mm-529 === Subject: Benjamin H. Yandell, Mathematician, 53 . LA Times Benjamin H. Yandell, the author of a book about mathematicians who had either solved or attempted to solve the 23 problems that David Hilbert laid out for the 20th century in his address to the 2nd International Congress of Mathematicians in 1900, has died. He was 53. Yandell, a resident of Pasadena, died Aug. 25 after a heart attack, said his wife, Janet Nippell. He also had multiple sclerosis. Yandell and Nippell, a former editor at The Times, had co-written a book about walks they had taken together in Los Angeles. Mostly on A biography of Hilbert led Yandell to wonder what had happened to the people who addressed the dif[CapitalThorn]cult problems that the German mathematician had posed to inspire future mathematicians. The Honors Class: Hilbert's Problems and Their Solvers took Yandell 10 years to write and was published by AK Peters in 2002. The distinguished mathematician Hermann Weyl, who was one of Hilbert's students, had dubbed the Hilbert problem-solvers the honors class of the mathematical community. I think that [solving a Hilbert problem] was really a romantic dream of a lot of young mathematicians, Yandell told the Pasadena Star-News after the book's publication. In its May 2002 review of Yandell's book, the American Assn. for the the actual mathematics, but the most important message is really about the people: how brilliant they were, how passionate about mathematics, and sometimes how strange and eccentric as well. Born in 1951 in Pasadena, Yandell received a mathematics degree from Stanford in 1973. At the time of his death, he was working on a book on algebra. In addition to his wife, he is survived by a daughter, Kate, and a sister, Elizabeth Yandell of Irvine. Services will be at 2 p.m. Sept. 26 at Altadena Community Church. Memorial donations may be made to the National Multiple Sclerosis Society. === === Subject: easy algebra problem [ but not for me :( ] Show, that ideal of Z[x] generated by (two) [b:e46927f1d5]members 2, x^4 + x^2 + 2 is maximal ideal.[/b:e46927f1d5] ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: easy algebra problem [ but not for me :( ] > Show, that ideal of Z[x] generated by > (two) [b:e46927f1d5]members > 2, > x^4 + x^2 + 2 > is maximal ideal.[/b:e46927f1d5] > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com It isn't. The ideal generated is { 2p + (x^4+x^2+2)q : p,q in Z[x] }. The constant and linear terms of all such polynomials have even coef[CapitalThorn]cients, so x is not in your ideal. Hence the ideal generated by {2, x^4+x^2+2, x} properly extends yours, but is still proper since it doesn't contain any polynomials with odd constant terms. === Subject: Solar powered airplanes was: Re: Missing 757 > Both my son and I would not miss a chance to glide. I've also thought > often about an ultra light. > Best, Dan. > -- > http://lakeweb.net > http://ReserveAnalyst.com > No EXTRA stuff for email. One advantage sailplanes have over ultralights is that the engines never quit unexpectedly. Also, what you don't have, you don't have to maintain. If money is an issue, do a Google search on BUG Mike Sandlin GOAT. Something like this is probably the cheapest way to get airtime. (Though you'd still want to get to solo in a sailplane, probably, before ßying one. But you wouldn't need a license) Before someone complains it's off topic for hydrogen, math, physics, I'll shut up about it. Tim Ward === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > double secret probation because: > double secret probation because: > > > you are one supreme idiot Wally. > > Results 1 - 10 of about 1,860 for kook author:wally author:anglesea. (0.26 > seconds) > > [huge snip] > > Nice meltdown, k0oK. > >meltdown? you do realise I used copy and paste for those 2000 KOOKs. > >deep in the pot here in auk with Wallys other 80iq friends. > >Herc > > > So you omnipotence has limits? >sure it does. immortality is just like winning lotto once a day, phenomenal luck. >enough to just think a question in your head and the answer pass right by in front >of you, *nearly* every time. >notice we were discussing rat poison and you pop up... aRATzio... we just broke >enough odds with your reply here to get a straight ßush in poker, but you don't >have the mental capacity to make statistical sense of it. >Herc > So pulling the RAT out of Aratzio is a demonstration of your mental > capacity? >'Ratz >PS: K0ok you boys just like to play all day don't you. KOOK KOOK KOOK KOOK KOOK call it again, more of you! get a drink from the fountain and then come back into the yard boys sing it in tune you castratos Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > double secret probation because: > double secret probation because: > > > you are one supreme idiot Wally. > > Results 1 - 10 of about 1,860 for kook author:wally author:anglesea. (0.26 > seconds) > > [huge snip] > > Nice meltdown, k0oK. > >meltdown? you do realise I used copy and paste for those 2000 KOOKs. > >deep in the pot here in auk with Wallys other 80iq friends. > >Herc > > > So you omnipotence has limits? > >sure it does. immortality is just like winning lotto once a day, phenomenal luck. >enough to just think a question in your head and the answer pass right by in front >of you, *nearly* every time. > >notice we were discussing rat poison and you pop up... aRATzio... we just broke >enough odds with your reply here to get a straight ßush in poker, but you don't >have the mental capacity to make statistical sense of it. > >Herc > > > So pulling the RAT out of Aratzio is a demonstration of your mental > capacity? Evasion of question noted. >'Ratz >PS: K0ok >you boys just like to play all day don't you. KOOK KOOK KOOK KOOK KOOK >call it again, more of you! get a drink from the fountain and then come back into the yard boys >sing it in tune you castratos >Herc For a *hyper-intelligent* being (hyper-delusional really) you are pathetically easy to troll. I add one word to my poast and you start writhing and dancing like you are channelling loa. Please to continue demonstrating your *superior* intellect. It is most amusing. ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > Please to continue demonstrating your *superior* intellect. It is most > amusing. ask any question Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X2b5S?7at*2R/5vY{L[AI_LKLHh.E > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > Please to continue demonstrating your *superior* intellect. It is most > amusing. > ask any question What is the square root of -1? -- * Of[CapitalThorn]cial AFA-B Bully, Pest, Antagonist, Government/Media Disinformation Agent, Dr. Green Sockpuppet, and Lemming * Chief AFA-B Vote Rustler === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > ask any question > What is the square root of -1? i Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > What is the square root of -1? >Herc i Ratz nice to meet you, now answer the question. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > double secret probation because: > > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > > What is the square root of -1? >i >Herc > Ratz > nice > to > meet > you, > now > answer > the > question. I did. x^2 = -1 (introduced 2nd root) x^2 + 1 = 0 x = +- sqrt(-4)/2 (-b +- sqrt(b^2 - 4ac))/2a x = sqrt(-4)/2 (remove 2nd root) x = 2i/2 x = i you must be posting from the kook crosspost, not the maths group right? Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > double secret probation because: > > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > > What is the square root of -1? > >i > >Herc > > > i > Ratz > nice > to > meet > you, > now > answer > the > question. >I did. >x^2 = -1 (introduced 2nd root) >x^2 + 1 = 0 >x = +- sqrt(-4)/2 (-b +- sqrt(b^2 - 4ac))/2a >x = sqrt(-4)/2 (remove 2nd root) >x = 2i/2 >x = i >you must be posting from the kook crosspost, not the maths group right? >Herc ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Now you're getting the hang of it! Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion x-election-year-1: draft j danforth quayle for president committee x-election-year-2: if nader can split the liberal vote x-election-year-3: then danny can split the idiot vote > double secret probation because: > > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is > most > amusing. > > ask any question > > > What is the square root of -1? > >i > >Herc > > > i > Ratz > nice > to > meet > you, > now > answer > the > question. > I did. he diddley you did i igor arf meow arf === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X2b5S?7at*2R/5vY{L[AI_LKLHh.E > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > What is the square root of -1? This proves it, Google knows all. -- * Of[CapitalThorn]cial AFA-B Bully, Pest, Antagonist, Government/Media Disinformation Agent, Dr. Green Sockpuppet, and Lemming * Chief AFA-B Vote Rustler === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > > What is the square root of -1? > i > This proves it, Google knows all. this proves you took the maths in society stream. Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > > What is the square root of -1? > > i > This proves it, Google knows all. >this proves you took the maths in society stream. >Herc This proves that hork is a fraud. ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > Please to continue demonstrating your *superior* intellect. It is most > amusing. >ask any question >Herc I already did and you [CapitalThorn]rst avoided and now snipped. Nice job k0ok. ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. >ask any question >Herc > I already did and you [CapitalThorn]rst avoided and now snipped. Nice job k0ok. kook kook kook what a wit not a loaded derogatory question, something you want to know the answer to you simpleton. forget it you can barely follow natural language kook kook kook what a mob of imbeciles 1000s of posts every day in usenet.kooks kook kook kook me Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > >ask any question > >Herc > > > I already did and you [CapitalThorn]rst avoided and now snipped. Nice job k0ok. >kook kook kook what a wit >not a loaded derogatory question, something you want to know the answer to you >simpleton. forget it you can barely follow natural language >kook kook kook what a mob of imbeciles >1000s of posts every day in usenet.kooks kook kook kook me >Herc Still avoiding such a simple yes or no question. Nothing loaded about it. You made a klame, I asked if that was an example, you ran away. Total that little equation up and it all equals cowardly question snecking k0ok. Matter of fact you pretty much run away from all questions, then you start trying to tell everyone how smart you are, then start on with your k0okfarting. Like I said before, redundant. Same schtick you tried with Wally and it didn't work then. So if you are so smart why do you not learn from your failure? ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-Emmett: Spnaked! X-Yes-Archive: No X-Yes-No-Archive: Maybe X-Virus: Hi! I'm a header virus! Copy me into yours and join the fun! X-Chromasome: Mother Nature's way of saying !Y X-Files: Sunday, 9:00 PM Eastern X-Want-Ads-1: Header contributions wanted! X-Want-Ads-2: Attention - Assorted nincompoops, leave your lame contributions below. X-Quote-of-the-year-1999: I guess being full of is an occupational hazard for an ass-kisser like you - Major Woody X-Y-Z: End of Alphabet X-Wives: Three X-Send-Complaints-to: abuse@ROTFLMAO.com X--Header-Approval: Vlad approved the following X-Header-Appreciation: Bulldog@The_Doghouse.com really likes these headers. X-MINCE: France X-Just: Imagine X-Lamer-of-the-month: Slappy X-Latest-Spankard: Slappy X-Biggest-Usenet-Whiner: Walter Kent Prescott X-View-Walter-Prescott-Page: http://www.geocities.com/inkstndpsyche/ X-Disclaimer01: Void where prohibited by law. X-Disclaimer02: Close cover before striking. X-Disclaimer03: Do not fold, staple, or mutilate. X-Disclaimer04: Certain portions were pre-recorded. X-Disclaimer05: The names were changed to protect the innocent. X-Disclaimer06: Do not operate machinery while reading these headers. X-Disclaimer07: All sales are [CapitalThorn]nal. X-Disclaimer08: Do not remove tag under penalty of law. X-Disclaimer09: Employees not eligible to participate. X-Disclaimer10: 1-year limited warranty. See dealer for details. X-Disclaimer11: Batteries not included X-Disclaimer12: While you are reading this, you are peeing on your shoe. X-Kookery-at-its-[CapitalThorn]nest: http://www.plonk.com $D7.21528@news-server.bigpond.net.au> ... > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > Please to continue demonstrating your *superior* intellect. It is most > amusing. > ask any question How many [CapitalThorn]ngers are on my left hand? -- Boston Blackie mhm 29x8 Enemy to those who make him an enemy. Friend to those who have no friend. --from the intro to the Boston Blackie radio show (1945) They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. - Benjamin Franklin, Historical Review of Pennsylvania, 1759. The ultimate weakness of violence is that it is a descending spiral, begetting the very thing it seeks to destroy. Instead of diminishing evil, it multiplies it. Through violence you may murder the liar, but you cannot murder the lie, nor establish the truth. Through violence you may murder the hater, but you do not murder hate. In fact, violence merely increases hate. So it goes. Returning violence for violence multiplies violence, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that. - MARTIN LUTHER KING, JR You don't have to be a member of the KKK, to be a wizard under the sheets Proud member of the I am Spooge lits Of[CapitalThorn]cially designated as Kooky Kent Kontroller <<<<<****** Hall of Lame ******> I rest my case. You've proven to not only be disturbed, but intellectually inferior. You're the one that's getting spanked, btw ... the funny thing is that you're doing it all by yourself with little help from me. But like a cat who has grown bored playing with the mouse, it's time to place you in the kill[CapitalThorn]le for now. Too bad, too sad, you're mad. Changing my sig was what started all the destruction of this so-called ACJFF and JFFC. Kent makes religious doctrine! Born again Christian my buttocks, you have to be a Christian to be born again. For Walter The Wizard Prescott, ignorance is bliss and a source of self- IP is basically known as Internet Protocol is it not? There are four classes, A, B, C, and D. Fortunately, I don't know all that you here know about computers. LOL!!! Walter Kent