mm-1007 === Subject: Limit of a decreasing function ? Given a function f(x) >= 0 which has a decreasing derivative (f(x) <0). Under what conditions can we conclude that eventually f(x) -> L and therefore f(x) -> 0 (where L is a limit). Do we need to have constraints on the norm of f(x) or f(x) ? How can we prove that? Also, where can i find literature about this specific subject (what field of calculus or diff. eq.)? === Subject: Re: Limit of a decreasing function ? > Given a function f(x) >= 0 which has a decreasing derivative (f(x) <0). > Under what conditions can we conclude that eventually f(x) -> L and > therefore f(x) -> 0 (where L is a limit). Do we need to have constraints > on the norm of f(x) or f(x) ? How can we prove that? Also, where can i > find literature about this specific subject (what field of > calculus or diff. eq.)? Heres a constraint you might want to try: Let {a_n} be a monotonically increasing sequence where ever f is defined. Then, if the infinite sum of f(a_n) converges, then f(x) converges. I doubt this is as general as youd like, but its a start. analysis would help you answer this sort of thing without too much difficulty. Check out Rudins Principles of Mathematical Analysis. Its a classic. Ôcid Ôooh === Subject: Re: Limit of a decreasing function ? > Given a function f(x) >= 0 which has a decreasing derivative (f(x) <0). > Under what conditions can we conclude that eventually f(x) -> L and > therefore f(x) -> 0 (where L is a limit). Do we need to have constraints > on the norm of f(x) or f(x) ? How can we prove that? Also, where can i > find literature about this specific subject (what field of > calculus or diff. eq.)? decreasing derivative and decreasing are different things. Presumably, since you specify a negative derivative, you want f to be decreasing. Also eventually f(x) -> L and therefore f(x) -> 0 sneaks in a conclusion (that f(x) -> 0) which doesnt follow from the premise. f can be VERY negative, as long as it is so for VERY SHORT stretches, without disturbing the existence of the limit of f. Decreasing nets of real numbers always have a limit in the extended real numbers. If theyre bounded, then the limit is in the reals. So the only hypothesis you need (assuming youre interested in x -> infinity) is that f be bounded below. --Ron Bruck === Subject: Re: Limit of a decreasing function ? > Decreasing nets of real numbers always have a limit in the extended > real numbers. If theyre bounded, then the limit is in the reals. So > the only hypothesis you need (assuming youre interested in x - infinity) is that f be bounded below. I remember back in first term calculus I learned this (sequence a(n) is increasing/decreasing and bounded from above/below, then a(n) converges.) I think there was a common name for this theorem, but I dont recall what it is. Im thinking Bolzano-Weierstrauss, but I think that just dealt with a bounded sequence and said that there is a convergent subsequence. Its been almost 8 years... I cant believe I forgot :-( I think the name was something crazily simple like upper-bound convergence theorem, but thats not quite it. J === Subject: Re: Limit of a decreasing function ? >> Decreasing nets of real numbers always have a limit in the extended >> real numbers. If theyre bounded, then the limit is in the reals. So >> the only hypothesis you need (assuming youre interested in x -> infinity) is that f be bounded below. > I remember back in first term calculus I learned this (sequence a(n) is >increasing/decreasing and bounded from above/below, then a(n) converges.) >I think there was a common name for this theorem, but I dont recall what >it is. > Im thinking Bolzano-Weierstrauss, but I think that just dealt with a >bounded sequence and said that there is a convergent subsequence. Yes. > Its been almost 8 years... I cant believe I forgot :-( > I think the name was something crazily simple like upper-bound >convergence theorem, but thats not quite it. There may have been a name for it in your calculus book, but I doubt theres a standard name for it - people dont think of this as a theorem so much as an obvious fact. ************************ David C. Ullrich === Subject: Re: Limit of a decreasing function ? convergence theorem, but thats not quite it. > There may have been a name for it in your calculus book, but I > doubt theres a standard name for it - people dont think of this > as a theorem so much as an obvious fact. Ah, I remember what we called it... the Monotone Covergence Theorem... sure, its pretty obvious, but to a freshman in first term calculus, proving it was a suitable midterm question. J === Subject: Re: Limit of a decreasing function ? >> I think the name was something crazily simple like upper-bound >>convergence theorem, but thats not quite it. >> There may have been a name for it in your calculus book, but I >> doubt theres a standard name for it - people dont think of this >> as a theorem so much as an obvious fact. > Ah, I remember what we called it... the Monotone Covergence Theorem... Right. Ive seen it called that in another book. Thats definitely not a standard name for this fact (it _is_ the name of a standard result in measure theory - I remember paging through that book one day, seeing a reference to the MCT, and wondering how in the world that got into a book on calculus...) >sure, its pretty obvious, but to a freshman in first term calculus, >proving it was a suitable midterm question. ************************ David C. Ullrich === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless Can someone answer me a question about technology of wireless. It seems to me that the USA had wireless technology way back to around say 1950. So my question would be why did it take 50 years before it could be commercially applied? Did we have to wait for computer technology to commercialize on wireless? We had walkie-talkies during WW2, so I just wonder why wireless never arose simultaneously with wire telephony after WW2? Granted that the wireless phones would have been huge. I post to sci.engr in hopes that some expert can answer the question. Whether the answer is that wireless had to wait for key technologies or perhaps wireless was prohibited purely by legallaws otherwise it would have sprung into existence circa 1950 instead of 1990s. In my mind, I am imagining that wireless appeared on the commercial scene rather later than what it realistically could have appeared as early as 1950 rather than 1990s. With Cingular and Vodafone bidding for AT&T wireless. I wonder if anyone has a reasonable timetable where wireless is as reliable and stable a feed as is fixed wire? Because the bidding for AT&T at around $35 billion is too generous because with the competition will knock out the smaller ones. And the price only justified because Cingular does not want to spend the time to out-compete in order to be in first place. Then again it can be argued that Cingular has amassed this $35 billion and must spend it somewhere. If I had the $35 billion I would think the fixed assets of Qwests fixed wire is more valueable than the ephemeral assets of AT&T wireless. (Even though I sold all of my Qwest shares). Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless > Can someone answer me a question about technology of wireless. It > seems to me that the USA had wireless technology way back to around > say 1950. So my question would be why did it take 50 years before it > could be commercially applied? Did we have to wait for computer > technology to commercialize on wireless? We had walkie-talkies during > WW2, so I just wonder why wireless never arose simultaneously with > wire telephony after WW2? Granted that the wireless phones would have > been huge. That is the point. The walkie-talkie were bulky and their range was very limited. Solid state technology with low power consumption is what turned the trick. Bob Kolker === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless > That is the point. The walkie-talkie were bulky and their range was very > limited. Solid state technology with low power consumption is what > turned the trick. > Bob Kolker I kind of thought that wireless could have naturally arrived in the 1970s instead of the 1990s but due to the Cold War that the government just did not want to have the air filled with talking machines and interfere with govt wireless walkie talkies. Has anyone traced the size of walkie talkies from 1950 onwards of the government and did they come in small size? Does anyone know if it is inherently unable for wireless to ever become as secure as wired computer? Wireless is inherently less stable than wired due to the nature of the medium of air compared to the wire. But what about the security issue? Archimedes Plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless > That is the point. The walkie-talkie were bulky and their range was very > limited. Solid state technology with low power consumption is what > turned the trick. > Bob Kolker > I kind of thought that wireless could have naturally arrived in the > 1970s instead of the 1990s but due to the Cold War that the government > just did not want to have the air filled with talking machines and > interfere with govt wireless walkie talkies. Has anyone traced the > size of walkie talkies from 1950 onwards of the government and did > they come in small size? > Does anyone know if it is inherently unable for wireless to ever > become as secure as wired computer? Wireless is inherently less > stable than wired due to the nature of the medium of air compared to > the wire. But what about the security issue? All these issues are tractable. People offer 802.11 service all over the place, and 802.16 type service has been offered, although 802.16 has yet to really made it to the market. > Archimedes Plutonium > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies -- Les Cargill === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless Speaking of wireless and size........... I believe the govt now has video and audio transmitters that replicate ßoating dust. An operative can enter a room, brush the dust from his Truth but Scary! > That is the point. The walkie-talkie were bulky and their range was very > limited. Solid state technology with low power consumption is what > turned the trick. > Bob Kolker > I kind of thought that wireless could have naturally arrived in the > 1970s instead of the 1990s but due to the Cold War that the government > just did not want to have the air filled with talking machines and > interfere with govt wireless walkie talkies. Has anyone traced the > size of walkie talkies from 1950 onwards of the government and did > they come in small size? > Does anyone know if it is inherently unable for wireless to ever > become as secure as wired computer? Wireless is inherently less > stable than wired due to the nature of the medium of air compared to > the wire. But what about the security issue? > Archimedes Plutonium > whole entire Universe is just one big atom where dots > of the electron-dot-cloud are galaxies === Subject: Re: OptimalStrategy via VonNeumann Gametheory on Playing the Stock Market; wireless >Can someone answer me a question about technology of wireless. It >seems to me that the USA had wireless technology way back to around >say 1950. So my question would be why did it take 50 years before it >could be commercially applied? Switching. === Subject: expectation problem BBC Radio2s Drivetime show has a phone-in guess the voice competition. When a new voice starts, the prize is 30 quid, and it goes up 10 quid each day nobody wins (each day two new players get one guess each) One day this week, the DJ said something implying that if it was a hard voice to guess, the BBC bosses preferred it as it saved them money. My instant gut reaction was that this couldnt be right, but I was driving and too busy with traffic to give it much thought. Im going to scribble some stuff on the back of an old envelope as soon as my 4 month old goes to sleep for the night (well, part of the night), but Id like to see what the experts here have to say as well. In a nutshell: To keep total prize money paid to a minimum, how fast do you want the voice guessed? Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc Oh yes: the prize is capped at 300 If the first guesser of the day gets it, 2nd guesser gets a brand new voice to have a go at, at the baseline 30 quid prize. === Subject: Re: expectation problem > BBC Radio2s Drivetime show has a phone-in guess the voice > competition. > When a new voice starts, the prize is 30 quid, and it goes up 10 quid > each day nobody wins (each day two new players get one guess each) > One day this week, the DJ said something implying that if it was a hard > voice to guess, the BBC bosses preferred it as it saved them money. > My instant gut reaction was that this couldnt be right, but I was > driving and too busy with traffic to give it much thought. > Im going to scribble some stuff on the back of an old envelope as soon > as my 4 month old goes to sleep for the night (well, part of the night), > but Id like to see what the experts here have to say as well. > In a nutshell: > To keep total prize money paid to a minimum, how fast do you want the > voice guessed? > Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc > Oh yes: the prize is capped at 300 > If the first guesser of the day gets it, 2nd guesser gets a brand new > voice to have a go at, at the baseline 30 quid prize. Max Cost = 60 perday min cost 300 paid once Unguess cost 10 perday for 27 days then 0 0, 10 < 30 If 2 prizes perday 10*2 ==> 20 20 < 30 cost per day unguessed 20 if guessed cost per day 30 -- If its Monday then I am a fool but not ignorant. === Subject: Re: expectation problem === >Subject: expectation problem >Message-id: competition. >When a new voice starts, the prize is 30 quid, and it goes up 10 quid >each day nobody wins (each day two new players get one guess each) >One day this week, the DJ said something implying that if it was a hard >voice to guess, the BBC bosses preferred it as it saved them money. >My instant gut reaction was that this couldnt be right, but I was >driving and too busy with traffic to give it much thought. >Im going to scribble some stuff on the back of an old envelope as soon >as my 4 month old goes to sleep for the night (well, part of the night), >but Id like to see what the experts here have to say as well. >In a nutshell: >To keep total prize money paid to a minimum, how fast do you want the >voice guessed? >Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >Oh yes: the prize is capped at 300 >If the first guesser of the day gets it, 2nd guesser gets a brand new >voice to have a go at, at the baseline 30 quid prize. If the voices are so easy that everyone gets it right, then the BBC bosses pay out 300 quid every week. If the voice is so hard that only the last caller on Friday gets it, they pay out 70 quid, saving the bosses 230 quid. Sure the prize keeps going up, but the fewer winners more than makes up for it. Suppose you spend $10000 to make 1000 trinkets for sale at the local fair. You know that you could sell all 1000 if you priced them at $10, but then you would make no profit. So the higher you price them the more money you make. But there is a catch. For every $1 you raise the price, you lose 50 sales. What do you price them at to maximize your profit? -- Mensanator Ace of Clubs === Subject: Re: expectation problem >In a nutshell: >To keep total prize money paid to a minimum, how fast do you want the >voice guessed? >Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >Oh yes: the prize is capped at 300 >If the first guesser of the day gets it, 2nd guesser gets a brand new >voice to have a go at, at the baseline 30 quid prize. This depends on how long the station intends to run the contest. Let: n = number of days between winners d = days in total Assume all voices are guessed after the same number of days and somebody wins as the contest ends. Then the payout for the station is: f(n, d) = 30 * (d / n) + 10 n this is minimized when 30 * d (-1/n^2) + 10 = 0 => n = Sqrt(3d) So lets say the contest runs for 60 days. Then the optimal difficulty is Sqrt(180) ~= 13,4 days between guesses, with a payout of £268. If we include the £300 maximum single payout, it gets trickier if the contest draws out. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: expectation problem >>In a nutshell: >>To keep total prize money paid to a minimum, how fast do you want the >>voice guessed? >>Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >>Oh yes: the prize is capped at 300 >>If the first guesser of the day gets it, 2nd guesser gets a brand new >>voice to have a go at, at the baseline 30 quid prize. >This depends on how long the station intends to run the contest. Let: >n = number of days between winners >d = days in total >Assume all voices are guessed after the same number of days and >somebody wins as the contest ends. Then the payout for the station is: >f(n, d) = 30 * (d / n) + 10 n Of course, this is wrong. The optimal strategy is to make the questions as hard as possible. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: expectation problem Toni Lassila wibbled: >In a nutshell: >To keep total prize money paid to a minimum, how fast do you want the >voice guessed? >Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >Oh yes: the prize is capped at 300 >If the first guesser of the day gets it, 2nd guesser gets a brand new >voice to have a go at, at the baseline 30 quid prize. > This depends on how long the station intends to run the contest. Let: We can assume its indefinitely. > n = number of days between winners > d = days in total > Assume all voices are guessed after the same number of days and > somebody wins as the contest ends. Then the payout for the station is: > f(n, d) = 30 * (d / n) + 10 n > this is minimized when > 30 * d (-1/n^2) + 10 = 0 > => n = Sqrt(3d) > So lets say the contest runs for 60 days. Then the optimal difficulty > is Sqrt(180) ~= 13,4 days between guesses, with a payout of £268. If > we include the £300 maximum single payout, it gets trickier if the > contest draws out. -- Where oh where is my 32? === Subject: Re: expectation problem === >Subject: Re: expectation problem nutshell: >>To keep total prize money paid to a minimum, how fast do you want the >>voice guessed? >>Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >>Oh yes: the prize is capped at 300 >>If the first guesser of the day gets it, 2nd guesser gets a brand new >>voice to have a go at, at the baseline 30 quid prize. >This depends on how long the station intends to run the contest. Let: >n = number of days between winners >d = days in total >Assume all voices are guessed after the same number of days and >somebody wins as the contest ends. Then the payout for the station is: >f(n, d) = 30 * (d / n) + 10 n >this is minimized when > 30 * d (-1/n^2) + 10 = 0 >=> n = Sqrt(3d) >So lets say the contest runs for 60 days. Then the optimal difficulty >is Sqrt(180) ~= 13,4 days between guesses, with a payout of £268. If >we include the £300 maximum single payout, it gets trickier if the >contest draws out. So in 60 days there are 4 winners each collecting 268 quid, so the BBC is out 1072 quid. How can that be considered a minimum when they would only pay 300 quid if the contest ran the full 60 days without a winner? >-- >Im not interested in mathematics that might have anything >to do with reality. -- Russell Easterly, in sci.math -- Mensanator Ace of Clubs === Subject: Re: expectation problem === >>Subject: Re: expectation problem nutshell: >To keep total prize money paid to a minimum, how fast do you want the >voice guessed? >Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >Oh yes: the prize is capped at 300 >If the first guesser of the day gets it, 2nd guesser gets a brand new >voice to have a go at, at the baseline 30 quid prize. >>This depends on how long the station intends to run the contest. Let: >>n = number of days between winners >>d = days in total >>Assume all voices are guessed after the same number of days and >>somebody wins as the contest ends. Then the payout for the station is: >>f(n, d) = 30 * (d / n) + 10 n >>this is minimized when >> 30 * d (-1/n^2) + 10 = 0 >>=> n = Sqrt(3d) >>So lets say the contest runs for 60 days. Then the optimal difficulty >>is Sqrt(180) ~= 13,4 days between guesses, with a payout of £268. If >>we include the £300 maximum single payout, it gets trickier if the >>contest draws out. > So in 60 days there are 4 winners each collecting 268 quid, > so the BBC is out 1072 quid. How can that be considered a > minimum when they would only pay 300 quid if the contest > ran the full 60 days without a winner? >>-- >>Im not interested in mathematics that might have anything >>to do with reality. -- Russell Easterly, in sci.math > -- > Mensanator > Ace of Clubs Accounting 60 * 60 = 3600 total liablity of contest each day the contest runs the liablity is reduced by 20 (& 30 after 27days) or 0 if guessed Carl -- If its Monday then I am a fool but not ignorant. === Subject: Re: expectation problem === >Subject: Re: expectation problem >Message-id: === >Subject: Re: expectation problem nutshell: >>To keep total prize money paid to a minimum, how fast do you want the >>voice guessed? >>Recap: Guess 1&2 win 30; 3&4 win 40; 5&6 win 50... etc >>Oh yes: the prize is capped at 300 >>If the first guesser of the day gets it, 2nd guesser gets a brand new >>voice to have a go at, at the baseline 30 quid prize. >This depends on how long the station intends to run the contest. Let: >n = number of days between winners >d = days in total >Assume all voices are guessed after the same number of days and >somebody wins as the contest ends. Then the payout for the station is: >f(n, d) = 30 * (d / n) + 10 n >this is minimized when > 30 * d (-1/n^2) + 10 = 0 >=> n = Sqrt(3d) >So lets say the contest runs for 60 days. Then the optimal difficulty >is Sqrt(180) ~= 13,4 days between guesses, with a payout of £268. If >we include the £300 maximum single payout, it gets trickier if the >contest draws out. >> So in 60 days there are 4 winners each collecting 268 quid, >> so the BBC is out 1072 quid. How can that be considered a >> minimum when they would only pay 300 quid if the contest >> ran the full 60 days without a winner? >-- >Im not interested in mathematics that might have anything >to do with reality. -- Russell Easterly, in sci.math >> -- >> Mensanator >> Ace of Clubs >Accounting >60 * 60 = 3600 total liablity of contest >each day the contest runs the liablity is reduced by 20 (& 30 after >27days) or 0 if guessed So do you agree that to keep total prize money paid to a minimum, how fast do you want the voice guessed? was a stupid question because you dont want the voice guessed at all? If there was a local minimum, then the question would make sense, but the prize schedule has no local minimum. To see an example of such a phenomena, suppose that once the prize reaches 100 quid, the daily increments are upped to 100 and the maximum is set at 1000 (keeping the rulle that if no one gets it at the 1000 quid level, the question is withdrawn). Then you would have this: number right wrong max savings 1 30 30 30 0 2 30 30 60 30 3 30 40 90 50 4 30 40 120 80 5 30 50 150 100 6 30 50 180 130 7 30 60 210 150 8 30 60 240 180 9 30 70 270 200 10 30 70 300 230 11 30 80 330 250 12 30 80 360 280 13 30 90 390 300 14 30 90 420 330 15 30 100 450 350 16 30 100 480 380 17 30 200 510 310 18 30 200 540 340 19 30 300 570 270 20 30 300 600 300 21 30 400 630 230 22 30 400 660 260 23 30 500 690 190 24 30 500 720 220 25 30 600 750 150 26 30 600 780 180 27 30 700 810 110 28 30 700 840 140 29 30 800 870 70 30 30 800 900 100 31 30 900 930 30 32 30 900 960 60 33 30 1000 990 -10 34 30 1000 1020 20 In this schedule, there is clearly a point that minimizes the liability to the BBC. They would want the questions hard enough such that the audience needs 16 tries to get it right, but no more than 16. With the original schedule, there is no motivation to moderate the difficulty. >Carl >-- >If its Monday then I am a fool but not ignorant. -- Mensanator Ace of Clubs === Subject: Re: expectation problem === >>Subject: Re: expectation problem >>Message-id: === >>Subject: Re: expectation problem nutshell: >To keep total prize money paid to a minimum, how fast do you want the >voice guessed? Never But this would be a promotional gimmick the cost of witch is fixed. the comp would stop when the budget is use. > In this schedule, there is clearly a point that minimizes > the liability to the BBC. They would want the questions > hard enough such that the audience needs 16 tries to > get it right, but no more than 16. the goal of the BBC is promotion the cost is FIXED > With the original schedule, there is no motivation to > moderate the difficulty. >>Carl >>-- >>If its Monday then I am a fool but not ignorant. > -- > Mensanator > Ace of Clubs This kind of thing is promotion not something to minmize, the cost is in a sense fixed ie no more than 60 per day. the promoter has to balance reward against excitment, -- If its Monday then I am a fool but not ignorant. === Subject: Re: expectation problem Mensanator wibbled: > So in 60 days there are 4 winners each collecting 268 quid, > so the BBC is out 1072 quid. How can that be considered a > minimum when they would only pay 300 quid if the contest > ran the full 60 days without a winner? Ah. My gf has told me the caveat to the 300quid cap: after 2 failed guesses at that point it becomes a write-in, and a new voice goes out for the phone-in. === Subject: Re: The Question is: What is The Question? > http://www.holidays.net/passover/question.html > Contextual background: > Orthodox micro-quantum theory in the Gell-Mann/Hartle decoherent > consistent histories interpretation, taken naively to the macro-realm > as in conventional quantum cosmology and quantum gravity with the > Wheeler-Dewitt timeless eq. HPSI(Universe) = 0 in Wheelers > superspace of 3D space-like geometries in the canonical 3+1 split > (ADM), has many BIT thought-like parallel universes (attractor basins > sans system point in Susskinds non-Bohmian Landscape that is subject > to the Darwinian natural selection of the Weak Anthropic Principle (WAP) > for the arbitrary parameters in the standard models of lepto-quarks > and chaotic inßationary cosmology): > These universes differ not only in the answers that the theory gives to > questions, but by the questions that are asked. Everetts relative-state > form of the theory, must therefore be supplemented by a theory of why > what we observe corresponds to the answers of certain questions, and not > to an infinite number of other questions. p. 43 Why you get answers to any question is known. Its been known since before ever an idiot named Newton arrived, with his pathetic tribe of Quantum mathematoids who couldnt think their way into a discount Anthropology class, nevermind a computer class. === Subject: Re: The Question is: What is The Question? http://www.holidays.net/passover/question.html > Contextual background: > Orthodox micro-quantum theory in the Gell-Mann/Hartle decoherent > consistent histories interpretation, taken naively to the macro-realm > as in conventional quantum cosmology and quantum gravity with the > Wheeler-Dewitt timeless eq. HPSI(Universe) = 0 in Wheelers > superspace of 3D space-like geometries in the canonical 3+1 split > (ADM), has many BIT thought-like parallel universes (attractor basins > sans system point in Susskinds non-Bohmian Landscape that is subject > to the Darwinian natural selection of the Weak Anthropic Principle (WAP) > for the arbitrary parameters in the standard models of lepto-quarks > and chaotic inßationary cosmology): These universes differ not only in the answers that the theory gives to > questions, but by the questions that are asked. Everetts relative-state > form of the theory, must therefore be supplemented by a theory of why > what we observe corresponds to the answers of certain questions, and not > to an infinite number of other questions. p. 43 > Why you get answers to any question is known. Its been known > since before ever an idiot named Newton arrived, > with his pathetic tribe of Quantum mathematoids > who couldnt think their way into a > discount Anthropology class, nevermind > a computer class. This _has_ to be computer generated. Ôcid Ôooh === Subject: Can BOIDS (AI life) algorithm be formulated in mathematics? Id be interested in formulating this algorithm in mathematics, e.g., in a Non-linear programming form. === Subject: Knuth Toom-Cook, exercise 4.3.3-(4) Has anyone implemented the n-way Toom-Cook multiplication algorithm from Knuth 4.3.3-T with the improvements from exercise 4.3.3-(4)? I want pointers to how to implement the Toom-Cook algorithm with evaluation points at $-r,..0..r$ or at $1^2,2^2,4^2,8^2,..,(2^r)^2$. Basically, the Toom Cook algorithm evaluates $W(x)=U(x) V(x)$ at arbitrary points amounting to $2 r + 1$ evaluations. Next it solves for the coefficients of W(x) by linear algebra with those evaluations of $W(x)$. The linear algebra is efficient, especially for large values of $r$, because of special properties between the values $W(p_{i+1}) - W(p_i)$ and the coefficients of $W(x)$. Can anyone help me implement the Toom-Cook algorithm at the evaluation points named above? P.S. I would also like to know of efficient ways to exactly divide large numbers by numbers of the form $2^k (2^j - 1)$. === Subject: Re: Knuth Toom-Cook, exercise 4.3.3-(4) > Has anyone implemented the n-way Toom-Cook multiplication algorithm from > Knuth 4.3.3-T with the improvements from exercise 4.3.3-(4)? I want pointers > to how to implement the Toom-Cook algorithm with evaluation points at > $-r,..0..r$ or at $1^2,2^2,4^2,8^2,..,(2^r)^2$. > Basically, the Toom Cook algorithm evaluates $W(x)=U(x) V(x)$ at arbitrary > points amounting to $2 r + 1$ evaluations. Next it solves for the > coefficients of W(x) by linear algebra with those evaluations of $W(x)$. The > linear algebra is efficient, especially for large values of $r$, because of > special properties between the values $W(p_{i+1}) - W(p_i)$ and the > coefficients of $W(x)$. > Can anyone help me implement the Toom-Cook algorithm at the evaluation > points named above? > P.S. I would also like to know of efficient ways to exactly divide large > numbers by numbers of the form $2^k (2^j - 1)$. LibTomMath [http://math.libtomcrypt.org] has an implementation of 2 and 3-way Toom-Cook and in the textbook I discuss polynomial basis multiplication (note: that section needs a lot of revision but the basic ideas are there) Tom === Subject: Re: easy...analysis problem.... > 2. let any a in E , n-derivative f^(n) (a) exist. > express f^(n) (a) by limit of f(x) ???? I didnt understand this second problem. Could you clarify, please? Artur === Subject: Re: easy...analysis problem.... Another hint for the first problem: since f(a) exists, f(a) exists > and f(a+h) = f(a) + h f(a) + (h2/2)f(a) + o(h^2). > Is that true? Taylors theorem only guarantees that > f(a+h) = f(a) + h f(a) + h^2/2 f(z) > for some z between a and a+h; but were NOT told that f is > continuous, only that it exists throughout the open set. > That was why I suggested LH. Im a bit surprised that this result is not so known as I thought it was. But actually, theres a theorem that says: Suppose f is defined on an open interval I and its n_th derivative exists at a point a of I. Then, for every h such that (a-h, a+h) is contained in I, we have f(a+h) = f(a) + hf(a)....+ h^n/(n!) f_n(a) + o (h^n). The only assumption is the existence of f_n(a). We dont need to assume f_n exists, let alone is continuous, in a neighborhood of a. This is not Taylors Theorem, which does require f_n to exist in a neighborhood of a. The proof is by induction on n (at least the classical proof Im aware of). First, we see that, if n=1, then the very definition of derivative implies that f(a+h) = f(a) + h f(a) + o(h). This is not the Mean Value Theorem, which is the particular case of Taylors Theorem for n=1. This comes just from the definition of derivative and all thats required is that f exists at a. Nothing more. This sets the base for the induction process. To make things simpler, we can admit, WLOG, that a=0 and f(a), as well as all of the first n derivatives of f, vanish at a. The very existence of f(a) implies the existence of f in a neighborhood of a, which, in turn, implies continuity of f in a neighborhood of a, say the interval (-d, d). If h is in (0,d), then the Mean Value Theorem holds in [0,h] and says there is a y between 0 and h such that f(h)/h = (f(h) - f(0))/(h-0) = f(y). Since y is in (0,d), the existence of f(0) implies that f(y) = f(0) + y f(0) + g(y) = g(y), where g(y)/y ->0 as y->0. Therefore, f(h)/h = g(y). Since y->0 as h ->0, we can write f(h)/h = o(h), which is the same as f(h) = o(h^2). If y is in (-d,0), we apply the same arguments. If n >2 and we assume the condition holds for n-1, then all we have to do is observe that f_n-1 is the derivative of f_n-2 and f_n is the second derivative of f_n-2. Exactly the same arguments show f(h) = o(h^n). The reason why its no loss of generality to assume a=0 and f(a) = f(a) = f_n(a)=0 is that we can always define g by g(x) = f(x) - (f(0) + x f(0)...+...x^n/(n!) f_n(0)), so that g(0) = g(0) = ... = g_n(0). If we apply the theorem to g, we get g(h) = o(h^n), hence f(h) = f(0) + h f(0)...