to bean explanation of the url content ?st please.In 2000, veri?d by google, I show that Jim Carrey and Laurie Holdenare associated through a real version of the ?truman show'I predict their association before it went public in year 2002 in Majestic.on topic comments appreciated.100,000 people there all know its true. I've even had 4 post so in aus.tv.little bit of detective work and you'll be suprised, maybe be the beginningof the end of my cage.Herc === no comment ?>its a simple question in propositional logic, circumstantial evidence I'm the Truman or not!You're mentally ill, dude. Please get help. === >no comment ?>its a simple question in propositional logic, circumstantial evidence I'm the Truman or not!> You're mentally ill, dude. Please get help.1/ stop verbally abusing me, show your medical quali?ations before you label me.2/ as I asked indicate you have checked the URLs ?st before replying3/ its a proposition type question, your answer was not in the form [True / False]Circumstantial evidence I'm the Truman or not?http://tinyurl.com/fuf8http://tinyurl.com/ fuf2TrueFalselets have a vote, just tick one.Herc === >no comment ?>its a simple question in propositional logic, circumstantial evidence I'm the Truman or not!> You're mentally ill, dude. Please get help.>1/ stop verbally abusing me, show your medical quali?ations before you label me.No one is verbally abusing you. I have simply pointed out, as doeseveryone else who bothers responding to you, that your posts screamout classic schizophrenia. Get help.>2/ as I asked indicate you have checked the URLs ?st before replyingYes. They don't help your cause. Get help.>3/ its a proposition type question, your answer was not in the form [True / False]>Circumstantial evidence I'm the Truman or not?>http://tinyurl.com/fuf8>http://tinyurl.com/fuf2>True> False>lets have a vote, just tick one.OK, I opt for schizophrenia is treatable, seek competent professionalhelp. Perhaps not what you wanted to hear, but what you *need* tohear. Get help. === =|-|erc warmed at our ?e and told this tale:>no comment ?>its a simple question in propositional logic, circumstantial evidence I'm the Truman or not!> You're mentally ill, dude. Please get help.>1/ stop verbally abusing me, show your medical quali?ations before you label me.I dated a schizophrenic for years. when she got off her meds, shestarted sounding like you. Seeki medical help. You can live a normallife.We are not abusing you. You are making ridiculous statments.>2/ as I asked indicate you have checked the URLs ?st before replyingIf read some of it. You ramble and make no sense at all.>3/ its a proposition type question, your answer was not in the form [True / False]>Circumstantial evidence I'm the Truman or not?>http://tinyurl.com/fuf8>http://tinyurl.com/fuf2>True> FalseFalse.--Douglas Berry gridlore@mindspring.comhttp:// gridlore.home.mindspring.comAtheist #2147, Atheist Vet #5Ezekiel 13:20 Wherefore thus saith the Lord GOD; Behold, I am against your pillows === =You need to get out more.> 1998 Truman released starring Jim Carrey, man who is on camera all his> life seeking a lady 2002 Majestic released starring Jim Carrey and> Laurie Holden, a romance. TRUMAN 2 > Its not PROOF I know, but answer this question:> If you were informed there is a real Truman, then do these 2 URLS> identify him accurately? > http://tinyurl.com/fuf8> 2000 quote : she looks exactly like Laurie Holden maybe I could see> her again > http://tinyurl.com/fuf2> 2000 quote : treat my loungeroom like a hollywood basement> government has spied on me so long> I.E. The truman show is based on me === > If complex numbers are allowed, you can use smaller matrices.> > I get it. About using matrices over complex numbers, does this give us> any extra feasibility? [I know one, you have all the solutions of the> characteristic polynomial within the ?ld]Look up Pauli Matrices. === =Say I have two sets A={a,b,c,d} and B={x,y,z}I need to ?d all combinations of the items in both sets, given thefollowing constraints:1) You must select at least one element from each set.2) The ?st item in the combination (or permutation?) must be fromset A.So examples include:a,xa,b,c,d,x,y,zb,x,yb,c,y,zd,x,zHow do I enumerate this in general??? === =2^4=number of objects in the power set of A (including the empty set)2^3=number of objects in the power set of B (including the empty set)(2^4-1)*(2^3-1)=(16-1)*(8-1)=15*7=105 is the result that you are lookingfor.> Say I have two sets A={a,b,c,d} and B={x,y,z}> I need to ?d all combinations of the items in both sets, given the> following constraints:> 1) You must select at least one element from each set.> 2) The ?st item in the combination (or permutation?) must be from> set A.> So examples include:> a,x> a,b,c,d,x,y,z> b,x,y> b,c,y,z> d,x,z> How do I enumerate this in general???> === I need to ?d all combinations of the items in both sets, given the> following constraints:> 1) You must select at least one element from each set.> 2) The ?st item in the combination (or permutation?) must be from> set A.I'm not sure about my answer.I suppose that one does not consider the order of the elements in thecombination (for example, the combination a,b,x,y would be the same thatb,a,y,x, where A = {a,b,c,d}, B = {x,y,z} and a,b,c,d are distinct elementsand x,y,z are distinct elements). But your point 2) refers a ?st item...Suppose that we don't consider the order. Also, lets suppose that we can'trepeat elements (for example, we can't have the combination a,a,b,x,y).The combination required can be build this way: ?st, we choose acombination of elements of A, then we choose a combination of elements of Band ?ally we add the combination together.For example, we choose the combination a,b from A, then we choose thecombination x,y from B and ?ally we add them together and get a,b,x,y.So, if we have n combinations of elements of A, and m combinations ofelements of B, then for each combination of elements of A, we can add mcombinations of elements of B. Therefore, we have nm ways of combinations ofelements in both sets.We only need to know how do compute n and m. We use this: if a set S as kelements, then the number of combinations (with at least one element) ofelements of S is 2^k - 1.For example, if A = {a,b} as 2 elements and B = {x,y,z} as 3 elements, thenn = 2^2 - 1 = 3 and m = 2^3 - 1 = 7, therefore the required number ofcombinations is mn = 21. They are: 1. a,x 2. a,y 3. a,z 4. a,x,y 5. a,x,z 6. a,y,z 7. a,x,y,z 8. b,x 9. b,y 10. b,z 11. b,x,y 12. b,x,z 13. b,y,z 14. b,x,y,z 15. a,b,x 16. a,b,y 17. a,b,z 18. a,b,x,y 19. a,b,x,z 20. a,b,y,z 21. a,b,x,y,zSorry my english. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com === 2^4=number of objects in the power set of A (including the empty set)> 2^3=number of objects in the power set of B (including the empty set)> (2^4-1)*(2^3-1)=(16-1)*(8-1)=15*7=105 is the result that you are looking> for.This doesn't deal with the implied ordering.>Say I have two sets A={a,b,c,d} and B={x,y,z}>I need to ?d all combinations of the items in both sets, given the>following constraints:>1) You must select at least one element from each set.>2) The ?st item in the combination (or permutation?) must be from>set A.>So examples include:>a,x>a,b,c,d,x,y,z>b,x,y>b,c,y,z>d,x,z>How do I enumerate this in general???>To count them do the following:Choices for 1st element: 4You can now think of the remaining two sets as one large set and ?d out how many orderings of those there are looking at all possible subsets of (A-{?st item}) union B.Since you must include B, some of the above will EXCLUDE all of B. That's the orderings of subsets of (A-{?st item}).Now, assemble the above results appropriately.To enumerate them??? I'd write a program.-- Will Twentyman === =I don't seem to be able to compose the appropriate search engine queryto ?d comparisons (summary tables) of computability concepts.Basically, what I would like to ?d is a table that would list thecorresponding levels such aselementary / regular language / deterministic ?iteautomatonprimitive recursive / context-free language / pushdown automatonrecursive / unrestricted language / Turing machineFor example, how to ? the rules of logical reasoning to theselevels? Or lambda calculus? Also, does the coupling of levels occurelsewhere than with generative grammars and their accepting automata? === =There is an oddity in the way complex numbers are introduced. Although acomplex number and its conjugate behave exactly the same, one half of theimaginary axis is thought of as positive and the other is thought of asnegative. -1 has two square roots, which are conjugates of each other, butonly one of them is given its own symbol. Why pretending there is asymmetrywhen there is none?If we accept that there is complete symmetry, then we are faced with astrange consequence: we cannot name any single non-real complex number. Wecan easily name real numbers -- -16, 21, pi, to take some examples -- butwith complex numbers we can only name pairs of numbers, such as the pairthat satis?s the equation x^2+1=0.One could go on and argue that it makes no more sence to talk about /thecomplex plane/ than to talk about /the plane/ (as if only one planeexisted).Comments?Mattias === > There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. No, they are not thought of as positive and negative--at least theyshouldn't be.GC> -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there is asymmetry> when there is none?> If we accept that there is complete symmetry, then we are faced with a> strange consequence: we cannot name any single non-real complex number. We> can easily name real numbers -- -16, 21, pi, to take some examples -- but> with complex numbers we can only name pairs of numbers, such as the pair> that satis?s the equation x^2+1=0.> One could go on and argue that it makes no more sence to talk about /the> complex plane/ than to talk about /the plane/ (as if only one plane> existed).> Comments?> Mattias === There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there isasymmetry> when there is none?