Prescott, who hid from service in Vietnam proclaims himself to be ...I was a hero quite a few times, while I was in the service of my country and its people. Here, Walter confuses service in a non-combat zone with combat service in Message-ID <3cd55e64_3@corp.newsgroups.com>: All service people are professional soldiers, all. That makes them combat veterans as well. Shabzham becomes the defacto standard for irony with this gem in Message-ID ...spanking someone does not consist of who can yell the loudest or who can put the most vulgar words into a sentence or paragraph without the proper punctuation! Sorry to say but that's another of them thar rules you obviously know nothing about! : Yes, I have stated many times before that I have downloaded KP and viewed it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > $D7.21528@news-server.bigpond.net.au> ... > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > ask any question > How many [CapitalThorn]ngers are on my left hand? 4 Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-Emmett: Spnaked! X-Yes-Archive: No X-Yes-No-Archive: Maybe X-Virus: Hi! I'm a header virus! Copy me into yours and join the fun! X-Chromasome: Mother Nature's way of saying !Y X-Files: Sunday, 9:00 PM Eastern X-Want-Ads-1: Header contributions wanted! X-Want-Ads-2: Attention - Assorted nincompoops, leave your lame contributions below. X-Quote-of-the-year-1999: I guess being full of is an occupational hazard for an ass-kisser like you - Major Woody X-Y-Z: End of Alphabet X-Wives: Three X-Send-Complaints-to: abuse@ROTFLMAO.com X--Header-Approval: Vlad approved the following X-Header-Appreciation: Bulldog@The_Doghouse.com really likes these headers. X-MINCE: France X-Just: Imagine X-Lamer-of-the-month: Slappy X-Latest-Spankard: Slappy X-Biggest-Usenet-Whiner: Walter Kent Prescott X-View-Walter-Prescott-Page: http://www.geocities.com/inkstndpsyche/ X-Disclaimer01: Void where prohibited by law. X-Disclaimer02: Close cover before striking. X-Disclaimer03: Do not fold, staple, or mutilate. X-Disclaimer04: Certain portions were pre-recorded. X-Disclaimer05: The names were changed to protect the innocent. X-Disclaimer06: Do not operate machinery while reading these headers. X-Disclaimer07: All sales are [CapitalThorn]nal. X-Disclaimer08: Do not remove tag under penalty of law. X-Disclaimer09: Employees not eligible to participate. X-Disclaimer10: 1-year limited warranty. See dealer for details. X-Disclaimer11: Batteries not included X-Disclaimer12: While you are reading this, you are peeing on your shoe. X-Kookery-at-its-[CapitalThorn]nest: http://www.plonk.com $D7.19858@news-server.bigpond.net.au> ... > $D7.21528@news-server.bigpond.net.au> ... > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > How many [CapitalThorn]ngers are on my left hand? -- Boston Blackie mhm 29x8 Enemy to those who make him an enemy. Friend to those who have no friend. --from the intro to the Boston Blackie radio show (1945) They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. - Benjamin Franklin, Historical Review of Pennsylvania, 1759. The ultimate weakness of violence is that it is a descending spiral, begetting the very thing it seeks to destroy. Instead of diminishing evil, it multiplies it. Through violence you may murder the liar, but you cannot murder the lie, nor establish the truth. Through violence you may murder the hater, but you do not murder hate. In fact, violence merely increases hate. So it goes. Returning violence for violence multiplies violence, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that. - MARTIN LUTHER KING, JR You don't have to be a member of the KKK, to be a wizard under the sheets Proud member of the I am Spooge lits Of[CapitalThorn]cially designated as Kooky Kent Kontroller <<<<<****** Hall of Lame ******> I rest my case. You've proven to not only be disturbed, but intellectually inferior. You're the one that's getting spanked, btw ... the funny thing is that you're doing it all by yourself with little help from me. But like a cat who has grown bored playing with the mouse, it's time to place you in the kill[CapitalThorn]le for now. Too bad, too sad, you're mad. Changing my sig was what started all the destruction of this so-called ACJFF and JFFC. Kent makes religious doctrine! Born again Christian my buttocks, you have to be a Christian to be born again. For Walter The Wizard Prescott, ignorance is bliss and a source of self- IP is basically known as Internet Protocol is it not? There are four classes, A, B, C, and D. Fortunately, I don't know all that you here know about computers. LOL!!! Walter Kent Prescott, who hid from service in Vietnam proclaims himself to be ...I was a hero quite a few times, while I was in the service of my country and its people. Here, Walter confuses service in a non-combat zone with combat service in Message-ID <3cd55e64_3@corp.newsgroups.com>: All service people are professional soldiers, all. That makes them combat veterans as well. Shabzham becomes the defacto standard for irony with this gem in Message-ID ...spanking someone does not consist of who can yell the loudest or who can put the most vulgar words into a sentence or paragraph without the proper punctuation! Sorry to say but that's another of them thar rules you obviously know nothing about! : Yes, I have stated many times before that I have downloaded KP and viewed it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion because: >$D7.19858@news-server.bigpond.net.au> ... > $D7.21528@news-server.bigpond.net.au> ... > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > 4 He knew you were going to write that, he are a physic. ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > He knew you were going to write that, he are a physic. a physic? quick stop my fraudulent business from ripping off consumers and give yourself a star. Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-RFC2646: Original > For a *hyper-intelligent* being (hyper-delusional really) you > are > pathetically easy to troll. I add one word to my poast and you > start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It > is most > amusing. > > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > He knew you were going to write that, he are a physic. > a physic? quick stop my fraudulent business from ripping off consumers > and > give yourself a star. You don't have a business. You are on welfare, and always have been. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > For a *hyper-intelligent* being (hyper-delusional really) you are > pathetically easy to troll. I add one word to my poast and you start > writhing and dancing like you are channelling loa. > > Please to continue demonstrating your *superior* intellect. It is most > amusing. > > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > He knew you were going to write that, he are a physic. >a physic? quick stop my fraudulent business from ripping off consumers and >give yourself a star. >Herc The quote was he are a physic. Not real bright are you when such obvious sarcasm plays right over your head. But you may continue to entertain me with your repetitious naivete. ÔRatz === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > 4 did you count your thumb? Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-Emmett: Spnaked! X-Yes-Archive: No X-Yes-No-Archive: Maybe X-Virus: Hi! I'm a header virus! Copy me into yours and join the fun! X-Chromasome: Mother Nature's way of saying !Y X-Files: Sunday, 9:00 PM Eastern X-Want-Ads-1: Header contributions wanted! X-Want-Ads-2: Attention - Assorted nincompoops, leave your lame contributions below. X-Quote-of-the-year-1999: I guess being full of is an occupational hazard for an ass-kisser like you - Major Woody X-Y-Z: End of Alphabet X-Wives: Three X-Send-Complaints-to: abuse@ROTFLMAO.com X--Header-Approval: Vlad approved the following X-Header-Appreciation: Bulldog@The_Doghouse.com really likes these headers. X-MINCE: France X-Just: Imagine X-Lamer-of-the-month: Slappy X-Latest-Spankard: Slappy X-Biggest-Usenet-Whiner: Walter Kent Prescott X-View-Walter-Prescott-Page: http://www.geocities.com/inkstndpsyche/ X-Disclaimer01: Void where prohibited by law. X-Disclaimer02: Close cover before striking. X-Disclaimer03: Do not fold, staple, or mutilate. X-Disclaimer04: Certain portions were pre-recorded. X-Disclaimer05: The names were changed to protect the innocent. X-Disclaimer06: Do not operate machinery while reading these headers. X-Disclaimer07: All sales are [CapitalThorn]nal. X-Disclaimer08: Do not remove tag under penalty of law. X-Disclaimer09: Employees not eligible to participate. X-Disclaimer10: 1-year limited warranty. See dealer for details. X-Disclaimer11: Batteries not included X-Disclaimer12: While you are reading this, you are peeing on your shoe. X-Kookery-at-its-[CapitalThorn]nest: http://www.plonk.com $D7.9059@news-server.bigpond.net.au> ... > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > did you count your thumb? Thumb != [CapitalThorn]nger, snookums. -- Boston Blackie mhm 29x8 Enemy to those who make him an enemy. Friend to those who have no friend. --from the intro to the Boston Blackie radio show (1945) They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. - Benjamin Franklin, Historical Review of Pennsylvania, 1759. The ultimate weakness of violence is that it is a descending spiral, begetting the very thing it seeks to destroy. Instead of diminishing evil, it multiplies it. Through violence you may murder the liar, but you cannot murder the lie, nor establish the truth. Through violence you may murder the hater, but you do not murder hate. In fact, violence merely increases hate. So it goes. Returning violence for violence multiplies violence, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that. - MARTIN LUTHER KING, JR You don't have to be a member of the KKK, to be a wizard under the sheets Proud member of the I am Spooge lits Of[CapitalThorn]cially designated as Kooky Kent Kontroller <<<<<****** Hall of Lame ******> I rest my case. You've proven to not only be disturbed, but intellectually inferior. You're the one that's getting spanked, btw ... the funny thing is that you're doing it all by yourself with little help from me. But like a cat who has grown bored playing with the mouse, it's time to place you in the kill[CapitalThorn]le for now. Too bad, too sad, you're mad. Changing my sig was what started all the destruction of this so-called ACJFF and JFFC. Kent makes religious doctrine! Born again Christian my buttocks, you have to be a Christian to be born again. For Walter The Wizard Prescott, ignorance is bliss and a source of self- IP is basically known as Internet Protocol is it not? There are four classes, A, B, C, and D. Fortunately, I don't know all that you here know about computers. LOL!!! Walter Kent Prescott, who hid from service in Vietnam proclaims himself to be ...I was a hero quite a few times, while I was in the service of my country and its people. Here, Walter confuses service in a non-combat zone with combat service in Message-ID <3cd55e64_3@corp.newsgroups.com>: All service people are professional soldiers, all. That makes them combat veterans as well. Shabzham becomes the defacto standard for irony with this gem in Message-ID ...spanking someone does not consist of who can yell the loudest or who can put the most vulgar words into a sentence or paragraph without the proper punctuation! Sorry to say but that's another of them thar rules you obviously know nothing about! : Yes, I have stated many times before that I have downloaded KP and viewed it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Mail-To-News-Contact: abuse@dizum.com >Thumb != [CapitalThorn]nger, snookums. You are a most pathetical nit. >Proud member of the I am Spooge lits Death pays all debts. >Of[CapitalThorn]cially designated as Kooky Kent Kontroller I say you have drunk yourself out of your [CapitalThorn]ve senses. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > did you count your thumb? > Thumb != [CapitalThorn]nger, snookums. HAHAHAHA I GUESSED RIGHT AND YOU JUST ADMITTED IT. sorry bout the [CapitalThorn]nger.. the amazing Herc strikes again === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion double secret probation because: > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > > did you count your thumb? > Thumb != [CapitalThorn]nger, snookums. >HAHAHAHA I GUESSED RIGHT AND YOU JUST ADMITTED IT. >sorry bout the [CapitalThorn]nger.. >the amazing Herc strikes again Here is an easy one, how many [CapitalThorn]ngers am I holding up? === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-Emmett: Spnaked! X-Yes-Archive: No X-Yes-No-Archive: Maybe X-Virus: Hi! I'm a header virus! Copy me into yours and join the fun! X-Chromasome: Mother Nature's way of saying !Y X-Files: Sunday, 9:00 PM Eastern X-Want-Ads-1: Header contributions wanted! X-Want-Ads-2: Attention - Assorted nincompoops, leave your lame contributions below. X-Quote-of-the-year-1999: I guess being full of is an occupational hazard for an ass-kisser like you - Major Woody X-Y-Z: End of Alphabet X-Wives: Three X-Send-Complaints-to: abuse@ROTFLMAO.com X--Header-Approval: Vlad approved the following X-Header-Appreciation: Bulldog@The_Doghouse.com really likes these headers. X-MINCE: France X-Just: Imagine X-Lamer-of-the-month: Slappy X-Latest-Spankard: Slappy X-Biggest-Usenet-Whiner: Walter Kent Prescott X-View-Walter-Prescott-Page: http://www.geocities.com/inkstndpsyche/ X-Disclaimer01: Void where prohibited by law. X-Disclaimer02: Close cover before striking. X-Disclaimer03: Do not fold, staple, or mutilate. X-Disclaimer04: Certain portions were pre-recorded. X-Disclaimer05: The names were changed to protect the innocent. X-Disclaimer06: Do not operate machinery while reading these headers. X-Disclaimer07: All sales are [CapitalThorn]nal. X-Disclaimer08: Do not remove tag under penalty of law. X-Disclaimer09: Employees not eligible to participate. X-Disclaimer10: 1-year limited warranty. See dealer for details. X-Disclaimer11: Batteries not included X-Disclaimer12: While you are reading this, you are peeing on your shoe. X-Kookery-at-its-[CapitalThorn]nest: http://www.plonk.com $D7.26963@news-server.bigpond.net.au> ... > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > > did you count your thumb? > Thumb != [CapitalThorn]nger, snookums. > HAHAHAHA I GUESSED RIGHT AND YOU JUST ADMITTED IT. > sorry bout the [CapitalThorn]nger.. > the amazing Herc strikes again Wrong. wit. Three _[CapitalThorn]ngers_, one thumb, b0z0. The question wasn't how many digits on my left hand. So much