+...h^n/(n!) f_n(0) + o(h^n). Maybe, from a strictly mathematical point of view, its a loss of generality to zero f(a) and its derivatives like I did. Maybe the most precise way to carry out the proof is to first prove the theorem for the simplified case and then extend the result to the general case. The important thing here is all we have to assume is the existence of f_n(a). This allows us to prove those theorems about relative maximum and minimum of f at a point a without assuming continuity or even existence of f_n in a neighborhood of a. For instance, if n is even and f, ....f_n-1 vanish at a and f_n(a)>0, then the equation f(a+h) - f(a) = (h^n)/(n!) + o(h^n) holds and clearly tells us f has a relative minimum at a. Except for f_n(a) we dont need any information about the existence, behavior or whatever of f in a neighborhhod of a. Im a bit surprised that I was aware of this theorem and Ron Bruck was not. I was introduced to it in Luenbergers book about non-Linear Programming. He used it to prove optmality conditions for constrained problems, those based on the Hessian and the gradient of functions from R^n to R. Artur PS. Of course, I dont have the slightest intention of teaching anything to any of you. Im just an engineer who likes math, and the difference between your knowledge and mine is greater than any M>0 arbitrarily chosen... === Subject: Re: hyperbolas > I would like to know how to find the gradient of the tangent at the > turning point of a hyperbola that doesnt have asymptotes going > straight up and straight across. Finding the turning point of a nice > simple hyperbola is easy since the gradient is 1 or -1, thats > obvious, but Im not quite sure how to find them for other The easiest method will depend a good bit on the form of equation giving the hyperbola. If it is known that the center of symmetry is at origin, with both axes of symmetry passing through origin, one can look for the two symmetrically placed points of the hyperbola which are closest to origin. That is, minimize the value of x^2 + y^2 subject to the requirement that (x,y) be on the hyperbola. More generally: The general xy-form of equation for an arbitrary conic section is A*x^2 + B*x*y + C*y^2 + D*x + E*y + F = 0, for suitable constants A,B,C,D,E,and F. Your simple case with vertical and horizontal asymptotes is one with A = C = 0 If the conic is known to be a hyperbola, one can show that B^2 > 4*A*C. You might check that this is the case with your equation. If B is not zero, the usual procedure for analysing such conics is to start with a rotation of coordinates to make a new axis parallel to an axis of symmetry of the conic, i.e., substituting for x and y with x = x*cos(t) - y*sin(t) and y = x*sin(t) + y*cos(t), where (A-C)/B = cotan(2*t). This is a tedious process at best, and shold be a last resort. === Subject: Probability of finding a Prime Number I have placed a paper on my web site at http://www.guffy.net/primes.pdf. In it I prove the following: Given the prime numbers 3 to p, the probability of finding the next prime is exactly: Product((p-1)/p) The paper is 3 pages and contains 4 equations. I would appreciate any comments. David === Subject: Re: Probability of finding a Prime Number > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) That statement is meaningless. > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. Meaningless statements dont cause me to run off and fetch papers on the same subject by the same author, so not read. Phil -- Unpatched IE vulnerability: Notepad popups Description: Opening popup windows without scripting Reference: http://computerbytesman.com/security/notepadpopups.htm Followup: http://msgs.securepoint.com/cgi-bin/get/bugtraq0308/55.html === Subject: Re: Probability of finding a Prime Number I do think that I have stated things correctly. Yes, the probability of finding a prime greater than p is 1. However, the probability of finding the next prime is not 1. You can think of it by asking how far past p do you have to go before you get to the next prime? If the probability is 1/10 then you can expect to find a prime somewhere between p and p+10. If the probability if 1/100 then you can expect to find a prime somewhere between p and p+100. Exactly where cant be determined. What I am fairly certain I have discovered is that this probability depends only on the primes from 3 to p and equals the product((p-1)/p). David > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. > David === Subject: Re: Probability of finding a Prime Number > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. > David Check this: http://mathworld.wolfram.com/MertensTheorem.html Leroy Quet === Subject: Re: Probability of finding a Prime Number > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. I could write the same paper, with one page and zero equations, and get a more accurate answer. Given a finite number of primes, the probability of finding the next prime is exactly: 1 (100%). Doug === Subject: Re: Probability of finding a Prime Number > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next > prime is > exactly: > Product((p-1)/p) > The paper is 3 pages and contains 4 equations. I would appreciate any > comments. > I could write the same paper, with one page and zero equations, and get a > more accurate answer. Given a finite number of primes, the probability of > finding the next prime is exactly: 1 (100%). You need to add Ôcontiguous and Ôfrom 2, or similar. e.g. from {7, 97, Phi(6,1463635^8192) } you cant find the next prime (whatever that might mean). Phil -- Unpatched IE vulnerability: mhtml wecerr CAB ßip Description: Delivery and installation of an executable Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0305/48.html === Subject: Re: Probability of finding a Prime Number more accurate answer. Given a finite number of primes, the probability of > finding the next prime is exactly: 1 (100%). Obviously, the author is not expressing himself accurately and perhaps maybe he doesnt have a good understanding of what probability is. But if you read the paper, it is obvious what he is trying to say. Given some n, the probability that is is not a multiple of 2 is 1/2, not a multiple of 3 is 2/3, not a multiple of 5 is 4/5, etc, so it is the product of (p-1)/p for all primes less than n. Then he basically goes on to say the some sort of thing when choosing an odd number. And then he starts comparing the product of (p-1)/p to the analytic results of the prime number theorem. J === Subject: Re: Probability of finding a Prime Number This is pseudo mathematical gibberish. It isnt even wrong. It is semantically void. If I (or anyone else) has a list of all the primes from 3 to p, the probability that I (or anyone else) can find the next time is 1. I can ALWAYS find the next prime. You can lead a horses ass to knowledge, but you cant make him think. === Subject: Re: Probability of finding a Prime Number I found this in one of your posts (theyre easy to find given that you sign everything with that horses ass comment): The probability that a large integer N is prime is the probability that it is not divisible by 2,3,5,7,11...sqrt(N). This probability is (1-1/2) * (1-1/3) * (1-1/5) .... all the primes up to N. My reasons for doing so are in the paper. Admittedly, stopping at sqrt(N) is the correct approach). So, the conclusion I came to is nothing new. However I think the approach I describe in the paper is relatively novel -- at least I enjoyed discovering it. On a personal note, I am an amateur who enjoys reading about math and trying to make some serious attempts at solving some problems. Your response (as well as some others) to my oringinal post was very discouraging. David > This is pseudo mathematical gibberish. > It isnt even wrong. It is semantically void. > If I (or anyone else) has a list of all the primes > from 3 to p, the probability that I (or anyone else) > can find the next time is 1. I can ALWAYS > find the next prime. > You can lead a horses ass to knowledge, but you cant make him think. === Subject: Re: Probability of finding a Prime Number > The probability that a large integer N is prime is the probability that > it is not divisible by 2,3,5,7,11...sqrt(N). This > probability is (1-1/2) * (1-1/3) * (1-1/5) .... > using > all the primes up to N. My reasons for doing so are in the paper. > Admittedly, stopping at sqrt(N) is the correct approach). ... > On a personal note, I am an amateur who enjoys reading about math and > trying > to make some serious attempts at solving some problems. Your response (as > well as some others) to my oringinal post was very discouraging. I thought Bob Silverman was a little harsh, although I found your first posting difficult to interpret. But if you test whether a number is divisible by any prime up to N, then surely you wouldnt find any new primes near N? Youd have to go up to N^2 at least. -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Probability of finding a Prime Number Yes, the probability that you will find a next prime is 1. What I am addressing is how far past p you have to go. If you go to infinity, yes you will certainly find another prime. What is the probability that a number in the neighborhood of n is prime? According to the prime number theorem it is 1/log(n). I argue that it is product((p-1)/p) where p takes on the prime values between 3 and n. In other words it only depends on the primes less than n and not on n itself. So, if you know the first m primes, then you can calculate the probability of finding the next prime. This is what I attempt to explain in my paper and I am pretty certain the arguments are sound. David > This is pseudo mathematical gibberish. > It isnt even wrong. It is semantically void. > If I (or anyone else) has a list of all the primes > from 3 to p, the probability that I (or anyone else) > can find the next time is 1. I can ALWAYS > find the next prime. > You can lead a horses ass to knowledge, but you cant make him think. === Subject: Re: Probability of finding a Prime Number > Yes, the probability that you will find a next prime is 1. What I am > addressing is how far past p you have to go. If you go to infinity, yes you > will certainly find another prime. > What is the probability that a number in the neighborhood of n is prime? > According to the prime number theorem it is 1/log(n). I argue that it is > product((p-1)/p) where p takes on the prime values between 3 and n. In > other words it only depends on the primes less than n and not on n itself. > So, if you know the first m primes, then you can calculate the probability > of finding the next prime. This is what I attempt to explain in my paper > and I am pretty certain the arguments are sound. I bet youve assumed that divisibility by each of the primes is independent of each other. However, the number 0 puts a big hole in that assumption -- its divisible by all primes. I.e. the primes are not entirely unrelated to each other, theyve contrived a divisibility coincidence at zero. (The mere fact that each one is defined in terms of properties of smaller ones is a dead giveaway to their non-indepenedence too.) This changes the probabilities noticably. Go research Eulers Gamma and Mertens Theorem. Phil -- Unpatched IE vulnerability: WMP local file bounce Description: Switching security zone, arbitrary command execution, automatic email-borne command execution Reference: http://www.ntbugtraq.com/default.asp?pid=36&sid=1&A2=ind0307& L=ntbugtraq&F=P& S=&P=6783 Exploit: http://www.malware.com/once.again!.html === Subject: Re: Probability of finding a Prime Number > What is the probability that a number in the neighborhood of n is prime? > According to the prime number theorem it is 1/log(n). I argue that it > is product((p-1)/p) where p takes on the prime values between 3 and n. The point the prime number theorems analytic results is that it is supposed to capture (an approximation to) the distribution of primes (or at least give some information about the distributions of primes up to some n). To calculate the product (p-1)/p over all p, you obviously need to know all p less than n, while no such requirements exist to calculate 1/log(n). > In other words it only depends on the primes less than n and not on n > itself. Well, thats obvious, but - again - the point to the prime number theorem is to give some idea of what PROD((p-1)/p) looks at at any specific n. > So, if you know the first m primes, then you can calculate the > probability of finding the next prime. This is what I attempt to > explain in my paper and I am pretty certain the arguments are sound. They are sound, and its all correct, but nothing new. I would think that any curious highschooler fiddling with primes would come across a similar notion. J === Subject: Re: Probability of finding a Prime Number > I have placed a paper on my web site at http://www.guffy.net/primes.pdf. > In it I prove the following: > Given the prime numbers 3 to p, the probability of finding the next prime is > exactly: > Product((p-1)/p) Let p = 5. Then the probability of finding the next prime (7) is exactly 4/5. What does this mean? Gib === Subject: Re: Involutionary Calculus > > Consider the following definitions: 1)The hyperderivative of a function f at a point x is defined as the limit as h approaches 0,of (f(x+h)/f(x))^(1/h),provided it exists. 2)The hyperintegral(or more properly the productal) of a function f over an > interval [a,b] over which it is positive is defined as the limit as n approaches infinity,of the continued product as k varies from 0 > to > n-1,of [f(a+(b-a)k/n)^1/n],provided it exists. It turns out to be that the two operations are related through following > theorem: > Suppose that F has f as its hyperderivative over some interval (a,b) and > continuous over [a,b] then > then the hyperintegral of f over [a,b] is F(b)/F(a). The calculus thus cerated has a structure analogous to the known calculus > and many theorems have their analogues. > It remains true,however,that the hyperderivative and the hyperintegral can > be expressed in terms of the known derivatives and known integrals. > If this idea is so much beautiful why is not it popular in mathematical > circles? We call it taking logs. >> I give a damn to what you call it or in what manner can the result be >> established.If you cannot see the inherent aesthetic appeal behind the >> results you better keep your comments with you. >> Uh, right. All hail Ashurosh, the discoverer of the amazing >> hypercalculus. >> The point is that your assumption that hypercalculus is not >> popular is simply incorrect. Its just calculus plus logarithms, >> and logarithms are _very_ popular... >> ************************ >> David C. Ullrich > The theory exists independent of logarithms.Only the connection between them > becomes obvious through the use of logarithms. > Thats not the only thing that becomes obvious through the use of > logarithms - one other thing is that the theory is really no big deal. > ************************ > David C. Ullrich I add that I do not claim that the theory is really new in the sense that it can be used to solve problems that are insoluble by the known methods since it is obvious that all problems that are demonstrable through the use of these ideas can be described in terms of known symblism and operations. My point was only to emphasize the aesthetical aspect of the theory. === Subject: Re: Involutionary Calculus >> The theory exists independent of logarithms.Only the connection between them >> becomes obvious through the use of logarithms. >> Thats not the only thing that becomes obvious through the use of >> logarithms - one other thing is that the theory is really no big deal. >> The idea seems new, interesting and promising. At least for monotonic >> functions it may provide speed of computation. An application may be >> in Fractals.Firstly, one might ask if a table of hyper derivatives >> and hyper integrals has been compiled for some functions like x^n, >> sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may ask >> more questions. How are the results related to normal derivatives and >> integrals? [ Surely one is not the log of the other always.] Have they >> been graphically interpreted ? Any work about existence, uniqueness >> etc.? Any hyper differential equations have been set up? >> How can one be so judgmental so soon about the deal size? When there >> is not enough clarity on these or similar matters, avoidance of >> suggestion to general condemnatives could perhaps better serve the >> aims of the newsgroup. >> Narasimham G.L. > I want to add few remarks. > We know that any infinitely differentiable (differentiable in the usual > sense of the word) can be written as an infinite polynomial. > Thats wrong for two reasons. First, you mean to say that any > infinitely differentiable function can be written as a _power > series_ - theres no such thing as an infinite polynomial. That is the way I define an infinite polynomial;as a formal power series. The fact that a person who can understand what an infinitely differential function means,objects the use of the term infinite polynomial is strange to me.Probably you mean to say that it is not a standard term in your mathematical dictionary.Point noted. > More important, its _not_ _true_ that any infinitely differentiable > function can be written as a power series. > We have an analogous theorem for the hyper derivative case. > It runs as follows: > Any function whose hyper derivatives exist upto all orders can be represented > as an infinite product of the following form: f(x)= Product as r varies from 0 to infinity of (A(r)^((x-a)^r)) > where A(r) is determined by a simple expression involving hyper derivatives > of f at a and is independent of x. > This theorem is completely analogous to the taylors theorem. > It remains true that this theorem has a trivial proof through the use of > logarithms. > Really? Id like to see the proof. The proof with the complete statement of the theorem runs as follows: Statement of the theorem: Given any function that is positive over an open interval and whose taylor series expansion converges to it over the interval has a representation as an infinite product of the above mentioned form which is convergent to the value of this function over this interval. (Denote the nth hyper derivative of f at x as H(f(x),n) and nth normal derivative as D(f(x),n)). The A(r)s are determined as follows: A(r)= D(f(a),r)^(1/r!). Since f(x)=Lim as r tends to infinity[(D(f(a),r)*((x-a)^k)/r!] (this is the assumption, obviously). Now write the infinite product as limit as n tends to infinity [e^(sum as r varies from 0 to n of (log(H(f(a),r))*((x-a)^r)/r!))]. This gives limit as n tends to infinity [e^(sum as r varies from 0 to n of (D(log(f(a)),r)*((x-a)^r)/r!))];using the fact that log(H(f(x),n))=D(f(x),n). Thus it follows that the given limit of the product is e^log(f(x)) or f(x); using taylors theorem on log(f(x)) over this interval. (Make certain to include a proof > of the _false_ version of the standard Taylors theorem that > youre evidently using.) Since the above statement was incomplete there is no question of a proof. > Perhaps I need to add further remarks as for the usefulness of a mathematical > theory as I understand it. > There are certain mathematical theories that have a kind of aesthetic > appeal in them.These theories need not only be studied for the sake of > their immediate usefulness but also for the sake of this appeal.The question > of applicability is of no doubt primary importance but the fact that no > immediate application is obvious doesnt imply its non existence. > ************************ > David C. Ullrich === Subject: Re: Involutionary Calculus > I want to add few remarks. >> We know that any infinitely differentiable (differentiable in the >> usual sense of the word) can be written as an infinite polynomial. >> Thats wrong for two reasons. First, you mean to say that any >> infinitely differentiable function can be written as a _power >> series_ - theres no such thing as an infinite polynomial. That is the way I define an infinite polynomial;as a formal power > series. The fact that a person who can understand what an infinitely > differential function means,objects the use of the term infinite > polynomial is strange to > me.Probably you mean to say that it is not a standard term in your > mathematical dictionary.Point noted. Dunno whether its a standard term in any mathematical dictionary. Citations to any mathematical dictionary containing it would be welcome. Anyway, its not true that a C^infinity function is analytic --- it isnt even locally true. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Involutionary Calculus >> That is the way I define an infinite polynomial;as a formal power >> series. >Dunno whether its a standard term in any mathematical dictionary. >Citations to any mathematical dictionary containing it would be >welcome. I wonder how many theorems about polynomials in general would be invalidated or at least rendered trivial if one included power series as polynomials? Does a power series have infinitely many zeros according to the fundamental theorem of algebra, etc. -- Im not interested in mathematics that might have anything to do with reality. -- Russell Easterly, in sci.math === Subject: Re: Involutionary Calculus > That is the way I define an infinite polynomial;as a formal power > series. >>Dunno whether its a standard term in any mathematical dictionary. >>Citations to any mathematical dictionary containing it would be >>welcome. >I wonder how many theorems about polynomials in general would be >invalidated or at least rendered trivial if one included power series >as polynomials? Lots. Infinitely many, in fact (although most of them are not yet _known_ to be true for actual polynomials...) >Does a power series have infinitely many zeros >according to the fundamental theorem of algebra, etc. Certainly not. ************************ David C. Ullrich === Subject: Re: Involutionary Calculus > That is the way I define an infinite polynomial;as a formal power > series. >>Dunno whether its a standard term in any mathematical dictionary. >>Citations to any mathematical dictionary containing it would be >>welcome. > I wonder how many theorems about polynomials in general would be > invalidated or at least rendered trivial if one included power series > as polynomials? Does a power series have infinitely many zeros > according to the fundamental theorem of algebra, etc. What are the infinitely many zeroes of 1 + z + z^2/2 + z^3/3! + .... ? :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: Involutionary Calculus The theory exists independent of logarithms.Only the connection between them > becomes obvious through the use of logarithms. > Thats not the only thing that becomes obvious through the use of > logarithms - one other thing is that the theory is really no big deal. The idea seems new, interesting and promising. At least for monotonic > functions it may provide speed of computation. An application may be > in Fractals.Firstly, one might ask if a table of hyper derivatives > and hyper integrals has been compiled for some functions like x^n, > sin(x), log(x), exp(x), cosh(x) etc. Are there references? One may ask > more questions. How are the results related to normal derivatives and > integrals? [ Surely one is not the log of the other always.] Have they > been graphically interpreted ? Any work about existence, uniqueness > etc.? Any hyper differential equations have been set up? How can one be so judgmental so soon about the deal size? When there > is not enough clarity on these or similar matters, avoidance of > suggestion to general condemnatives could perhaps better serve the > aims of the newsgroup. Narasimham G.L. >> I want to add few remarks. >> We know that any infinitely differentiable (differentiable in the usual >> sense of the word) can be written as an infinite polynomial. >> Thats wrong for two reasons. First, you mean to say that any >> infinitely differentiable function can be written as a _power >> series_ - theres no such thing as an infinite polynomial. > That is the way I define an infinite polynomial;as a formal power series. > The fact that a person who can understand what an infinitely differential > function means,objects the use of the term infinite polynomial is strange to > me. A lot of things are strange to me... >Probably you mean to say that it is not a standard term in your > mathematical dictionary.Point noted. >> More important, its _not_ _true_ that any infinitely differentiable >> function can be written as a power series. By the way, you dont seem to have noticed this... >> We have an analogous theorem for the hyper derivative case. >> It runs as follows: >> Any function whose hyper derivatives exist upto all orders can be represented >> as an infinite product of the following form: >> f(x)= Product as r varies from 0 to infinity of (A(r)^((x-a)^r)) >> where A(r) is determined by a simple expression involving hyper derivatives >> of f at a and is independent of x. >> This theorem is completely analogous to the taylors theorem. >> It remains true that this theorem has a trivial proof through the use of >> logarithms. >> Really? Id like to see the proof. > The proof with the complete statement of the theorem runs as follows: > Statement of the theorem: Given any function that is positive over an open > interval and whose taylor series expansion converges to it over the interval > has a representation as an infinite product of the above mentioned form > which is convergent to the value of this function over this interval. Well I believe _this_ - its clear because if f > 0 and f is analytic then log(f) is also analytic. This is not the result you stated the other day. > (Denote the nth hyper derivative of f at x as H(f(x),n) and nth normal > derivative as D(f(x),n)). > The A(r)s are determined as follows: > A(r)= D(f(a),r)^(1/r!). > Since > f(x)=Lim as r tends to infinity[(D(f(a),r)*((x-a)^k)/r!] (this is the > assumption, obviously). > Now write the infinite product as limit as n tends to infinity > [e^(sum as r varies from 0 to n of (log(H(f(a),r))*((x-a)^r)/r!))]. > This gives limit as n tends to infinity > [e^(sum as r varies from 0 to n of (D(log(f(a)),r)*((x-a)^r)/r!))];using the > fact that log(H(f(x),n))=D(f(x),n). > Thus it follows that the given limit of the product is e^log(f(x)) or f(x); > using taylors theorem on log(f(x)) over this interval. No, its not taylors theorem. >(Make certain to include a proof >> of the _false_ version of the standard Taylors theorem that >> youre evidently using.) > Since the above statement was incomplete there is no question of a proof. >> Perhaps I need to add further remarks as for the usefulness of a mathematical >> theory as I understand it. >> There are certain mathematical theories that have a kind of aesthetic >> appeal in them.These theories need not only be studied for the sake of >> their immediate usefulness but also for the sake of this appeal.The question >> of applicability is of no doubt primary importance but the fact that no >> immediate application is obvious doesnt imply its non existence. >> ************************ >> David C. Ullrich ************************ David C. Ullrich === Subject: Algebra programs I was wondering if anyone could give me a list of some of the top graduate programs in algebr (commutative algebra specifically). === Subject: Problem For a given commutative binary operation over the set of natural numbers construct another binary operation in the following manner: Let * be the commutative operation over N.Define ^ over N as a binary operation as: If n is a natural number then n^1=n and n^(k+1)=(n^k)*n (1) n^(k+1)=n*(n^k) (2) where k is any natural number. I raise the following question: What are the restrictions on * so that ^ is also commutative over N ? Suppose * is any binary operation over N.One can derive in general two binary operations from * by choosing the first or the second definition,as quoted above. I ask the following somewhat vague and more general question: What properties are in general common in * and ^? Certainly not assosiativity or commutativity. I require general theorems relating the properties of * and ^. Under which field of Modern algebra thus the abstraction of above concepts appear? === Subject: Re: Problem > For a given commutative binary operation over the set of natural numbers > construct another binary operation in the following manner: > Let * be the commutative operation over N.Define ^ over N as a binary > operation as: > If n is a natural number then > n^1=n and > n^(k+1)=(n^k)*n (1) > n^(k+1)=n*(n^k) (2) where k is any natural number. > I raise the following question: > What are the restrictions on * so that ^ is also commutative over N ? You want n^1 = 1^n? so n = 1*1*1*...*1 (n times). Thus n*1 = n+1. You want n^2 = 2^n? Then n*n = 2*2*...*2. etc. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Help Please I cant quite seem to get this proof to come out exactly right. Can If a, g E Sn , prove that the permutations a and g-1ag have the same parity.(ie they are either both even or both odd). g-1 is the inverse of g E means that a and g exist in Sn === Subject: Re: Help Please > I cant quite seem to get this proof to come out exactly right. Can > If a, g E Sn , prove that the permutations a and g-1ag have the same > parity.(ie they are either both even or both odd). > g-1 is the inverse of g > E means that a and g exist in Sn Erm... I like giving my students homework, too. How about posting *your* effort that you Ôcant seem to get to come out right and then you might get some help on getting it to come out correctly. But without showing me/us that you can understand the problem *and* have put some sweat equity into trying to find an answer, I think that my participation in your education, save for the Ôroyal road drachma, is over. Drieux === Subject: Bound of a sum I admit I am no math genius, but I need to find some way to bound k-1 n sum sum 2/(j-i+1) i=1 j=k+1 1 <= k<= n. I think (stupid empirical experiment) that the sum is bounded by about 0.7n for any value of k .. but I really cant figure how to make something looking like a sound proof. Any upper bound that is some constant by n is OK. Any ideas ? Soren -- Fjern de 4 bogstaver i min mailadresse som er indsat for at hindre s... Remove the 4 letter word meaning junk mail in my mail address. === Subject: Re: Bound of a sum >I admit I am no math genius, but I need to find some way to bound >k-1 n >sum sum 2/(j-i+1) >i=1 j=k+1 >1 <= k<= n. >I think (stupid empirical experiment) that the sum is bounded by about >0.7n for any value of k .. but I really cant figure how to make >something looking like a sound proof. >Any upper bound that is some constant by n is OK. Write the sum as S = sum_{r=3}^n f(r,k)/r where f(r,k) = min(k-1,n-r+1) - max(0,k-r+1). For r <= k we have f(r,k) <= (k-1) - (k-r+1) = r - 2, so sum_{r=3}^k f(r,k)/r < k. For r > k we use f(r,k) <= k - 1, so sum_{r=k+1}^n f(r,k)/r <= (k-1) sum_{r=k+1}^n 1/r <= (k-1) int_k^n 1/t dt = (k-1) ln(n/k) Thus S < k + (k-1) ln(n/k). As a function of k on [1,n] this is increasing (its derivative is 1/k + ln(n/k)), and so its maximum is at k=n, where its value is n. So S < n. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Bound of a sum > I admit I am no math genius, but I need to find some way to bound > k-1 n > sum sum 2/(j-i+1) > i=1 j=k+1 > 1 <= k<= n. > I think (stupid empirical experiment) that the sum is bounded by about > 0.7n for any value of k .. but I really cant figure how to make > something looking like a sound proof. This will be a little sketchy. Fact: There exists a constant C such that 1/k + 1(k+1) + ... + 1/n <= C*ln(n/k) for all n, 1 <= k <= n; this is easy to show. So sum(j=k+1,n) 1/(j-i) <= C*ln[(n-i)/(k+1-i)]. So we just need to consider C*sum (i=1,k) ln[(n-i)/(k+1-i)]. But that sum of logs is the log of the product. And that ends up looking a lot like ln(n choose k). Recall Pascals triangle shows (n choose k) is maximized when k = n/2 (at least for even n). So we need a bound on ln(n!/[(n/2)!]^2). But n!/[(n/2)!]