> [snip]> with complex numbers we can only name pairs of numbers, such as the pair> that satis?s the equation x^2+1=0.When real equations such as x^2+1=0 have complex solutions, they always havesuch solutions as conjugate pairs. This does not apply if the equationsinclude complex coef?ients.Example: x^2 - i = 0 has two roots (like any quadratic), namely thesquare roots of i: x1 = 1/sqr2 + i/sqr2 x2 = -1/sqr2 - i/sqr2These are negatives of each other but not conjugates.For a more complex example, try x^2 - (1+i)x + i = 0x^2 - (1+i)x + i = 0b^2 = (1+i)^2 = 2ib^2-4ac = -2isqr(b^2-4ac) = 1-ix1 = [1+i + (1-i)]/2 = 1x2 = [1+i - (1-i)]/2 = iNot conjugates, not even negatives. === =Mattias-> There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there isasymmetry> when there is none?Well, it's improper to call the half of the axis positive and the othernegative.You still bring up an elegant symmetry: If you replace i by -i everywhere,complex algebra is still consistent. Thus, the choice of which square rootof -1 you denote as i and which you call -i is in that sense arbitrary. Butas soon as you're working with more than one complex number, you just needto make sure that your choice for all the complex nubmers you're using usethe same choice for which root you call i.For example,(1+2i)(3+4i)=-5+10i(1+2(-i))(3+4(-i))=-5-10iSo, the mapping i->-i preserves multiplication (and addition, etc.). Butthat's only true because our choice of i was consistent between the twomultiplied numbers. So which you pick doesn't really matter; just as longas you stick with your choice.Travis === Message-id: There is an oddity in the way complex numbers are introduced. Although a>complex number and its conjugate behave exactly the same, one half of the>imaginary axis is thought of as positive and the other is thought of as>negative.And one half of the real axis is thought of as positive and the other half isthought of as negative.> -1 has two square roots, which are conjugates of each other, but>only one of them is given its own symbol. The symbol refers to the axis.>Why pretending there is asymmetry>when there is none?What asymmetry are you refering to? negative realpositive realnegative imaginarypositive imaginaryLooks symmetrical to me.>If we accept that there is complete symmetry, then we are faced with a>strange consequence: we cannot name any single non-real complex number.Who's we? You got a turd in your pocket? Just because you can't ?ure itout, doesn't mean nobody can.> We can easily name real numbers -- -16, 21, pi, to take some examplesIn the context of the complex plane, there are no real numbers. The examplesyou gave are complex numbers whose imaginary component is 0. In other words-16 is actually -16 + 0*i21 is actually 21 + 0*ipi is actually pi + 0*iAnd pi*i would be a non-real number. And in the context of the complex plane,it would be a complex number whose real part is 0, i.e. 0 + pi*i> -- but with complex numbers we can only name pairs of numbers, such as thepair>that satis?s the equation x^2+1=0.They come in pairs because that is a 2nd degree polynomial. Third degreepolynomials have triplets that satisfy the equation.>One could go on and argue that it makes no more sence to talk about /the>complex plane/ than to talk about /the plane/ (as if only one plane>existed).And one would be wrong when so arguing.>Comments?When you don't understand something, the problem is most likely with you andnot the system. Don't assume everything is wrong just because you can'tunderstand it. Go ahead and ask questions, that's what the newsgroup is herefor. But don't listen to James S. Harris, there are no fundamental errors beingtaught in math. When your understanding seems at odds with what is taught, askfor help in ?uring out why your understanding is wrong.>Mattias--Mensanator2 of Clubs http://members.aol.com/mensanator666/2ofclubs/2ofclubs.htm === There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there is asymmetry> when there is none?> We are not pretending there is asymmetry. What we are doing is arbitarily picking one of the roots of -1 and calling it i (or j). Then the other one has to be -i. But there is no asymetry here, because we don't know which of the two roots of -1 we called i.-- Stephen Montgomery-Smithstephen@math.missouri.eduhttp:// www.math.missouri.edu/~stephen === Mattias-> There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half ofthe> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other,but> only one of them is given its own symbol. Why pretending there is> asymmetry> when there is none?> Well, it's improper to call the half of the axis positive and the other> negative.> You still bring up an elegant symmetry: If you replace i by -ieverywhere,> complex algebra is still consistent. Thus, the choice of which squareroot> of -1 you denote as i and which you call -i is in that sense arbitrary.But> as soon as you're working with more than one complex number, you just need> to make sure that your choice for all the complex nubmers you're using use> the same choice for which root you call i.> For example,> (1+2i)(3+4i)=-5+10i> (1+2(-i))(3+4(-i))=-5-10i> So, the mapping i->-i preserves multiplication (and addition, etc.). But> that's only true because our choice of i was consistent between the two> multiplied numbers. So which you pick doesn't really matter; just as long> as you stick with your choice.> Travis>The need to remain consistent in choice of i versus -i is brought out incomplex analysis also. Suppose that f:C-->C is entire. One might thinkthat f:C-->C-->C would also be entire where the second map is complexconjugation, but that is not the case. Similiarly C-->C-->C wherethe ?st map is complex conjugation and the second is given by f is alsonot entire. (The Cauchy Riemann equations suf?e to see that thesemaps are generally not entire.) However C-->C-->C-->C where themiddle map is f and the other 2 are complex conjugation does give anentire function! This may be somewhat clearer if one makes a commutativesquare:C-->C| |/ /C-->CHorizontal arrows are f and vertical arrows are complex conjugation.Reiterating, all of this may be interpreted as saying that the choiceof i versus -i must be consistent. Thus I presume that an entire functionmust be viewed as mapping C to _itself_ rather than to some copy of C!Alternately, if one starts with two copies of C, then an entire functionfrom one copy to the other identi?s the two copies!So I vote Travis's explanation as adequate reason for giving the two squareroots of -1 distinct symbols. They are needed for consistency ofexposition.David C. Harris === There is an oddity in the way complex numbers are introduced....> -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretend there is asymmetry> when there is none?> If we accept that there is complete symmetry, then we are faced with a> strange consequence: we cannot name any single non-real complex number....> with complex numbers we can only name pairs of numbers, such as the pair> that satis?s the equation x^2+1=0....perhaps evaded the real issue. I agree with your main concern, and Ithink the answer lies in the connection between mathematics and ourphysical diagrams. As you suggest, complex conjugation is an automorphism of the complex?ld, leaving invariant every real number, but swapping i with -i. That means that most algebraic statements we make about complex numbersremain true if i and -i are transposed, so how do we know which iswhich? The mathematics assures us that the equation z^2 + 1 = 0 has twosolutions, but it treats them even-handedly, giving us no way to stick aspecial label on one of them rather than the other. The down-to-earth everyday answer is that our diagrams represent thereal line horizontally with positive numbers to the right, and theimaginary axis perpendicular to it with i above and -i below. Thatdepends upon our notions of right and left (or clockwise andanti-clockwise), which we acquired as small children by being shownsomething *physical*. You can't de?e right or left mathematically. Amathematical argument (perhaps about real vector spaces) can tell you thatthere are two sides, but it treats them even-handedly, giving us no way tostick a special label on one of them rather than the other. I've repeatedthose words, to suggest that your question about i versus -i is prettyclose to the more basic question of right versus left. If you're doing really pure mathematics, then calling certain complexnumbers i and -i isn't going to cause any trouble. You never *need*to decide which is which. But if you're a normal human being who uses aphysical plane diagram to help in thinking about complex numbers, then youput 1 on the right, -1 on the left, i above and -i below. It'sas humdrum as that. Ken Pledger. === There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there is asymmetry> when there is none?Complex numbers can be viewed as some special 2x2 matrices withreal entries.For more on that, you could take a look at the following Web page:http://www.sosmath.com/matrix/complex/complex.htmlDavid Bernier === =|There is an oddity in the way complex numbers are introduced. Although a|complex number and its conjugate behave exactly the same, one half of the|imaginary axis is thought of as positive and the other is thought of as|negative. -1 has two square roots, which are conjugates of each other, but|only one of them is given its own symbol. Why pretending there is asymmetry|when there is none?This is an issue that I've seen raised in philosopher's discussionsof the concept of mathematical structuralism, which is the ideathat the identity of an element of a structure (such as the complexnumbers) depends only on it's position in the structure, which isgiven only up to isomorphism.A number of more-or-less correct answers have been given, but I wantto put it a little differently.The exact same behavior you describe is for certain structuresde?ed for the complex numbers. There are various layers of structurewe could mention. The complex numbers are a topological space, forexample. They're a metric space. There's an af?e structure on them.They're a ?ld.But why stop there? I would say that the complex numbers are morethan just the complex numbers taken as a ?ld with a norm on it(which has the symmetry under complex conjugation you mention).They also have a real part function and an imaginary partfunction de?ed on them. This allows us to identify them withpoints (x,y) on the plane. With this additional structure, complexconjugation is not a complete symmetry.Considered as a topological space or as a metric space, the complexnumbers have symmetries that take any point to any other point. Wewouldn't say that because of that, there's a problem with consideringthem distinguished from each other because of other properties. Thesame thing goes when going from the complex numbers as a ?ld witha metric to the complex numbers as the complex numbers. In a sensewhat seems funny is that it's such a small additional bit of structure,but obviously we want to have it (part of the time).Euclidean space has symmetries that move any point to any other point,but it's very popular to consider Eulidean space after a choice ofcoordinate system has been made, i.e., as R^2 or R^3. One thing thatmight seem different there is that usually the way one shows thatEulidean space exists is by showing that R^2 or R^3 with the rightmetric on it is an example. (We wouldn't have to do it like that,of course.) But the complex numbers are the same way. If you considerthe proof that the real numbers have an algebraic closure, hiddeninside is the fact that R[x]/(x^2+1) is such an algebraic closure.This construction gives us an obvious basis {1,x}. So as we constructthe complex numbers, we've already constructed them as elements of theform a+bx where x^2 is identi?d with -1. That's already got thisextra structure that we get by picking one of the square roots of -1to be i (except it gets called something else, like x). So in bothexamples, the construction showing that the more symmetrical structureexists goes by ?st showing that the more structured object exists,and then removing the structures that we may not care about.Keith Ramsay === There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative.Its true to say that the the conjugate is equal to the negative, only forcomplex numbers of the form (0+bi) or (0, b), where b is a real number.i.e. conj(0+bi) = (0-bi), and neg(0+bi) = (0-bi)But, conj(a+bi)=neg(a+bi) fails when a is not equal to 0.e.g. conj(a+bi)=a-bi and neg(a+bi)=-a-bi.