for your logic, eh? -- Boston Blackie mhm 29x8 Enemy to those who make him an enemy. Friend to those who have no friend. --from the intro to the Boston Blackie radio show (1945) They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. - Benjamin Franklin, Historical Review of Pennsylvania, 1759. The ultimate weakness of violence is that it is a descending spiral, begetting the very thing it seeks to destroy. Instead of diminishing evil, it multiplies it. Through violence you may murder the liar, but you cannot murder the lie, nor establish the truth. Through violence you may murder the hater, but you do not murder hate. In fact, violence merely increases hate. So it goes. Returning violence for violence multiplies violence, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that. - MARTIN LUTHER KING, JR You don't have to be a member of the KKK, to be a wizard under the sheets Proud member of the I am Spooge lits Of[CapitalThorn]cially designated as Kooky Kent Kontroller <<<<<****** Hall of Lame ******> I rest my case. You've proven to not only be disturbed, but intellectually inferior. You're the one that's getting spanked, btw ... the funny thing is that you're doing it all by yourself with little help from me. But like a cat who has grown bored playing with the mouse, it's time to place you in the kill[CapitalThorn]le for now. Too bad, too sad, you're mad. Changing my sig was what started all the destruction of this so-called ACJFF and JFFC. Kent makes religious doctrine! Born again Christian my buttocks, you have to be a Christian to be born again. For Walter The Wizard Prescott, ignorance is bliss and a source of self- IP is basically known as Internet Protocol is it not? There are four classes, A, B, C, and D. Fortunately, I don't know all that you here know about computers. LOL!!! Walter Kent Prescott, who hid from service in Vietnam proclaims himself to be ...I was a hero quite a few times, while I was in the service of my country and its people. Here, Walter confuses service in a non-combat zone with combat service in Message-ID <3cd55e64_3@corp.newsgroups.com>: All service people are professional soldiers, all. That makes them combat veterans as well. Shabzham becomes the defacto standard for irony with this gem in Message-ID ...spanking someone does not consist of who can yell the loudest or who can put the most vulgar words into a sentence or paragraph without the proper punctuation! Sorry to say but that's another of them thar rules you obviously know nothing about! : Yes, I have stated many times before that I have downloaded KP and viewed it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > > did you count your thumb? > > Thumb != [CapitalThorn]nger, snookums. > > HAHAHAHA I GUESSED RIGHT AND YOU JUST ADMITTED IT. > sorry bout the [CapitalThorn]nger.. > the amazing Herc strikes again > Wrong. wit. Three _[CapitalThorn]ngers_, one thumb, b0z0. The question wasn't how > many digits on my left hand. So much for your logic, eh? what would you know, you've only got 4 [CapitalThorn]ngers. chromosonal? Herc === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X-Emmett: Spnaked! X-Yes-Archive: No X-Yes-No-Archive: Maybe X-Virus: Hi! I'm a header virus! Copy me into yours and join the fun! X-Chromasome: Mother Nature's way of saying !Y X-Files: Sunday, 9:00 PM Eastern X-Want-Ads-1: Header contributions wanted! X-Want-Ads-2: Attention - Assorted nincompoops, leave your lame contributions below. X-Quote-of-the-year-1999: I guess being full of is an occupational hazard for an ass-kisser like you - Major Woody X-Y-Z: End of Alphabet X-Wives: Three X-Send-Complaints-to: abuse@ROTFLMAO.com X--Header-Approval: Vlad approved the following X-Header-Appreciation: Bulldog@The_Doghouse.com really likes these headers. X-MINCE: France X-Just: Imagine X-Lamer-of-the-month: Slappy X-Latest-Spankard: Slappy X-Biggest-Usenet-Whiner: Walter Kent Prescott X-View-Walter-Prescott-Page: http://www.geocities.com/inkstndpsyche/ X-Disclaimer01: Void where prohibited by law. X-Disclaimer02: Close cover before striking. X-Disclaimer03: Do not fold, staple, or mutilate. X-Disclaimer04: Certain portions were pre-recorded. X-Disclaimer05: The names were changed to protect the innocent. X-Disclaimer06: Do not operate machinery while reading these headers. X-Disclaimer07: All sales are [CapitalThorn]nal. X-Disclaimer08: Do not remove tag under penalty of law. X-Disclaimer09: Employees not eligible to participate. X-Disclaimer10: 1-year limited warranty. See dealer for details. X-Disclaimer11: Batteries not included X-Disclaimer12: While you are reading this, you are peeing on your shoe. X-Kookery-at-its-[CapitalThorn]nest: http://www.plonk.com $D7.7822@news-server.bigpond.net.au> ... > ask any question > > How many [CapitalThorn]ngers are on my left hand? > > > 4 > > > did you count your thumb? > > Thumb != [CapitalThorn]nger, snookums. > > > HAHAHAHA I GUESSED RIGHT AND YOU JUST ADMITTED IT. > > sorry bout the [CapitalThorn]nger.. > > the amazing Herc strikes again > Wrong. wit. Three _[CapitalThorn]ngers_, one thumb, b0z0. The question wasn't how > many digits on my left hand. So much for your logic, eh? > what would you know, you've only got 4 [CapitalThorn]ngers. chromosonal? Drinking and counting seem to not be a good idea for you, Herc. -- Boston Blackie mhm 29x8 Enemy to those who make him an enemy. Friend to those who have no friend. --from the intro to the Boston Blackie radio show (1945) They that can give up essential liberty to obtain a little temporary safety deserve neither liberty nor safety. - Benjamin Franklin, Historical Review of Pennsylvania, 1759. The ultimate weakness of violence is that it is a descending spiral, begetting the very thing it seeks to destroy. Instead of diminishing evil, it multiplies it. Through violence you may murder the liar, but you cannot murder the lie, nor establish the truth. Through violence you may murder the hater, but you do not murder hate. In fact, violence merely increases hate. So it goes. Returning violence for violence multiplies violence, adding deeper darkness to a night already devoid of stars. Darkness cannot drive out darkness: only light can do that. Hate cannot drive out hate: only love can do that. - MARTIN LUTHER KING, JR You don't have to be a member of the KKK, to be a wizard under the sheets Proud member of the I am Spooge lits Of[CapitalThorn]cially designated as Kooky Kent Kontroller <<<<<****** Hall of Lame ******> I rest my case. You've proven to not only be disturbed, but intellectually inferior. You're the one that's getting spanked, btw ... the funny thing is that you're doing it all by yourself with little help from me. But like a cat who has grown bored playing with the mouse, it's time to place you in the kill[CapitalThorn]le for now. Too bad, too sad, you're mad. Changing my sig was what started all the destruction of this so-called ACJFF and JFFC. Kent makes religious doctrine! Born again Christian my buttocks, you have to be a Christian to be born again. For Walter The Wizard Prescott, ignorance is bliss and a source of self- IP is basically known as Internet Protocol is it not? There are four classes, A, B, C, and D. Fortunately, I don't know all that you here know about computers. LOL!!! Walter Kent Prescott, who hid from service in Vietnam