^2 <= 2^n for even n, which is easily verified by induction. So were getting ln(2^n) = nln(2), which is looking good. I think the above can be made into a real proof. === Subject: Re: Bound of a sum > I admit I am no math genius, but I need to find some way to bound > k-1 n > sum sum 2/(j-i+1) > i=1 j=k+1 > 1 <= k<= n. > I think (stupid empirical experiment) that the sum is bounded by about > 0.7n for any value of k .. but I really cant figure how to make > something looking like a sound proof. > Any upper bound that is some constant by n is OK. > Any ideas ? > Soren An overly complicated _start_ to a solution, hopefully. I assume the double sum is, as it appears, a function of both k and n. (But k is bounded by n.) Your double-sum can be written as: s(k,n) = 2* sum{j=k+1 to n} (H(j) -H(j-k+1)), where H(m) = sum{j=1 to m} 1/j. So, s(k,n) = 2 * ((n+1)H(n) -n -(k+1)H(k) +k -(n-k+2)H(n-k+1) +n-k+1 -1), since: sum{k=1 to m} H(k) = (m+1)*H(m) - m. So we have: s(k,n) = 2 * ((n+1)H(n) -(k+1)H(k) -(n-k+2)H(n-k+1)). But H(n) ~ ln(n) + gamma +1/(2n) - O(1/n^2), where gamma is Eulers constant. So, s(k,n) = 2(n+1)(ln(n) +gamma) +(1+1/n) -2(k+1)(ln(k) +gamma) -(1+1/k) -2(n-k+2)(ln(n-k+1) +gamma) -(1-1/(n-k+1)) - O(1/n^2) - O(1/k^2)-O(1/(n-k+1)^2) And the O()s make sense in terms of n, k, and n-k+1 all increasing to infinity. Hopefully I did not make an error. Leroy Quet === Subject: Re: Bound of a sum > I admit I am no math genius, but I need to find some way to bound > k-1 n > sum sum 2/(j-i+1) > i=1 j=k+1 > 1 <= k<= n. > I think (stupid empirical experiment) that the sum is bounded by about > 0.7n for any value of k .. but I really cant figure how to make > something looking like a sound proof. oops 1.4 n, or maybe really 2ln2. > Any upper bound that is some constant by n is OK. > Any ideas ? > Soren Soren -- Fjern de 4 bogstaver i min mailadresse som er indsat for at hindre s... Remove the 4 letter word meaning junk mail in my mail address. === Subject: Re: Algebraic integers, simple result Correction to my previous post on this ... >What I discovered was an easy demonstration by looking at >(1+sqrt(-167) since (1+sqrt(-167))/2[(1-sqrt(-167))/2] = -7 This calculation by JSH is incorrect. The right side should be +42 = 7*6. >you might assume that (1+sqrt(-167)) has non-unit factors in common >with 7 in the ring of algebraic integers, but it does not. The way to quickly and easily see that is to consider >z = (1+sqrt(-167))/r_1 >where r_1 is presumed to be an algebraic integer factor of 7 that is >also a factor of (1+sqrt(-167)) so that z is an algebraic integer. >However, you find that if r_1 is not a unit, then z is the root of a >non-monic polynomial of order equal to 2 times the order of the >minimum polynomial with integer coefficients which has r_1 as a root. >And you know that polynomial which has z as a root, cant have a >rational root because neither (1+sqrt(-167))/r_k, nor >(1-sqrt(-167))/r_k, where k is some counting number equal to or >greater than 1 that is less than or equal to the order of the >polynomial, can be rational. >The polynomial must be non-monic because even if r_1 were a factor of >(1+sqrt(-167)) it wouldnt be a factor of (1-sqrt(-167)), and all of >the other roots cant be a factor of both either. >I thought Id use r_1 just to simplify things slightly. >Oh yeah, so there doesnt exist an algebraic integer r_1 that is a >non-unit factor of 7 where z can be an algebraic integer. >Its a fascinatingly simple to show result. Clearly, if your argument were correct, it also applies to (1 - sqrt(-167)). Therefore what you have shown is that the following theorem is false: [note: the following is corrected also] Theorem. If A and B are algebraic integers and A * B = 7*6 Then at least one of A or B must have algebraic integer factors in common with 7. Perhaps I should give the proof of the theorem so you can tell us all where it contains an error. Proof: Suppose that A does not have an algebraic integer factor in common with 7. Then A is coprime to 7. We therefore want to show that B is not coprime to 7. The fact that A is coprime to 7 means that there exist algebraic integers s and t such that A*s + 7*t = 1. Multiplying both sides by B, I have: A*B*s + 7*B*t = B. Since A*B = 7*6, this implies that 7*6*s + 7*B*t = B, or 7*(6*s + B*t) = B. That is, 7 itself is a factor of B. QED. Now: would you also like to know where your fascinatingly simple reasoning has gone wrong? Nora B. === Subject: Re: Algebraic integers, simple result > What I discovered was an easy demonstration by looking at > (1+sqrt(-167) since (1+sqrt(-167))/2[(1-sqrt(-167))/2] = -7 > you might assume that (1+sqrt(-167)) has non-unit factors in common > with 7 in the ring of algebraic integers, but it does not. We might, but must we? We *might* also assume that 7 divides (1-sqrt(-167))/2 We might assume many things. > The way to quickly and easily see that is to consider > z = (1+sqrt(-167))/r_1 > where r_1 is presumed to be an algebraic integer factor of 7 that is > also a factor of (1+sqrt(-167)) so that z is an algebraic integer. > However, you find that if r_1 is not a unit, then z is the root of a > non-monic polynomial of order equal to 2 times the order of the > minimum polynomial with integer coefficients which has r_1 as a root. > And you know that polynomial which has z as a root, cant have a > rational root because neither (1+sqrt(-167))/r_k, nor > (1-sqrt(-167))/r_k, where k is some counting number equal to or > greater than 1 that is less than or equal to the order of the > polynomial, can be rational. where have these r_k come from? So it has no rational root, and? > The polynomial must be non-monic because even if r_1 were a factor of > (1+sqrt(-167)) it wouldnt be a factor of (1-sqrt(-167)), and all of > the other roots cant be a factor of both either. > I thought Id use r_1 just to simplify things slightly. > Oh yeah, so there doesnt exist an algebraic integer r_1 that is a > non-unit factor of 7 where z can be an algebraic integer. > Its a fascinatingly simple to show result. > James Harris So, z is a root of some polynomial, that you assure is is non-monic (a brief proof would have been nice), and how exactly have you shown this to be irreducible? Somehow I dont think anyone will trust your intuition about these things. === Subject: Re: Algebraic integers, simple result >What I discovered was an easy demonstration by looking at >(1+sqrt(-167) since (1+sqrt(-167))/2[(1-sqrt(-167))/2] = -7 >you might assume that (1+sqrt(-167)) has non-unit factors in common >with 7 in the ring of algebraic integers, but it does not. The way to quickly and easily see that is to consider >z = (1+sqrt(-167))/r_1 >where r_1 is presumed to be an algebraic integer factor of 7 that is >also a factor of (1+sqrt(-167)) so that z is an algebraic integer. >However, you find that if r_1 is not a unit, then z is the root of a >non-monic polynomial of order equal to 2 times the order of the >minimum polynomial with integer coefficients which has r_1 as a root. >And you know that polynomial which has z as a root, cant have a >rational root because neither (1+sqrt(-167))/r_k, nor >(1-sqrt(-167))/r_k, where k is some counting number equal to or >greater than 1 that is less than or equal to the order of the >polynomial, can be rational. >The polynomial must be non-monic because even if r_1 were a factor of >(1+sqrt(-167)) it wouldnt be a factor of (1-sqrt(-167)), and all of >the other roots cant be a factor of both either. >I thought Id use r_1 just to simplify things slightly. >Oh yeah, so there doesnt exist an algebraic integer r_1 that is a >non-unit factor of 7 where z can be an algebraic integer. >Its a fascinatingly simple to show result. Clearly, if your argument were correct, it would also apply to (1 - sqrt(-167)). Therefore what you have shown implies that the following theorem is false: Theorem. If A and B are algebraic integers and A * B = 7, Then at least one of A or B must have algebraic integer factors in common with 7. Perhaps I should give the proof of the theorem so you can tell us all where it contains an error. Proof: Suppose that A does not have an algebraic integer factor in common with 7. Then A is coprime to 7. This means that there exist algebraic integers s and t such that A*s + 7*t = 1. Multiplying both sides by B, I have: A*B*s + 7*B*t = B. Since A*B = 7, this implies that 7*s + 7*B*t = B, or 7*(s + B*t) = B. That is, 7 itself is a factor of B. QED. Now: would you also like to know where your fascinatingly simple reasoning has gone wrong? Nora B. >James Harris === Subject: Re: Algebraic integers, simple result > Its a fascinatingly simple to show result. > James Harris And how many of these fascinatingly simple to show results, like JSHs vaunted fascinatingly simple to show proof of FLT, have slunk quietly out of sight when they proved to be fascinatingly false? === Subject: Re: Algebraic integers, simple result Apparently your argument was refuted in a previous thread. That is usually what prompts you to start a new one. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: JSH: how you can prove to the mathematicians you are right Adjunct Assistant Professor at the University of Montana. [.snip.] [.typo corrected in statement of Dedekinds Prague Theorem below.] >> DEDEKINDS PRAGUE THEOREM (Lemma 2 in Chapter 2, Section 5, of David >> Hilberts Zahlbericht) If the coefficients a1,a2,....,b1,b2,... of two >> polynomials in one variable >> F(x) = a1*x^r + a2*x^{r-1} + ... >> G(x) = b1*x^s + b2*x^{s-1} + ... >> are algebraic integers, and the coefficients of the product >> F(x)*G(x) = c1*x^{r+s} + c2*x^{r+s-1} + ... >> are all divisible by the rational integer w, then each of the numbers >> a1*b1, a1*b2, a1*b3,...,a2*b1, a2*b2,... is also divisible by w. >> (Also proven Kronocker, Mertens, and Hurwitz). [.snip.] >I was interested to note that the statment of the theorem >requires w to be a rational integer. (are there any known >counterexamples if w is an algebraic integer but not >at rational integer?) I think the result is true if we replace the condition that w be a rational integer with the condition that it be an arbitary algebraic integer, and by basically the same argument. So we want to prove that: THEOREM. If the coefficients a1, a2, ..., b1, b2, ..., of two polynomials in one variable F(x) = a1*x^r + a2*x^{r-1} + ... G(x) = b1*x^s + b2*x^{s-1} + ... are algebraic integers, and the coefficients of the product are all divisible by the algebraic integer w, then each of the products a1*b1, a1*b2, ..., a2*b1, a2*b2,.... are divisible by w. (I will be using unique factorization of ideals in number fields, which may be why this was not proven in this generality by Dedekind: the Prague Theorem is one of the key steps ->towards<- establishing unique factorization into prime ideals; and yes, Im almost certainly doing this in a roundabout way...). Proof. Let K be the field of definition of F and G, and let R be its ring of integers. Define the content of a polynomial with coefficients in R to be the ideal generated by the coefficients. We show first that the content of the product is equal to the product of the contents. Using Prop. 8.6 (Factoring out the content) in our preprint, http://www.math.umt.edu/magidin/preprints/gauss.pdf pp. 24, there is an extension L of K where we can write each of the polynomials as F(x) = c_f*f(x) G(x) = c_g*g(x) F(x)G(x) = c_h*h(x) where c_f, c_g, c_h are algebraic integers, (c_f) = content of F (lifted to the ring of integers of L), and likewise (c_g)=content(G), (c_h)=content(F*G); and f(x), g(x), and h(x) are each primitive with algebraic integer coefficients. By Gausss Lemma for Number Fields (Theorem 8.4, pp. 23 of the preprint), we must have (c_f*c_g)=(c_h), and f(x)*g(x) primitive and equal to h(x), up to multiplication by a unit. Now taking the ideals and intersecting with R, we have that content(F*G) = (c_h)cap R = (c_f*c_g)cap R = [(c_f)cap R]*[(c_h)cap R] = content(F)*content(G). Now consider the ideal (w); since w divides every coefficient of F*G, then (w) divides the content of F*G, which equals the content of F times the content of G. Factor (w), cont(F), and cont(G) into prime ideals, and we can write (w) = P1*...*Pr*Q1*...*Qs where each of Pi, Qj are prime ideals, P1*...*Pr divides cont(F), and Q1*...*Qs divides cont(G). In particular, every ai is in P1*...*Pr, and every bj is in Q1*...*Qs; therefore, each product ai*bj is in P1*...*Pr*Q1*...*Qs = (w), which is equivalent to w dividing each product ai*bj, as desired. QED -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: how you can prove to the mathematicians you are right <87znbpo5p5.fsf@phiwumbda.org> <87smhgokhq.fsf@phiwumbda.org> <87ad3ooy8l.fsf@phiwumbda.org> Discussion, linux) > Thats not what Im saying. He would decide that his argument is > correct AND his formalism is correct, but that the proof-checker, > and the whole theory of proof-checkers, had a fundamental ßaw. I dont know why I missed that obvious prediction. Nice list of honors for JSH. I think you missed the line of James Harris original scents. Maybe some sneakers, too. -- Jesse F. Hughes What you call reasonable is suspect since youve proven yourself to be an enemy of mathematics. -- James S. Harris defends the cause. === Subject: Re: Please help me >I need to solve this equation step by step. >y(x) = y(x)/x + 3(y(x))^2 >Could someone help me? I think that this is Bernoullis equation, But not sure. Some algebraic manipulation brings this to y - xy ------- = -3x y^2 or alternately, let u = y/x so that y = ux, then the equation becomes ux + u = u + 3 u^2 x^2 whence u/u^2 = 3x Thats as much of a hint as I can give without giving the whole thing away. Rob Johnson take out the trash before replying === Subject: Re: Please help me > I need to solve this equation step by step. > yÇ(x) = y(x)/x + 3(y(x))^2 Set y(x) = u(x)*x: y = u*x + u => u*x + u = u + 3*x^2*u^2 u = 3*x*u^2 This can now be solved by separation of variables. -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: The New Lenny-Maxist Dialectic Jack, you say: Ô... dark energy and dark matter ... are both virtual exotic vacuum states of w = -1 residual zero point energy density of negative and positive quantum pressure respectively .... That statement is oversimplistic, at least, and wrong if taken literally. We had an extensive e-mail discussion about that point some time ago. One of the distinguishing points between DE and DM in astrophysical observations is that for DE, w = -1 and for DM, w = 0. Yes, exactly, but perhaps I have given you credit for what I thought was your idea? That idea is quite obvious physically. Imagine for example a huge sphere, i.e. the galactic halo of attractive exotic vacuum whose interior has w = -1 with positive pressure. To the distant outside observer it is indistinguishable from w = 0 CDM holding the galaxies in check. I thought you had that idea? Please clarify for the record in my book since I dont want to give you credit for a really good idea if I invented it and not you? :-) The e-mail exchange was so fast that I was not sure who got the idea? Its like an entangled thought BIT pilot waves. Of course, we also need signal nonlocality beyond micro-quantum theory. If w = -1 for both of them, the evolution curves for our universe would NOT be consistent with observation, as shown in Tegmarks paper at http://xxx.lanl.gov/abs/arXiv:astro-ph/0207199 and in particular a figure therein, attached to this message, that clearly shows that all matter, ordinary or dark, should obey the w = 0 equation of state in order to be consistent with observation. This not a relevant argument as I showed above. Of course a sphere of w = -1 in the interior looks like w = 0 CDM for the exterior observer in terms of its gravity effect on orbiting masses like galaxies. The math is very simple. Suppose, in a toy model, the galactic halo is a uniform sphere of radius R with a constant /zpf < 0 ~ zero point energy density of positive quantum pressure with interior w = -1. The effective w = 0 CDM equivalent mass seen by the distant observer using light signals is obviously M(CDM w = 0) ~ c^2/zpfR^3/G As I recall, our discussion got around this difficulty in your model by making dark matter things to be like bon-bons, with a w = -1 interior and a surface coating of w = 0 stuff, so that your dark matter could act cosmologically like w = 0. Yes, exactly what I just said. Was that not your idea originally or was it mine? Therefore, it seems to me to be misleading for you to say that DM has w = -1 unless you go on to qualify the statement by saying that your DM lives as bon-bon bits with outer coatings of w = 0 stuff Well, obviously thats what I mean. Lets not get too nit-picky legalistic here. Your legal training is showing. :-) since your bon-bon structure of DM will be needed to be taken into account when you do DE/DM engineering. Right, so tell me. Who got the idea first? Or was it truly collective which is the way I am telling it in Super Cosmos. Please clarify what your memory is. We can always go back to the e-mails I suppose. Let some graduate student do it 50 years from now. :-) Jack, you mention ÔThe whole idea of metric engineering is that as Men Who Would Be Gods we can engineer any damn Universe we like ... and you have referred to Tegmarks Parallel Universes and to Susskinds Landscapes. Correct. force constant ratios can vary (other than by renormalization with energy level), I disagree because I can calculate model is in reasonably good agreement with experimental observation. You may be correct and I and the others may be wrong. We cannot decide that yet. I do not understand what your input assumptions are and how you make those calculations. Until you can communicate more clearly and effectively on what the key ideas are there the issue is moot. I think Paul Steinhardt at Princeton agrees at least with your philosophy on this. As far as I can see, the 17 or is it 23 epicycle fudge factors in the gauge force model are contingent. Until you can give me some compelling heuristic organizing idea such as I have in my model, i.e. vacuum coherence quenches the random zero point micro-quantum vacuum ßuctuations etc. Until you have a grand notion like that, easily expressed in ordinary language, like Einsteins equivalence principle, like Heisenbergs uncertainty principle, like the principle of relativity, like More is different, like The Question is: What is The Question? like Law without law, Mass without mass, Charge without charge, Spin without spin, Universe as a self-excited circuit etc. until you have some compelling images like that which express a significant homomorphic compressed faithful image of the original more detailed idea, like a jpeg of a tiff , until you have that + a toy model back of the envelope calculation of a mass and a coupling, no one will be able to tell if you have the right stuff or not. Problem is every one is too busy with their own ideas and cannot take off a couple of years to read all the papers you have read. Although there have been some recent papers that advocate a variable fine structure constant, a recent PRL paper by Ashenfelter, Mathews, and Olive, at http://link.aps.org/abstract/PRL/v92/e041102 says Ô... We show that the synthesis of 25,26Mg at the base of the convective envelope in low-metallicity asymptotic giant branch stars can produce the isotopic ratios needed to explain the low-z subset (with z<1.8) of the many-multiplet data from quasar absorption systems without invoking a time variation of the fine structure constant. ... Fine, but that is not the idea. Its not that alpha is or is not constant in cosmic time, its that alpha may be different in different universes next door. Thats what the Susskind Landscape Idea is in sense of WAP and parallel Hubble sphere universes on a single post-inßation bubble in a cosmic champagne bubbly in hyperspace. No wonder Rashi de Troyes made Champagne 1000 years ago in the South of France Cathar Country of the Priory of Sion and the Grail Da Vinci Code. How fitting! :-) In other words, as far as experimental results go, the fine structure constant is still OK to be constant, which is in agreement with my model in which it is calculated. You are using the ergodic theorem here unconsciously. The Landscape idea is that there is an ensemble of parallel universes with all the fudge factors randomly distributed and what we measure is because we are here to measure it in the sense of Darwinian natural selection as explained for example by Lee Smolin. Now this is a clash of world pictures of monumental proportions I agree. But first we must never forget The Question is: What is The Question. Wheeler must have gotten that idea from watching Abbott and Costello do the shtick Whos on first. :-) http://www.baseball-almanac.com/humor4.shtml Of course, since the Landscapes etc of Lenny and Max ... Maybe we should call it Maxism? :-) Lenin and Marx? Ill be Trotsky? :-) do not make any predictions, other than to say anthropomorhically that whatever we observe we observe because we are here, which to me seems like meaningless sophistry, they are not falsifiable Not true. It is falsifiable I think. But that is a major discussion. and you and they are as entitled to believe and publish about that, just as creationists are entitled to believe and publish that the universe was created by G-d in 4004 BC, with fossils etc all being put in place then. Definitely my variation on the Lenny-Maxist Doctrine is Popper-falsifiable in many ways. Seriously I do not think your argument here is correct. A whole book will need to be written why. Jack, you say, about the WMAP ratios, that I need to say why I ... think they are static .... I do NOT think that they are static, and my paper at http://www.innerx.net/personal/tsmith/WMAPpaper.pdf clearly says that. Yes, but the problem here is that I am not able to follow the argument of your paper here. Do you know anyone who can and can help you explain it better? The whole point of the paper is that there is a ratio 67% : 27%: 6% for 2 billion years ago when the expansion curve of our universe last intersected with a linear expansion line, and that ratio has evolved to the present day. Explain how you get this set of numbers and the other? Its all a dense fog to me. My calculated WMAP ratio of 75.3% : 20.2% : 4.5% comes from evolution of the 2 billion year ago ratio through the past 2 billion years. The WMAP ratio is NOT static. It evolves, and that one of the things that my paper is about. Glad to hear that, but you need to explain how exactly that comes about in a clearer way. Jack, you say ... dark energy and dark matter ... are both virtual exotic vacuum states of w = -1 residual zero point energy density of negative and positive quantum pressure respectively .... That statement is oversimplistic, at least, and wrong if taken literally. We had an extensive e-mail discussion about that point some time ago. One of the distinguishing points between DE and DM in astrophysical observations is that for DE, w = -1 and for DM, w = 0. Yes, exactly, but perhaps I have given you credit for what I thought was your idea? That idea is quite obvious physically. Imagine for example a huge sphere, i.e. the galactic halo of attractive exotic vacuum whose interior has w = -1 with positive pressure. To the distant outside observer it is indistinguishable from w = 0 CDM holding the galaxies in check. I thought you had that idea? Please clarify for the record in my book since I dont want to give you credit for a really good idea if I invented it and not you? :-) The e-mail exchange was so fast that I was not sure who got the idea? Its like an entangled thought BIT pilot waves. Of course, we also need signal nonlocality beyond micro-quantum theory. If w = -1 for both of them, the evolution curves for our universe would NOT be consistent with observation, as shown in Tegmarks paper at http://xxx.lanl.gov/abs/arXiv:astro-ph/0207199 and in particular a figure therein, attached to this message, that clearly shows that all matter, ordinary or dark, should obey the w = 0 equation of state in order to be consistent with observation. This not a relevant argument as I showed above. Of course a sphere of w = -1 in the interior looks like w = 0 CDM for the exterior observer in terms of its gravity effect on orbiting masses like galaxies. The math is very simple. Suppose, in a toy model, the galactic halo is a uniform sphere of radius R with a constant /zpf < 0 ~ zero point energy density of positive quantum pressure with interior w = -1. The effective w = 0 CDM equivalent mass seen by the distant observer using light signals is obviously M(CDM w = 0) ~ c^2/zpfR^3/G As I recall, our discussion got around this difficulty in your model by making dark matter things to be like bon-bons, with a w = -1 interior and a surface coating of w = 0 stuff, so that your dark matter could act cosmologically like w = 0. Yes, exactly what I just said. Was that not your idea originally or was it mine? Therefore, it seems to me to be misleading for you to say that DM has w = -1 unless you go on to qualify the statement by saying that your DM lives as bon-bon bits with outer coatings of w = 0 stuff, Well, obviously thats what I mean. Lets not get too nit-picky legalistic here. Your legal training is showing. :-) since your bon-bon structure of DM will be needed to be taken into account when you do DE/DM engineering. Tony Right, so tell me. Who got the idea first? Or was it truly collective which is the way I am telling it in Super Cosmos. Please clarify what your memory is. We can always go back to the e-mails I suppose. Let some graduate student do it 50 years from now. :-) Jack, you mention The whole idea of metric engineering is that as Men Who Would Be Gods we can engineer any damn Universe we like ... and you have referred to Tegmarks Parallel Universes and to Susskinds Landscapes. Correct. force constant ratios can vary (other than by renormalization with energy level), I disagree because I can calculate model is in reasonably good agreement with experimental observation. You may be correct and I and the others may be wrong. We cannot decide that yet. I do not understand what your input assumptions are and how you make those calculations. Until you can communicate more clearly and effectively on what the key ideas are there the issue is moot. I think Paul Steinhardt at Princeton agrees at least with your philosophy on this. As far as I can see, the 17 or is it 23 epicycle fudge factors in the gauge force model are contingent. Until you can give me some compelling heuristic organizing idea such as I have in my model, i.e. vacuum coherence quenches the random zero point micro-quantum vacuum ßuctuations etc. Until you have a grand notion like that, easily expressed in ordinary language, like Einsteins equivalence principle, like Heisenbergs uncertainty principle, like the principle of relativity, like More is different, like The Question is: What is The Question? like Law without law, Mass without mass, Charge without charge, Spin without spin, Universe as a self-excited circuit etc. until you have some compelling images like that which express a significant homomorphic compressed faithful image of the original more detailed idea, like a jpeg of a tiff , until you have that + a toy model back of the envelope calculation of a mass and a coupling, no one will be able to tell if you have the right stuff or not. Problem is every one is too busy with their own ideas and cannot take off a couple of years to read all the papers you have read. Although there have been some recent papers that advocate a variable fine structure constant, a recent PRL paper by Ashenfelter, Mathews, and Olive, at http://link.aps.org/abstract/PRL/v92/e041102 says ... We show that the synthesis of 25,26Mg at the base of the convective envelope in low-metallicity asymptotic giant branch stars can produce the isotopic ratios needed to explain the low-z subset (with z<1.8) of the many-multiplet data from quasar absorption systems without invoking a time variation of the fine structure constant. .... Fine, but that is not the idea. Its not that alpha is or is not constant in cosmic time, its that alpha may be different in different universes next door. Thats what the Landscape Idea is in sense of WAP and parallel Hubble sphere universes on a single post-inßation bubble in a cosmic champagne bubbly in hyperspace. No wonder Rashi de Troyes made Champagne 1000 years ago in the South of France Cathar Country of the Priory of Sion and the Grail Da Vinci Code. How fitting! :-) In other words, as far as experimental results go, the fine structure constant is still OK to be constant, which is in agreement with my model in which it is calculated. You are using the ergodic theorem here unconsciously. The Landscape idea is that there is an ensemble of parallel universes with all the fudge factors randomly distributed and what we measure is because we are here to measure it in the sense of Darwinian natural selection as explained for example by Lee Smolin. Now this is a clash of world pictures of monumental proportions I agree. But first we must never forget The Question is: What is The Question. Wheeler must have gotten that idea from watching Abbott and Costello do the shtick Whos on first. :-) http://www.baseball-almanac.com/humor4.shtml Of course, since the Landscapes etc of Lenny and Max Maybe we should call it Maxism? :-) Lenin and Marx? Ill be Trotsky? :-) do not make any predictions, other than to say anthropomorhically that whatever we observe we observe because we are here, which to me seems like meaningless sophistry, they are not falsifiable Not true. It is falsifiable I think. But that is a major discussion. and you and they are as entitled to believe and publish about that, just as creationists are entitled to believe and publish that the universe was created by G-d in 4004 BC, with fossils etc all being put in place then. Definitely my variation on the Lenny-Maxist Doctrine is Popper-falsifiable in many ways. Seriously I do not think your argument here is correct. A whole book will need to be written why. Jack, you say, about the WMAP ratios, that I need to say why I ... think they are static .... I do NOT think that they are static, and my paper at http://www.innerx.net/personal/tsmith/WMAPpaper.pdf clearly says that. Yes, but the problem here is that I am not able to follow the argument of you paper here. Do you know anyone who can and can help you explain it better? The whole point of the paper is that there is a ratio 67% : 27%: 6% for 2 billion years ago when the expansion curve of our universe last intersected with a linear expansion line, and that ratio has evolved to the present day. Explain how you get this set of numbers and the other? Its all a dense fog to me. My calculated WMAP ratio of 75.3% : 20.2% : 4.5% comes from evolution of the 2 billion year ago ratio through the past 2 billion years. The WMAP ratio is NOT static. It evolves, and that one of the things that my paper is about. Glad to hear that, but you need to explain how exactly that comes about in a clearer way. === Subject: Re: The New Lenny-Maxist Dialectic line news reader for the Linux platform NewsFleX homepage: http://www.home.zonnet.nl/panteltje/newsßex/ and ftp download ftp://sunsite.unc.edu/pub/linux/system/news/readers/ The way putin is going about it, maybe we should be worried a bit about 2 much of that stuff. Or is this what they want you to think ;-)? === Subject: Re: Ttttttt....two unknowns for an Inverse Kinemaniac > For the solution of an inverse kinematics problem > I need some help. I have two equations of the form > a sin(q) + b sin(p) + c sin(q) + d cos(p)cos(q) + e cos(p)sin(q) + f=0 > g sin(q) + h sin(p) + i sin(q) + j cos(p)cos(q) + k cos(p)sin(q) + l=0 > so two equations with two unknowns. Is there a general explicit solution > for these? Id solve these as a pair of linear equations in, say, sin(q) and cos(q) and then use sin^2(q) + cos^2(q) = 1 to obtain a single condition in p, then in the latter replace sin(p) and cos(p) by functions of t = tan(p/2) to obtain a polynomial in t. Not exactly explicit, but it should solve the problem. ------------------------------------------------------------- -------------- John R Ramsden (jr@adslate.com) ------------------------------------------------------------- -------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Continuum hypotheses and Lebesgue measure > Hi all, Consider this statement: If a Lebesgue-measurable subset M of the set R of real numbers has > cardinal less than the cardinal of R then the Lebesgue measure of > M is zero. This is trivial if we assume the continuum hypothsis. And what if we > dont? Its easy to see that if there were subsets of R whose cardinal > was in between the cardinals of N (the naturals) and R, then, among > those sets, there would be Lebesgue-measurable sets with Lebesgue > measure zero. But would there be necessarily, among those sets, > Lebesgue-measurable sets with Lebesgue measure greater than zero? (*) Every subset of R with cardinal < c is Lebesgue measurable > [and therefore, as Ullrich noted, of measure zero] > (*) is independent of ZFC. (*) is a consequence of Martins Axiom. > But there are also models of ZFC + notCH where (*) fails: > were there are subsets of cardinal strictly between aleph_0 and c that > have positive Lebesgue outer measure but (of course) zero inner > measure. As I recall (It has been a long time...) the random reals > type models do this. Random reals? Ive never heard of these. Could you provide some information or sources of information? Ôcid Ôooh === Subject: Checking ring of algebraic integers It turns out that theres an incredibly simple way to check the ring of algebraic integers and see that theres a problem with how its currently taught. You can take two quadratics: x^2 - x + 42 = 0 and y^2 - by - 7 = 0 where by how algebraic integers are currently taught, youd think that there exists an algebraic integer b, such that a root of the second quadratic is a factor of a root of the first. Intriguingly, there does not, and its easy to show. Now x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 which is z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) where now the question is, can an algebraic b exist such that z is an algebraic integer? The square roots dont tell me much, but working to eliminate them adds solutions, as a first step consider: 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 where I have *two* solutions, where they are z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) and z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) and working still further I get 196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0 and I can divide both sides by 28 to finally get 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 where I have four solutions, which are z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). (Those not sure can simply multiply out (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.) Now then our question was, could what is now z_1 be an algebraic integer for an algebraic integer b? And now an important theorem comes into play, as an algebraic integer cannot be the root of a non-monic primitive with integer coefficients that is irreducible over Q. Going back to x^2 - x + 42, I know that the sum of its roots is 1, so the roots are coprime to each other. So looking at the four solutions and assuming that z_1 is an algebraic integer, that leaves z_4 as a second possibility that is also an algebraic integer. But (Z-z_1)(Z-z_4) = [Z - (1 + sqrt(-167))/(b + sqrt(b^2 + 28))][Z - (1 + sqrt(-167))/(b + sqrt(b^2 + 28))], which is Z^2 + (b - sqrt(-167) sqrt(b^2 + 28))Z/14 + 6 = 0, where the middle coefficient cannot be an integer with algebraic integer b, as that would require that b^2 + 28 = -167n^2, with n an integer, which means that b cant be real. If you use that irrational b then isolate the irrational term you have sqrt(-167n^2 - 28) Z/14 = -Z^2 + 167n^2 Z/14 - 6 which when you square both sides to get read of the square root just gives you a non-monic quartic again! Therefore, z cannot be an algebraic integer. Therefore, in the ring of algebraic integers, the roots of x^2 - x + 42 = 0 and y^2 - by - 7 = 0 never share non-unit factors in the ring of algebraic integers. Notice that the key theorem here is that one that prevents an algebraic integer from being the root of a non-monic primitive with integer coefficients irreducible over Q. Take away the condition that you have an algebraic integer factor, and youre ok, which proves that there must be a ring beyond algebraic integers, where they *can* share non-unit factors. I have called the more inclusive ring, the ring of objects, or the Object Ring. The Object Ring is a commutative ring that includes all numbers such that -1 and 1 are the only members that are both a unit and an integer, where no non-unit member is a factor of any two integers that are coprime. One of the most surprising results that follows quickly from understanding the limitations of the ring of algebraic integers, and having the fuller definition, is a proof of Fermats Last Theorem of around two to three pages. James Harris === Subject: Re: Checking ring of algebraic integers > It turns out that theres an incredibly simple way to check the ring > of algebraic integers and see that theres a problem with how its > currently taught. Is your problem now with the TEACHING of the subject? Can you be more specific? === Subject: Re: Checking ring of algebraic integers > It turns out that theres an incredibly simple way to check the ring > of algebraic integers and see that theres a problem with how its > currently taught. Is your problem now with the TEACHING of the subject? Can you be more > specific? My assertion is that mathematicians teach that given x2 - x + 42 = 0 the roots, which are algebraic integers, have non-unit algebraic integer factors in common with 7 in the ring of algebraic integers. It turns out they do not, and Im working out the argument--drafts first and then refining--as Ive done before. This thread is now the key one, as I take the step of eliminating sqrt(-167) without multiplying it out which goes to the splitting field of x2 - x + 42 = 0 for those who wonder what splitting field Im discussing. Does that make sense to you? James Harris === Subject: Re: Checking ring of algebraic integers > This thread is now the key one, as I take the step of eliminating > sqrt(-167) without multiplying it out which goes to the splitting > field of > x2 - x + 42 = 0 > for those who wonder what splitting field Im discussing. Do you know what field that is? Q(sqrt(-167)), the field of numbers of the form r + s.sqrt(-167), with r and s rational. The ring of integers in this field is formed by all numbers of the form a + b.sqrt(-167) with a and b both integer or both one half of an odd integer. This ring of integers contains a number that is a common factor of [(1 + sqrt(-167))/2]^11 and 7^11, say p. The splitting field of x^11 - p contains a common factor of (1 + sqrt(-167))/2 and 7. (The power 11 comes from the fact that the class number of the field is 11, and so it is also not a UFD.