> -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there isasymmetry> when there is none?'The' square root of x, only has sense in the domains of: positive realnumbers, positive integers, natural numbers.eg. sqrt(16)=4.'The' square root of x does not exist in R or C or H etc. It is not unique,ie, we can talk about ?a' square root of x but we cannot talk about ?the'square root unless we specify ?the positive' square root, or ?the negative'square root.In the domain of complex numbers:The positive square root of (-1) exists and is equal to +i.The negative square root of (-1) exists and is equal to -i.> If we accept that there is complete symmetry, then we are faced with a> strange consequence: we cannot name any single non-real complex number.There are just as many non-real complex numbers as there are real numbers.(bi) is a non-real complex number for all real values of b.> We> can easily name real numbers -- -16, 21, pi, to take some examples -- but> with complex numbers we can only name pairs of numbers, such as the pair> that satis?s the equation x^2+1=0.Not so. the number pair that satis?s x^2+1=0 is (+i, -i), but, that pairdoes not express a complex number of the form (a, b) or (a+bi).--- (16)i, (21)i, (pi)i, are examples of non-real complex numbers.Owen> One could go on and argue that it makes no more sence to talk about /the> complex plane/ than to talk about /the plane/ (as if only one plane> existed).> Comments?> Mattias> === > There is an oddity in the way complex numbers are introduced. Although a> complex number and its conjugate behave exactly the same, one half of the> imaginary axis is thought of as positive and the other is thought of as> negative. -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretending there is asymmetry> when there is none?>Complex numbers can be viewed as some special 2x2 matrices with>real entries.A fascinating and sometimes useful fact, but I don't see how ithas anything to do with the question. Among those matrices thereare two that have square equal to -I; why do we decide one of themis i and the other is -i?>For more on that, you could take a look at the following Web page:>http://www.sosmath.com/matrix/complex/complex.html> David BernierDavid C. Ullrich === There is an oddity in the way complex numbers are introduced. Although a>complex number and its conjugate behave exactly the same, one half of the>imaginary axis is thought of as positive and the other is thought of as>negative. -1 has two square roots, which are conjugates of each other, but>only one of them is given its own symbol. Why pretending there is asymmetry>when there is none?Many replies seem to have missed the point.What a few people have said is what's correct: The symmetry exists,and there is _no_ reason why one of those complex numbers shouldbe called i and the other one -i. We just picked one and called it i.(And no, there's no way to say _which_ of the two is the one wedecided to call i.)>If we accept that there is complete symmetry, then we are faced with a>strange consequence: we cannot name any single non-real complex number. We>can easily name real numbers -- -16, 21, pi, to take some examples -- but>with complex numbers we can only name pairs of numbers, such as the pair>that satis?s the equation x^2+1=0.>One could go on and argue that it makes no more sence to talk about /the>complex plane/ than to talk about /the plane/ (as if only one plane>existed).>Comments?>Mattias> David C. Ullrich === =bQ.10472@news01.bloor.is.net.cable.rogers.com...[ snip]> Its true to say that the the conjugate is equal to the negative, only for> complex numbers of the form (0+bi) or (0, b), where b is a real number.[snip]Apparently you de?e i as the ordered pair (0,1), and a + bi as the orderedpair (a, b). With this de?ition one can indeed name complex numbers, butthis is not how complex numbers arose historically. I am sure Cardano didnot de?e a complex number as an ordered pair. I think he must haveintroduced i as a number that satis?s the equation x^2+1=0. The problem isthat there are /two/ such numbers.Mattias === =$Mc.1023455@newsread1. prod.itd.earthlink.net...> Mattias-> There is an oddity in the way complex numbers are introduced. Althougha> > complex number and its conjugate behave exactly the same, one half of> the> imaginary axis is thought of as positive and the other is thought ofas> negative. -1 has two square roots, which are conjugates of each other,> but> > only one of them is given its own symbol. Why pretending there is> asymmetry> when there is none?> Well, it's improper to call the half of the axis positive and the other> negative.> You still bring up an elegant symmetry: If you replace i by -i> everywhere,> complex algebra is still consistent. Thus, the choice of which square> root> of -1 you denote as i and which you call -i is in that sense arbitrary.> But> as soon as you're working with more than one complex number, you justneed> to make sure that your choice for all the complex nubmers you're usinguse> the same choice for which root you call i.> For example,> (1+2i)(3+4i)=-5+10i> (1+2(-i))(3+4(-i))=-5-10i> So, the mapping i->-i preserves multiplication (and addition, etc.).But> that's only true because our choice of i was consistent between the two> multiplied numbers. So which you pick doesn't really matter; just aslong> as you stick with your choice.> Travis> The need to remain consistent in choice of i versus -i is brought out in> complex analysis also. Suppose that f:C-->C is entire. One might think> that f:C-->C-->C would also be entire where the second map is complex> conjugation, but that is not the case. Similiarly C-->C-->C where> the ?st map is complex conjugation and the second is given by f is also> not entire. (The Cauchy Riemann equations suf?e to see that these> maps are generally not entire.) However C-->C-->C-->C where the> middle map is f and the other 2 are complex conjugation does give an> entire function! This may be somewhat clearer if one makes a commutative> square:> C-->C> | |> / /> C-->C> Horizontal arrows are f and vertical arrows are complex conjugation.> Reiterating, all of this may be interpreted as saying that the choice> of i versus -i must be consistent. Thus I presume that an entire function> must be viewed as mapping C to _itself_ rather than to some copy of C!> Alternately, if one starts with two copies of C, then an entire function> from one copy to the other identi?s the two copies!This was helpful.Mattias === > There is an oddity in the way complex numbers are introduced....> -1 has two square roots, which are conjugates of each other, but> only one of them is given its own symbol. Why pretend there is asymmetry> when there is none?> If we accept that there is complete symmetry, then we are faced with a> strange consequence: we cannot name any single non-real complexnumber....> with complex numbers we can only name pairs of numbers, such as the pair> that satis?s the equation x^2+1=0....> perhaps evaded the real issue. I agree with your main concern, and I> think the answer lies in the connection between mathematics and our> physical diagrams.> As you suggest, complex conjugation is an automorphism of the complex> ?ld, leaving invariant every real number, but swapping i with -i.> That means that most algebraic statements we make about complex numbers> remain true if i and -i are transposed, so how do we know which is> which? The mathematics assures us that the equation z^2 + 1 = 0 has two> solutions, but it treats them even-handedly, giving us no way to stick a> special label on one of them rather than the other.Exactly!> The down-to-earth everyday answer is that our diagrams represent the> real line horizontally with positive numbers to the right, and the> imaginary axis perpendicular to it with i above and -i below. That> depends upon our notions of right and left (or clockwise and> anti-clockwise), which we acquired as small children by being shown> something *physical*. You can't de?e right or left mathematically. A> mathematical argument (perhaps about real vector spaces) can tell you that> there are two sides, but it treats them even-handedly, giving us no way to> stick a special label on one of them rather than the other. I've repeated> those words, to suggest that your question about i versus -i is pretty> close to the more basic question of right versus left.True.Mattias === |There is an oddity in the way complex numbers are introduced. Although a> |complex number and its conjugate behave exactly the same, one half of the> |imaginary axis is thought of as positive and the other is thought of as> |negative. -1 has two square roots, which are conjugates of each other,but> |only one of them is given its own symbol. Why pretending there isasymmetry> |when there is none?> This is an issue that I've seen raised in philosopher's discussions> of the concept of mathematical structuralism, which is the idea> that the identity of an element of a structure (such as the complex> numbers) depends only on it's position in the structure, which is> given only up to isomorphism.> A number of more-or-less correct answers have been given, but I want> to put it a little differently.> The exact same behavior you describe is for certain structures> de?ed for the complex numbers. There are various layers of structure> we could mention. The complex numbers are a topological space, for> example. They're a metric space. There's an af?e structure on them.> They're a ?ld.> But why stop there? I would say that the complex numbers are more> than just the complex numbers taken as a ?ld with a norm on it> (which has the symmetry under complex conjugation you mention).> They also have a real part function and an imaginary part> function de?ed on them. This allows us to identify them with> points (x,y) on the plane. With this additional structure, complex> conjugation is not a complete symmetry.This is a good answer to my question, but if complex numbers have this extrastructure, I do not see that complex numbers arise naturally in mathematics.