proclaims himself to be ...I was a hero quite a few times, while I was in the service of my country and its people. Here, Walter confuses service in a non-combat zone with combat service in Message-ID <3cd55e64_3@corp.newsgroups.com>: All service people are professional soldiers, all. That makes them combat veterans as well. Shabzham becomes the defacto standard for irony with this gem in Message-ID ...spanking someone does not consist of who can yell the loudest or who can put the most vulgar words into a sentence or paragraph without the proper punctuation! Sorry to say but that's another of them thar rules you obviously know nothing about! : Yes, I have stated many times before that I have downloaded KP and viewed it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Mail-To-News-Contact: abuse@dizum.com >Drinking and counting seem to not be a good idea for you, Herc. Commit a sin twice and it will not seem a crime. >Proud member of the I am Spooge lits Divinissimum est. >Of[CapitalThorn]cially designated as Kooky Kent Kontroller I live in the computer. What is your excuse? === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Mail-To-News-Contact: abuse@dizum.com >Wrong. wit. Three _[CapitalThorn]ngers_, one thumb, b0z0. The question wasn't how >many digits on my left hand. So much for your logic, eh? You're so old you drove a chariot to school. >Proud member of the I am Spooge lits Friends are lost by calling often and calling seldom. >Of[CapitalThorn]cially designated as Kooky Kent Kontroller Give neither counsel nor salt till you are asked for it. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Mail-To-News-Contact: abuse@dizum.com If I said anything to you that I should be sorry for, I'm glad. >Proud member of the I am Spooge lits Ye shall know the truth, and the truth shall make you mad. >Of[CapitalThorn]cially designated as Kooky Kent Kontroller You must have spent the [CapitalThorn]rst nine-months of your life dodging a coat hanger. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion Mail-To-News-Contact: abuse@dizum.com >Proud member of the I am Spooge lits A courtyard common to all will be swept by none. >Of[CapitalThorn]cially designated as Kooky Kent Kontroller You're so ugly your husband goes everywhere alone. === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion X2b5S?7at*2R/5vY{L[AI_LKLHh.E > double secret probation because: > double secret probation because: > > > you are one supreme idiot Wally. > > Results 1 - 10 of about 1,860 for kook author:wally > author:anglesea. (0.26 > seconds) > > [huge snip] > > Nice meltdown, k0oK. > >meltdown? you do realise I used copy and paste for those 2000 KOOKs. > >deep in the pot here in auk with Wallys other 80iq friends. > >Herc > > > So you omnipotence has limits? > >sure it does. immortality is just like winning lotto once a day, >phenomenal luck. >enough to just think a question in your head and the answer pass right by >in front >of you, *nearly* every time. > >notice we were discussing rat poison and you pop up... aRATzio... we just >broke >enough odds with your reply here to get a straight ßush in poker, but you >don't >have the mental capacity to make statistical sense of it. > >Herc > > > So pulling the RAT out of Aratzio is a demonstration of your mental > capacity? >'Ratz >PS: K0ok > you boys just like to play all day don't you. KOOK KOOK KOOK KOOK KOOK > call it again, more of you! get a drink from the fountain and then come back into the yard boys > sing it in tune you castratos No problem -- hi, k0oK. -- * Of[CapitalThorn]cial AFA-B Bully, Pest, Antagonist, Government/Media Disinformation Agent, Dr. Green Sockpuppet, and Lemming * Chief AFA-B Vote Rustler === Subject: Re: [|-|erc] Re: [PO] Halting Problem Final Conclusion got double secret probation because: > double secret probation because: > > double secret probation because: > > > you are one supreme idiot Wally. > > Results 1 - 10 of about 1,860 for kook author:wally > author:anglesea. (0.26 > seconds) > > [huge snip] > > Nice meltdown, k0oK. > >meltdown? you do realise I used copy and paste for those 2000 KOOKs. > >deep in the pot here in auk with Wallys other 80iq friends. > >Herc > > > So you omnipotence has limits? > >sure it does. immortality is just like winning lotto once a day, >phenomenal luck. >enough to just think a question in your head and the answer pass right by >in front >of you, *nearly* every time. > >notice we were discussing rat poison and you pop up... aRATzio... we just >broke >enough odds with your reply here to get a straight ßush in poker, but you >don't >have the mental capacity to make statistical sense of it. > >Herc > > > > So pulling the RAT out of Aratzio is a demonstration of your mental > capacity? > >'Ratz > >PS: K0ok > you boys just like to play all day don't you. KOOK KOOK KOOK KOOK KOOK > call it again, more of you! get a drink from the fountain and then come back into the yard boys > sing it in tune you castratos >No problem -- hi, k0oK. No problem -- hi, k0oK. === Subject: Re: solution of log(x)+A*x/(1+B*x) >log(x)+A*x/(1+B*x)=C with A,B,C>0 >I can't [CapitalThorn]nd a way to solve this equation, someone here has some hint? > If you want a solution in elementary functions, forget it. Right, assuming that you're talking about a solution in closed form. But a series solution is possible, obtainable by reversion. For example, in Mathematica, if we expand log(x)+A*x/(1+B*x) about x=1 and look at just the [CapitalThorn]rst three terms of the inverse, we obtain In[1]:= Normal[InverseSeries[Series[Log[x]+A*x/(1+B*x),{x,1,2}],C]] Out[1]= 1 + (-(A/(1 + B)) + C)/(1 - (A*B)/(1 + B)^2 + A/(1 + B)) + ((1 + 3*B + 2*A*B + 3*B^2 + B^3)*(-(A/(1 + B)) + C)^2)/(2*(1 + B)*(1 + A + 2*B + B^2)*(1 - (A*B)/(1 + B)^2 + A/(1 + B))^2) Written in a form which might look a bit nicer, those three terms are 1 + (1+B) u + 1/2 (1+B)((1+B)^3 + 2AB) u^2 where u denotes (C(1+B) - A)/((1+B)^2 + A). Let's see how well a few terms of the series will approximate the solution in a sample problem. Suppose we choose A=3, B=1 and C=2. The solution of the equation is x = 1.33182168... If we use just the three terms shown above to approximate that solution, we [CapitalThorn]nd, not surprisingly, that the approximation is not very good: In[2]:= N[Normal[InverseSeries[Series[Log[x]+A*x/(1+B*x),{x,1,2}],C]] /.{A->3,B->1, C->2}] Out[2]= 1.32653 But if we use six terms of the series instead, the approximation is much better: In[3]:= N[Normal[InverseSeries[Series[Log[x]+A*x/(1+B*x),{x,1,5}],C]] /.