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Checking ring of algebraic integers > My assertion is that mathematicians teach that given > x2 - x + 42 = 0 > the roots, which are algebraic integers, have non-unit algebraic > integer factors in common with 7 in the ring of algebraic integers. > It turns out they do not, and Im working out the argument--drafts > first and then refining--as Ive done before. > This thread is now the key one, as I take the step of eliminating > sqrt(-167) without multiplying it out which goes to the splitting > field of > x2 - x + 42 = 0 > for those who wonder what splitting field Im discussing. Why not respond to those who wonder why you keep posting an argument which has been proven false? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Checking ring of algebraic integers Your argument has already been proven wrong. Why repost the same thing? Are you a troll -- or just an idiot? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Checking ring of algebraic integers >It turns out that theres an incredibly simple way to check the ring >of algebraic integers and see that theres a problem with how its >currently taught. >[...] >And now an important theorem comes into play, as an algebraic integer >cannot be the root of a non-monic primitive with integer coefficients >that is irreducible over Q. >[...] >Notice that the key theorem here is that one that prevents an >algebraic integer from being the root of a non-monic primitive with >integer coefficients irreducible over Q. >Take away the condition that you have an algebraic integer factor, and >youre ok, which proves that there must be a ring beyond algebraic >integers, where they *can* share non-unit factors. Youre a fascinating guy. I remember a while back when you spent a few months explaining that this extremely well-known result was wrong, posting erroneous counterexample after erroneous counterexample. Finally decided it was correct, eh? Just curious: You decided its correct because you know how to prove it, or are you just taking the word of the evil mathematical establishment? >I have called the more inclusive ring, the ring of objects, or the >Object Ring. >The Object Ring is a commutative ring that includes all numbers such >that -1 and 1 are the only members that are both a unit and an >integer, where no non-unit member is a factor of any two integers that >are coprime. >One of the most surprising results that follows quickly from >understanding the limitations of the ring of algebraic integers, and >having the fuller definition, is a proof of Fermats Last Theorem of >around two to three pages. Heh-heh. >James Harris ************************ David C. Ullrich === Subject: Re: Checking ring of algebraic integers Adjunct Assistant Professor at the University of Montana. [.snip.] >>Notice that the key theorem here is that one that prevents an >>algebraic integer from being the root of a non-monic primitive with >>integer coefficients irreducible over Q. >>Take away the condition that you have an algebraic integer factor, and >>youre ok, which proves that there must be a ring beyond algebraic >>integers, where they *can* share non-unit factors. >Youre a fascinating guy. I remember a while back when you spent >a few months explaining that this extremely well-known result >was wrong, posting erroneous counterexample after erroneous >counterexample. Finally decided it was correct, eh? >Just curious: You decided its correct because you knowmathematical establishment? Took some doing, but he went over the proof and agreed to it. -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Checking ring of algebraic integers > [.snip.] >Notice that the key theorem here is that one that prevents an >algebraic integer from being the root of a non-monic primitive with >integer coefficients irreducible over Q. >Take away the condition that you have an algebraic integer factor, and >youre ok, which proves that there must be a ring beyond algebraic >integers, where they *can* share non-unit factors. >>Youre a fascinating guy. I remember a while back when you spent >>a few months explaining that this extremely well-known result >>was wrong, posting erroneous counterexample after erroneous >>counterexample. Finally decided it was correct, eh? >>Just curious: You decided its correct because you knowhow to prove it, or are you just taking the word of the evil >>mathematical establishment? >Took some doing, but he went over the proof and agreed to it. Well yes, thats more or less what he _said_ happened, but it doesnt really answer my question - the fact that he says he agrees a proof is correct doesnt prove that he knows how to prove the result... ************************ David C. Ullrich === Subject: Tea with Sarfatti Tea with Sarfatti at Kensington Palace Gardens TSAs I recall, our discussion got around this difficulty in your model by making dark matter things to be like bon-bons, with a w = -1 interior and a surface coating of w = 0 stuff, so that your dark matter could act cosmologically like w = 0. Yes, exactly what I just said. Was that not your idea originally, or was it mine? TS: Therefore, it seems to me to be misleading for you to say that DM has w = -1 unless you go on to qualify the statement by saying that your DM lives as bon-bon bits with outer coatings of w = 0 stuff Aha! I see. No your bon bon analogy is not needed at all. So the idea was mine after all. I did not look closely enough at your wording! I donÍt need no w = 0 layer at the surface of the w = -1 glob. I donÍt need no damn bon bons! Who ordered that? Not me. Take it back, it ainÍt kosher. There is a ßy in that soup. The proper distinction is interior w = -1 mimics w = 0 when observed at a distance with light signals or gravity waves or neutrinos or anything since gravity is universal. === Subject: Re: Tea with Sarfatti > Tea with Sarfatti at Kensington Palace Gardens > TSAs I recall, our discussion got around this difficulty in your model > by making dark matter things to be like bon-bons, with a w = -1 interior > and a surface coating of w = 0 stuff, so that your dark matter could act > cosmologically like w = 0. > Yes, exactly what I just said. Was that not your idea originally, or was > it mine? > TS: Therefore, it seems to me to be misleading for you to say that DM > has w = -1 unless you go on to qualify the statement by saying that your > DM lives as bon-bon bits with outer coatings of w = 0 stuff > Aha! I see. No your bon bon analogy is not needed at all. So the idea > was mine after all. I did not look closely enough at your wording! I > donÍt need no w = 0 layer at the surface of the w = -1 glob. I donÍt > need no damn bon bons! Who ordered that? Not me. Take it back, it ainÍt > kosher. There is a ßy in that soup. The proper distinction is > interior w = -1 mimics w = 0 when observed at a distance with light > signals or gravity waves or neutrinos or anything since gravity is > universal. So Sarfatti has graduated from having conversations with himself to having tea with himself, I see. -E === Subject: Re: Tea with Sarfatti > Tea with Sarfatti at Kensington Palace Gardens > TSAs I recall, our discussion got around this difficulty in your model > by making dark matter things to be like bon-bons, with a w = -1 interior > and a surface coating of w = 0 stuff, so that your dark matter could act > cosmologically like w = 0. > Yes, exactly what I just said. Was that not your idea originally, or was > it mine? > TS: Therefore, it seems to me to be misleading for you to say that DM > has w = -1 unless you go on to qualify the statement by saying that your > DM lives as bon-bon bits with outer coatings of w = 0 stuff > Aha! I see. No your bon bon analogy is not needed at all. So the idea > was mine after all. I did not look closely enough at your wording! I > don.89t need no w = 0 layer at the surface of the w = -1 glob. I don.89t > need no damn bon bons! Who ordered that? Not me. Take it back, it ain.89t > kosher. There is a ßy in that soup. Looks more like a hot, steaming bowl full of bullshit to me, Jack. Better ask the waiter to take it back to the kitchen before it stinks up the restaurant. === Subject: Re: Sequence of Diophantine Equations by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1F04bk15222; Im thinking about lots of 1s, lots of 2s, and lots of (x-1)s. Also, let us observe that, if z has this property and y_i=2 for some i, then z+3 has the same property, too... Ady. >> Hi! >> Is there some (huge) positive integer M with the following >> property: for any z>M, there exist positive integers >> x,y_1,y_2,...,y_z such that >> x^x=(y_1)^(y_1)+(y_2)^(y_2)+...+(y_z)^(y_z) ? >> (Please remark that the ys are >=1 and need not to be >> necessarily distinct). >> Ignorantly Yours, >> Ady. >A partial result??:... > x^x/(x-1)^(x-1) ~ x*e. >So, (for x not y; z >1) >z has to be *at least* in the _vicinity_ of x*e, >if all ys = (x-1). >(I hope this clue {or clue} does not lead you astray.) >Leroy >Quet === Subject: Re: Ring of algebraic integers, comparison check by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1F04cr15279; A minor misprint below at [*****] - >> CORRECTION POST: >> It turns out that theres an incredibly simple way to check the ring >> of algebraic integers and see that theres a problem with how its >> currently taught. >> You can take two quadratics: >> x^2 - x + 42 = 0 >> and >> y^2 - by - 7 = 0 >> where by how algebraic integers are currently taught, youd think that >> there exists an algebraic integer b, such that a root of the second >> quadratic is a factor of a root of the first. Intriguingly, there >> does not, and its easy to show. >> Now >> x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and >> y^2 - by - 7 = 0, has as one of its roots >> (b + sqrt(b^2 + 28))/2. >> So I can simply introduce z, where >> (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2 >> which is >> z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >> where now the question is, can an algebraic b exist such that z is an >> algebraic integer? >> The square roots dont tell me much, but working to eliminate them >> adds solutions, as a first step consider: >> 28z^2 + 2(1+sqrt(-167))bz - (1+sqrt(-167))^2 = 0 >> where I have *two* solutions, where they are >> z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >> and >> z_2 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) >> and working still further I get >> 196 z^4 + 28b z^3 + (168b^2 + 2324) z^2 - 168bz + 7056 = 0 >> and I can divide both sides by 28 to finally get >> 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 >> where I have four solutions, which are >> z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >> z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) >> z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) >> z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). >> (Those not sure can simply multiply out >> (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.) >> Now then our question was, could what is now z_1 be an algebraic >> integer for an algebraic integer b? >> And now an important theorem comes into play, as an algebraic integer >> cannot be the root of a non-monic primitive with integer coefficients >> that is irreducible over Q. >> So integer bs are eliminated as a possibility immediately. > OK. >> There are several ways to show that none of the the zs can be a >> fraction for an integer b, > I thought you just eliminated integer bs. >> but some of you may wonder why theyd need >> to be, like why couldnt they be irrational algebraic integers? >> Well, unless >> 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 >> is reducible over Q with that integer b, the theorem I mentioned >> requires that *none* of its roots can be an algebraic integer, as its >> non-monic and primitive. >> Therefore, even if you had an irrational algebraic integer root, the >> polynomial would have to be reducible over Q, meaning that youd need >> at least one rational root. > You must be assuming that a polynomial that is reducible over >Q must have at least one rational root. That is false. Consider >3*x^4 - 15*x^2 + 5. [*****] This should be 3*x^4 - 16*x^2 + 5. N.B. > Plus, if b is not an ordinary integer but it is an algebraic >integer, it must be irrational. A polynomial with irrational >coefficients cannot be reducible over Q. >> If b is irrational--remember its an algebraic integer--then you get a >> polynomial of degree equal to the order of the monic polynomial with >> integer coefficients for which b is a root times 4. >> However, for any such polynomial, you run into the same problem that >> itd need at least one root thats rational. > You have not shown that such a polynomial would have to be reducible >over Q. Even if it were, that does not imply that it has a rational root. >> But you can characterize every possible root by >> z_j = (1 + sqrt(-167))/W_j >> z_k = (1 - sqrt(-167))/W_k >> where j and k are counting numbers, and W_j and W_k are algebraic >> integers. >> The problem is that (1+sqrt(-167)/U where U is an algebraic integer >> cannot be rational unless U has (1+sqrt(-167) itself as a factor. >> Therefore, in the ring of algebraic integers, the roots of >> x^2 - x + 42 = 0 >> and >> y^2 - by - 7 = 0 >> never share non-unit factors in the ring of algebraic integers. >> Notice that the key theorem here is that one that prevents an >> algebraic integer from being the root of a non-monic primitive with >> integer coefficients irreducible over Q. > A true theorem, but you are assuming something else: that a polynomial >reducible over Q has to have at least one rational root: clearly >a false assumption. > Nora B. >> Take away the condition that you have an algebraic integer factor, and >> youre ok, which proves that there must be a ring beyond algebraic >> integers, where they *can* share non-unit factors. >> I have called the more inclusive ring, the ring of objects, or the >> Object Ring. >> The Object Ring is a commutative ring that includes all numbers such >> that -1 and 1 are the only members that are both a unit and an >> integer, where no non-unit member is a factor of any two integers that >> are coprime. >> One of the most surprising results that follows quickly from >> understanding the limitations of the ring of algebraic integers, and >> having the fuller definition, is a proof of Fermats Last Theorem of >> around two to three pages. >> James Harris === Subject: Re: Metamath Axiom of Choice >>While looking at the Metamath site, and specifically at their statement of >>the Axiom of Choice, I note that their statement of the Axiom of Choice is >>effectively the statement that for any set x, there exists a set y such >>that for any nonempty element w of x, there is at least one element t of y >>such that w is an element of t, and that w has exactly one element in >>common with the union of all elements of y which have w as an element, >>i.e. there is a set y such that w is an element of some element t of y, >>and if s = {t in y : w in t}, then w n (Us) has exactly one element, where >>A n B denotes the intersection of A and B, and where Us is the union of >>the elements of s. >>That this implies the Axiom of Choice is obvious (if w is a nonempty >>element of x, let f(w) be the unique element of w n (Us), where >>s = {t in y : w in t}). The question is how can one prove the Metamath >>statement of the Axiom of Choice in ZFC? >> How does this sound for a proof. Let f : x-{empty set} -> Ux be a choice >> function (so that f(v) is an element of v for all nonempty elements v of x). >> Let y = {{v,f(v)} : v in x, v nonempty}. The first thing to note is that >> if w is a nonempty element of x, then there is an element of y which has w >> as an element, specifically, {w,f(w)}. Let w be a nonempty element of x, >> then s = {t in y : w in t} = {{w,f(w)}} u {{v,w} : v in x, v nonempty, >> f(v) = w}, where A u B denotes the union of A and B. It follows that >> Us = {w,f(w)} u {v in x : v nonempty, f(v) = w}, and so the elements of >> Us are w, f(w), and nonempty elements v of x such that f(v) = w. Since >> w is an element of v for all nonempty elements v of x such that f(v) = w, >> then the elements of Us are w, f(w) and some sets which have w as an >> element. By the axiom of regularity, the only common element of w and Us >> is f(w), completing the proof of the Metamath statement of the Axiom of >> Choice. >> David McAnally >> At the moment, they (the Time Lords) are far from being all-powerful. >> Thats why its been left up to me and me and me. >> quote by: Patrick Troughton in The Three Doctors >> ------- >Uhm... the Axiom of Choice is an _axiom_. For specificity, it is a >wff of the FOL of set theory which is (roughly speaking) stipulated >true, and can thus be used in deductions. If you decide that you >dont like the axiom of choice, you can choose to not work with FO >languages which include it. But even then, its hard to deny its use >in axiom systems which _do_ use it. (Since, by construction, youre >setting up a conditional... If the axiom of choice is true, and >...blah..., then ...blah...) >In any case, there is nothing to prove. As I stated above, the Metamath formulation of the Axiom of Choice is effectively the statement that given any set x, there exists a set y such that for all nonempty elements w of x, w n (Us) has exactly one element, where s = {t in y : w in t}. This is not amongst the formulations of the Axiom of Choice of which I was previously aware. The question of the equivalence of the Metamath formulation with the more standard versions under ZF naturally arises, as such an equivalence is necessary if the Metamath set theoretic axioms are to be taken as axioms of ZFC. In spite of what you might think, I was NOT trying to prove the Axiom of see that I initially gave a proof of a more standard version of the Axiom of Choice from ZF *and* the Metamath version of the Axiom of Choice, and that I was wondering about the possibility of a proof of the Metamath version of the Axiom of Choice from ZFC (which I believe that I supplied in the subsequent posting). In fact, I cant tell where you could have got any idea that I was trying to prove the Axiom of Choice from ZF. Your condescendingly pointing out that the Axiom of Choice is an axiom, and your final statement that there is nothing to prove demonstrate that you have completely missed the point of what I was trying to do, i.e. to demonstrate that the axioms adopted by Metamath do in fact form a set of axioms for ZFC, a fact which was not immediately obvious to me. Both Jesse Hughes and David Ullrich picked up what I was trying to do. Why couldnt you? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: Metamath Axiom of Choice >>Uhm... the Axiom of Choice is an _axiom_. For specificity, it is a >>wff of the FOL of set theory which is (roughly speaking) stipulated >>true, and can thus be used in deductions. If you decide that you >>dont like the axiom of choice, you can choose to not work with FO >>languages which include it. But even then, its hard to deny its use >>in axiom systems which _do_ use it. (Since, by construction, youre >>setting up a conditional... If the axiom of choice is true, and >>...blah..., then ...blah...) >>In any case, there is nothing to prove. >As I stated above, the Metamath formulation of the Axiom of Choice is >effectively the statement that given any set x, there exists a set y >such that for all nonempty elements w of x, w n (Us) has exactly one >element, where s = {t in y : w in t}. This is not amongst the >formulations of the Axiom of Choice of which I was previously aware. >The question of the equivalence of the Metamath formulation with the >more standard versions under ZF naturally arises, as such an equivalence >is necessary if the Metamath set theoretic axioms are to be taken as >axioms of ZFC. Also, the motivation behind proving the equivalence of the Metamath formulation of the Axiom of Choice and more standard formulations is the same as the motivation behind proving the equivalence of the Axiom of Choice and Zorns Lemma, and the equivalence of the Axiom of Choice and the Well-Ordering Theorem. I presume that you have seen the proofs of these last two mentioned equivalences, Acid Pooh. David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e*pi is irrational >> e*pi _is_ irrational, so why the need for a question mark? >G. A. Edgar cited MathWorld. So he did *not* put a question >mark. Instead, he gave the reference and asked for comments. >He could as well have written to the editors, hinting at >this non-sequitur and asking them to remove it. >> But (how) does that follow from the Gelfond-Schneider theorem? >Thats the point :-) >When David C. Ullrich called e*pi is irrational an error, >then indeed he didnt mean it (as I jokingly supposed), but >he called the implication an error. No, I meant that I thought it was unknown whether pi*e was irrational - I thought that was clear from my reply, I guess not. I wasnt certain, but I thought it was unknown for years, until I today when I saw Nat Silver post a reference saying it was known to be irrational. Since then hes posted another reference saying its unknown... >The fact that at least one of e+pi and e*pi must be irrational >is so funny that I think I can get it from my memory: >(x-e)(x-pi) = 0 has no rational roots, OK? >The left side is x^2 - (e+pi)*x + e*pi. Assuming e+pi and e*pi >both rational you have x^2 - Ax + B = 0 with rational >A=e+pi and B=e*pi. But then x is rational. Contradiction. >Rainer Rosenthal >r.rosenthal@web.de ************************ David C. Ullrich === Subject: Re: e*pi is irrational >> Um... I thought it was stated (but not proved) in Hardy & Wright, >> but now I cant find it. Sorry if Ive misremembered. >> It seems that I did remember correctly, see Nat Silvers contribution. >> H&W give no reference (in the 4th ed) but they mention 1929 which >> was the date of Gelfonds paper. >> The e + pi or e*pi result refers to transcendentality >> (transcendence?). >in 1991, Old and New Unsolved Problems in Plane Geometry and Number >Theory, Stan Wagon and Victor Klee write on page 23: >For example, the second problem posed above is equivalent to asking >whether the real number pi/e is rational. Equally difficult unsolved >problems >ask whether any of the following are rational: pi+e, pi*e, pi^e, e^e, pi^pi. So in particular Wagon and Klee said it was not known in 1991 whether pi*e was rational? I _thought_ that was unknown. Of course this contradicts what you cited from Hardy&Wright. Maybe its a typo? >Such combinations are expected to be transcendental numbers, but proving >transcendence of individual numbers is difficult. ************************ David C. Ullrich === Subject: Re: e*pi is irrational > So in particular Wagon and Klee said it was not known in 1991 > whether pi*e was rational? I _thought_ that was unknown. Of > course this contradicts what you cited from Hardy&Wright. > Maybe its a typo? It is a typo that survived into the fourth and fifth editions. Unfortunately for Eric Weisstein, he probably lifted his information from H&W. Anyway, Klee and Wagon point out that as well as knowing about the pair, pi+e and pi*e, it is also true that at least one, in the following pairs, is irrational: a) pi+e and pi-e b) pi*e and pi/e === Subject: Re: e*pi is irrational > I dont have a proof. I may well be wrong, but my recollection > is that its known that one of e + pi or e*pi must be irrational > but its not known which. I certainly hope one at least of them is, otherwise e and pi must both be at most quadratic irrationalities, which makes a nonsense of Lindemanns proof of the transcendence of pi (and Hermites proof of the same for e ?) (Apologies if I missed some irony there.) ------------------------------------------------------------- -------------- John R Ramsden (jr@adslate.com) ------------------------------------------------------------- -------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: e*pi is irrational > H&W give no reference (in the 4th ed) but they mention > 1929 which was the date of Gelfonds paper. As far as I can see, in 1929, C. Siegel (Germany) showed 2^sqrt(2) is transcendental. Gelfond/Schneider, independently using different methods, found their more general result in 1934. Showing 2^sqrt(2) is transcendental was important in the context of the time. In 1919, Hilbert asserted that the Riemann hypothesis would probably be proved before he died and that Fermats Last Theorem might be proved before the end of the century but that the transcendence of 2^sqrt(2) would turn out to be more trouble- some than the others. Theorem: A real algebraic number of degree n is not approximable to any order greater than n. This theorem was improved upon successively by Thue, Siegel, Dyson, and Gelfond. Finally, in 1955, Roth showed that no irrational algebraic number is approximable to order greater than 2. === Subject: Re: e*pi is irrational > Um... I thought it was stated (but not proved) in Hardy & Wright, > but now I cant find it. Sorry if Ive misremembered. > It seems that I did remember correctly, see Nat Silvers contribution. > H&W give no reference (in the 4th ed) but they mention 1929 which > was the date of Gelfonds paper. > The e + pi or e*pi result refers to transcendentality > (transcendence?). in 1991, Old and New Unsolved Problems in Plane Geometry and Number Theory, Stan Wagon and Victor Klee write (on page 243): For example, the second problem posed above is equivalent to asking whether the real number pi/e is rational. Equally difficult unsolved problems ask whether any of the following are rational: pi+e, pi*e, pi^e, e^e, pi^pi. Such combinations are expected to be transcendental numbers, but proving transcendence of individual numbers is difficult. === Subject: Re: Math Too Advanced For Mainstream Economists Ive been asking two questions, the first a simple yes-or-no question, the second asks for basic empirical support. : Question 1: Does the originator of this thread believe that a : well-behaved supply and demand model is never appropriate for : explaining how labor markets function? : Question 2: Give than there are many types of models for explaining : labor markets, what observed aspect of labor market behavior does : this approach explain better than other established models? > My URL suggests that explaining wages and employment by the supply > and demand for labor is problematic: > is sometimes reasonable. This is true, I do believe that the supply and demand model is sometimes reasonable. Yes or no, do you believe that the supply and demand model is never a reasonable description of what we observe in the world? Here, Ill make it easy for you. ___ I, Robert Vienneau, believe that the supply and demand model is sometimes a reasonable way of modelling the world we observe. ___ I, Robert Vienneau, believe that the supply and demand model is never reasonable way of modelling the world we observe. Please put your X in the appropriate space. > But he cannot and will not present any > argument whatsoever that addresses my argument (which merely > echoes well-established results in the literature). > Instead he tries to shift the burden of proof: > Question: Does the originator of this thread believe that a > well-behaved supply and demand model is never appropriate for > explaining how labor markets function? > This raise an empirical question. Does poor Mark Witte realize > he has no point? I think it was clear early on that there was never any point to this thread beyond it being merely a strange polemic. Of course, if someone would present an interesting model that did a good job of describing something that we see, I could be persuaded otherwise. But...thats just not going to happen, is it? > -- > Try http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/ Bukharin.html > To solve Linear Programs: .../LPSolver.html > r c A game: .../Keynes.html > v s a Whether strength of body or of mind, or wisdom, or > i m p virtue, are found in proportion to the power or wealth > e a e of a man is a question fit perhaps to be discussed by > n e . slaves in the hearing of their masters, but highly > @ r c m unbecoming to reasonable and free men in search of > d o the truth. -- Rousseau === Subject: Re: Math Too Advanced For Mainstream Economists > Poor Mark Witte seems to believe that the supply and demand model > is sometimes reasonable. > This is true, I do believe that the supply and demand model is > sometimes reasonable. Yes or no, do you believe that the supply and > demand model is never a reasonable description of what we observe in > the world? Here, Ill make it easy for you. > ___ I, Robert Vienneau, believe that the supply and demand model is > sometimes a reasonable way of modelling the world we observe. > ___ I, Robert Vienneau, believe that the supply and demand model is > never reasonable way of modelling the world we observe. > Please put your X in the appropriate space. I dont think this will work. At least it hasnt in the past. How about we try the affirmative silence gambit? Say, Robert, if you believe the supply and demand model is never a reasonable approach, dont respond to this question. === Subject: easy...analysis....simple confirm... hello.......genius...... let function f : R->R and g : R->R is continuous. let f(r) = g(r) for all r in rational. show that f(x) = g(x) ---------------------------------- this is my solving process first......method...... any r in rational, if for each e>0 there is a d, |x-r| |f(x)-f(r)|0, |x-c| |h(x)-h(c)|, on >but since Ive completed the most basic high-school maths, You will be shocked once you realize the extent to which your HS shortchanged you. There is very little Mathematics in contemporary HS Mathematics. >Is mathematics a far too great field to summarize into one reference >book? Yes. Is there a good library near you? Id advise checking out library books rather than buying your own copies, at least until you have a better feel for your direction. Number Theory may be a relatively painless area to start with. Euclidean Geometry is also something that can be covered without assuming any other Mathematics; avoid books that have an analytic (Cartesian) rather than synthetic (axiomatic) presentation unless you have already learned some Real Analysis. >Of course, the best way to learn math is to solve problems and >exercises, Yes. Thats important for learning the material, and some of the problems are interesting in their own right. >I might add that I wish to learn mathematics up to, >lets say, a university level. Thats really the starting point these days. The curriculum has been greatly watered down from what it was in the early 20th century. >My wallet-size isnt infinite, so one or two major works is about >what I can buy. One or two major works isnt enough. Have you considered using some online resources? Is there a good university near you? Maybe theyd let you use their library. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: The role of infinity in math <402a6518_3@newsfeed.slurp.net> at 11:23 PM, rusin@vesuvius.math.niu.edu (Dave Rusin) said: >D. There are aleph-1 of them or the set of them is countably >infinite. > Look up the definition of countably infinite and tell me this > isnt nearly a tautology! It isnt a tautology. In fact, its false. There are only Aleph-0 integers. >I dont believe thats a useful concept in itself. Well, one and two point compactifications are useful, and the term infinity is used productively in Projective Geometry. >Use infinite, the definable adjective, whenever possible instead >of an undefined noun infinity. Id say that they are equally meaningful in context and equally meaningless out of context. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: The role of infinity in math at 10:39 PM, Martin Johansen said: >What is the role of infinity in math: The word is used for a number of very different things. >How is it defined? In what context? >Why is it needed? Like any other technical term, it allows you to express common phrases more concisely. >At what precition does math work? None. Mathematics is not the same as Engineering. If you need to do a calculation, your objective will dictate what precision is appropriate. Note, however, that there is a branch of Mathematics called Numerical Analysis that will help you determine what precision you need for a calculation in a particular context. at 11:02 PM, Martin Johansen said: >It is practical, yes. But why use it instead of a high number, Because a high number would be totally irrelevant to the uses to which Mathematicians put the term infinity. Thats like asking a farmer wy he feeds his cattle corn instead of pumice; theyre not remotely similar. >Again all measurement work with presition, but what about math in >itself? The concept of precision is inapplicable to most of Mathematics. Even where it is relevant, the Mathematics has no precision, but rather proves results about the precision of calculations. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Limit Of # Of Coprime Integers >.... > You are right, I was *WRONG*!... I discovered I was wrong this morning when I noted that n*6/pi^2 - sum{k=3 to n+1} 1/zeta(j) is negative for all high enough ns, ...obviously wrong! And I did not even notice myself that I had an incorrect > 1/m^2, where I should have had a 1/m^3. But, in any case, I did finally get the n=2 case: limit{m-> oo} > (1/m^3) *sum{k=1 to m} sum{j=1 to m} H(m;k,j) = (6/pi^2) *product{p=primes} (1 -1/(p(p+1))), which is what you got. Also, the above limit (6/pi^2 *product) can be rewritten as: (zeta(3)*36/pi^4) * product{p=primes} (1- 1/(p(p+1))^2), which converges, obviously, faster than my original product. Did I do my math correct? (I multiplied product by product{p=primes} (1 +1/(p(p+1))), which I assume is zeta(2)/zeta(3); then divided constant coefficent by this.) Leroy Quet (I seem to remember seeing somewhere that the product in my > representation is equal to some famous constant,...or perhaps that > was with (p(p-1)) in the denominator of the fraction instead...) > With a (p(p-1)), the product would have been Artins constant: > http://www.worldwideschool.org/library/books/sci/math/ MiscellaneousMathematic alConstants/chap8.html > Leroy As for n > 2, I have not figured that out, nor will I investigate this > any more for now. Leroy Quet ... === Subject: Over There Jack, you say ... interior w = -1 mimics w = 0 when observed at a distance with light signals or gravity waves or neutrinos or anything since gravity is universal. .... If the interior of your DM ball is w = -1, but your DM ball looks from the exterior like w = 0 then is that saying that your DM ball looks like a black hole? No obviously not necessarily. It looks like a dark cold mass M with w = 0 where M ~ c^2|/zpf|R^3/G provided that R > |/zpf|R^3 Of course if /zpf gets too strongly negative you get an event horizon and in fact there are black holes at the centers of some galaxies. I got it all Bhubba. I got The Right Stuff and the problem is that the Ed Wittens and the Brian Greenes et-al are shining strong lights in the wrong part of the dark room. :-) Some of what they are doing is OK though. Indeed, I use it shamelessly. :-) If so, then is there any way to distinguish your model from the many proposals that DM might be made up of black holes? I just showed you. You can certainly make artificial black holes as a WMD, but I would rather make Star Gates. Such proposals sometimes use primordial black holes that were created in the early universe and are sometimes either sufficiently massive as not to have yet decayed or are taken to be Planck-mass with the idea that the Planck mass is a minimum mass for a black hole, so that it wont decay. Sure, but I take the active view of the Observer-Participator, The Pisher-Paysher my Grandfather (who worked for the Army) called it. :-) We are the Corps of Metric Engineers, the Masters of Hyperspace. Thats what the UFOs are all about. The Endless Frontier is opening up again. Its our Manifest Destiny Matrix. And we wont come back till were finished Over There. http://www.lib.virginia.edu/speccol/exhibits/music/patriotic_ overthere.html === Subject: Re: e is transcendental (was: classes of transcendental numbers ? > I,have a reason , with my due respect. > Panagiotis Stefanides Mus be the snow. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: e is transcendental (was: classes of transcendental numbers ? So? The real part of exp(i pi) is cos(pi), and its imaginary part >>is sin(pi), so all you are saying is that cos(pi) = -1 and >>sin(pi) = 0, and we were already aware of these facts. There is >>NO reason to conclude that exp(i pi) = 0. > I,have a reason , with my due respect. > Panagiotis Stefanides Yes, but you DONT tell us what your reason is. You cant expect us to accept your claims without giving support for those claims. So what possible reason could you have for expecting us to agree with your claim that exp(i pi) = 0? David McAnally At the moment, they (the Time Lords) are far from being all-powerful. Thats why its been left up to me and me and me. quote by: Patrick Troughton in The Three Doctors ------- === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1F04bl15237; >.93 Eur Ing Panagiotis Stefanides [NonBreakingSpace].8b.96.87À.8c .97.99.95 .93fi.92.9b.93.87 >> I,have a reason , with my due respect. >> Panagiotis Stefanides This is Your opinion! At least the following may be appropriate as an answer regarding my respect andcontributions and respect to the Greek Science : http://www.stefanides.gr/why_logarithm.htm This gives the reasoning behind the widely use word of LOGARITHM Gives relationships to LOGARITHMIC SPIRALS WITH ARCHIMEDES SPIRAL AND LOGARITMS etc. etc. http://www.stefanides.gr/plato.htm http://www.stefanides.gr/logarithm.htm http://www.stefanides.gr/nautilus.htm http://www.stefanides.gr/platostriangle.htm Panagiotis Stefanides http://www.stefanides.gr === Subject: Re: JSH: What it would take > Now its interesting to see what it would take for > z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > to be an algebraic integer for an algebraic integer b. > First off notice that for an algebraic integer b, (b + sqrt(b^2 + 28)) > is an algebraic integer, so it fits into the format I used in my > previous post where I just said r_1. > Next, its kind of obvious from demonstration that what makes it > impossible is whats required to get a polynomial with integer > coefficients. > You need a monic quadratic with integer coefficients for z, but cant > get one so its not an algebraic integer. Wrong again. Why do you need a *quadratic* with integer coefficients? > Next, its kind of obvious from demonstration that what makes it > impossible is whats required to get a polynomial with integer > coefficients. No, you can not get a quadratic with integer coefficients, this does not prove that z is not an algebraic integer. > You need a monic quadratic with integer coefficients for z, but cant > get one so its not an algebraic integer. Why do you need a quadratic? > Now consider if you have something like z = sqrt(2)sqrt(7)/sqrt(2), > then obviously you can divide off sqrt(2), and get that monic > quadratic with integer coefficients. Yup, and when you have z = sqrt(2)(sqrt(3) + sqrt(7))/sqrt(2), you will *not* get a quadratic with integer coefficients. And when you have z = sqrt(2)cbrt(7)/sqrt(2), you will not get that quadratic. And when you have z = p^(1/11)(b + sqrt(b^2 + 28))/(b + sqrt(b^2 + 28)) (as is the case), you will not get a quadratic. So what is your point? > What Ive shown in a rather simple way is that only such simple and > direct results will work, so if you dont *see* the non-unit factors > between two irrational algebraic integers, then you dont have any in > the ring of algebraic integers! A completely new view, to be sure. > Thats fascinating to me as for a long time Ive been trying to figure > out how to show the limitations of the ring, but kept bumping into > what you cant physically see, but here now I get to use it to show my > point. I do not see it. What is the common factor of 3 and (1 + sqrt(-5))? (They have a common algebraic integer factor.) -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: What it would take >Now its interesting to see what it would take for >z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >to be an algebraic integer for an algebraic integer b. >First off notice that for an algebraic integer b, (b + sqrt(b^2 + 28)) >is an algebraic integer, so it fits into the format I used in my >previous post where I just said r_1. Yes ... similarly for (b - sqrt(b^2 + 28)). See below. >Next, its kind of obvious from demonstration that what makes it >impossible is whats required to get a polynomial with integer >coefficients. >You need a monic quadratic with integer coefficients for z, but cant >get one so its not an algebraic integer. Let c be any algebraic integer divisor of 7. Let b = c - 7/c. Then c satisfies the equation c^2 - b*c - 7 = 0. so c = (b + sqrt(b^2 + 28))/2 or c = (b - sqrt(b^2 + 28))/2. So what you claim to have shown above is essentially that (1 + sqrt(-167)) cannot have an algebraic integer factor in common with 7. If your argument is correct, it also works for (1 - sqrt(-167)). However, [(1 + sqrt(-167))/2]*[(1 - sqrt(-167))/2] = 7*6. Therefore you have two algebraic integers whose product is divisible by 7, but neither of them has any algebraic integer factors in common with 7. Right? That is, 7 | A * B, but both A and B are coprime to 7. Are you sure you dont want to think about that some more ? Nora B. PS: And what exactly do you think is wrong with the Keith Ramsay factor of (1 + sqrt(-167))/2 ??? >Now consider if you have something like z = sqrt(2)sqrt(7)/sqrt(2), >then obviously you can divide off sqrt(2), and get that monic >quadratic with integer coefficients. >What Ive shown in a rather simple way is that only such simple and >direct results will work, so if you dont *see* the non-unit factors >between two irrational algebraic integers, then you dont have any in >the ring of algebraic integers! >Thats fascinating to me as for a long time Ive been trying to figure >out how to show the limitations of the ring, but kept bumping into >what you cant physically see, but here now I get to use it to show my >point. >James Harris === Subject: Re: JSH: What it would take >Now its interesting to see what it would take for >z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >to be an algebraic integer for an algebraic integer b. >First off notice that for an algebraic integer b, (b + sqrt(b^2 + 28)) >is an algebraic integer, so it fits into the format I used in my >previous post where I just said r_1. > Yes ... similarly for (b - sqrt(b^2 + 28)). See below. Fuck off you stupid dumbshit. Yes Im back to cursing at you Nora Baron. FUCK OFF!!! James Harris === Subject: Re: JSH: What it would take > **** off you stupid dumb****. > Yes Im back to cursing at you Nora Baron. > **** OFF!!! > James Guttersnipe Harris Because shes proven you wrong _again_, naturally. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: JSH: What it would take >Now its interesting to see what it would take for z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) to be an algebraic integer for an algebraic integer b. First off notice that for an algebraic integer b, (b + sqrt(b^2 + 28)) >is an algebraic integer, so it fits into the format I used in my >previous post where I just said r_1. > Yes ... similarly for (b - sqrt(b^2 + 28)). See below. Fuck off you stupid dumbshit. > Yes Im back to cursing at you Nora Baron. > FUCK OFF!!! > James Harris Harris, you such an anus. You can try again. Appended are my comments that you deleted, plus some additional. > Yes ... similarly for (b - sqrt(b^2 + 28)). See below. >Next, its kind of obvious from demonstration that what makes it >impossible is whats required to get a polynomial with integer >coefficients. >You need a monic quadratic with integer coefficients for z, but cant >get one so its not an algebraic integer. > Let c be any algebraic integer divisor of 7. Let b = c - 7/c. > Then c satisfies the equation > c^2 - b*c - 7 = 0. > so > c = (b + sqrt(b^2 + 28))/2 or > c = (b - sqrt(b^2 + 28))/2. > So what you claim to have shown above is essentially that > (1 + sqrt(-167)) > cannot have an algebraic integer factor in common with 7. > If your argument is correct, it also works for > (1 - sqrt(-167)). > However, > [(1 + sqrt(-167))/2]*[(1 - sqrt(-167))/2] = 7*6. > Therefore you have two algebraic integers whose product > is divisible by 7, but neither of them has any algebraic integer > factors in common with 7. Right? > That is, > 7 | A * B, > but both A and B are coprime to 7. > Are you sure you dont want to think about that some more ? Heres why you should. If A is coprime to 7, then by definition there are algebraic integers s and t such that A*s + 7*t = 1. Multiply both sides by B: A*B*s + 7*B*t = B. Since 7 | A*B, let A*B = 7*k for some algebraic integer k, and subsitute: 7*k*s + 7*B*t = B and factor: 7*(k*s + B*t) = B, which implies 7 | B. Therefore if A is coprime to 7, B most certainly is not. Overall conclusion: if 7 | A*B, one or the other of A and B must not be coprime to 7. Even if you dont understand the argument, it makes *intuitive* sense. In this one respect, algebraic integers are like ordinary integers. Nora B. > Nora B. > PS: And what exactly do you think is wrong with the Keith Ramsay > factor of (1 + sqrt(-167))/2 ??? === Subject: Re: JSH: What it would take >>Now its interesting to see what it would take for >>z = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) >>to be an algebraic integer for an algebraic integer b. >>First off notice that for an algebraic integer b, (b + sqrt(b^2 + 28)) >>is an algebraic integer, so it fits into the format I used in my >>previous post where I just said r_1. >> Yes ... similarly for (b - sqrt(b^2 + 28)). See below. >Fuck off you stupid dumbshit. >Yes Im back to cursing at you Nora Baron. >FUCK OFF!!! Giggle. You _really_ think that when she says things about the math and your only reply is to tell her to fuck off its not going to be obvious that the reason is you see shes right (or at least fear she may be right and dont want to think about it?) HINT: This is another one of those things where even if someone knew nothing whatever about the math he would decide on the basis of your behavior that you must be wrong. Not a good thing if youre trying to convince people. >James Harris ************************ David C. Ullrich === Subject: Re: JSH: What it would take > Thats fascinating to me as for a long time Ive been trying to figure > out how to show the limitations of the ring, but kept bumping into > what you cant physically see, but here now I get to use it to show my > point. > James Harris In polite circles, it is called ßashing. === Subject: Re: Countable perfect set |Maybe Im not seeing this properly (perhaps have a definition wrong). | |What if you only have one point x in your space X={x}? Give it the |obvious topology ({} and {x}=X are open). Make your metric be d(x,y) |= 0 since x must be equal to y. I think x is the only possible |accumulation point. Your point x is an isolated point of the set, not an accumulation point. A point x is in the closure of a set X if for each epsilon>0, there exists a point y in X (not necessarily different from x) such that d(x,y)0 such that the only element y of X satisfying d(x,y)0, there are infinitely many values w such that e^w=z. If w is one of them, the others differ from it by multiples of 2*pi*i, so that the possibilities for log z are ..., w-4*pi*i, w-2*pi*i, w, w+2*pi*i, w+4*pi*i, .... Choosing a branch means choosing one of these for each z to be the value of log z. This would not be so bad if it were possible to make this choice in a continuous way. You can picture log z as being like a ramp in a parking garage, where each each time you go around you wind up at the next level of the garage. If you start on one branch of log z and follow it continuously as you take z around the origin, you arrive at a different branch of log z. Suppose for example you just went around the unit circle going counterclockwise, starting with log 1 = 0. When you get to z=i, you have log(i)=pi*i/2. When you get to z=-1, you have log(-1)=pi*i. Proceeding on around to z=-i, you have log(-i)=3*pi*i/2. And finally when you get back to z=1, you have arrived at the branch which has log(1)=2*pi*i. Its essentially the same thing as the problem of angles being defined only up to multiples of 2*pi radians. Another analogous situation is the way that in order to define time zones, there has to be a date line somewhere. As you go around the world to the east, you keep having to set your clock forward. For consistency, however, there has to be some place (somewhat arbitrarily chosen) where as you go east, you get switched back to the previous date. Choosing a branch of log is like choosing where to put the international date line. Or, if you think of log z as being like a spiral ramp, choosing a branch is like chopping out just one turn of the ramp. Typically, one chooses a connected region in the plain not containing 0 and not containing any closed curve passing around 0, and for some point z in the region chooses a value of w satisfying e^w=z. Often the region consists of all the points not on a particular ray from 0, called a branch cut. Then value at 1 is chosen; the choice log(1)=0 is called the principal branch. Sometimes the value of log is defined on the ray as well, to be equal to the limit as one approaches the ray from one side. If you do that, though, the function is discontinuous because the limit as you approach the line from the other side differs from the value on the line by 2*pi*i or -2*pi*i. However you do it, you either have to leave log z undefined for some values of z<>0, or you get a discontinuous function. Its not necessary to make the branch cut a straight line. You can draw some kind of curious non-self-intersecting curve from 0 out to infinity and define a branch of log which is continuous everywhere in the complex plane except on that curve. The same kind of issue comes up for defining the square root of complex numbers or inverse trig functions. If you continuously follow a branch of the complex square root as you go around 0, when youve taken one complete trip around 0 you arrive at the other branch, the one with the opposite sign. Keith Ramsay === Subject: Re: Help to verify an open mapping >f: R^n{0} -> S^(n-1) >x |-> x/||x|| >where ||.|| is the euclidean norm. >How to prove, analytically, that f is an open mapping? Hint: any neighbourhood of x <> 0 in R^n contains a neighbourhood of x in the sphere of radius ||x||. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. Yup, that is easy. Arturo already provided an easy proof. But take the smallest field that contains all three. It will be the splitting field of some polynomial. On the other hand, you are assuming (I think) that if a and b are in the same splitting field that their non-unit algebraic integer in common also should be there. That is false. > (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) What is the difference between a field and a field of rationals? Something is a field when it satisfies the requirements of a field. Do you know them? > Thats a nifty and powerful result. Why am I the one who had to > discover it? Not so very powerful. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. > (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) > Thats a nifty and powerful result. Why am I the one who had to > discover it? Thats the simplest way to explain, which is accessible to the mathematicians out there familiar with Galois Theory. The question now is, will they sit back and hope that its harder for me to explain in detail? I have reason to think they may try, unfortunately, which means Ill keep at working out a simple explanation that doesnt require you know advanced mathematics, or trust mathematicians who do but behave in such an egregious manner. James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) Thats a nifty and powerful result. Why am I the one who had to > discover it? > As with all the other of JSHs nifty and powerful results, one should avoid using it to build bridges, or anything else except castles in air. > Thats the simplest way to explain, which is accessible to the > mathematicians out there familiar with Galois Theory. Simple, and false. A field is a field is a field. That JSH is so unawre of what a field is that he could make such statements is typical of him. > The question now is, will they sit back and hope that its harder for > me to explain in detail? The question is, will they all sit back chortling and waiting for the newsgroup clowns next pratfall. === Subject: Re: JSH: Splitting field, algebraic integer factors > Thats a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Because youre so special in Gods eyes. === Subject: Re: JSH: Splitting field, algebraic integer factors Thats a nifty and powerful result. Why am I the one who had to > discover it? > James Harris > Because youre so special in Gods eyes. The math isnt that complicated. Consider that Im using x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2. So I can simply introduce z, where (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. Now it doesnt take a math genius to see that you cant multiply by an algebraic integer z, and manage that sqrt(-167) on the left side without having b^2 + 28 = -167n^2, which is why I brought up the splitting field. But to see replies in this thread youd think I brought up the Tooth Fairy. Its like rank stupidity on an awesome scale as if mathematicians are dimwits. Why is it so wacky for me to point out the obvious? Why are you all so stupid? James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors >> Thats a nifty and powerful result. Why am I the one who had to >> discover it? >> James Harris >> Because youre so special in Gods eyes. >The math isnt that complicated. Consider that Im using >x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and >y^2 - by - 7 = 0, which has as one of its roots >(b + sqrt(b^2 + 28))/2. >So I can simply introduce z, where >(1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. >Now it doesnt take a math genius to see that you cant multiply by an >algebraic integer z, and manage that sqrt(-167) on the left side >without having b^2 + 28 = -167n^2, which is why I brought up the >splitting field. >But to see replies in this thread youd think I brought up the Tooth >Fairy. Uh, part of the reason is that we all know that you have no idea what a splitting field actually is. I mean youve made it clear many times that you dont even know the definition of the word ring... when you refer to splitting fields as things which are not true fields you induce gales of laughter - thats just the way it works. >Its like rank stupidity on an awesome scale as if mathematicians are >dimwits. >Why is it so wacky for me to point out the obvious? >Why are you all so stupid? And its not just us here on sci.math! The world-famous mathematicians youve contacted are _also_ all too stupid to see youre right about any of this. Every single one of them. Hard to believe, isnt it? (Well, _most_ people would find it hard to believe...) >James Harris ************************ David C. Ullrich === Subject: Re: JSH: Splitting field, algebraic integer factors Thats a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Because youre so special in Gods eyes. > The math isnt that complicated. Consider that Im using > x^2 - x + 42 = 0, which has (1 + sqrt(-167))/2 as on of its roots, and > y^2 - by - 7 = 0, which has as one of its roots > (b + sqrt(b^2 + 28))/2. > So I can simply introduce z, where > (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2. > Now it doesnt take a math genius to see that you cant multiply by an > algebraic integer z, and manage that sqrt(-167) on the left side > without having b^2 + 28 = -167n^2, which is why I brought up the > splitting field. > But to see replies in this thread youd think I brought up the Tooth > Fairy. You brought up the equation 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0, derived from (1 + sqrt(-167))/2 = (b + sqrt(b^2 + 28))z/2, after several false starts, and claime that that equation is irreducible over Q for any integer b, for some simple reasons . But for b = 6, which was an integer last time I looked, 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 7z^4 + 6z^3 + 299z^2 - 36z + 252 = (z^2 + z + 42)(7z^2 - z + 6). which certianly fits my definition of being reducible. Which brings up teh question of what your definition of irreducible versus reducible must look like. > Its like rank stupidity on an awesome scale as if mathematicians are > dimwits. It appears to be rank stupidity on an awsome scale, but not by any mathematicians. > Why is it so wacky for me to point out the obvious? Was it nature or nurture that gave you all that wackyness? > Why are you all so stupid? Just lucky, I guess. I really prefer our sort of stupidity to your sort of intelligence. Mine tends to get me to correct answers to mathematical questions much more often that your way gets you to them. > James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. Err, *the* same splitting field? the 2nd and 4th root of 2, say, are not coprime, yet they lie in different splitting fields for different polynomials (over Q). Is there some magic polynomial that we ca call THE polynomial, and hence THE splitting field? > (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) > Thats a nifty and powerful result. Why am I the one who had to > discover it? > James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. Err, *the* same splitting field? the 2nd and 4th root of 2, say, are > not coprime, yet they lie in different splitting fields for different > polynomials (over Q). Is there some magic polynomial that we ca call > THE polynomial, and hence THE splitting field? Ok, more correctly either you can directly see the factors between irrational integers, like with 2^{1/4} being a factor of 2^{1/2}, or they have to be members of the same splitting field. James Harris === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. > Err, *the* same splitting field? the 2nd and 4th root of 2, say, are > not coprime, yet they lie in different splitting fields for different > polynomials (over Q). Is there some magic polynomial that we ca call > THE polynomial, and hence THE splitting field? Ok, more correctly either you can directly see the factors between > irrational integers, like with 2^{1/4} being a factor of 2^{1/2}, or > they have to be members of the same splitting field. > James Harris Look at Arturos post, there is always a splitting field in which any two (or finitely many) algebraic integers will lie. I still dont see how you are using the word Ôthe, without telling us how to find *the* polynomial. At the moment I dont see anyway of forcing any meaningful uniqueness. === Subject: Re: JSH: Splitting field, algebraic integer factors Adjunct Assistant Professor at the University of Montana. [.snip.] >Err, *the* same splitting field? the 2nd and 4th root of 2, say, are >not coprime, yet they lie in different splitting fields for different >polynomials (over Q). Is there some magic polynomial that we ca call >THE polynomial, and hence THE splitting field? Well, any finite number of algebraic numbers lie in a unique minimal splitting field over Q, surely; namely, if r1,...,rk are algebraic numbers, then the normal closure of Q(r1,...,rk) is the minimal splitting field over Q which contains r1, r2,...,rk. Not much of an observation, granted. (The polynomial whose splitting field we are talking about would be the product of the minimal polynomials of r1, or r2,...., and of rk). -- Its not denial. Im just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: JSH: Splitting field, algebraic integer factors > [.snip.] >Err, *the* same splitting field? the 2nd and 4th root of 2, say, are >not coprime, yet they lie in different splitting fields for different >polynomials (over Q). Is there some magic polynomial that we ca call >THE polynomial, and hence THE splitting field? > Well, any finite number of algebraic numbers lie in a unique minimal > splitting field over Q, surely; namely, if r1,...,rk are algebraic > numbers, then the normal closure of Q(r1,...,rk) is the minimal > splitting field over Q which contains r1, r2,...,rk. > Not much of an observation, granted. (The polynomial whose splitting > field we are talking about would be the product of the minimal > polynomials of r1, or r2,...., and of rk). Which is true irrespective of the coprimality of the r_i. He would need to state more surely. As in his reply he accepted that silly counter example, it cant be that he is thinking of the normal closure of the extension by the two alg. integers, since the counter example falls under the heading of Ôeasily spotted; although he hasnt spotted x^2-2 splits in any extension where x^4-2 splits, and surely that would be an easy example in which to explain what he meant. I mean, how would he use it in the cases 1. a=sqrt(2) b=sqrt(3) 2. a=sqrt(6) b=sqrt(10) I cant see that the would be anyway of disinguishing those cases with a splitting field argument. But they do fall into the easily spotted category. === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. > (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) > Thats a nifty and powerful result. Why am I the one who had to > discover it? > James Harris Dont you want to rephrase that like an anti-Gertrude: A field is not a field is not a field ? Discoveries like this are what liven up the game of James watching. === Subject: Re: JSH: Splitting field, algebraic integer factors > Previously I posted that if you cant *see* the factors between > irrational algebraic integers then theyre not there, but more > correctly the situation is that two algebraic integers have to be > members of the same splitting field to have a non-unit algebraic > integer in common. > (I say more correctly as there may be some terminology issues here > because what mathematicians currently call a splitting field is not > a true field, but something close, like the field of rationals. But > thats another issue for another time.) > Thats a nifty and powerful result. Why am I the one who had to > discover it? You should be asking yourself why you are so often convinced that you are correct when you are wrong. > James Often in error, but never in doubt. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: universal set with 3 valued logic >It just doesnt make sense to introduce a strange new system of logic just >to salvage the Universal Set. Is there anything really wrong with the usual >2-value logic (T or F)-- other than prohibiting the existence of the >Universal Set, I mean? >Dan > i never said there was anything Ôwrong with either binary logic or > ZFC. i dont even know what Ôright and Ôwrong mean in math other > than consistency. Quite right. to me, the prohibition of a universal set is > Ôwrong, if anything is, but maybe thats just where you and i differ. > to me, this is better than making the object {x|x=x} a proper class, > whatever that is but maybe thats just me. An interesting idea. While the universal set is a troublesome construct in some ways, it does seem a bit counterintuitive to say that it does not exist. What do you make of Cantors power set theorem that for any set s and its power set p, p is larger than s? Does the universal set u have a power set in your system? Is it larger than u? Dan === Subject: Re: universal set with 3 valued logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1F1JLK20417; >But there are some three-step self-referential set definitions >that are still contradictory. So you can introduce more truth-values >for them. === Subject: Re: Ring of algebraic integers, comparison check >>You can take two quadratics: >>x^2 - x + 42 = 0 >>and >>y^2 - by - 7 = 0 >>where by how algebraic integers are currently taught, youd think that >>there exists an algebraic integer b, such that a root of the second >>quadratic is a factor of a root of the first. Can you explain why? It doesnt seem obvious to me. > Let r1 and r2 be the two roots of x^2-x+42 = 0. Since 7|r1*r2=42, > there is a factorization of 7 into algebraic integers, 7=s*t, such > that s|r1 and t|r2. > The polynomial y^2 - (s-t)y - 7 = (y-s)(y+t) has roots -t and s. s > divides the first root of x^2-x+42, t divides the second root of > x^2-x+42. Setting b = s-t gives the result. > So where does that leave James argument? It shows that it is clearly false. Unfortunately, this will have no effect on James. Arturos argument starts with the fact that if algebraic integers a, b are such that p|ab, then there must be algebraic integers s and t such that st=p, s|a and t|b. James has refused to accept this result. Also, James claims that proofs dont fight. In order to attack one of James claims it is not enough to present an argument with a contardictory conclusion. [1] Rather, you must show where James argument is incorrect. James will most likely ignore this argument. -William Hughes [1] Like many of James proscriptions (use of square roots, ad hominem arguments etc.) this does not appy to James. James will happily brand an argument false, based on contradictory results from a completely different approach. === Subject: Re: Ring of algebraic integers, comparison check > >You can take two quadratics: > >x^2 - x + 42 = 0 > >and > >y^2 - by - 7 = 0 > >where by how algebraic integers are currently taught, youd think that > >there exists an algebraic integer b, such that a root of the second > >quadratic is a factor of a root of the first. > > Can you explain why? It doesnt seem obvious to me. > That is not difficult. Lets have p a divisor of 7, so 7 = p.q. > (And let p and q be algebraic integers). We have: > (y - p)(y + q) = y^2 - (p - q).y - 7 > so set b = (p - q) and we have p a root of the polynomial in y above. > So *any* factor of 7 is the root of such a polynomial. > On the other hand, a root r of x^2 - x + 42 is not co-prime to 7, and > so there is an algebraic integer factor of x, such that it is also a > factor of 7. And by the previous paragraph it is a root of the > polynomial in y. > But we know that common factor due to work by Keith Ramsay. > Set a = (44444 - 111.sqrt(-167))^(1/11), and a its complex conjugate. > Now b = a - a will do the trick. > (The reason is that with > z = ([(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2])^(1/11) > (a.z)^11 = [(1 + sqrt(-167))/2]^11, so with the proper choice of > the eleven possibilities for z, a.z = (1 + sqrt(-167))/2, a root > of the polynomial in x, and (a.a)^11 = 7^11, so a.a = 7.) Ive been puzzling over that now and I realize why Im so sick of you people. Do I want to push an argument thats wrong? No. So Im using you yahoos to check stuff, and half the time youre wrong, and the rest of the time youre too damn incoherent for me to tell! I need some new people. Oh well, theyll grow up and get on sci.math eventually, but in the meantime, I just have this crappy crew so here goes. The problem I see with what you have Dik Winter, is that I can take y^2 - by - 7 = 0, which has as one of its roots (b + sqrt(b^2 + 28))/2. Trouble is, that sqrt(b^2 + 28) there, as it needs to be sqrt(-167) or some multiple of it. But that means its b^2 + 28 = -167n^2, where n is some integer. Now, that follows rather basically, and the problem is that the splitting field for y^2 - by - 7, is a lot different than the one for x^2 - x - 42, because one has a -42 on the end and the other has a -7. Now you can babble all you want about Keith Ramsay did this or that, but Im sick of you people because Im finding it harder and harder to use you and get something coherent out of you!!! Maybe your brains are fried. Whens the next generation up? James Harris === Subject: Re: Ring of algebraic integers, comparison check > Do I want to push an argument thats wrong? No. So Im using you > yahoos to check stuff, and half the time youre wrong, and the rest of > the time youre too damn incoherent for me to tell! What James means here is that he tries to get real mathematicians to clean up and correct his phony math so he can take credit for their work. Half the time he cant steal their results because they prove him wrong, and the rest of the time he doesnt understand them well enough to steal from them. Hint for James: The things you dont understand are also proof that youre wrong. -- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvise fwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: Ring of algebraic integers, comparison check > Now you can babble all you want about Keith Ramsay did this or that, > but Im sick of you people because Im finding it harder and harder to > use you and get something coherent out of you!!! > Maybe your brains are fried. Whens the next generation up? > James Harris JSH has, as usual, got thing backwards. The incoherence and fried brains are at his end of the pipeline. Note how he ignores his own error in the folloeinf 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > is irreducible over Q for any integer b, for some simple reasons Try b = 6. Then 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 7z^4 + 6z^3 + 299z^2 - 36z + 252 = (z^2 + z + 42)(7z^2 - z + 6). This error on JSHs part clearly establishes whose brains are fried. === Subject: Re: Ring of algebraic integers, comparison check ... > (The reason is that with > z = ([(298-23.sqrt(-167))[(-3-7.sqrt(-167))/2])^(1/11) > (a.z)^11 = [(1 + sqrt(-167))/2]^11, so with the proper choice of > the eleven possibilities for z, a.z = (1 + sqrt(-167))/2, a root > of the polynomial in x, and (a.a)^11 = 7^11, so a.a = 7.) > Ive been puzzling over that now and I realize why Im so sick of you > people. Because you are not willing to check what we write? There are p, q and r such that p^11.q^11 = [(1 - sqrt(-167))/2]^11 and p^11.r^11 = 7^11. And you are still maintaining that (1 - sqrt(-167)/2 and 7 do not share the factor p? Bizarre. > Do I want to push an argument thats wrong? No. So Im using you > yahoos to check stuff, and half the time youre wrong, and the rest of > the time youre too damn incoherent for me to tell! Darn, into curse mode again. > The problem I see with what you have Dik Winter, is that I can > take > y^2 - by - 7 = 0, which has as one of its roots > (b + sqrt(b^2 + 28))/2. > Trouble is, that sqrt(b^2 + 28) there, as it needs to be sqrt(-167) or > some multiple of it. Why? You simply assert and do not give reasons. > Now, that follows rather basically, and the problem is that the > splitting field for y^2 - by - 7, is a lot different than the one for > x^2 - x - 42, because one has a -42 on the end and the other has a -7. Yes, so what? (You may note in passing that the splitting field of y^2 - by - 7 depends quite a bit on b. For instance, when b = -6, the splitting field is just the rationals.) > Now you can babble all you want about Keith Ramsay did this or that, > but Im sick of you people because Im finding it harder and harder to > use you and get something coherent out of you!!! I thought Keith Ramsay had been some use because he provided an explicit algebraic integer factor common to (1 + sqrt(-167))/2 and 7. Of course, that use would not advocate your problem. > Maybe your brains are fried. Whens the next generation up? Eh? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Ring of algebraic integers, comparison check > CORRECTION POST: > It turns out that theres an incredibly simple way to check the ring > of algebraic integers and see that theres a problem with how its > currently taught. > You can take two quadratics: > x^2 - x + 42 = 0 > and > y^2 - by - 7 = 0 > where by how algebraic integers are currently taught, youd think that > there exists an algebraic integer b, such that a root of the second > quadratic is a factor of a root of the first. Because JSH might think so, it does not follow that anyone else might. JSHs mental processes are notoriously atypical. Intriguingly, there > does not, and its easy to show. > Now > x^2 - x + 42 = 0 has (1 + sqrt(-167))/2 as on of its roots, and > y^2 - by - 7 = 0, has as one of its roots > (b + sqrt(b^2 + 28))/2. Is this to be a proof that (a) there is a particular root of the second equation not an algebraic integer divisor of a particular root of the first, or that (b) no root of the second is such a divisor of any root of the first? They are quite different statements. weaker statement than (b), and is of no use, by itself, in establishing his other claims. Since such a proof, even if valid, is irrelevant, I have clipped it. === Subject: Re: Ring of algebraic integers, comparison check ... > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) > z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) > z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). ( z_1 and z_4 are algebraic integers for a proper value of b, z_2 and z_3 are not for that value.) > So integer bs are eliminated as a possibility immediately. Remember this, it is true! > Well, unless > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 > is reducible over Q with that integer b, the theorem I mentioned > requires that *none* of its roots can be an algebraic integer, as its > non-monic and primitive. But this does not show that integer bs are eliminated. They are eliminated only if you can indeed show that the quartic is indeed irreducible. You have not done that. > Therefore, even if you had an irrational algebraic integer root, the > polynomial would have to be reducible over Q, meaning that youd need > at least one rational root. This is wrong. 3x^4 + 4x^3 + 7x^2 + 4x + 3 is reducible, but it has no rational root. > If b is irrational--remember its an algebraic integer--then you get a > polynomial of degree equal to the order of the monic polynomial with > integer coefficients for which b is a root times 4. This makes no sense as written. With a proper choice of b the polynomial in z splits as: (Z - z_1)(Z - z_4) and 7(Z - z_2)(Z - z_4) both with algebraic integer coefficients. > However, for any such polynomial, you run into the same problem that > itd need at least one root thats rational. This is wrong, as shown above. It is true if you can split off linear terms, not when you can split off quadratics or higher order polynomials. > But you can characterize every possible root by > z_j = (1 + sqrt(-167))/W_j > z_k = (1 - sqrt(-167))/W_k > where j and k are counting numbers, and W_j and W_k are algebraic > integers. > The problem is that (1+sqrt(-167)/U where U is an algebraic integer > cannot be rational unless U has (1+sqrt(-167) itself as a factor. It is *not* necessary that you will get a rational number. > Therefore, in the ring of algebraic integers, the roots of > x^2 - x + 42 = 0 > and > y^2 - by - 7 = 0 > never share non-unit factors in the ring of algebraic integers. Silently disregarding the proof that such numbers indeed do exist. > Notice that the key theorem here is that one that prevents an > algebraic integer from being the root of a non-monic primitive with > integer coefficients irreducible over Q. Not so. > Take away the condition that you have an algebraic integer factor, and > youre ok, which proves that there must be a ring beyond algebraic > integers, where they *can* share non-unit factors. Ah, yes. Up to now the algebraic integers suffice, there is a proven common algebraic non-unit factor. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Ring of algebraic integers, comparison check > CORRECTION POST: Though it pains me to admit it the poster Nora Baron was actually right about something as I had a mistake in thinking below. 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 > where I have four solutions, which are > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) > z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) > z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)). > (Those not sure can simply multiply out > (Z - z_1)(Z - z_2)(Z - z_3)(Z - z_4) to verify.) > Now then our question was, could what is now z_1 be an algebraic > integer for an algebraic integer b? > And now an important theorem comes into play, as an algebraic integer > cannot be the root of a non-monic primitive with integer coefficients > that is irreducible over Q. > So integer bs are eliminated as a possibility immediately. > There are several ways to show that none of the the zs can be a > fraction for an integer b, but some of you may wonder why theyd need > to be, like why couldnt they be irrational algebraic integers? > Well, unless > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 0 > is reducible over Q with that integer b, the theorem I mentioned > requires that *none* of its roots can be an algebraic integer, as its > non-monic and primitive. > Therefore, even if you had an irrational algebraic integer root, the > polynomial would have to be reducible over Q, meaning that youd need > at least one rational root. Thats not true, as the poster Nora Baron pointed out you *can* have irrational roots for a polynomial reducible over Q, but its a minor error. The correct conclusion is that 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 is irreducible over Q for any integer b, for some simple reasons, but its easier to just look at each of its roots: z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)) and realize that for it to be reducible, youd have to have a monic quadratic with integer coefficients with two of those as its roots. James Harris === Subject: Re: Ring of algebraic integers, comparison check > The correct conclusion is that > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > is irreducible over Q for any integer b, for some simple reasons Try b = 6. Then 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 = 7z^4 + 6z^3 + 299z^2 - 36z + 252 = (z^2 + z + 42)(7z^2 - z + 6). Something is simple here, but it is not reasons. Looks like its back to the drawing board for old JSH! === Subject: Re: Ring of algebraic integers, comparison check ... > The correct conclusion is that > 7z^4 + bz^3 + (6b^2 + 83)z^2 - 6bz + 252 > is irreducible over Q for any integer b, for some simple reasons, but > its easier to just look at each of its roots: Yes, but I thought that b could be an algebraic integer. > z_1 = (1 + sqrt(-167))/(b + sqrt(b^2 + 28)) > z_2 = (1 - sqrt(-167))/(b + sqrt(b^2 + 28)) > z_3 = (1 + sqrt(-167))/(b - sqrt(b^2 + 28)) > z_4 = (1 - sqrt(-167))/(b - sqrt(b^2 + 28)) > and realize that for it to be reducible, youd have to have a monic > quadratic with integer coefficients with two of those as its roots. Why? You only need a monic quadratic with *algebraic* integer coefficients. The roots of x^2 + sqrt(2).x + 1 are algebraic integers. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Ring of algebraic integers, comparison check > Though it pains me to admit it the poster Nora Baron was actually > right about something as I had a mistake in thinking below. Of course it pains you. It does so because you are not, and never have been, in search of the truth. > James The truth hurts. Harris -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Ring of algebraic integers, comparison check > Consider the polynomial: > 3z^4 + 4z^3 + 7z^2 + 4z + 3 > it is not monic, not all of the coefficients are divisible by 3, > nevertheless, it has algebraic integer roots. To wit, it can be > written as: > (x^2 + x + 1)(3x^2 + x + 3). > Now set b = 4, we have: > 3z^3 + bz^3 - (b^2 - 23)z + bz + 3. > That 23 is coprime to 3 is no reason for it to have no algebraic > integer roots for any b. > Yes Dik Winter, there are indeed non-monic polynomials with integer > coefficients that have rational roots! NB: Dik is talking about the existence of algebraic integer roots. JSH is talking about the existence of rational roots. > Any readers who needed you to show them that probably arent > well-versed in mathematics. Any readers who cannot distinguish between rational roots and algebraic integer roots probably arent well-versed in reading, much less mathematics. > James Harris Who does not distinguish between rational and algebraic integer === Subject: how to calculate the degrees of an angle? Knowing the sides of the triangle, and knowing that its a right triangle, how do you find out the degrees of each angles? (One is 90, I know that.) For example, if I got a triangle with sides (0,0), (1,0), (0,2) what are the angles? Please dont tell me to find out the tangent and look it up on a table. If thats the answer, then how do I calculate the values of that table? Alex === Subject: Re: how to calculate the degrees of an angle? > Knowing the sides of the triangle, and knowing that its a right > triangle, how do you find out the degrees of each angles? (One is 90, > I know that.) > For example, if I got a triangle with sides (0,0), (1,0), (0,2) what > are the angles? > Please dont tell me to find out the tangent and look it up on a > table. If thats the answer, then how do I calculate the values of > that table? Google for CORDIC on calculating circular functions. There is also the McLaurin expansion. -- G.C. === Subject: Re: how to calculate the degrees of an angle? > Knowing the sides of the triangle, and knowing that its a right > triangle, how do you find out the degrees of each angles? (One is 90, > I know that.) > For example, if I got a triangle with sides (0,0), (1,0), (0,2) what > are the angles? > Please dont tell me to find out the tangent and look it up on a > table. If thats the answer, then how do I calculate the values of > that table? > Alex I can imagine a triangle with (0,0), (1,0) and (0,2) as vertices, but my imagination does not run to having a triangle with (0,0), (1,0) and (0,2) as sides. === Subject: Re: how to calculate the degrees of an angle? > Knowing the sides of the triangle, and knowing that its a right > triangle, how do you find out the degrees of each angles? (One is 90, > I know that.) > For example, if I got a triangle with sides (0,0), (1,0), (0,2) what > are the angles? > Please dont tell me to find out the tangent and look it up on a > table. If thats the answer, then how do I calculate the values of > that table? > Alex I would find the lengths of all three sides and then apply the law of cosines. David Moran === Subject: Re: how to calculate the degrees of an angle? > Knowing the sides of the triangle, and knowing that its a right > triangle, how do you find out the degrees of each angles? (One is 90, > I know that.) > For example, if I got a triangle with sides (0,0), (1,0), (0,2) what > are the angles? > Please dont tell me to find out the tangent and look it up on a > table. If thats the answer, then how do I calculate the values of > that table? > Alex > I would find the lengths of all three sides and then apply the law of > cosines. > David Moran If it is a right triangle, then you can just use right triangle trigonometry. David Moran === Subject: intersections of compact sets someone asked me this question, and i cant find a counterexample, nor do i know if its true : is the arbitrary intersection of compact sets a compact set? If there is a counterexample, please post. === Subject: Re: intersections of compact sets >someone asked me this question, and i cant find a counterexample, nor >do i know if its true : is the arbitrary intersection of compact sets >a compact set? If there is a counterexample, please post. In Hausdorff spaces, yes: In such spaces, compact sets are closed. Hence, an arbritrary intersection thereof is closed, and closed subsets of compact sets are compact. For a counterexample (from Moore T.O., Elementary General Topology), consider N where the topology consists of N and all sets disjoint from {0,1}. Then N {0} and N {1} are compact, but their intersection is not. The space in this counterexample is not even T0. Can we come up with a T1 counterexample? -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: intersections of compact sets > The space in this counterexample is not even T0. Can we come up with a > T1 counterexample? Let E=[0,1[ with standard topologie. Let A= E U {1_a} U {1_b} define a topologie on A such E U {1_a} ans E U {1_b} both homeomorphs to [0,1]. But neighborohood of 1_a does not contain 1_b and vice-versa. (Uou can also obtain A, with [-1,1] ans identify x with -x forall x in ]0,1[) E is locally compact-hausdorf hence T1. E U {1_a} and E U {1_b} are compact but not E. === Subject: Looking for page 145 of Hardy and Wright I need to have a look at page 145 of the introduction to theory of numbers of Hardy and Wright. I went to the library in the science department of my town, but the book was not there !!! Could someone make a quick and dirty scan of that page (maybe 144-146 ???) and sent it to me. Any resolution will be welcome: it is not for printing or for keeping but only in order to check a referece. e-mail: thomas.baruchel@laposte.net -- .82 nous devons agir comme si la chose qui peut-.90tre ne sera pas devait .90tre é (Kant, M.8etaphysique des moeurs, doctrine du droit, II conclusion) Thomas Baruchel === Subject: Re: Looking for page 145 of Hardy and Wright Thomas Baruchel escribi.97: > I need to have a look at page 145 of the introduction to theory of > numbers of Hardy and Wright. I went to the library in the science > department of my town, but the book was not there !!! Could someone > make a quick > and dirty scan of that page (maybe 144-146 ???) and sent it to me. > Any resolution will be welcome: it is not for printing or for > keeping but only in order to check a referece. > e-mail: thomas.baruchel@laposte.net In the fifth edition it contains the prove of Theorem 177: The continued fraction which represents a quadratic surd is periodic. The prove, inicied in page 144, conclue at the end of this page. ËIs that on what you are interested? -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Looking for page 145 of Hardy and Wright > I need to have a look at page 145 of the introduction to theory of numbers > of Hardy and Wright. I went to the library in the science department of my > town, but the book was not there !!! Could someone make a quick and dirty > scan of that page (maybe 144-146 ???) and sent it to me. You should probably specify an edition, since there are five of them, and not all libraries will have the latest, or, better yet, specify what particular material you are looking for so people can make sure they get the right part from their edition. -- --Tim Smith === Subject: Re: Graph Theory: Number of maximal Matchings >thank you for your hints. Unfortunately my bipartite graphs are not >regular. Well, the same upper bound will work in the case where > the maximum degree is k (since removing edges can only decrease the > number of possible matchings). > Im not sure I understood the underlying argument: > Are you assuming that a graph with maximum degree k can be always > identified as a subgraph of a k-regular graph on the same vertices? > Because thats not true. Oops, youre right of course. Stupid of me. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Anybody want to play with a new math of infinities? > Consider the question of what happens when we > divide 1 by 0? > A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? > A simpler solution is to use the symbology: > The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. > Since the / has a pre-existing meaning, let us say, > The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. > When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. > Does this lead to any inconsistency? Yes it does: 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. ---- David === Subject: Re: Anybody want to play with a new math of infinities? > Consider the question of what happens when we > divide 1 by 0? A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? A simpler solution is to use the symbology: The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. Since the / has a pre-existing meaning, let us say, The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. Does this lead to any inconsistency? > Yes it does: > 1= 1-0*0 = 1-0*(0+0) = 1-0*0 + 1-0*0 = 1+1 = 2. > ---- David > 1= 1~0*0 = 1~0*(0+0) = 1~0*0 + 1~0*0 = 1+1 = 2. You are right! Can we outlaw something here? === Subject: Re: Anybody want to play with a new math of infinities? > You are right! > Can we outlaw something here? YES! We can outlaw division by zero. === Subject: Re: Anybody want to play with a new math of infinities? In sci.math, bhanwara divide 1 by 0? > A lot of work assumes there is an answer that > correlates to infinity, which is problematic, > because what happens when you multiply this > infinity by 0 again? > A simpler solution is to use the symbology: > The result of dividing 1 by 0 is 1/0. > The result of dividing 2 by 0 is 2/0. etc. > Since the / has a pre-existing meaning, let us say, > The result of dividing 1 by 0 is 1~0. > The result of dividing 2 by 0 is 2~0. etc. > When you multiply 1~0 by 0, you get 1. > But when you multiple 2~0 by 0, you get 2. > Does this lead to any inconsistency? > What about comparisons? > It would seem 1~0 should be greater than 2~0, which > should be greater than 3~0. 0~0 would be greather than 1~0. > Are there theorems than can be proved using > this new symbology? One has the minor problem of evaluating 0~0, which is the result of dividing 0/0. This can be gotten by e.g. subtracting 1~0 - 1~0, and could be any real at all. Of course, for normal real values r, r - r = 0. Also, if one multiplies 1~0 * 1~0, what does one get? Ditto for 1~0 / 1~0 and r / 1~0, which may or may not be 0. Whatever it is, its not a field... :-) -- #191, ewill3@earthlink.net Its still legal to go .sigless. === Subject: Re: Anybody want to play with a new math of infinities? > What about comparisons? > It would seem 1~0 should be greater than 2~0, which > should be greater than 3~0. 0~0 would be greather than 1~0. Oops, the other way... We can also do the whole lot of normal arithmetical operations with standard props, as long as we keep the ~0 there.. === Subject: . Zenos paradox . Hi John Schoenfeld, You state, I am interested in knowing how transfinite numbers solved Zenos paradox , By either stoping after the numbers dip below a given scale, or by assuming that the infinite series would converge. ( Because such well tested assumptions are quite useful ) But nature cant be limited by such prejudices. For example: If one could somehow change the scale he uses for things like heat, mass-energy, time, and space, one could forever approach, yet never reach, the beginning of the big bang. And because the accelerated expansion of the universe means that its Always been cooling, even scientifically, theres no beginning. http://click.exacttarget.com/?fe911170706d007976- fe3016707360067c711779 It simply states that as far back as we can extrapolate, the cosmos has been expanding, thinning out and cooling down. The big bang is best thought of not as a singular event but as an ongoing process, a gradual molding of order out of chaos. Yet, no matter how ones brain defines the word Order , nature will always seem like this : You have a consumption rate. First nature gives you everything for free, and then, in the end, it takes it all back ... in between it seems like a semi-bizarre casino. And this is true for Both physical and biological systems. === Subject: Is logical induction a valid form of proof in math? The closest I can think of is Mathematical Induction. Take the example of summing the x first odd numbers. sum 2n-1 from n=1 to n=x : 1 + 3 + 5 + ... First, it is possible to deduce the sum. But I can also induce a formula x^2 from looking at this section of the sequence, but that is clearly not a proof. The (mathematical) inductive hypothesis is a deduced form of proof, but is deducing an induced formula to be true, logical induction? Or more generally is logical induction a valid form of proof in math? === Subject: Re: Is logical induction a valid form of proof in math? >The closest I can think of is Mathematical Induction. >(...) >Or more generally is logical induction a valid form of proof in math? If you mean by logical induction the induction argument as used in natural science using experimental Ôevidence, the answer is definitely no, by all math. standards. BTW the term math. induction is more or less misleading. In German it is often called complete induction - better would be complete>d< induction because it completes inductive reasoning based on looking at what happens for n=0,1,2,...m for some (in general very) small m so that you get a valid math. argument - this is valid because the possibility of math. induction is at the bottom of the nature of natural numbers - see e.g. the Peano axioms. In axiomatic set theory one can >prove< math. induction, i.e. that a certain set (of things then called natural numbers) - with an additional structure - satisfies Peanos axioms of which the one stating (the base of) math. induction is the only one naively (not quite) trivial ... === Subject: Re: Stuck on two combinatorics problems >Anyone got ideas on these two problems ? >1)A is a rectangle consisting of n * 2 numbered squares. Compute a_n, the >number of ways to partition A into domino bricks ( that is 2 x 1 >rectangles). For instance a_1 = 1, a_2 = 2 Each nx2 arrangement can be broken down into one of two cases. 1. the rightmost column consists of a single vertical brick. In this case, there are a_{n-1} arrangements of the remaining n-1 columns. 2. the rightmost column consists of two horizontal bricks. In this case, there are a_{n-2} arrangements of the remaining n-2 columns. Thus, a_n = a_{n-1} + a_{n-2}. >2) Show that if p and q are odd primes then >2^(pq+1) [equivalance symbol] 2^(p+q) (mod pq) Use the fact that 2^p = 2 mod p for all primes p <> 2. Rob Johnson take out the trash before replying === Subject: Re: Stuck on two combinatorics problems yes i forgot to the condition p =/= . Sloppy of me. > Valentino Berti escribi.97: > Anyone got ideas on these two problems ? > 1)A is a rectangle consisting of n * 2 numbered squares. Compute > a_n, the number of ways to partition A into domino bricks ( that is 2 > x 1 rectangles). For instance a_1 = 1, a_2 = 2 > You can make a n*2 rectangle adding a two horixontal dominos a one (n-2)*2 > rectangle, or adding a vertical domino to a (n-1)*2 rectangle, then ... > 2) Show that if p and q are odd primes then > 2^(pq+1) [equivalance symbol] 2^(p+q) (mod pq) > If p = q odd prime it fails. Did you forgotten the condition p =/= q? > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: Stuck on two combinatorics problems What i meant to say was: yes i forgot to the condition p =/= q. Sloppy of me(again) :) > yes i forgot to the condition p =/= . Sloppy of me. > Valentino Berti escribi.97: > Anyone got ideas on these two problems ? 1)A is a rectangle consisting of n * 2 numbered squares. Compute > a_n, the number of ways to partition A into domino bricks ( that is 2 > x 1 rectangles). For instance a_1 = 1, a_2 = 2 > You can make a n*2 rectangle adding a two horixontal dominos a one (n-2)*2 > rectangle, or adding a vertical domino to a (n-1)*2 rectangle, then ... > 2) Show that if p and q are odd primes then > 2^(pq+1) [equivalance symbol] 2^(p+q) (mod pq) > If p = q odd prime it fails. Did you forgotten the condition p =/= q? > -- > Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: A series for the inverse sine cardinal function >The sine cardinal function, > ( 1 if x = 0 > sinc(x) = ( > ( sin(x)/x otherwise, >arises in several applications, and so its inverse should also be of >interest. This note gives a series expansion for the inverse of sinc(x), >0 <= x <= x0, where x0 (approx. 