> Considered as a topological space or as a metric space, the complex> numbers have symmetries that take any point to any other point. We> wouldn't say that because of that, there's a problem with considering> them distinguished from each other because of other properties. The> same thing goes when going from the complex numbers as a ?ld with> a metric to the complex numbers as the complex numbers. In a sense> what seems funny is that it's such a small additional bit of structure,> but obviously we want to have it (part of the time).Personally I would prefer not to have it.> Euclidean space has symmetries that move any point to any other point,> but it's very popular to consider Eulidean space after a choice of> coordinate system has been made, i.e., as R^2 or R^3.I do not like when people do this, because the extra structure makes it hardto see what one is really doing.Mattias === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> Its true to say that the the conjugate is equal to the negative, onlyfor> complex numbers of the form (0+bi) or (0, b), where b is a real number.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as theordered> pair (a, b). With this de?ition one can indeed name complex numbers, but> this is not how complex numbers arose historically. I am sure Cardano did> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. The problemis> that there are /two/ such numbers.There is no problem at all.+i=(the positive square root of -1).-i=(the negative square root of -1).Where do you ?d a problem with this concept?> Mattias> === >There is an oddity in the way complex numbers are introduced. Although a>complex number and its conjugate behave exactly the same, one half of the>imaginary axis is thought of as positive and the other is thought of as>negative. -1 has two square roots, which are conjugates of each other, but>only one of them is given its own symbol. Why pretending there is asymmetry>when there is none?>Complex numbers can be viewed as some special 2x2 matrices with>real entries.> A fascinating and sometimes useful fact, but I don't see how it> has anything to do with the question. Among those matrices there> are two that have square equal to -I; why do we decide one of them> is i and the other is -i?I read Keith Ramsay's reply. It got me thinking about the question: How did/would Bourbaki de?e the structure C used in complex analysis?[ I say structure because it's known that the ?ld C has about 2^continuum ?ld automorphisms. That's related to transcendence degrees, for example as in Thomas Hungerford's _Algebra_ . So I think a C ?ld automorphism need not preserve |z| i.e. if phi: C->C autom., |phi(z)| = |z| isn't guaranteed)What I've got so far in my attempt to de?e C in an intrinsic way(draft) is:(1) The elements of C are all R-linear maps from R^2 to R^2(2) for any mapping f: R^2 -> R^2, f is in C _only if_ = K* for some real K independent of u, v in R^2(3) [ I assume here that for an R-linear f: R^2->R^2, det(f) is intrinsic; anyway, it won't depend on the basis chosen for R^2 so I think det(f) is a basis-free concept] For any f in C, det(f)>=0 .(4) [de?ition of the norm of an f in C] |f| = sqrt(det(f)) .(5) [de?ition of multiplication] For f, g in C, f*g := f little_circle g , f o g.(6) [de?ition of addition] f + g is de?ed by: (f+g)(u) = f(u) + g(u), u in R^2(7) [C becomes a metric space] distance(f,g) = |f - g|(8) [add a topology ] C gets its natural topology because we've made C into a metric space(9) Obviously, C has a multiplicative identity, 1 := id: R^2 -> R^2, id(u) = u, any u in R^2(10) If all goes well, C should now be a ?ld..(11) What is R as embedded in C? The new R is the closure in C of the ?ld generated by 1, where 1:= id: R^2-> R^2 .(12) De?ing the conjugation map conjug: f|-> f^{bar} : conjug is the identity on new R conjug o conjug = id_C conjug =/= id_C conjug is a ?ld automorphism conjug is an isometry from C-> C(13) De?ing i: with all the structure so far, I don't see how to de?e i in an intrinsic way...(14) a musing: We need to give an orientation to R^2, hopefully, C will inherit an intrinsic orientation from the orientation given to R^2, and then , with the orientation, we could hopefully de?e arg:C{0} -> R, and then i is the complex number which has norm 1 and arg = pi/2 .David Bernier === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> > Its true to say that the the conjugate is equal to the negative, only> for> complex numbers of the form (0+bi) or (0, b), where b is a realnumber.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers,but> this is not how complex numbers arose historically. I am sure Cardanodid> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. Theproblem> is> that there are /two/ such numbers.> There is no problem at all.> +i=(the positive square root of -1).> -i=(the negative square root of -1).> Where do you ?d a problem with this concept?De?e one of the concepts on the following list, and I can de?e all theothers:1a) i1b) -i2a) Arg(z)2b) -Arg(z)3a) Im(z)3b) -Im(z)4a) PositiveSquareRoot(z)4b) NegativeSquareRoot(z)The problem is that I cannot de?e any of the concepts above unless one ofthem has already been de?ed (for example, above you show how to de?e iif positive square roots have been de?ed).(I can, however, de?e the following things:1) {i, -i}2) {Arg(z), -Arg(z)}3) {Im(z), -Im(z)}4) {PositiveSquareRoot(z), NegativeSquareRoot(z)})Mattias === =Correction:> 4a) PositiveSquareRoot(z)> 4b) NegativeSquareRoot(z)I meant:4a) PositiveSquareRoot(z)4b) Conjugate(PositiveSquareRoot(z))5a) NegativeSquareRoot(z)5b) Conjugate(NegativeSquareRoot(z))> 4) {PositiveSquareRoot(z), NegativeSquareRoot(z)}I meant:4) {PositiveSquareRoot(z), Conjugate(PositiveSquareRoot(z))}5) {NegativeSquareRoot(z), Conjugate(NegativeSquareRoot(z))} === =Mattias,[snip]> De?e one of the concepts on the following list, and I can de?e all the> others:> 1a) i> 1b) -i> 2a) Arg(z)> 2b) -Arg(z)> 3a) Im(z)> 3b) -Im(z)> 4a) PositiveSquareRoot(z)> 4b) NegativeSquareRoot(z)1a) Pick i to be either square root of -1. It doesn't matter which one youpick (i.e., the math will always work the same) given that you apply yourchoice consistently. Then, as you say, you can de?e the other concepts.A word of caution, though: In the context of square roots of complexnumbers, ?positive' and ?negative' are convenient terms, but there's nomathematical basis for identifying one imaginary number as one or the other.Yes, once we've de?ed i to be one root or the other, we can de?e afunction that maps complex numbers to reals, ImaginaryPart(a+bi)=b (where a,b real), and we can associate positive and negative *reals* with givenimaginary numbers--but, this strictly is a matter of convention. Therefore,associating one imaginary number with ?positive' while associating itsadditive inverse with ?negative' is arbitrary in exactly the same sense as(and in fact, is a direct consequence of) which root of -1 we de?e to bei.The choice of i as the ?positive' square root of -1 is certainly not theonly place where an arbitrary mathematical decision is made to keep aconcept consistent: Think right-hand rule.[snip]Travis === .... > 1a) Pick i to be either square root of -1.... How? That's the question. Ken Pledger. === > > .... > 1a) Pick i to be either square root of -1....> How? That's the question.> Ken Pledger.This is of course seen in one way quite a reasonable question. Thereis an automorphism of C taking any of the square roots to any other,i.e. there are no properties that one can distinguish. However here are of course lots (2^|C|) of automorphisms of C (as a?ld).For example let s and t be trancendental. Then there is anautomorphism of C taking s to t. In particular there is an orbit ofAut(C) whose complement is countable (the algebraic numbers).On the other hand, refuting that it is reasonable to pick a member ofa ?ite set will not allow you to do much mathematics.Also, one need not mention i at all when we consider C as a ?ld. Theimport of most mathematical equations that state i has property P(which maybe mentions -i aswell) is that for any x with x^2=-1 wehave that x has property p (with respect to -x).The algebriac (?st order) theory of C is quite well understood(decidable) and i is not a 0-de?able (that is the point of the ?stparagraph). The second paragraph of course says that almost allelements are not de?able or even algebraic. This is much more of aproblem, surely, from your point of view. === I am sure Cardano did> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. The problem is> that there are /two/ such numbers.> MattiasABSOLUTELY!My researches into the Gamma Function (http://www.wehner.org/euler/ )lead to the GAMMA RIDDLE -which I tell you now is INSOLUBLE.The Euler equation begins with Nature having REAL numbers, and thenbeing shown the IMAGINARY by Man - with Exp(iX).Nature obliges by saying If you insist on imagining, then you goround in circles. Hence the CosX +iSinX .When the left of the Gamma curve is multiplied by the right, itsgammaness vanishes, leaving only a spiral of magnitude Pi.This spiral is -1 to the power of X. That is, it is REAL to the powerof REAL.We offer this up to Nature, and Nature says That's a LEFT-RIGHThanded spiral.We can see the SHADOW of a spiral on the XY REAL plane.We struggle to understand LEFT-RIGHT. Is it left OR right?Is it left AND right?We choose a left-handed spiral, and construct it for NEGATIVE valuesof the factorials (Gamma below 1). Then we use the factorial law toshift this spiral to the RIGHT along the X axis (increasing X).As X becomes positive, half-loops of spiral intrude where they do notbelong.We choose a right-handed spiral, and ......... ditto.So logical OR is ruled out.Cardano may have been confused into thinking he had a choice.In fact, with the Gamma function the spiral in complex space is left(logical AND) right.How can a spiral be left-handed and right-handed? AMBIDEXTROUS?Impossible.Therefore in the context of the Gamma function, the negation function-1 to the X CANNOT be complex.So the Gamma is incompatible with complex maths.Either the Gamma is an invention of Man, or the Complex Maths is.Further reasoning is that, in creating the Gamma function we fedFACTORIALS in.These are in the SET of REAL.There are no IMAGINARY components in the set of real.Charles Douglas Wehner === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> > Its true to say that the the conjugate is equal to the negative, only> for> complex numbers of the form (0+bi) or (0, b), where b is a real number.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers, but> this is not how complex numbers arose historically. I am sure Cardano did> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. The problem> is> that there are /two/ such numbers.> There is no problem at all.> +i=(the positive square root of -1).> -i=(the negative square root of -1).> Where do you ?d a problem with this concept?> I ?d a problem with positive and negative applied to _any_ complexnumbers. They're ok applied to reals, put not complex numbers.GC> Mattias> === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> > Its true to say that the the conjugate is equal to the negative,only> for> complex numbers of the form (0+bi) or (0, b), where b is a realnumber.