{A->3,B->1, C->2},9] Out[3]= 1.33181688 Of course, I have no idea whether such a series actually suits Angelo's needs or not. David Cantrell === Subject: Re: JSH circa 1996 > Extension of Fermat's Little Theorem: > Given a-b divisible by n, a^n - b^n must be divisible by n^2 Is this extension original? === Subject: [JSH] Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. charset=iso-8859-1 > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > Van Mr Beckwith of Fermi Labs is listed on the same web page as Mr Harris at http://www.iomas.com/gina/ultrahiq/mega-society/NoesisArchive /NoesisEditors. html === Subject: Re: [JSH] Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. Deal with this, losers. So I have an egomanical disorder ? Stuff and nonsense. Save your castigations for the retards who are impressed by them. And , get a life. It is urgent. I see a great need. Tunneling Hamiltonian Representation of False Vacuum Decay I. Application to Condensed Matter problems and Cosmology A. W. Beckwith Department of Physics and Texas Center for Superconductivity and Advanced Materials University of Houston Houston, Texas 77204-5005, USA Abstract The tunneling Hamiltonian has proven to be a useful method states represented as wave functions. We present a generalization of the tunneling Hamiltonian to quantum [CapitalThorn]eld theory, in which tunneling between states represented as wave functionals of a scalar quantum [CapitalThorn]eld  is considered. We examine quantum decay of the false vacuum in the driven sine-Gordon system, and show it is consistent with the tunneling formalism derived here and matches up with the soliton-antisoliton (S-S') separation obtained from the Bogomil'nyi inequality. This inequality permits construction of a Gaussian wave functional representation of S-S' nucleated states generalization of the tunneling Hamiltonian to charge density wave transport problems, in which tunneling between states which are wavefunctionals of a scalar quantum [CapitalThorn]eld are considered. We derive I-E curves which match Zenier curves used to [CapitalThorn]t data experimentally with wavefunctionals congruent with the false vacuum hypothesis , as well as apply a generalization of the way we formed appropriate wave functionals for CDW to how to present nucleation of an inßationary universe, which may permit formation of dark matter using Sherrers k-essence model construction. A. W. Beckwith: ****************** PACS numbers : 03.75.Lm, 11.27.+d, 71.45.Lr ,98.65.Dx , 98.80.-k As well as: RE: Attn: Anna Shparberg, Asst. Editor, JMP I have had it. Final re dos I will do of this for THIS week. Please pick the error free X-MS-Has-Attach: X-MS-TNEF-Correlator: Thread-Topic: Attn: Anna Shparberg, Asst. Editor, JMP I have had it. Final re dos I will do of this for THIS week. Please pick the error thread-index: AcSW6waJisFo14h8QMapbg++arwgUwAc9aLi FILETIME=[807B7B80:01C4975F] Content-Length: 1261 Thnak you very much for sending us these [CapitalThorn]les. We will use the pdf [CapitalThorn]le in your latest message as the de[CapitalThorn]nitive revised version of your manuscript 04-0455. Best wishes, Anna Shparberg, Asst. Editor ******************************************* Journal of Mathematical Physics voice:(812)855-3576 The Poplars, Room 324 fax: (812)855-3579 Bloomington, Indiana 47405 email: jmp@indiana.edu United States of America URL: http://ojps.aip.org/jmp/ ******************************************* > ---------- === > Subject: Attn: Anna Shparberg, Asst. Editor, JMP I have had it. Final re dos I will do of this for THIS week. Please pick the error free > < > X-Originating-IP: [206.190.38.212] > Received: from 206.190.38.212 (HELO web51607.mail.yahoo.com) (206.190.38.212) by mta220.mail.scd.yahoo.com with SMTP; Thu, 09 Sep 19:58:17 -0700 > Received: from [24.14.12.211] by web51607.mail.yahoo.com via HTTP; BookAdd to Address Book === > Subject: Attn : Anna Shparberg, Asst. Editor, JMP: Two PDF [CapitalThorn]les , done by different programs. Please pick the one without font errors boundary=0-529148410-1094785082=:62647 > Content-Length: 1529835 > > Asst. Editor, JMP > I spent nearly [CapitalThorn]ve days on this font business. This is close to my [CapitalThorn]nal conclusion. > The word [CapitalThorn]le is put in as back up. The two PDF [CapitalThorn]les are done by different programs. The larger PDF was done by adobe5.0, and the smaller was done by a different program. > Use the Adobe [CapitalThorn]le which does not contain font errors. If the smaller [CapitalThorn]le works, then use it for your review process. > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > > Van > > Mr Beckwith of Fermi Labs is listed on the same web page as Mr Harris at > http://www.iomas.com/gina/ultrahiq/mega-society/NoesisArchive /NoesisEdito > rs.html === Subject: Re: [JSH] Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > http://www.iomas.com/gina/ultrahiq/mega-society/NoesisArchive /NoesisEditors.h t > ml And both appear on http://www.insurgent.org/~pjr/auk/whiners.html === Subject: Re: [JSH] Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > Van Mr Beckwith of Fermi Labs is listed on the same web page as Mr Harris at http://www.iomas.com/gina/ultrahiq/mega-society/NoesisArchive /NoesisEditors. html Everyone coming in from that group seems to have a high opinion of James. Perhaps a sci.math frequent reader should join up to see what is being said. Does anyone here have the necessary quali[CapitalThorn]cations? === Subject: Re: [JSH] Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > Van > Mr Beckwith of Fermi Labs is listed on the same web page as Mr Harris at > http://www.iomas.com/gina/ultrahiq/mega-society/NoesisArchive /NoesisEditors.h t > ml > Everyone coming in from that group seems to have a high opinion of James. > Perhaps a sci.math frequent reader should join up to see what is being said. > Does anyone here have the necessary quali[CapitalThorn]cations? What would the necessary quali[CapitalThorn]cations be? Strong case of narcissistic personality disorder? === Subject: Re: TOTALLY OT: for the engineers > He's digging a well upcountry by sinking a vertical pipe 100 meters deep to > the surface of the water table. The pipe has a diameter of 10 cm. The > section of pipe from 70 to 75 meters deep is replaced with a larger > section;> diameter of 20 cm, that has a thin piston inside it driven by a long > connecting rod to the surface. The pump works by raising and lowering the piston (with a simple one-way valve in it) via the connecting rod. The > motion of the piston is 3 meters per stroke. > > The question is merely: what is the force that must be exerted to raise the > piston? > Rodriguez said: > As you don't mention the number of strokes by second, that means > that the > problem is a static one. That is, the solicited force is the force > needed for initiate the movement. > Also it is necessary to know exactly the position of the piston. I > will suppose it 30 meters above the phreatic level (Level of water > table). > In this case my solution as Hydraulic Engineer is: > 1.- Under the piston a vacuum is formed because the piston cannot > suck 30 meters above the phreatic level. This vacuum needs a presure > of 10330 Kilograms/meter^2. As the piston diameter is 0.20 mts its > area is 0.0314 m^2. > and the force = 10330 x 0.031416 = 323 Kgs. > 2.