4.4934) denotes the positive value of x >at which sinc(x) reaches its absolute minimum. >[Besides posting this to sci.math, I have also posted it to >sci.math.num-analysis and comp.dsp in the hope that some people might be >able to give references to previous mathematical treatments of the inverse >sinc function. In particular, Id be interested to know if the series given >here is already known.] >With f(x) = 2*x + 3*x^3/10 + 321*x^5/2800 + 3197*x^7/56000 + > 445617*x^9/13798400 + 1766784699*x^11/89689600000 + > 317184685563*x^13/25113088000000 + > 14328608561991*x^15/1707689984000000 + > 6670995251837391*x^17/1165411287040000000 + > 910588298588385889*x^19/228420612259840000000 + ..., >it can be shown that the desired inverse, abbreviated as Asinc here, is >given by > Asinc(x) = Sqrt(3/2) * f(Sqrt(1 - x)) [example snipped] >The radius of convergence of the above series for f is slightly larger than >1.1 . As expected, when x is near that value, convergence is very slow. sinc(x) >= -0.21723362821 (minimum at x = +/-4.49340945791). Therefore, sqrt(1-sinc(x)) <= 1.10328311335, hopefully the radius of convergence. >The series for f was obtained by reversion of series, etc. To be more >specific, for anyone interested, in Mathematica the series may be obtained >using >Simplify[Normal[InverseSeries[1 - Series[Sin[x]/x, {x, 0, n}]]/Sqrt[3/2] >/. x -> x^2], x > 0] Another way to do this is using the Taylor series for x = sinc(y). We want to find y in terms of x where y^2 y^4 y^6 x = 1 - --- + --- - ---- + ... 6 120 5040 Define f(z) by z^2 z^3 f(z) = z - --- + --- - ... 20 840 oo --- k-1 k 6 = > (-1) z ------- --- (2k+1)! k=1 so that 1 x = 1 - - f(y^2) 6 and -1 y = sqrt(f (6-6x)) For the inverse of f, Mathematica gives Normal[InverseSeries[Series[Sum[-(-x)^k 6/(2k+1)!,{k,1,10}],{x,0,10}],x]] 2 3 4 5 6 x 2 x 13 x 4957 x 58007 x x + -- + ---- + ----- + --------- + ----------- + 20 525 37800 145530000 16216200000 7 8 9 1748431 x 4058681 x 5313239803 x ------------- + -------------- + ------------------- + 4469590125000 92100645000000 1046241656460000000 10 2601229460539 x ---------------------- 4365681093774000000000 Although this doesnt yield a simple formula for the general term of the series, at least the first 40 terms all have positive coefficients. I think there is a good chance that all the terms may share this property, but I need to look into this a bit more. Rob Johnson take out the trash before replying === Subject: Rudins proof of the open mapping theorem Hi all, The open mapping theorem mentioned at the subject of this post is the one concerning continuous linear maps L from a topological vector space E into a topological vector space F. Walter Rudin proves the theorem in two of his textbooks, Real and Complex Analysis and Functional Analysis. On the first of these textbooks, the hypothesis are that E and F are Banach spaces and that L is surjective and the conclusion is equivalent to the assertion that L is an open mapping; I have no problem with Rudins proof here. On Functional Analysis, the theorem is stated in a much more general setting. Rudin defines an F-space as a topological vector space (BTW, for Rudin all topological vector spaces are T1 spaces) such that its topology is induced by an invariant metric (that is, a metric d such that, for v, w, and u in E we have d(u + v,u + w) = d(v,w)) with respect to which E is a complet metric space. The hypothesis of the theorem are: a) E is an F-space; b) L(E) is of second category. The conclusions are: 1) L is surjective; 2) L is an open mapping; 3) F is an F-space. Conclusion 2) is the most important one; Rudin shows how the other two conclusions follow from it. In order to prove 2), Rudin proceeds as follows: he starts with a neighborhood V of 0 in E and then attempts to prove that L(V) is a neighborhood of 0 in F. He chooses r > 0 such that {x in E | d(x,0) < r} is a subset of V and he defines V_n as {x in E : d(x,0) < r/2^n}, for each non-negative integer n. He then proves that the closure of L(V_1) is a neighborhood of 0 in F and he asserts that the closure of L(V_1) is contained in L(V). In order to prove that assertion, he shows that, if you fix y_1 in the closure of L(V_1), there is some sequence (x_n)_n of vectors of E and some sequence (y_n)_n of vectors of F such that: a) each x_n belongs to V_n; b) each y_n belongs to the closure of L(V_n); c) y_{n+1} = y_n - L(x_n). Then Rudin defines x = x_1 + x_2 + x_3 + ...; this makes sense, because E is complete and the sequence (x_1 + x_2 + ... + x_n)_n is a Cauchy sequence. Then we have, L(x_1 + ... + x_n) = L(x_1) + ... + L(x_n) = y_1 - y_{n+1}. If we knew that lim_n y_n = 0, it would follow that L(x) = y_1. Rudin says that lim_n y_n = 0 by continuity of L and I suspect this this argument is not correct. Heres one reason why I think that. Take for E the real numbers, with its usual topology and structure of real vector space. Take for F the real numbers with its usual structure of real vector space, but this time the topology is the one for which the closed subsets are the finite subsets (except for the whole space, of course). Let L be the identity. Then L is linear and its image is of second category. Furthermore, if we define V_n as ]-1/2^n,1/2^n[, it is true that the closure of each L(V_n) is a neighborhood of 0 (since its equal to F!). However, it is not true that the closure of L(V_1)( = F) is contained in L(V_0)( = ]-1,1[). Besides, if you take y_1 = 2 and try to get sequences (x_n)_n and (y_n)_n as above, youll see that you wont get lim_n y_n = 0. Of course, this is not really a counter-example: with the topology defined above, F is *not* a topological vector space; neither the sum from F x F into F nor the product from R x F into F are continuous functions. On the other hand, it is still true that F is T1, that the sum is continuous on each variable and that the product is continuous enough. So, does anyone disagrees with my criticism? And, if it is correct, is the open mapping theorem, as stated in Rudins Functional Analysis, true? I have looked up in other books on topological vector spaces, specially those written by authors who try to state their theorems in the more general possible way (specially Bourbaki and Koethe) and it seems that nobody else tries to prove the open mapping theorem assuming only that F is a topological vector space. Jose Carlos Santos === Subject: Re: Rudins proof of the open mapping theorem >Hi all, >The open mapping theorem mentioned at the subject of this post is >the one concerning continuous linear maps L from a topological vector >space E into a topological vector space F. >Walter Rudin proves the theorem in two of his textbooks, Real and >Complex Analysis and Functional Analysis. On the first of these >textbooks, the hypothesis are that E and F are Banach spaces and that >L is surjective and the conclusion is equivalent to the assertion that >L is an open mapping; I have no problem with Rudins proof here. >On Functional Analysis, the theorem is stated in a much more general >setting. Rudin defines an F-space as a topological vector space (BTW, >for Rudin all topological vector spaces are T1 spaces) such that its >topology is induced by an invariant metric (that is, a metric d such >that, for v, w, and u in E we have d(u + v,u + w) = d(v,w)) with >respect to which E is a complet metric space. The hypothesis of the >theorem are: >a) E is an F-space; >b) L(E) is of second category. >The conclusions are: >1) L is surjective; >2) L is an open mapping; >3) F is an F-space. >Conclusion 2) is the most important one; Rudin shows how the other two >conclusions follow from it. In order to prove 2), Rudin proceeds as >follows: he starts with a neighborhood V of 0 in E and then attempts >to prove that L(V) is a neighborhood of 0 in F. He chooses r > 0 such >that {x in E | d(x,0) < r} is a subset of V and he defines V_n as >{x in E : d(x,0) < r/2^n}, for each non-negative integer n. He then >proves that the closure of L(V_1) is a neighborhood of 0 in F and he >asserts that the closure of L(V_1) is contained in L(V). In order to >prove that assertion, he shows that, if you fix y_1 in the closure of >L(V_1), there is some sequence (x_n)_n of vectors of E and some >sequence (y_n)_n of vectors of F such that: >a) each x_n belongs to V_n; >b) each y_n belongs to the closure of L(V_n); >c) y_{n+1} = y_n - L(x_n). >Then Rudin defines x = x_1 + x_2 + x_3 + ...; this makes sense, >because E is complete and the sequence (x_1 + x_2 + ... + x_n)_n is a >Cauchy sequence. Then we have, > L(x_1 + ... + x_n) = L(x_1) + ... + L(x_n) = y_1 - y_{n+1}. >If we knew that lim_n y_n = 0, it would follow that L(x) = y_1. Rudin >says that lim_n y_n = 0 by continuity of L and I suspect this this >argument is not correct. The fact that y_n is in the closure of L(V_n) shows that y_n -> 0. where I waved my hands is exactly what goes wrong in the not-quite-counterexample below. Were set if we can show that in any TVS, if O is a neighborhood of 0 then there exists another neighborhood of 0, V, such that the closure of V is contained in O, right? This must be true... Aha. It follows from the (joint!) continuity of the operations that the origin has a neighborhood V such that V - V = {x - y : x, y in V} is a subset of O. Suppose that x is in the complement of O. Then x + V is a neighborhood of x, and V intersect (x + V) is empty, hence x is not in the closure of V. >Heres one reason why I think that. Take for >E the real numbers, with its usual topology and structure of real >vector space. Take for F the real numbers with its usual structure of >real vector space, but this time the topology is the one for which the >closed subsets are the finite subsets (except for the whole space, of >course). Let L be the identity. Then L is linear and its image is of >second category. Furthermore, if we define V_n as ]-1/2^n,1/2^n[, it >is true that the closure of each L(V_n) is a neighborhood of 0 (since >its equal to F!). However, it is not true that the closure of L(V_1)( >= F) is contained in L(V_0)( = ]-1,1[). Besides, if you take y_1 = 2 >and try to get sequences (x_n)_n and (y_n)_n as above, youll see that >you wont get lim_n y_n = 0. >Of course, this is not really a counter-example: with the topology >defined above, F is *not* a topological vector space; neither the sum >from F x F into F nor the product from R x F into F are continuous >functions. On the other hand, it is still true that F is T1, that the >sum is continuous on each variable and that the product is continuous >enough. >So, does anyone disagrees with my criticism? And, if it is correct, is >the open mapping theorem, as stated in Rudins Functional Analysis, >true? >I have looked up in other books on topological vector spaces, >specially those written by authors who try to state their theorems in >the more general possible way (specially Bourbaki and Koethe) and it >seems that nobody else tries to prove the open mapping theorem >assuming only that F is a topological vector space. You gotta get up pretty early in the morning to catch Rudin. >Jose Carlos Santos ************************ David C. Ullrich === Subject: Re: Rudins proof of the open mapping theorem > The fact that y_n is in the closure of L(V_n) shows that y_n -> 0. I wish I knew why... > where I waved my hands is exactly what goes wrong in the > not-quite-counterexample below. Were set if we can show that > in any TVS, if O is a neighborhood of 0 then there exists another > neighborhood of 0, V, such that the closure of V is contained > in O, right? This must be true... > Aha. It follows from the (joint!) continuity of the operations > that the origin has a neighborhood V such that V - V > = {x - y : x, y in V} is a subset of O. Suppose that x is in > the complement of O. Then x + V is a neighborhood of x, > and V intersect (x + V) is empty, hence x is not in the > closure of V. Well, I already knew that, for topological groups, T_0 <=> T_3. But I do not see how to deduce from that fact that lim_n y_n = 0. Besides, to which neighborhoods of 0 do you apply this? The closures of L(V_n) are already closed neighborhoods. I suppose that you apply it to the interiors of the closures of L(V_n), but in my fake counter-example these are equal to the whole space. Jose Carlos Santos === Subject: unit ball is closed? Let X be a bounded subset of R^n, i.e. m(X) f (in L_1), that is, INT |f_n - f| --> 0. I need to show that f is in B, i.e., that INT f^2 <= 1. Ive tried doing things like breaking up the region of integration into the regions where |f| <= 1 and |f| > 1 and applying some (obvious) inequalities, but that hasnt gotten me anywhere. I also know INT (f_n)^2 <= 1, so it would work to show INT |f_n - f|^2 --> 0, but if I could show that then I wouldnt be posting here... Any hints? I also got the itch to try to use dominated convergence (or one of its spinoffs), but no luck on that front either... === Subject: Re: unit ball is closed? >Let X be a bounded subset of R^n, i.e. m(X)ball of L_2(X) is a closed subset of L_1(X) ? >Call the ball B. Then B = { f | INT f^2 <= 1}. Now, suppose f_n is a sequence >of functions in B and f_n --> f (in L_1), that is, INT |f_n - f| --> 0. I need >to show that f is in B, i.e., that INT f^2 <= 1. A standard result says that some subsequence converges to f a.e. Now Someones Lemma shows that ||f||_2 <= 1. >Ive tried doing things like breaking up the region of integration into the >regions where |f| <= 1 and |f| > 1 and applying some (obvious) inequalities, >but that hasnt gotten me anywhere. I also know INT (f_n)^2 <= 1, so it would >work to show INT |f_n - f|^2 --> 0, but if I could show that then I wouldnt be >posting here... >Any hints? I also got the itch to try to use dominated convergence (or one of >its spinoffs), but no luck on that front either... ************************ David C. Ullrich === Subject: Re: unit ball is closed? For extra credit ( :-) ) show that the unit ball of L_2(X), when viewed as a subset of L_1(X), has empty interior. >Let X be a bounded subset of R^n, i.e. m(X)ball of L_2(X) is a closed subset of L_1(X) ? >Call the ball B. Then B = { f | INT f^2 <= 1}. Now, suppose f_n is a sequence >of functions in B and f_n --> f (in L_1), that is, INT |f_n - f| --> 0. I need >to show that f is in B, i.e., that INT f^2 <= 1. > A standard result says that some subsequence converges to f a.e. > Now Someones Lemma shows that ||f||_2 <= 1. >Ive tried doing things like breaking up the region of integration into the >regions where |f| <= 1 and |f| > 1 and applying some (obvious) inequalities, >but that hasnt gotten me anywhere. I also know INT (f_n)^2 <= 1, so it would >work to show INT |f_n - f|^2 --> 0, but if I could show that then I wouldnt be >posting here... >Any hints? I also got the itch to try to use dominated convergence (or one of >its spinoffs), but no luck on that front either... > ************************ > David C. Ullrich === Subject: Re: unit ball is closed? Hmm...is this true? Suppose f is an interior point. Then there is an e>0 s.t. I := { g | INT |f-g| < e } contained in B. This generates a contradition if its true that functions which square-integrate to a value > 1 are dense in L_1, but I scarcely know how to show that... === Subject: Re: unit ball is closed? >Hmm...is this true? >Suppose f is an interior point. Then there is an e>0 s.t. >I := { g | INT |f-g| < e } contained in B. >This generates a contradition if its true that functions which >square-integrate to a value > 1 are dense in L_1, but I scarcely know how to >show that... Suppose that f = a * 1_[0,b], that is, a times the characteristic function of [0,b], where b is small. What are the L_1 and L_2 norms of f? Is sqrt(b) smaller than b or larger? How much smaller or larger? ************************ David C. Ullrich === Subject: proof of the dimension theorem of linear algebra Hi all, I know this will sound like a homework. Part of the effort will be to convince you that it is not. Suppose I am Ôthe mathematician, I have just heard from a friend that certain things called vector spaces are of interest, and I have just discovered the importance of the idea of linear (in)dependence, representation of a vector as a linear combination, et cetera. Now my task is the following: Theorem: All bases of a vector space are equinumerous. My textbook says many things, I will describe ... but the point is, they are interested only in the finite case, I am not [I know it will be harder]. Proof: [my textbook] { Suppose B_1 and B_2 are two bases of vector space V. As B_1 spans V, B_2 is linearly independent, |B_1| >= |B_2| As B_2 spans V, B_1 is linearly independent, |B_2| >= |B_1| } Of course, this proof assumes another theorem, which becomes my main concern from now on. But there are two things I will like to know (if you would be kind enough to explain), 1) Do I have to be in debt to the total ordering of cardinalities? 2) Can I somehow arrange a shortcut that establishes a bijection between B_1 and B_2 from which the theorem follows at once? I can *see* a natural choice of course [the invertible basis transformation ... but I should not know about it at this stage, and certainly to justify that, I have to prove this theorem first :-)] Now, the question boils down to: Theorem: If S is a spanning set, and L is a linearly independent set of a vector space V, then |S| >= |L|. Proof: [my textbook] { I can smell there should be a proof by contradiction, and the obvious point to hit is independence. Anyway, my book constructs, for w_i in S and v_j in L, v_j = sum a_ij w_i to see that this leads to a contradiction. } But all these countings ... we have just pushed them into the domain of common sense [my book reads, we must end up with at least one row of zeroes]. This amounts to, IMHO, the same thing that we are supposed to prove. Can you guys provide me with an elegant proof of this fundamental theorem, using rigorous arguments only, extending to infinite cardinalities? Please. I beg you. Too hard for me. :-D === Subject: Re: proof of the dimension theorem of linear algebra >Hi all, >I know this will sound like a homework. Part of the effort will be to >convince you that it is not. >Suppose I am Ôthe mathematician, I have just heard from a friend that >certain things called vector spaces are of interest, and I have just >discovered the importance of the idea of linear (in)dependence, >representation of a vector as a linear combination, et cetera. Now my >task is the following: >Theorem: All bases of a vector space are equinumerous. >My textbook says many things, I will describe ... but the point is, >they are interested only in the finite case, I am not [I know it will >be harder]. >Proof: [my textbook] { >Suppose B_1 and B_2 are two bases of vector space V. >As B_1 spans V, B_2 is linearly independent, |B_1| >= |B_2| >As B_2 spans V, B_1 is linearly independent, |B_2| >= |B_1| >Of course, this proof assumes another theorem, which becomes my main >concern from now on. But there are two things I will like to know (if >you would be kind enough to explain), >1) Do I have to be in debt to the total ordering of cardinalities? >2) Can I somehow arrange a shortcut that establishes a bijection >between B_1 and B_2 from which the theorem follows at once? I can >*see* a natural choice of course [the invertible basis transformation >... but I should not know about it at this stage, and certainly to >justify that, I have to prove this theorem first :-)] >Now, the question boils down to: >Theorem: If S is a spanning set, and L is a linearly independent set >of a vector space V, then |S| >= |L|. >Proof: [my textbook] { >I can smell there should be a proof by contradiction, and the obvious >point to hit is independence. Anyway, my book constructs, for w_i in >S and v_j in L, >v_j = sum a_ij w_i >to see that this leads to a contradiction. >But all these countings ... we have just pushed them into the domain >of common sense [my book reads, we must end up with at least one row >of zeroes]. This amounts to, IMHO, the same thing that we are >supposed to prove. Yes, that is what you need to prove. Thats why the proof is complete at that point, because this has been proved! >Can you guys provide me with an elegant proof of >this fundamental theorem, using rigorous arguments only, extending to >infinite cardinalities? I doubt it - my impression is that the argument in infinite dimensions is different. >Please. I beg you. Too hard for me. :-D ************************ David C. Ullrich === Subject: which is the most efficient method to solve 0-1 integer programming? Can you give me some suggestion? Or where can I find papers in this topic? Chonghai Hu === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > (I am posting this for fun, for it is probably easily solved by those > with some background in number theory.) > Let us say we have 2 spherical dice which somehow each represent all > positive integers, each integer represented exactly once per die, on > their countably-infinite number of sides. And each integer is equally > likely to be rolled on both dice. > So, the dice-pair is rolled an infinite number of times. > What is the closed-form representation of the probability that > for *at least one* roll of the dice-pair > the m_th rolling of the pair produces integers j and k where > GCD(j,k) = m? (later versions noted) Maybe one way to try and look at the problem is to imagine a plane lattice Z^2, then draw unit squares (or unit-area hexagons) centred on each point (x, y) where both x and y are multiples of an integer m say, and then finally stereographically project the plane onto a hemisphere of unit total area and work out the total area, a_m, on the hemisphere of all the projected squares or hexagons. Then the probability of your pair of spherical dice landing on two Ôsides with GCD at least m is presumably (1 - a_m)^2. Elaborating this kind of argument might give an answer of sorts to your question. Trouble is, I doubt if youd be happy with an infinite dice (die, to be pedantic) that was constructed in this way, as it would be loaded in favour of small numbers. But in a hypothetical infinite spherical die of unit area with faces equally distributed, the sum of the area of all faces with values dividible by m would presumably be simply 1/m. ------------------------------------------------------------- -------------- John R Ramsden (jr@adslate.com) ------------------------------------------------------------- -------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Let us say we have 2 spherical dice which somehow each represent all >positive integers, each integer represented exactly once per die, on >their countably-infinite number of sides. And each integer is equally >likely to be rolled on both dice. >So, the dice-pair is rolled an infinite number of times. >What is the closed-form representation of the probability that >for *at least one* roll of the dice-pair >the m_th rolling of the pair produces integers j and k where >GCD(j,k) = m? >(ie. there is at least one m, where m = GCD(j,k) on the m_th rolling >of the dice-pair.) As has been noted elsewhere, an infinite-sided die is impossible, but let us pick an n and assume that on roll m, we roll two nm-sided dice. We can then let n tend to infinity. For any n, the probability of both dice having a factor of m is 1/m^2. However, for a large enough n, the probability of there being no other common factors is 6/pi^2. Thus, the probability of the GCD of the dice being m is 1/m^2 6/pi^2. Therefore, the probability of at least one match is oo --- 1 6 1 - | | ( 1 - --- ---- ) m=1 m^2 pi^2 Using Stirlings Formula, it is fairly simple to show that oo --- a^2 | | ( 1 - --- ) m=1 m^2 Gamma(m+a) Gamma(m-a) Gamma(1)^2 = lim ---------- ---------- ---------- m->oo Gamma(1+a) Gamma(1-a) Gamma(m)^2 1 = --------------------- Gamma(1+a) Gamma(1-a) sin(pi a) = --------- pi a = sinc(pi a) Using a = sqrt(6)/pi, we get the probability of at least one match to be 1 - sinc(sqrt(6)), which is about .7394732364. Furthermore, the average number of matches is oo --- 1 6 > --- ---- --- m^2 pi^2 m=1 = 1 Rob Johnson take out the trash before replying === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs Again (as in my other reply), here is the new-and-improved statement of the problem,...(!) but this time with an example: Official (as of now) version: Let us say we have 2 spherical dice which somehow each represent all positive integers, each integer represented exactly once per die, on their countably-infinite number of sides. And each integer is equally likely to be rolled on both dice. So, the dice-pair are rolled repeatedly *until* (if ever) on the m_th roll, GCD(k,j) = m, where j and k are the integers of the dice. What is the probability that we will ever stop rolling the dice (assuming we are eternal and need to keep rolling until GCD(j,k) =m on the m_th roll)? For example: roll 1: j=3; k = 1026; GCD(3,1026) = 3, not 1. roll 2: 25 ; 23534545; GCD = 5, not 2. roll 3: 63 ; 2121; GCD = 21, not 3. roll 4: 36 ; 20; GCD= 4, which DOES = 4. So we CAN stop rolling now! But we will not necessarily be able to ever stop! By the way, as I calculate, the probability is *between* 0 and 1... Leroy Quet === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >(...) >Let us say we have 2 spherical dice which somehow each represent all >positive integers, each integer represented exactly once per die, on >their countably-infinite number of sides. And each integer is equally >likely to be rolled on both dice. As no answer until now states it explicitely, Ill mention it. At least with standard probability theory based on standard measure theory, the last quoted sentence is an impossible assumption. Here is why: Let k be the supposed probability that each pos. integer has to appear on one of the dices (ie. I consider here only one dice). The sum of Prob.(number n appears on the dice) summed over all n in N (N=the set of natural numbers) is - because measure/probability is (not only finitely, but) countably additive -, the prob. that >some< integer n appears, which is the certain event, so the sum must be 1. But every term of the sum is supposed to be equal to k. If k>0, the sum is +oo (infinity); if k=0, this sum is 0 - in both cases, we have a contradiction ! I can add: in virtue of the theory of infinite sums, the probability for an n in N to appear >must< tend to 0 for n->oo and even sufficiently rapidly for the infinite sum to remain finite ! (and the sum must be =1) === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs > [H]ere is the new-and-improved statement of the problem positive integers, each integer represented exactly once per die, on > their countably-infinite number of sides. And each integer is equally > likely to be rolled on both dice. > So, the dice-pair are rolled repeatedly *until* (if ever) > on the m_th roll, > GCD(k,j) = m, > where j and k are the integers of the dice. > What is the probability that we will ever stop rolling the dice[?] Now, speaking as the voice of common sense and low education, Id say this is impossible. You cant make *every* natural number equally likely to come up -- it just doesnt work! What you *can* do, though, as has been suggested, is take a limit of n-sided dice as n *goes* to infinity. Problem is, on what quantity do you take the limit? If n is finite and greater than 1, then there *will* be sequences of rolls that never stop. So taking the limit of average time to stop rolling doesnt work. Nor does anything else I can think of. So my conclusion is that the problem as stated has no meaningful solution, and I dont see any way to re-state it so as to permit one. Sorry. -Arthur === Subject: Re: puzzle: GCDs of Infinite Set of Integer Pairs >Im inclined to go with Chuck. The whole infinite thing is kind of >hinky. Especially the countably-infinite part. Also, somehow >makes me doublecheck my wallet. But, knowing nothing about number >theory...and upon rereading the problem, I say 100% is the >probability. But I dont know what a closed-form representation is. > I dont think the puzzle as stated has a well-defined answer. I think > to turn it in to a wel-defined puzzle, you reinterpret it as follows: > imagine you have two n sided dice, which you will roll > n times. What is the probability that on the mth roll the GCD of > the number of the two dice is m? Find the limit as n approaches > infinity. > I expect the answer is probably zero, but I cant proe it. > George OK, except... You are right about the limit-as-n-approaches-infinity part. But I must restate this sentence of the problem: >What is the closed-form representation of the probability that >for *at least one* roll of the dice-pair >the m_th rolling of the pair produces integers j and k where >GCD(j,k) = m? Here is the new-and-improved statement of the problem,...