> [snip]> > Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers,but> this is not how complex numbers arose historically. I am sure Cardanodid> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. Theproblem> is> that there are /two/ such numbers.> There is no problem at all.> +i=(the positive square root of -1).> -i=(the negative square root of -1).> Where do you ?d a problem with this concept?> I ?d a problem with positive and negative applied to _any_ complex> numbers. They're ok applied to reals, put not complex numbers.> GCWhy do you have a problem with negatives?neg(a+bi) =df (-a)+(-b)i.In general: neg(a+bi+cj+dk ...) = (-a-bi-cj-dk - ...)Where a and b are real (possitive and negative) numbers.(a+bi) = (a, b), by de?ition: an ordered pair of reals.i.e. -(a+bi) = (-a, -b).(a+bi)+(c+di) = (a+c) + (b+d)i.(a+bi)-(c+di) = (a-c) + (b-d)i.And, (a+b(+i)) = (a + b(-i)) is false, for all b<>0.+i = -i, is contradictory.We cannot deal with complex numbers without negation, imo.Owen> Mattias> === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> Its true to say that the the conjugate is equal to the negative, onlyfor> complex numbers of the form (0+bi) or (0, b), where b is a real number.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as theordered> pair (a, b). With this de?ition one can indeed name complex numbers, but> this is not how complex numbers arose historically. I am sure Cardano did> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. The problemis> that there are /two/ such numbers.Both +sqrt(-1) and -sqrt(-1), work. The ?st is positive (+i) and thesecond is negative (-i).We can assume that (i)=(-i), instead of (i)=(+i), but there would be nodifference once the de?itions are clear. That is, i^2 = (-1) is true foreach.> Mattias> === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> Its true to say that the the conjugate is equal to the negative,> only> for> complex numbers of the form (0+bi) or (0, b), where b is a real> number.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers,> but> this is not how complex numbers arose historically. I am sure Cardano> did> not de?e a complex number as an ordered pair. I think he must have> > introduced i as a number that satis?s the equation x^2+1=0. The> problem> is> that there are /two/ such numbers.> There is no problem at all.> +i=(the positive square root of -1).> -i=(the negative square root of -1).> Where do you ?d a problem with this concept?> I ?d a problem with positive and negative applied to _any_ complex> numbers. They're ok applied to reals, put not complex numbers.> GC> Why do you have a problem with negatives?> neg(a+bi) =df (-a)+(-b)i.I have no problem with multiplying a complex number z by (-1): z(-1) =z, but I do have have a problem with describing either z or -z aspositive or negative. Note: _not_ one is the negative of the otherbut one is negative unconditionally.This is the way one introduces an order into a ring R:(i) specify a non-empty subset P of R such that:(ii) every element x in R is such that just one of x = 0 x in P -x in P holds,(iii) if x, y in P then x + y, x * y in P.(iv) de?e x < y as y - x in P.Now, if -x in P, x is called negative, if x in P, x is calledpositive.Also one can redo all the above by specifying properties for anunde?ed < and de?ing P from it.So, with R = complex numbers, how will you de?e P or bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> > Its true to say that the the conjugate is equal to the negative, only> for> complex numbers of the form (0+bi) or (0, b), where b is a real number.> [snip]> Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers, but> this is not how complex numbers arose historically. I am sure Cardano did> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. The problem> is> that there are /two/ such numbers.> Both +sqrt(-1) and -sqrt(-1), work. The ?st is positive (+i) and the> second is negative (-i).Positive numbers are greater than negative, so -i < +ian inequality can be multiplied through by a positive number withoutchanging its sense, so (+i)(-i) < (+i)(+i) 1 < -1Happy?GC> We can assume that (i)=(-i), instead of (i)=(+i), but there would be no> difference once the de?itions are clear. That is, i^2 = (-1) is true for> each.> Mattias> > === =,> Both +sqrt(-1) and -sqrt(-1), work. The ?st is positive (+i) and the> second is negative (-i).Minor nit picked:Non-real complex numbers are neither positive nor negative in themselves, but they may be positive or negative multiples of some other complex number. What you may mean is that +i is a positive multiple of i and the -i is a negative multiple of i, where the multipliers are the real numbers +1 and -1, respectively. === > bQ.10472@news01.bloor.is.net.cable.rogers.com...> [snip]> > Its true to say that the the conjugate is equal to the negative,only> for> complex numbers of the form (0+bi) or (0, b), where b is a realnumber.> [snip]> > Apparently you de?e i as the ordered pair (0,1), and a + bi as the> ordered> pair (a, b). With this de?ition one can indeed name complex numbers,but> this is not how complex numbers arose historically. I am sure Cardanodid> not de?e a complex number as an ordered pair. I think he must have> introduced i as a number that satis?s the equation x^2+1=0. Theproblem> is> that there are /two/ such numbers.> Both +sqrt(-1) and -sqrt(-1), work. The ?st is positive (+i) and the> second is negative (-i).> Positive numbers are greater than negative, so> -i < +iYes.> an inequality can be multiplied through by a positive number without> changing its sense, so> (+i)(-i) < (+i)(+i)> 1 < -1This contradiction shows that your assumption is false.It is true when multiplied by a positive real number but it obviously failsfor positive complex multipliers.It is also false to say that positive numbers multiplied by positive numbersare alway positive, and so on.A de?ition of greater and less etc., for complex numbers:1. (a+bi) > (c+di), de?ed, (a>c and b>d) or (a>c and b=d) or (a=c andb>d).2. (a+bi) < (c+di), de?ed, (a< (c+di), de?ed, (a>c and b (c+di), de?ed, (ad).For real numbers a, b, it's true that: a>b or a(c+di) or (a+bi)<(c+di) or (a+bi)=(c+di) or(a+bi)><(c+di) or (a+bi)<>(c+di).Owen> Happy?> GC> We can assume that (i)=(-i), instead of (i)=(+i), but there would be no> difference once the de?itions are clear. That is, i^2 = (-1) is truefor> each.> Mattias> === [snip...> I ?d a problem with positive and negative applied to _any_ complex> numbers. They're ok applied to reals, put not complex numbers.> GC>Why do you have a problem with negatives?>neg(a+bi) =df (-a)+(-b)i.>In general: neg(a+bi+cj+dk ...) = (-a-bi-cj-dk - ...)>Where a and b are real (possitive and negative) numbers.>(a+bi) = (a, b), by de?ition: an ordered pair of reals.>i.e. -(a+bi) = (-a, -b).>(a+bi)+(c+di) = (a+c) + (b+d)i.>(a+bi)-(c+di) = (a-c) + (b-d)i.>And, (a+b(+i)) = (a + b(-i)) is false, for all b<>0.>+i = -i, is contradictory.>We cannot deal with complex numbers without negation, imo.>Owen>He didn't say he had a problem with negatives (or negation). He has aproblem with ?positives' and ?negatives', which really make senseonlly when there is an order. When you say ?a is postive' (in thereals), you mean ?a>0'. What do you mean by ?a is positive' in thecomplex ?ld?Larry(this space unintentially left blank ..... === =|This is a good answer to my question, but if complex numbers have this extra|structure, I do not see that complex numbers arise naturally in mathematics.How would you prove that the complex numbers (characterized in someother way) exist, without at some stage constructing the complexnumbers as I've characterized them (i.e., having enough additionalstructure to keep complex conjugation from being an isomorphism)?Keith Ramsay === |This is a good answer to my question, but if complex numbers have thisextra> |structure, I do not see that complex numbers arise naturally inmathematics.> How would you prove that the complex numbers (characterized in some> other way) exist, without at some stage constructing the complex> numbers as I've characterized them (i.e., having enough additional> structure to keep complex conjugation from being an isomorphism)?First approach. Start with a Euclidean plane P. Let G be the group ofparallel displacements of P. Let T be the set of tranformations of G thatare rotations combined with magni?ations. T forms a model of the complexnumbers. The problem with this approach is that we ?st need models ofEuclidean geometry.Second approach. Let S be a set with two elements. The complex numbers arenow modelled as triples (m, a, s), where m is a non-negative numberrepresenting the modulus, a is a number in the interval [0, pi] representingthe size of the argument, and s is an element in S representing thedirection of the argument. (Some numbers have more than one representation,but this problem can be solved by considering equivalence classes ofrepresentations.)Mattias === > [snip...> I ?d a problem with positive and negative applied to _any_complex> numbers. They're ok applied to reals, put not complex numbers.> GC>Why do you have a problem with negatives?>neg(a+bi) =df (-a)+(-b)i.> >In general: neg(a+bi+cj+dk ...) = (-a-bi-cj-dk - ...)> >Where a and b are real (possitive and negative) numbers.> >(a+bi) = (a, b), by de?ition: an ordered pair of reals.> >i.e. -(a+bi) = (-a, -b).>(a+bi)+(c+di) = (a+c) + (b+d)i.>(a+bi)-(c+di) = (a-c) + (b-d)i.>And, (a+b(+i)) = (a + b(-i)) is false, for all b<>0.>+i = -i, is contradictory.>We cannot deal with complex numbers without negation, imo.>Owen> He didn't say he had a problem with negatives (or negation). He has a> problem with ?positives' and ?negatives', which really make sense> onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?positive(A) <-> A>0, for all A.(for: real, complex, and hypercomplex numbers)positive(a) <-> a>0positive(a+bi) <-> (a+bi)>0positive(a+bi) <-> (a+bi)>(0+0i)positive(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).(a+bi) is positive, if, a>0 and b>0.(a+bi) is positive, if, a>0 and b=0.(a+bi) is positive, if, a=0 and b>0.positive(a+0i) <-> (a>0 and b=0).positive(0+bi) <-> (a=0 and b>0).Similarly, negative(a+bi) <-> (a+bi)<0.negative(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).negative(a+0i) <-> a<0 and b=0.negative(0+bi) <-> a=0 and b<0.Owen> Larry> (this space unintentially left blank ..... === > [snip...> I ?d a problem with positive and negative applied to _any_> complex> numbers. They're ok applied to reals, put not complex numbers.> GC>Why do you have a problem with negatives?>neg(a+bi) =df (-a)+(-b)i.>In general: neg(a+bi+cj+dk ...) = (-a-bi-cj-dk - ...)>Where a and b are real (possitive and negative) numbers.>(a+bi) = (a, b), by de?ition: an ordered pair of reals.>i.e. -(a+bi) = (-a, -b).>(a+bi)+(c+di) = (a+c) + (b+d)i.> >(a+bi)-(c+di) = (a-c) + (b-d)i.>And, (a+b(+i)) = (a + b(-i)) is false, for all b<>0.>+i = -i, is contradictory.>We cannot deal with complex numbers without negation, imo.>Owen> He didn't say he had a problem with negatives (or negation). He has a> problem with ?positives' and ?negatives', which really make sense> onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?> positive(A) <-> A>0, for all A.> (for: real, complex, and hypercomplex numbers)> positive(a) <-> a>0> positive(a+bi) <-> (a+bi)>0> positive(a+bi) <-> (a+bi)>(0+0i)> positive(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).> (a+bi) is positive, if, a>0 and b>0.> (a+bi) is positive, if, a>0 and b=0.> (a+bi) is positive, if, a=0 and b>0.> positive(a+0i) <-> (a>0 and b=0).> positive(0+bi) <-> (a=0 and b>0).> Similarly, negative(a+bi) <-> (a+bi)<0.> negative(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).(Correction)negative(a+bi) <->.(a<0 and b<0) or (a<0 and b=0) or (a=0 and b<0).Owen> negative(a+0i) <-> a<0 and b=0.> negative(0+bi) <-> a=0 and b<0.> Owen> Larry> (this space unintentially left blank .....> === Why pretending there is asymmetry>when there is none?> What a few people have said is what's correct: The symmetry exists,> and there is _no_ reason why one of those complex numbers should> be called i and the other one -i. We just picked one and called it i.> (And no, there's no way to say _which_ of the two is the one we> decided to call i.)> THIS IS A POINT I ALREADY COVERED in a posting in this thread.Is it not remarkable that when Nature delivers TWO, the mathematicianinstinctively thinks he has a CHOICE (a OR b)?That is begging the question. In the Gamma (factorial) function, Ihave shown that Nature delivers an AND function of iX and -iX:Is it left OR right?Is it left AND right?The negation function -1 to the power of X is (as I already said) REALto the power of REAL. We do not feed anything imaginary INTO themathematics - and wait for Nature to tell us what the imaginary partis. In the context of the Gamma function, the negation functionspirals left AND right at the same time.Accordingly, Nature says the spiral is NOT ALLOWED to spiral. NatureDOES NOT TAKE SIDES - and Nature ALLOWS NO CHOICE.This means that Man INVENTED the complex maths or Man INVENTED theGamma function.I used the Gamma function as a tool to investigate the complexmaths. Good mathematicians do not leave their tools behind. Thus,until it is proven that the negation function spirals BOTH ways in ALLapplications, one has to consider that it might be a special propertyof the Gamma function alone.However, I have established to my own satisfaction that the complexmaths is incompatible with the Gamma function - or with my fRactorials(fractional factorials).>If we accept that there is complete symmetry, then we are faced with a>strange consequence: we cannot name any single non-real complex number. We>can easily name real numbers -- -16, 21, pi, to take some examples -- but>with complex numbers we can only name pairs of numbers, such as the pair>that satis?s the equation x^2+1=0.>What is so information-poor about the complex maths is when you takeany polynomial and substitute a complex argument (see my FOUR-BOXALGORITH at http://wehner.org/euler ). I did this for the fractionalfactorials.I obtained the Fraccos, Fracsin, Fraccosh and Fracsinh for example -that is to say, the equivalent of Sine, Cos, Shine and Cosh but withthe Gamma polynomial replacing the Exponential polynomial.That pointless symmetry kept popping up - as if to say this researchis jejeune.>One could go on and argue that it makes no more sense to talk about /the>complex plane/ than to talk about /the plane/ (as if only one plane>existed).> >Sloppy! On the page of my website mentioned above, I show images inthe iX complex plane and in the iY complex plane. Those who do seriousmaths know that there is more than one complex plane.>Comments?>Mattias> David C. Ullrich> I have elaborated on your posting - but have not disagreed with a wordthat you or Mattias Wikstr.9am said.Charles Douglas Wehner === > ...> He didn't say he had a problem with negatives (or negation). He has a> problem with ?positives' and ?negatives', which really make sense> > onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?> positive(A) <-> A>0, for all A.> (for: real, complex, and hypercomplex numbers)So is i > 0 in the complex numbers? There is no useful order > inthe complex numbers.GC === > ...> He didn't say he had a problem with negatives (or negation). He has a> problem with ?positives' and ?negatives', which really make sense> > onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?> positive(A) <-> A>0, for all A.> (for: real, complex, and hypercomplex numbers)> So is i > 0 in the complex numbers? There is no useful order > in> the complex numbers.> GCOf course there is, ..why don't you pay attention? === > ...> He didn't say he had a problem with negatives (or negation). He has a> problem with ?positives' and ?negatives', which really make sense> onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?> positive(A) <-> A>0, for all A.> (for: real, complex, and hypercomplex numbers)> So is i > 0 in the complex numbers? There is no useful order > in> the complex numbers.> GC> Of course there is, ..why don't you pay attention?He means that the complex numbers are not an ordered ?ld.Try a Google search on ordered ?ld if you don't know the de?ition.-- Dave SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === > [snip...> I ?d a problem with positive and negative applied to _any_> complex> numbers. They're ok applied to reals, put not complex numbers.> GC>Why do you have a problem with negatives?>neg(a+bi) =df (-a)+(-b)i.>In general: neg(a+bi+cj+dk ...) = (-a-bi-cj-dk - ...)>Where a and b are real (possitive and negative) numbers.>(a+bi) = (a, b), by de?ition: an ordered pair of reals.>i.e. -(a+bi) = (-a, -b).>(a+bi)+(c+di) = (a+c) + (b+d)i.> >(a+bi)-(c+di) = (a-c) + (b-d)i.>And, (a+b(+i)) = (a + b(-i)) is false, for all b<>0.>+i = -i, is contradictory.>We cannot deal with complex numbers without negation, imo.>Owen> He didn't say he had a problem with negatives (or negation). He has a> > problem with ?positives' and ?negatives', which really make sense> onlly when there is an order. When you say ?a is postive' (in the> reals), you mean ?a>0'. What do you mean by ?a is positive' in the> complex ?ld?> positive(A) <-> A>0, for all A.> (for: real, complex, and hypercomplex numbers)> positive(a) <-> a>0> positive(a+bi) <-> (a+bi)>0> positive(a+bi) <-> (a+bi)>(0+0i)> positive(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).> (a+bi) is positive, if, a>0 and b>0.> (a+bi) is positive, if, a>0 and b=0.> (a+bi) is positive, if, a=0 and b>0.> positive(a+0i) <-> (a>0 and b=0).> positive(0+bi) <-> (a=0 and b>0).> Similarly, negative(a+bi) <-> (a+bi)<0.> negative(a+bi) <->. (a>0 and b>0) or (a>0 and b=0) or (a=0 and b>0).>(Correction)>negative(a+bi) <->.(a<0 and b<0) or (a<0 and b=0) or (a=0 and b<0).>Owen> negative(a+0i) <-> a<0 and b=0.> negative(0+bi) <-> a=0 and b<0.This doesn't describe a (total) order on C. What about 1-i? is it >0or <0.Now matter how hard you try, you can't de?e a ?reasonable' totalorder on C. > Owen> Larry> (this space unintentially left blank .....>Larry(this space unintentially left blank ..... === =Let s(m,n) = an unsigned Stirling number of the 1st kind.( s(0,0) =1, s(0,n) =0 if n not 0,s(m+1,n) = m*s(m,n) +s(m,n-1) )First, (this must be well-known)If m+n is odd, then:(m(m-1)/2), the (m-1)th triangular number,divides s(m,n).-Okay, this is less probably well-known:Let T(0,m,n,q) = s(m+q,n+q)/(m+q-1)!,for m, n = positive integers, and q = nonnegative integer.And, for r = positive integer:T(r,m,n,q) = sum{k=1 to m} T(r-1,k,n,q);then:(m+q-1)! T(r,m,n,q) is congruent to(-1)^(m+n) (m+q-1)! T(s,m,n,q) (mod{m+q+r+s}),for r and s = any nonnegative integers.(I am not even trying to get a closed-form for T().)Leroy Quet === =Let S(m,n) = an unsigned Stirling number of the 2nd kind.( S(0,0) =1, S(0,n) =0 if n not 0,S(m+1,n) = n*S(m,n) +S(m,n-1) )Let b(m) = sum{k=0 to ?m-1)/2)} S(m,2k+1) (-1)^kAnd let c(m) = sum{k=0 to ?/2)} S(m,2k) (-1)^kThen:For p = any prime,p divides (b(p+2) -b(p+1) +b(p))andp divides (c(p+2) -c(p+1) +c(p) +2)This is a direct consequence of the result:If a(m) = q*b(m) + r*c(m),then: a(0) = r, a(1) = q,and a(m+2) = a(m+1) - sum{k=0 to m} binomial(m,k) 2^(m-k) a(k).This a-sequence result is found at:http://mathforum.org/discuss/sci.math/a/t/389070Leroy Quet === =There is no such thing as a model containing all ordinals DUMBASSHow can we know V>L?stop equivocating on V George Greene has been arguing there that we can know V>L, sincethere are models of ZFC + V>L, and so the non-constructible sets fromsuch models show there really are non-constructible sets. Daryl McCullough has been arguing against this, noting it is possible forthe construction of ZFC + V>L models (under suitable consistency pemises)to be itself done in L, undermining the argument that there must really benon-constructible sets. I will write about further ways constructibility properties can becomplicated between models. Daryl points out it is possible to have a smaller M0 ZFC model with aset it believes to be non-constuctible, being a member of a larger M1 ZFCmodel making the set constructible. So non-constructibility can change toconstructibility when passing to a larger universe. For well-founded small models M0 this does not reverse. If x is a set inM0, and M0 |= x is constructible, and M0 is a member of a larger ZFCmodel M1, and M1 |= M0 is a standard transitive model, thenM1 also |= x is construcible. Namely the Godel construction of x is stillin M1, and M1 still recognizes this as such a construction. For M0 a non-well-founded ZFC model though, this can change. We needsome conventions to even make sense of the inclusions needed to state theseproperties. So we are considering models M0 of form , where N0 is a set(the universe of M0), and E is some binary relation on N0, to interpretthe epsilon symbol from the language of ZF, but we allow for thepossibility the E is not the true epsilon on N0. We want to talk about x a set in M0, its properties there, and thenconstrast them with x's properties in a larger M1 model. In general wecan't relate x as it sits in M0 with its E members, to the true set xwith its true epsilon members. We can make sense of this in a special case of how M0 sits in M1 though,which will be the case under discussion. Namely any M0 = ZFCmodel has a well founded part, namely those elements with E-transitiveclosure being well-founded with respect to E. So we will consider the case of M0 a member of a larger M1 ZFC with M1satisfying that M0 has the additional property I will call s-t-wf, namelystandard transitive wellfounded, which is that on the well-founded part ofM0 E is the true epsilon, and that well-founded part of N0 is a transitiveset with respect to either E or epsilon, same thing. Then we can consider x from the well-founded part of M0, as this isde?ed in M1, and for these we can