(!) Let us say we have 2 spherical dice which somehow each represent all positive integers, each integer represented exactly once per die, on their countably-infinite number of sides. And each integer is equally likely to be rolled on both dice. So, the dice-pair are rolled repeatedly *until* (if ever) on the m_th roll, GCD(k,j) = m, where j and k are the integers of the dice. What is the probability that we will ever stop rolling the dice (assuming we are eternal and need to keep rolling until GCD(j,k) =m on the m_th roll)? By the way, as I calculate, the probability is *between* 0 and 1... Leroy Quet My original post: >(I am posting this for fun, for it is probably easily solved by those >with some background in number theory.) >Let us say we have 2 spherical dice which somehow each represent all >positive integers, each integer represented exactly once per die, on >their countably-infinite number of sides. And each integer is equally >likely to be rolled on both dice. >So, the dice-pair is rolled an infinite number of times. >What is the closed-form representation of the probability that >for *at least one* roll of the dice-pair >the m_th rolling of the pair produces integers j and k where >GCD(j,k) = m? >(ie. there is at least one m, where m = GCD(j,k) on the m_th rolling >of the dice-pair.) >Hopefully the solution to this puzzle is not too well known... >Leroy Quet === Subject: Re: Analysis >> I saw this one before but I forgot the proof. Any hints will help. Prove >> that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1. >> Steven > I think you mean X>-1 [Broken formatting fixed...if you are going to top-post, at least do it right: dont put the quoted material in your signature. Geesh!] Why not X >= -1? At X = -1, it becomes 0^n >= 1-n which is true for n >= 1. Whether or not this holds for all natural numbers depends on whether or not you are in the 0 is a natural number school or not. (In fact, for those of the 0 is not a natural number school, X >= -1 is not the best that can be said. Its good for X >= -2, for instance). -- --Tim Smith === Subject: Re: Analysis >> I saw this one before but I forgot the proof. Any hints will help. Prove >> that (1+X)^n >= (1+nX), n is a natural #, X is a real number >= -1. >> Two questions: >> (1) Whats the linear approximation to (1+x)^n when x>-1? >> (2) Whats the concavity of the function (1+x)^2? >One question: whats wrong with induction on n? Not a damn thing. adam === Subject: Re: universal set with 3 valued logic by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1FEBrn12783; hi Dan An interesting idea. While the universal set is a troublesome construct in some ways, it does seem a bit counterintuitive to say that it does not exist. What do you make of Cantors power set theorem that for any set s and its power set p, p is larger than s? Does the universal set u have a power set in your system? Is it larger than u? some remarks. the subsets axiom has been changed but if one looks at the truth tables on pages 3 and 4, one sees that this is an extension of the subsets axiom; ie, if my subsets axiom were applied to crisp sets, then youd have the same results. therefore, cantors diagonal argument to show how no set x can be mapped onto its powerset has to be treated with some delicacy. when one does so, one finds out that the diagonal set that exists is a fuzzy set. ie, whether the truth value of the statement z E D_f, where D_f is the diagonal set, and z is the element of x mapped to D_f, *is the third truth value*. this is discussed on page 6. *what cantors argument proves is this: if there is an f that maps x onto P(x) then P(x) contains at least one fuzzy set. the contra- positive of that is this: if P(x) contains no fuzzy sets then there is no f that maps x onto P(x). this is in version 10-1 of my paper which can be found at the same address, just replace the 9 with 10-1. on page 4, i show how Russells set is fuzzy but its existence is not otherwise contradictory. on page 5, i prove that P(x)=U iff x=U; hence, U is inaccesible via powersetting a smaller set and the powerset of U is U (not just in bijection with). this goes along with the *-ed statement above for P(U)=U and P(U) does contain a fuzzy element, namely, the S in Russells paradox. --Brian === Subject: Re: e is transcendental (was: classes of transcendental numbers ? by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1FEBtj12860; >is sin(pi), so all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and we were already aware of these facts. There is >NO reason to conclude that exp(i pi) = 0. >> I,have a reason , with my due respect. >> Panagiotis Stefanides >Yes, but you DONT tell us what your reason is. You cant expect us to >accept your claims without giving support for those claims. So what >possible reason could you have for expecting us to agree with your claim >that exp(i pi) = 0? The reason is simply that exp(ipi0=-1 should be accompanied by the statement that this is the real part solution. Is it fair? Panagiotis Stefanides >David McAnally > >------- === Subject: Re: skyline matrix solver by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1FEBuk12879; >Please, > I need a skyline matrix solver algorithm. Where can I find one? >Cpplayer I have skyline matrix solver in fortran 90, do you want it ? === Subject: How fast is the Confinued Fraction factorization algorithm? Hi! Im factoring numbers n using the Continued Fraction algorithm, finding pairs (x,y) such that x^2==y mod n. The y values are factorized as y = product(p_i^e_i) * p, where p_i are primes<=307, p<307^2 is another prime, and e_i>=0. Im running a home-made program on a 1 GHz standard desktop computer, using the GNU MP library for all the basic integer functions. Im able to factor the 26-digit number n_26 in 90 seconds and the 28-digit number n_28 in 9 minutes, where n_26 = 42240587706227658402774373 = 4646464647359 * 9090909091547 and n_28 = 4224058769561941597796273857 = 46464646464983 * 90909090909479. Im somewhat disappointed at these time measurements. I was expecting speeds several order of magnitudes faster; considering that it is now possible to factor 70 digit numbers, albeit with a different algorithm. So, is it because of inefficiences in my program, or is it simply the limit of the algorithm? Has anybody else experimented with this algorithm? I can add that for n_26 a total of 6567 pairs (x,y) are considered, and for n_28 the count is 11368. Im considering putting an effort into optimizing the program, but I would like to hear other peoples experiences as to whether I can expect any significant speed-up. If not, then Ill switch over to the Elliptic Curve algorithm. I would like to hear peoples experiences with this algorithm too. -Michael. === Subject: Re: How fast is the Confinued Fraction factorization algorithm? > Im factoring numbers n using the Continued Fraction algorithm, finding > pairs (x,y) such that x^2==y mod n. The y values are factorized as y = > product(p_i^e_i) * p, where p_i are primes<=307, p<307^2 is another > prime, and e_i>=0. Isnt this usually called the Quadratic Sieve? Where do continued fractions come into it? Regarding the speed, there is a standard trick using logarithms to determine if a number is smooth (has small factors), using the fact that if it is not smooth there must be quite a large discrepancy. Do you use this? -- Timothy Murphy e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie tel: +353-86-2336090, +353-1-2842366 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: there is no such thing as infinity > @mozo.cc.purdue.edu: Very inefficient. I modified the program to count in increments > of M/10, and it was done very quickly. It turns out that > M= 11(M/10), which is embarrassing mostly because if wed have > thought of this sooner, we could have worked it out by hand. > Well, Im glad thats finally settled. (Hmmm...I wonder what > journal I should send this off to....) >> The two-line program that I posted previously, namely >> print *, huge(1) >> end >> prints 2147483647 on my machine. Huge is a standard intrinsic >> function. > Must be a glitch, since 11/10 * Huge isnt even an integer. > Sure it is. The statement > print *, 11/10 * huge(1) prints exactly the same number as > print *, huge(1) > and therefore M = huge(1). I rest my case. Yeah, huge(1). Sure. Thats what /all/ men think. ;P (...Starblade Riven Darksquall...) === Subject: Re: there is no such thing as infinity > You should take a look at Cantors theory of sets. In it, Cantor does > treat infinities as numbers, cardinals and ordinals. Try a Google > search on it. Cantor was a rusha. He confused generations of mathematicians by > telling them there are lots of infinities. > Whats so confusing about them? > It is impossible to get a > BS in math without admitting that his theories are correct. The diagonal argument was a sham because it presupposes the existence > of infinity without proving its existence. > I already told you that one assumes the existence of an infinite set, > such as the set of all integers. Do you accept the set of all > integers? If one assumes the existence of that set, one does not need > to then prove it. Do you accept the continuum of the real line of > numbers? > His theorem is proof that > infinity does not in fact exist, because if it did, then there would > be only one type, not aleph null, aleph one etc., because infinity by > definition is as far as one can go. > I dont know whose definition of infinity youre using, but it sure > isnt a definition that mathematicians use. In math, infinity is about > the cardinalities and ordinalities of abstract sets, not about > distances in physical space. > He even defiled the Hebrew > alphabet by putting subscripts next to aleph when the gematria of > aleph is not infinity or even M but one. > Im not up on Hebrew mysticism. His theory of infinities was in fact a precursor to the communist > revolution. Marx believed in thesis and antithesis, both infinite premise of > dialectical materialism is true. > I thought Marx was on about economics and the fair dispersal of > resources in a society. > Had Cantor never been alive, there would be world peace and for sure > the number M would have been determined. > There have been and still are many threats to world peace, most of > these threats have nothing to do with Marxism. > Define this number M. > One of the brilliant things Cantor did was to define an infinite set > as a set whose elements can be put into 1-1 correspondence with a > proper subset of itself (obviously something one cannot do with a > finite set). For instance, you can put the set of positive integers > into 1-1 correspondence with the set of positive even integers by the > correspondence n -> 2n It seems to go against common sense to say that these two sets have > the same cardinalities. Then again, we dont have any common > experience working with infinities. That is even more evidence for the ridiculousness of there being an > infinity. Any math that completely contradicts the senses is wrong > just as any scientific theory that contradicts observation is wrong. > You just cannot have an infinite set. Dr. Ben Zona > Mathematics is founded on self consistency, not on sensibility. > Science uses calculus, and calculus uses infinity. > Patrick It uses infinity as a limit, or rather, a lackthereof. They are using infinity as simply a gap to fill in the sentence Take the limit of y as x goes to so that they dont have to keep saying Take the limit of y as x increases/decreases without bound, so infinity works here only as a convenience, not as a thing that actually exists. (...Starblade Riven Darksquall...) === Subject: Re: there is no such thing as infinity Ive thought really hard about this one and came to the conclusion > that there is no scientific evidence of infinity existing. > Idiot. > . > / > / > ------- > / > / > ------------- > With the apex of the triangle as starting point, sequentially draw a > straight line to every point in the line forming the base of the > triangle. Compare the number of points in the line forming the base > to the number of points in the intermediary line which each straight > line must also intersect but is inarguably shorter than the base (in > this case, only half as long). > Now you have an infinity in hand infinitely larger than the number of > integers, which is also infinite - but countable. If it makes you > feel any better, git, use a smaller triangle. You stupid twit. You just cant divide by zero in this case. A point and a line are two different things. A line simply isnt composed of an infinite number of points. A line segment is exactly that: A _SEGMENT_. A point is just a point. Simply having two points does not give you a line segment. So clearly a line isnt just what you get when you add an infinite number of points. Dumbass. > Whats to say that eventually there is a number where it is impossible > to count higher than? > Nothing. Nothing at all. Try the trivial proof that there is no > largest prime number. Primes are a very small fraction of all > integers. Idiot. > I am currently running a computer program that will eventually find > this magic number > Idiot. IEEE 80-bit (long_double_precision) data type gives 64 bits in > the mantissa so: > 2^64 = 1.84x10^19 > is the biggest number in your box (unless you ran a Microsoft > compiler, which defaults double_long_precision source code to > long_precision executables). > 2^80 = 1.21x10^24 > BTW, idiot, (2^20996011)-1 is the 40th Mersenne prime. It has > 6,320,430 deciml digits. Get back to us when you count that high. > Here are the digits for you to compare with your work, > I told you, youve got to wait for Intel to come out with a 64 > bits! >Fortran says positive infinity = 2147483647 and negative infinity = >-2147483648. Weird thing is 2*(positive infinity) = 0. >Where should I publish my findings? Idiot: 2^31-1 = 2 billion, the number you quoted; the highest possible number a 32bit intel chip can recognize. If you wanted to say something bigger, maybe that there are 10^80 atoms in the known galaxy. Thats big, but not infinite. But, to get to the point, answer me this: we can only magnify about 10^-19 meters, far larger than the Planck length of 10^-34 meters. We can only see out in space as far as Hubble will let us. Obviouslly we can predict things smaller than we know that the galaxy is NOT-finite. And that is as acurate as we can get right now. To say the galaxy is finite when we cannot now see the end of it is ignorance. Of course you may argue the we just cant see the end yet, that is true. But all the data, all the math, points at a galaxy far larger than the human mind can conceive. And if we cant see mentally picture something that big, to our puny minds it is infinite. You quoted Fortran. Obviously you are familiar with the process of type casting and concactenation. The integer is being left shifted and type casted to an int so the computer recognizes it. In the process it is being concacted to fit in the register. Obviously a large portion of the number is lost. Same is true when we try to picture something as large as the universe. All we know now is that it is NOT-finite. === Subject: Re: Motion > I am interesting in knowing how transfinite numbers solved Zenos paradox. Transfinite Number http://mathworld.wolfram.com/TransfiniteNumber.html Zenos Paradoxes http://mathworld.wolfram.com/ZenosParadoxes.html One central element of the this theory of the transfinite numbers is a precise definition of when two infinite collections are the same size, and when one is bigger than the other -- with such a definition in hand it is then possible to order the infinite numbers just as the finite numbers are ordered. === Subject: Re: motion by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i1FEBuN12884; >A ball is thrown upward from the edg of a cliff in such a way that x >seconds later, it is y=-16x^2 + 64x + 80 feet above the ground at the >base of the cliff. When does the ball hit the ground? Solving for y=0 gives only one +ve value for x( 5 seconds).. thats the answer....... >James > === Subject: Re: What is a number?/What is not a number? > a number is the value of a function that assigns to a subset of all > objects a word indicating that the people who saw some reason to > create the subset also had some reason to quantify it. When the usual > way of counting the entities in the subset fails, these people hire > salesmen and women called mathematicians to invent a term such as Ôpi > or Ôe to convince the rest of us that numerosity prevails despite no > realistic quantification. Its been proven So everything thats obseved and told is a number? :0 Carl === Subject: Re: What is a number?/What is not a number? > a number is the value of a function that assigns to a subset of all > objects a word indicating that the people who saw some reason to > create the subset also had some reason to quantify it. When the usual > way of counting the entities in the subset fails, these people hire > salesmen and women called mathematicians to invent a term such as Ôpi > or Ôe to convince the rest of us that numerosity prevails despite no > realistic quantification. Its been proven So everything thats obseved and told is a number? :0 Carl === Subject: Re: What is a number?/What is not a number? >> What is a number?/What is not a number? >> Numbers seem to be simple things (objects) or a property of things. >> Yet the deeper you study numbers the more complex they become. >> As children understanding of numbers starts with the Naturals then the >> Intergers, the Rationals, the Reals, the Complex and beyond. >> Mathamatics later turns this all on its head > Whatdja mean dude? Doesnt mathematics go in the order Naturals = Integers => Rationals => Reals => Complex and beyond, in the sense of > defining each in terms of the previous one? So if you define the > naturals `a la von Neumann and youre happy with set theory, No Then you go to set theory and reconstruct it all youre > done. No it all starts again the logic knots up, the proof is the axiom, a tautology which can define everything :). Again: What is not a number? Carl === Subject: Re: What is a number?/What is not a number? > a number is the value of a function that assigns to a subset of all > objects a word indicating that the people who saw some reason to > create the subset also had some reason to quantify it. When the usual > way of counting the entities in the subset fails, these people hire > salesmen and women called mathematicians to invent a term such as Ôpi > or Ôe to convince the rest of us that numerosity prevails despite no > realistic quantification. Its been proven Yes I Can apply a function to anything, so anything has a property of a number. Again: What is not a number? Does the maping of an object produce a number or is the number a universal property Carl === Subject: Re: What is a number?/What is not a number? What is a number?/What is not a number? > Numbers seem to be simple things (objects) or a property of things. > Yet the deeper you study numbers the more complex they become. As children understanding of numbers starts with the Naturals then the > Intergers, the Rationals, the Reals, the Complex and beyond. Mathamatics later turns this all on its head > Whatdja mean dude? Doesnt mathematics go in the order Naturals = Integers => Rationals => Reals => Complex and beyond, in the sense of > defining each in terms of the previous one? So if you define the > naturals `a la von Neumann and youre happy with set theory, youre > done. I interpreted his comments as an ontological question... What are the substantives which we call Ônumbers? This is a tricky question, and quite likely intractable. I think the best answer is that numbers are the objects which satisfy the axioms of a given number system. Any other description just adds layers of unnecessary complication. Ôcid Ôooh === Subject: Re: What is a number?/What is not a number? >> What is a number?/What is not a number? >> Numbers seem to be simple things (objects) or a property of things. >> Yet the deeper you study numbers the more complex they become. >> As children understanding of numbers starts with the Naturals then >> the Intergers, the Rationals, the Reals, the Complex and beyond. >> Mathamatics later turns this all on its head >> Whatdja mean dude? Doesnt mathematics go in the order Naturals => Integers => Rationals => Reals => Complex and beyond, in the sense of >> defining each in terms of the previous one? So if you define the >> naturals `a la von Neumann and youre happy with set theory, youre >> done. > I interpreted his comments as an ontological question... What are > the substantives which we call Ônumbers? > This is a tricky question, and quite likely intractable. I think the > best answer is that numbers are the objects which satisfy the axioms > of a given number system. Any other description just adds layers of > unnecessary complication. > Ôcid Ôooh yea Carl === Subject: Re: What is a number?/What is not a number? > What is a number?/What is not a number? > Numbers seem to be simple things (objects) or a property of things. > Yet the deeper you study numbers the more complex they become. > As children understanding of numbers starts with the Naturals then the > Intergers, the Rationals, the Reals, the Complex and beyond. > Mathamatics later turns this all on its head Maybe not. Ill offer this definition: A number is an element of a number set. A number set is either the set of integers Z or any ring/field R that contains Z. Or if you prefer, a number set is either the set of integers Z or any ring/field R such that there exists a 1-1 ring homomorphism from Z into R. My apologies to Peano, but here we treat the Naturals as numbers because they are a subset of Z which forms a semigroup. Perhaps someone else can fix this defect of definition. By the definition I gave, Clifford algebras over fields of 0 characteristic and even GL(n,Z) are also number sets. Patrick === Subject: Re: mathworld peculiarity > I suppose these are too numerous to deserve mention. > But here: > [ http://mathworld.wolfram.com/ Lindemann-WeierstrassTheorem.html ] > we see the symbol Q_p, which is used for the p-adic numbers. > Presumably it should be Q, the rationals. Are you drawing Eric Weissteins attention to these errors (I assume thats what they are)? -- G.C. === Subject: limit of sequence of functions I must be missing something, because this problem seems way too easy: Let {p_n} be a sequence of continuous functions on S^1(-pi,pi) (i.e., the functions are 2pi -periodic). Suppose that for every continuous function f on S^1, the sequence f_n (x) = INT(-pi, pi) p_n (x-y) f(y) dy has the property that lim f_n (0) exists. Then, for some K, INT(-pi, pi) |p_n (x)| dx < = K, n = 1, 2, 3, 4, .... My solution: Since the p_n are continuous and 2pi -periodic, they must be bounded, say |p_n (x)| < = C. Then INT(-pi, pi) |p_n (x)| dx < = 2 pi C and we are done. What did I gloss over? I dont really see how/why to use the assumption. I guess if you were to use it, you could let f be 1 (its continuous on S^1) and then lim INT(-pi, pi) p_n (-y) dy exists, but this alone wouldnt mean those integrals are bounded for all n (I dont think)... === Subject: Re: limit of sequence of functions >I must be missing something, because this problem seems way too easy: >Let {p_n} be a sequence of continuous functions on S^1(-pi,pi) (i.e., the >functions are 2pi -periodic). Suppose that for every continuous function f on >S^1, the sequence >f_n (x) = INT(-pi, pi) p_n (x-y) f(y) dy >has the property that lim f_n (0) exists. Then, for some K, >INT(-pi, pi) |p_n (x)| dx < = K, n = 1, 2, 3, 4, .... >My solution: Since the p_n are continuous and 2pi -periodic, they must be >bounded, say |p_n (x)| < = C. Then >INT(-pi, pi) |p_n (x)| dx < = 2 pi C >and we are done. What did I gloss over? Each one is bounded: |p_n(x)| <= C_n. But you havent shown that the same C_n works for all n. (And in fact there need not exist a C such that |p_n(x)| <= C for all n.) >I dont really see how/why to use the >assumption. What course is this? In particular have you recently covered anything about Banach spaces? >I guess if you were to use it, you could let f be 1 (its >continuous on S^1) and then >lim INT(-pi, pi) p_n (-y) dy >exists, but this alone wouldnt mean those integrals are bounded for all n (I >dont think)... ************************ David C. Ullrich === Subject: Re: limit of sequence of functions I got this problem from an archive of functional analysis qualifying exam problems... === Subject: HELP: combinatorics straining my limited brain capacity Desperately need help on this one. DonÇt even know where to start. Here goes: 2*n different persons stand in row. The first n persons wear a red hat, the next n persons wear a blue hat. Hats are distinguishable(even those of the same color). Also, persons are distinguishable. The hats are thrown up in the air and land, one hat for each person. Calculate the number of ways the hats can land so that a) the n first persons still wear a red hat a) the n first persons wear a blue hat c) the color of the hat alternate among the persons, and the first person wears a red hat d) formulate an assumption that generalizes the result of a), b) and c) === Subject: Re: Got a speeding ticket and need to fight back Approximately f lamda = c f lamda = 186 000 mps f -> frequency in hertz h = 60min > = 60 * 60 sec f lamda = 186 000 > = 186 000 mps f -> 1000,000,000 s^-1 (gigahertz) lamda = .000186 miles Thats the radars wavelength approximately. Take the radars time > window, > it internally uses to compute the speed: 1/10 second ( the human eye > can > see about 30 frames / sec ). Assuming Reaction time of a driver = 1/10 > sec > approx 0.000186/ .1s = .00186 mps = 6.6 mph This has nothing to do with how these radar units actually work. Google > on ÔDoppler. And then understand that most traffic court judges have probably heard > them all. And thrown them all out of court. If you cant identify some > procedural error on the part of the officer, pay the fine. > Amid all the noise associated with this posting, I am > wondering why nobody has bothered to question what > computational time and human reaction time have to do > with a Doppler velocity measurement. > If your case is weak, try to bafße them with BS. > I will answer your question why reaction time is important > A) even for DSP analysis, you know there are going to be processing > delays. Your MP3 player for example, actually is always a few milli seconds > ahead, processing all the crappy songs you listen to. Delay in processing does not make the data taken for the reading any different nor does it change the computed results. > B) it gives a good accurate measure of spontaneous velocity. You point the > radar at some big stupid object. Internally, it(the object) is composed of > billions and billions of molecules. Even air, for example, could causes > interference to measurement. The way, you get rid of the error, is by > averaging, a lot of data, over a period of time. Which has nothing to do with reaction time. > C) Syncrozning with the Speedometer reading. Speedometer is usally analog. > If you are not that blind, you will notice that needle that does the > reading, is actually bumping up and down, around a certain value. That is > because the car, is actually vibrating at a very high angular velocity. I dont understand how bumping up and down has anything to do with angular velocity nor reaction time. > However, by averaging out the measurement and by slowing the speedometer > needle down to your reaction time, you have a steady-state reading. > If things were not measured around your reaction time, you know the speed of > the car can be anywhere from a million 1,000,000,000 mph to 0 mph > Learn some science first, before you spew your crap. I didnt state anything -- I merely asked a question. > To talk smart, you need to know how smart people talk. Ive been watching for an example from you. > Too bad, catch-22 for u. According to some two-bit school on the Hill in Boulder, I knew enough to sneak through their engineering program. I must have slept through more of those EE classes than I thought. I guess I should also refund the money Ive been paid me over the years to work on instrumentation systems. === Subject: Re: Got a speeding ticket and need to fight back Devanathan) >If you ever got a Ph.D in psychology, Why would one wanting to understand statistics get a degree in psychology? Statistics is Mathematics and there are REAL Ph.D.s in Statistics. === Subject: Re: Got a speeding ticket and need to fight back Devanathan) >The last i checked, I dictate the laws of physics. I am God, Obviously you are not God as God doesnt resort to your immature tactics or name calling. He doesnt need to. === Subject: Re: Got a speeding ticket and need to fight back Devanathan) >you know what increases with age? fratulence. For example: take you. Your list is truncated. What increases with age (at least most of the time): 1). Knowledge 2). Experience 3). Wisdom 4). Maturity 5). Knowing the limits 6). And yes gas if you eat to many carbohydrates or other irritants routinely. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. > The lower bound of the Hausdorff dimension of the graph of > f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 11. > is s. Is it equal to s? If anyone is still interested s >= HausDim(graph f) >= s - c/log(t) but I dont know what constant c is :-( -- G.C. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >> The lower bound of the Hausdorff dimension of the graph of >> f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 11. >> is s. Is it equal to s? >If anyone is still interested > s >= HausDim(graph f) >= s - c/log(t) Huh. GIve us a hint/sketch of how you prove that. >but I dont know what constant c is :-( ************************ David C. Ullrich === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >The lower bound of the Hausdorff dimension of the graph of >f(x)=sum^{infty}_{i=1}t^{(s-2)i}sin(t^ix) For 11. >is s. Is it equal to s? > If what you say is so and my scribblings are correct then yes. > But I wonder if you really meant to say that s is a lower bound; Youre right, I meant lower bound. I have Haus(graph f) <= s and Box(graph f) >= s. I cant prove that Haus(graph f) = s which squares with G A Edgar saying that the Hausdorff dimension is an open problem. > getting an upper bound is much easier, its hard for me to > imagine how a person could know a lower bound but be > unable to find an upper bound (its easier to cover the > graph by a bunch of little squares than to show it cant > be covered by an even smaller bunch of squares): > If I did the calculations right then f is in Lip_alpha for alpha = > 2 - s, and for any f in Lip_alpha the (2-alpha)-dimensional > measure is sigma-finite (the measure of the part of the > graph above a compact interval is finite.) Hence the > s-dimensional measure of the graph of this function > is sigma-finite, which shows that s is an _upper_ bound > for the dimension. > Do you really know that s is a lower bound? How? > (Years ago I read a paper on the dimension of these > graphs; the argument was not nearly as trivial as the > calculations I did to show that s was an upper bound...) > ************************ > David C. Ullrich -- G.C. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. > Hausdorff dimension <= Box dimension in general. Sorry, my last paragraph should have read Let S be the graph of f , then Box(S) >= s , and the Hausdorff dimension of f is <= s . But I cant prove more. Not Let S be the graph of f , then Box(S) >= s , and the Hausdorff dimension of f is >= s . But I cant prove more. -- G.C. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >> Hausdorff dimension <= Box dimension in general. >Sorry, my last paragraph should have read >Let S be the graph of f , then Box(S) >= s , and the Hausdorff >dimension of f is <= s . So as I conjectured, you really meant that s is an upper bound. >But I cant prove more. So the whole thing reads >Let N_x(S) be the smallest number of sets of diameter at most x >which can cover S and define >Box(S)=lim_{x -> 0} inf (-log N_x(S)/log x) >Let S be the graph of f , then Box(S) >= s , and the Hausdorff >dimension of f is <= s . But I cant prove more. I imagine the notation lim_{x -> 0} inf means the lim inf as How do you show that Box(S) >= s? What seems easy to me is that Box(S) <= s, which is how I showed that the Hausdorff dimension was <= s this morning. Of course getting a lower bound on Box should be easier than a lower bound on the Hausdorff dimension, but still... For that matter I dont see how Box(S) >= s is going to imply the Hausdorff dimension is <= s. Could it be that there were two typos in the original and you only corrected one? >Not >Let S be the graph of f , then Box(S) >= s , and the Hausdorff >dimension of f is >= s . But I cant prove more. ************************ David C. Ullrich === Subject: Re: Hausdorff dimension of graph of Weierstrass function. > ... Could it be that > there were two typos in the original and you only > corrected one? Work in progress. Ill return to it tomorrow, perhaps. -- G.C. === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >> ... Could it be that >> there were two typos in the original and you only >> corrected one? >Work in progress. Ill return to it tomorrow, perhaps. In the meantime I thought about it - actually it seems like getting a lower bound for _Box_ might not be that hard (its clearly much easier than for the Hausdorff dimension.) You just have to show the graph intersects a large number of boxes in a grid. So throw away the tail (where the absolute value of the tail is less then half the grid size) and show that whats left has a large derivative on a lot of long intervals... Of course that leaves out a few details, but I retract my skepticism about you having got a lower bound for Box of that graph. ************************ David C. Ullrich === Subject: Re: Hausdorff dimension of graph of Weierstrass function. ... Could it be that > there were two typos in the original and you only > corrected one? >>Work in progress. Ill return to it tomorrow, perhaps. >In the meantime I thought about it - actually it seems like >getting a lower bound for _Box_ might not be that hard >(its clearly much easier than for the Hausdorff dimension.) >You just have to show the graph intersects a large number >of boxes in a grid. So throw away the tail (where the >absolute value of the tail is less then half the grid size) >and show that whats left has a large derivative on a lot >of long intervals... Indeed, if t is large enough a very crude argument suffices to show Box = s. (The condition is t^(2-s) > c1 and t^(s-1) > c2; didnt bother to find c1 and c2 exactly, since the argument is not going to work all the way down to t > 1 anyway; presumably a slightly subtler argument works if just t > 1. If t is large enough and you look at a grid of side h = t^(-N) then the oscillation of the N-th term in each vertical column is 2*t^((s-2)N), while the oscillation of the sum of the other terms is less than half that; use the absolute value for some and the size of the derivative for the others and add some geometric series. It follows that the graph intersects about t^(sN) = h^(-s) squares in the grid for 0 < x < 1...) >Of course that leaves out a few details, but I retract my >skepticism about you having got a lower bound for Box >of that graph. >************************ >David C. Ullrich ************************ David C. Ullrich === Subject: Re: Hausdorff dimension of graph of Weierstrass function. >[...] >Indeed, if t is large enough a very crude argument >suffices to show Box = s. Having dusted off a few cobwebs in the lacunary-series section of my brain, its clear that Box = s for t > 1 and 1 < s < 2. The proof depends on the fact that the supremum of a lacunary series is larger than a constant times the sum of the absolute value of the coefficients. This is well-known, for example it can be proved using Riesz products (see Katznelson). We need a slightly sophisticated local version, the proof of which must be somewhere in the lacunary series folder in the filing cabinet in my office: Lemma. Suppose L > 1 and a > 0. There exists C = C(L, a) > 0 such that if f(x) = sum_1^N a_j cos(n_j x), where n_{j+1}/n_j > L for all j, and I is an interval of length a/n_1, then the sup of |f| on I is larger than C sum |a_j|. Assuming that, consider a square grid in the plane with side length h. Take the sum and split into the sum from 1 to N - 1 plus the tail, where N is chosen so that the oscillation of the sum from 1 to N - 1 on any interval of length h is less than h (use obvious estimates on the derivatives). If you choose N maximal with this property it turns out that h > a/t^N for some constant a > 0, so the lemma shows that the max of the tail minus the min of the tail on any interval of length h is large. Hence the same is true of f itself, and this shows that the graph hits a large number of squares in the grid, forcing Box > s. ************************ David C. Ullrich === Subject: Re: pentiamond rep-tiles >That leaves these two: > ____ __/ > /____ /___/ >I see no reason that these should not be rep-tilable. Neither do I see >a rep-tiling of either! Does anybody? I think I have a dissection of the pentabolo corresponding to the first, which is composed of copies of itself either of the original orientation or rotated by 180 degrees. This can be deformed to give a dissection of the pentiamond. However this is an ir-rep-tiling. -- Stewart Robert Hinsley http://www.meden.demon.co.uk/Fractals/fractals.html