relate the true x from M1 to how thatx is seen in M0. So with all this background, it is possible to have x a member of theweell-founded (in the sense of M1) part of M0, with M0 a s-t-wf ZFCmodel with nonstandard E membership relation, this M0 being a set in M1and having those properties in M1, such that M0 models x isconstructible and M1 models x is non-constreuctible. Ie the reversedirection from above, the one that couldn't happen for standardtransitive models. Namely, x is from the M1-wf part of M0, but the Godel construction of xmaking M0 think x is constuctible comes from the non-wf part of M0, so M1does not recognize that as a real construction, and in fact M1 provides noalternative constructions. Here is a construction of such examples M0 and M1 (under suitableconsistency assumptions). As I recall, there is a theorem given inBarwise's Admissible Sets and Structures, and application of the Barwisecompactness theorem. I don't have that book at hand, so I am going bymemory, but I also think I reconstructed a proof myself. Anyway, if I amremembering this wrong and my reconstruction is also wrong, this part will So here is the theorem. I will use strange looking numbers in thestatement to make things line up with how I will apply it in a moment. Before stating it I will de?e that for ZF models M-1 = and M0 = , M0 end extends M-1 is de?ed:N0 superset N-1 and E-1 = E0 restricted to N-1,and if m1 in N-1 and m0 in N0 and m0 E0 m1 then m0 in N-1 (ie m1 from the little model doesn't get any new memers in the big model). Suppose M-1 (ie M with subscript or whatever -1 = the integer minus1) suppose M-1 = is a countable model of ZF. (Yes I thinkwe can do this with just ZF, making this even more remarkable). Then thereis M0 = a model end extending M-1, with M0 |= ZFC + V=L . So ie, given any M-1 model, possibly V>L, possible ~AC even, by addingextra garbage to the top of the model, not changing what the original setswere, just creating new higher sets, you can make it look like a V=L model. So we apply that theorem for the orignal observation, from suitablepremises. Suppose we assume we have M1 a ZFC model with x a set in M1and M1 |= x is not constructible. (So we assume M1 |= V>L). Supposealso N-1 is a set in M1 and x is in N-1and letting M-1 = , suppose M1 |= M-1 |= ZFC andalso M1 |= N-1 is countable transitive. Then we apply the above theorem in M1, obtaining M0 containing xand |= V=L. So M0 |= x is constructible (since M0 thinks everything is). ButM1 |= x is not constructible. Using the same construction aove to replace half the members of asequence, having ?st preprocessed an assumed sequence to make all modelscountable in the next model, I think I have proved the result below, sayingit can be tricky how moving to bigger universes changes whether things seemconstructible. To state the result, I must de?e a certain type of sequence of ZFCmodels. By an accumulating tower of ZFC models I mean an ordinally indexedsequence of ZFC models, each s-t-wf, s.t. each model in the towerconstains a member of its well-founded part the restriction of the tower tobelow that location. So the main result:In ZFC, suppose also that x is hereditarily countable(ie every member of the transitive closure of {x} is countable).Suppose also that x is not constructible.Suppose (aleph_1)^L = (aleph_1)^L[x]Suppose there is asequence of ordinals in length (aleph_1)^L,such that each of those ordinals alpha has L_alpha[x] |= ZFC.Then there is an accumulating tower of countable ZFC models in length(aleph_1)^L, with x a member of the well-founded part of every model inthe tower, such that the question of whether x is constructible zigzagsthroughout the tower adjacently: ie even ordinals versus odd ordinals. So the tower changes its mind (aleph_1)^L many times, trying to decideif x is constructible. Namely the idea here is adjust the original sequence of L models, alwayspicking the smallest ordinal making the next model of ZFC making allprevious models countable, and making the initial segment of the sequencelength countable. Inductively, the next ordinal doing that is alwayscountable in L, (by a collapsing argument), so that keeps the inductiongoing. So we obtain a tower, where each part sees everything below as countable. Now on every second step, do that Barwise theorem, replacing the originalmodel by an extension making x look countable. Do this theorem inside thenext model in the tower. Replace those steps by their extensions. We use the sequence has length (aleph_1)^L to see that the originalstep could produce the nextt structure countable in L, so that stayscountable later to apply Barwise (or whoever proved it: I say Barwisebecause I saw it in that book). The Barwise result needs (as stated) that the starting structure iscountable. To make an example showing this, I will work inZFC + x is a non-construcible real + aleph_1 = (aleph_1)^L. By Cohen'smethods, this theory is consistent if ZFC is. So if a ZFC proof couldgeneralize Barwise, apply it in that theory and clash against my examplebelow, showing that theory inconsistent, and hence by Cohen ZFCinconsistent. So working in that theory, suppose the starting model M-1includes in its well-founded part all countable^L ordinals (ie so makingthis a case not claimed in the actual theorem). Then, if we could applythe theorem still, we would obtain an end extension M0 making xconstructible. By Godel's proof of CH^L, applied in M0,we would have some countable^M0 ordinal stage of L constructing x. Since x is not constructible in our working theory, the M0 constructionof it must come from the nonconstrucible part of M0. M0 end extends M-1,and M-1 contains all countable^L ordinals, so M0 has ordinals itbelieves countable beyond all the true countable^L ordinals. So M0 provides a map from omega with range including all the truecountable ordinals. Our outer working theory can extract that map from M0,and obtain a countable enumeration of all the true countable^L ordinals. But this contradicts aleph_1 = (aleph_1)^L from our working theory. That concludes the example showing Barwise needed the starting modelcountable. So my tower result also needs the tower is only length (aleph_1)^L, moreor less. Because the tower includes the sequence leading up to its levelas a member, at each stage. So the well-founded parts of the tower keepgetting longer. So if you continued a tower like this past (aleph_1)^L, the modelsappearing there would include (aleph_1)^L in their wellfounded part. So, depending on where you put the ones constructing x (even or odd), at (aleph_1)^L or (aleph_1)^L + 1, you would hit a structure saying x isconstructible, but at least in the special case above of aleph_1 =(aleph_1)^L, a model with such big well-founded part can't be mistakenunderlying theorem. So the same issues show the tower result has the same limitations.--David Libert (ah170@freenet.carleton.ca)1. I used to be conceited but now I am perfect.2. So self-quoting doesn't seem so bad. -- David Libert3. So don't be a morron. -- Marek Drobnik bd308 rhetorical salvo IRC sig === =converge uniformely to some functions. All of the examples of the in?ite series that I saw had very simple forms. For example,1) s(x)=1+x^2+x^3+...+x^n+....2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...When I was reading the de?ition of what power series are, the book said: if all of the coef?ients in the in?ite series are independent of x, the series is called a power series.If this is the case, how about the in?ite series like this one?s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3) }+{(x^5)/(1^3)}+{(x^6)/(2^4)}+...The series above has the following properties.1) whenever x^(even numbers, like 0,2,4..),then the denominators of each term is in the form: 2^(some number).2) whenever x^(odd numbers), then the denominators of each term is in the form: 1^(some numbers)Since all of the coef?ients of the terms in the in?ite series is independent of x, could I call this series a power series? If so is there any way for me to know what this power series converges to? The reason why I am asking this question is that it is relatively easy to make many kinds of in?ite series, but then I don't know which in?ite series I come up with is more useful than the others. I also would like to know if there is any general method to check that the in?ite series that I am interested in can be converged to some function of a simple form? When I was looking at the way book explained how some of the in?ite series can be changed to a simpler forms, it seems that they strongly depended on the form of the in?ite series. However, if there is some general way to tackle problems of simplifying in?ite series, that would help me very much.in my in?ite series. I could not use some math text editor to write more neatly than that. === =In?ite series can be thought of as polynomials that go on forever (noin?ite zero coef?ients)3x^5+2x^7-8x^16+.... is an in?ite series for example.> converge uniformely to some functions. All of the examples of the> in?ite series that I saw had very simple forms. For example,> 1) s(x)=1+x^2+x^3+...+x^n+....> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...> When I was reading the de?ition of what power series are, the book said:> if all of the coef?ients in the in?ite series are independent of> x, the series is called a power series.> If this is the case, how about the in?ite series like this one?>s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^ 3)}+{(x^5)/(1^3)}+{(x^6)/(2^4)}+...> The series above has the following properties.> 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each> term is in the form: 2^(some number).> 2) whenever x^(odd numbers), then the denominators of each term is in> the form: 1^(some numbers)> Since all of the coef?ients of the terms in the in?ite series is> independent of x, could I call this series a power series? If so is> there any way for me to know what this power series converges to?> The reason why I am asking this question is that it is relatively easy> to make many kinds of in?ite series, but then I don't know which> in?ite series I come up with is more useful than the others. I also> would like to know if there is any general method to check that the> in?ite series that I am interested in can be converged to some> function of a simple form? When I was looking at the way book explained> how some of the in?ite series can be changed to a simpler forms, it> seems that they strongly depended on the form of the in?ite series.> However, if there is some general way to tackle problems of simplifying> in?ite series, that would help me very much.> in my in?ite series. I could not use some math text editor to write> more neatly than that.> === =Yes,s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{( x^4)/(2^3)}+{(x^5)/(1^3)}+{(x^6)/(2^4)}+... is an in?ite seriess(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^ 3)}+{(x^5)/(1^3)}+{(x^6)/(2^4)}+...= (1/2)+(1/1)x+(1/2^2)x^2+(1/1^2)x^3+(1/2^3)x^4+...The point is, an in?ite series can be expressed as the sum and differenceof an in?te number of (combined) terms of the form-- a constant times xraised to a non-negative integer. If the meaning of integer, constant and/ornon-negative integer are not clear then you MUST look them up.Good luck> converge uniformely to some functions. All of the examples of the> in?ite series that I saw had very simple forms. For example,> 1) s(x)=1+x^2+x^3+...+x^n+....> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...> When I was reading the de?ition of what power series are, the book said:> if all of the coef?ients in the in?ite series are independent of> x, the series is called a power series.> If this is the case, how about the in?ite series like this one?>s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^ 3)}+{(x^5)/(1^3)}+{(x^6)/(2^4)}+...> The series above has the following properties.> 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each> term is in the form: 2^(some number).> 2) whenever x^(odd numbers), then the denominators of each term is in> the form: 1^(some numbers)> Since all of the coef?ients of the terms in the in?ite series is> independent of x, could I call this series a power series? If so is> there any way for me to know what this power series converges to?> The reason why I am asking this question is that it is relatively easy> to make many kinds of in?ite series, but then I don't know which> in?ite series I come up with is more useful than the others. I also> would like to know if there is any general method to check that the> in?ite series that I am interested in can be converged to some> function of a simple form? When I was looking at the way book explained> how some of the in?ite series can be changed to a simpler forms, it> seems that they strongly depended on the form of the in?ite series.> However, if there is some general way to tackle problems of simplifying> in?ite series, that would help me very much.> in my in?ite series. I could not use some math text editor to write> more neatly than that.> === converge uniformely to some functions. All of the examples of the > in?ite series that I saw had very simple forms. For example,> 1) s(x)=1+x^2+x^3+...+x^n+....> 2) s(x)=1+x/1+(x^2)/(2!)+(x^3)/(3!)+...+(x^n)/(n!)+...> When I was reading the de?ition of what power series are, the book said:> if all of the coef?ients in the in?ite series are independent of > x, the series is called a power series.> If this is the case, how about the in?ite series like this one?> s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{ (x^5)/(1^3)}+{(x^6)> /(2^4)}+...> The series above has the following properties.> 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each> term is in the form: 2^(some number).> 2) whenever x^(odd numbers), then the denominators of each term is in > the form: 1^(some numbers)> > Since all of the coef?ients of the terms in the in?ite series is > independent of x, could I call this series a power series?Yes> If so is > there any way for me to know what this power series converges to?Yes.Note: (1) 1/2 + x^2/2^2 + x^4/2^3 + x^6/2^4 +... =1/2(1 + (x^2/2)^1 + (x^2/2)^2 + (x^2/2)^3 +...) The series is geometricin x^2/2.Also (2) x + x^3 + x^5 +....=x(1 + (x^2) + (x^2)^2 +...). The series is also geometric in x^2.The two series both converge for -1 < x < 1, you know what theyconverge to and your series is the sum of (1) and (2).> The reason why I am asking this question is that it is relatively easy > to make many kinds of in?ite series, but then I don't know which > in?ite series I come up with is more useful than the others. That's backwards. You don't make up a series and then ?ure out whatit's good for; basically, you come up with a series when all elsefails. (I may get some argument about that.)> I also > would like to know if there is any general method to check that the > in?ite series that I am interested in can be converged to some > function of a simple form? Simple form? Well you know series for several common functions so youcould try to turn your series into one of those. As a rule, if you arelucky, ingenuity might do it. In general, it is a non-trivial problemand even though the power series may converge with non-zero radius ofconvergence, it doesn't converge to an elementary function.> When I was looking at the way book explained > how some of the in?ite series can be changed to a simpler forms, it > seems that they strongly depended on the form of the in?ite series. Sure.> However, if there is some general way to tackle problems of simplifying > in?ite series, that would help me very much.I'm afraid there is none. Also, to make matters worse, not all seriesare power series.[...]-- Paul SperryColumbia, SC (USA) === When I was reading the de?ition of what power series are, the book said:> if all of the coef?ients in the in?ite series are independent of> x, the series is called a power series.> If this is the case, how about the in?ite series like this one?> s(x)={1/2}+{x/1}+{(x^2)/(2^2)}+{(x^3)/(1^2)}+{(x^4)/(2^3)}+{ (x^5)/(1^3)}+{(x^6)/(2^4)}+...> The series above has the following properties.> 1) whenever x^(even numbers, like 0,2,4..),then the denominators of each> term is in the form: 2^(some number).> 2) whenever x^(odd numbers), then the denominators of each term is in> the form: 1^(some numbers)> Since all of the coef?ients of the terms in the in?ite series is> independent of x, could I call this series a power series? If so is> there any way for me to know what this power series converges to?Yes & yes.Try seperating the two parts into two series and see if each converges.> The reason why I am asking this question is that it is relatively easy> to make many kinds of in?ite series, but then I don't know which> in?ite series I come up with is more useful than the others. I also> would like to know if there is any general method to check that the> in?ite series that I am interested in can be converged to some> function of a simple form?Wishful thinking.> When I was looking at the way book explained> how some of the in?ite series can be changed to a simpler forms, it> seems that they strongly depended on the form of the in?ite series.> However, if there is some general way to tackle problems of simplifying> in?ite series, that would help me very much.>Yup, practice, experience, insight, repeat as necessary.> in my in?ite series. I could not use some math text editor to write> more neatly than that.>Don't bother, often they do not show up as you intend.x + (x^2)/(2^2) + ... is suf?ientx + x^2 / 2^2 + ... is loose tho possiblePlease use some spaces for easier reading. Compare: x^2=3y+7=27+6z x^2 = 3y+7 = 27+6z === The reason why I am asking this question is that it is relatively easy >to make many kinds of in?ite series, but then I don't know which >in?ite series I come up with is more useful than the others. I also >would like to know if there is any general method to check that the >in?ite series that I am interested in can be converged to some >function of a simple form? When I was looking at the way book explained >how some of the in?ite series can be changed to a simpler forms, it >seems that they strongly depended on the form of the in?ite series. >However, if there is some general way to tackle problems of simplifying >in?ite series, that would help me very much.To rephrase the question: given a convergent series, how can you ?d the sum in closed form (if indeed a closed form exists)?There is no completely general answer AFAIK. There are certain tricks, e.g.:1) Check for telescoping series, i.e. series that can be written in the form sum_n (f(n+1)-f(n)) for some function f.2) Write the series as a power series, or as a special case of a powerseries, i.e. if your series is sum_n f(n) look at sum_n f(n) z^n. Then a) recognize the series for some of the standard elementary functions,or even some of the non-elementary ones (e.g. hypergeometric seriescover a lot of territory) b) see if differentiation or integration will simplify the form of the series c) see if the series satis?s some differential equation that can be solved3) If the coef?ients are rational functions of n times an n'th power, use a partial fraction decomposition.4) If the sum is numeric, evaluate it numerically to high precision and see if it is recognized by the Inverse Symbolic Calculator at Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > The reason why I am asking this question is that it is relatively easy >to make many kinds of in?ite series, but then I don't know which >in?ite series I come up with is more useful than the others. I also >would like to know if there is any general method to check that the >in?ite series that I am interested in can be converged to some >function of a simple form? When I was looking at the way book explained >how some of the in?ite series can be changed to a simpler forms, it >seems that they strongly depended on the form of the in?ite series. >However, if there is some general way to tackle problems of simplifying >in?ite series, that would help me very much.> To rephrase the question: given a convergent series, how can you ?d the > sum in closed form (if indeed a closed form exists)?> There is no completely general answer AFAIK. There are certain tricks, > e.g.:> 1) Check for telescoping series, i.e. series that can be written in the > form sum_n (f(n+1)-f(n)) for some function f.> 2) Write the series as a power series, or as a special case of a power> series, i.e. if your series is sum_n f(n) look at sum_n f(n) z^n. Then> a) recognize the series for some of the standard elementary functions,> or even some of the non-elementary ones (e.g. hypergeometric series> cover a lot of territory)> b) see if differentiation or integration will simplify the form of the series> c) see if the series satis?s some differential equation that can be > solved> 3) If the coef?ients are rational functions of n times an n'th power, > use a partial fraction decomposition.> 4) If the sum is numeric, evaluate it numerically to high precision and > see if it is recognized by the Inverse Symbolic Calculator at > Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2> Also, take a look at A=B at http://www.cis.upenn.edu/~wilf/AeqB.htmlAbsolutely amazing stuff.Martin Cohen === =[...]> Also, take a look at A=B at http://www.cis.upenn.edu/~wilf/AeqB.html> Absolutely amazing stuff.> Martin Cohen>You can even download the whole book. === =How does one say critical point in French? A critical point is apoint, on the plot of a function, where the derivatve is zero. === How does one say critical point in French? A critical point is a> point, on the plot of a function, where the derivatve is zero.un point critique (google 7050/179) orun point stationnaire (google 349/20) orun point extr.8emal (google 30/4)First number is number of hits.have the words minimal and maximalExample: http://www.google.com/search?&q=%22point+critique%22+minimal+ maximal&lr=lang_frDirk Vdm