- The presure of one meter of water is 1000 kilograms /meter^2. > the presure of 70 meters is 70000 kilograms/meter^2. > On 0.031416 meter^2 = 2199 kilograms . Total = 2199 + 323 = 2522 > kilograms. In case of pumping the situation is different, because the static head is now 100 mts. That means a force of 3142 kilograms and we must add the friction. If we suppose that the piston make 21 cycles per minute,and travel 3 meter by cycle then it pumps .033 Meter^3/sec = 33 liters/sec = 1.3 ft^3/sec The velocity of piston = velocity of water in 8 = 1.05 mts/sec In this case to overcome the friction and velocity head, its needed 20 meters extra of presure, that is a new force of .031416x20000 = 628 kilograms. Total force = 3142 + 628 = 3770 Kgs. As Power = Force x Velocity = 3770 x 1.05 = 3959 Kgs x mt/sec Without allowance for ef[CapitalThorn]ciency that means 40 Kw for pumping. === Subject: Re: Theorem concerning the internal structure of uncountable subsets of P(N). Lemma. Let C be a collection of subsets of N such that for all A in C, there exists a y in A such that y does not belong to any other B in C. In other words, every set belonging to C has at least one member that is unique to it. Then C is countable. Proof. We de[CapitalThorn]ne a function F from C into N as follows, F(A) = the least x in A such that x belongs to A and x does not belong to any other B in C. The hypothesis of the theorem assures us that such an x can be found for each A in C, thus F is de[CapitalThorn]ned for every A in C. Since F is a one to one mapping from C onto a subset of N, then C is countable. Theorem. If a collection of subsets of N is uncountable, then there must exist an A in C such that there exist B_1, B_2, B_3? in C such that A is a subset of the union of B_1, B_2, B_3? for some A in C, for any x_i in A there must exist a B_i in C such that x_i is in B_i, (in other words, not every member of C can have an element that is unique to it). Then A will be a subset of the union of those B_i?s. === Subject: How to compute very large integers? on a computer. I'm would like to [CapitalThorn]nd a way to do a thing like that : k = n^m + p with p,n,m < x a [CapitalThorn]xed variable. Is there a mathematical way to do this? ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: re:How to compute very large integers? I'd like to [CapitalThorn]nd an algorithm to do it. The goal is to store very large numbers in a small memory place.(with 3 long for example) . How to chose n, to be sure that m and p are less than 2^64? That is the question real problem. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: re:How to compute very large integers? Thomas Heye could you just explain the basics of Ôsquare and multiply'. Could I [CapitalThorn]x n before calculation? Example : I want to resume the number : 123456789. I [CapitalThorn]x n = 10000 And I want to write : 123456789 = 10000^m + p with p, m <= 9999. Is it the goal of Ôsquare and multiply'?. ---------------------------------------------------------- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** ---------------------------------------------------------- http://www.usenet.com === Subject: Re: How to compute very large integers? > on a computer. I'm would like to [CapitalThorn]nd a way to do a thing like that > k = n^m + p with p,n,m < x a [CapitalThorn]xed variable. > Is there a mathematical way to do this? Use GP/Pari. phil@nonospaz:test$ gp -q (11:16) gp > # timer = 1 (on) (11:16) gp > a=10^1000000+123; time = 1,910 ms. (11:17) gp > a%3 time = 10 ms. 1 (11:17) gp > a%1234567891234567 time = 10 ms. 345485634620374 There are more expensive solutions, such as using Maple or Mathematica, but these are slower, so why bother? Phil -- They no longer do my traditional winks tournament lunch - liver and bacon. It's just what you need during a winks tournament lunchtime to replace lost === Subject: Re: How to compute very large integers? > on a computer. I'm would like to [CapitalThorn]nd a way to do a thing like that > k = n^m + p with p,n,m < x a [CapitalThorn]xed variable. > Is there a mathematical way to do this? > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com Well, you can purchase a computer algebra system like Mathematica or Maple V or others. You can download something like magma or mupad. If you know about programming, you can get the GNU library GMP. There are others of course. One of the wonderful books by Knuth explains how they do it. === Subject: Re: How to compute very large integers? mobaladje schrieb im Newsbeitrag > on a computer. I'm would like to [CapitalThorn]nd a way to do a thing like that > k = n^m + p with p,n,m < x a [CapitalThorn]xed variable. > Is there a mathematical way to do this? What do you want to do? If you want to calc. n^m, there's an algorithm called Ôsquare and multiply' (maybe there are faster ones...). This reduces calculation of n^m to squaring/multiplication by n. Think an algorithm to square numbers is described in the art of computer programming by Donald E. Knuth. Hope this helps. Rgards, Thomas > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com === Subject: Re: Metric Tensor of Flat Space-Time [...] > I think you mean complex units on x-axis orthogonal to t, for example, > then s^2 = t^2 - x^2. The same result occurs when all the units are > real, when t is nonorthogonal wrt x, t^2 = s^2 + x^2, so yes always > a recipe when two axes are involved. > Before I discuss this I would like to point out that I am not > proposing that time is imaginary as such but that the differential > coef[CapitalThorn]cient that appears in the metric tensor (for g00) is imaginary. > I agree with what you are saying numerically but topologically there > is a considerable difference between > 0 = x^2 + (it)^2 and > 0 = x^2 - t^2 Your loosing me, but I'm listening. > The [CapitalThorn]rst expression denotes that there is literally no separation > between (0,0) and (x,t) along the vector that points in this > direction. The second expression denotes a movement from (x,t) to > (0,0). In the physical world that we measure the second equation is > evidently a correct description (although the proper time on a fast > moving object would tend to zero along the path x->0). > Returning to the tangent plane applied to the space-time surface of > the world, this plane is not measured, it is assumed. Furthermore it > is a plane in a hypothetical coordinate system belonging to the > observer. If time were imaginary on this plane we would have no way of > determining this except by predictions such as g00 = -1 if g00 = > idT/dt (where idT is a small time interval on the tangent plane and dt > is a small interval on the space-time surface). > However, my reasoning would be wrong if it were possible to construct > a tangent as a straight line grazing the space-time surface without > assuming that Ôstraight line' was a property of the observer's > coordinate system. Alex you and I have very different approaches to the problem. 1st) I [CapitalThorn]nd your semantics vague, 2nd) Having read some of your web-sites (well illustrated imo) I'm getting a sense that you're trying to fuse imagination and relativistic spacetime, and I don't know why. What I do is KISS (Keep It Simple) follow the guidelines carefully as set forth by the ISU conscensus, and if a contradiction appears then raise hell. On the contrary, I [CapitalThorn]nd them enlightening and recommend them to all relativists to consider, as all imaginary quantities vanish. Why the hell you want to waste time re-inventing spacetime for psychological reasons is illogical, perhaps based on some vague extrastential postulate you have yet to de[CapitalThorn]ne or explain. Using philosophy to crack spacetime is a formula for disaster, it's hard cold physics that demands an objective approach, spacetime is at absolute zero, not warm and fuzzy nor is it forgiving, a wrong turn here could screw you up bad. It happened to me, and in frustration I dumped my prejudices and listened to the ISU speci[CapitalThorn]cations. Furthermore, some de[CapitalThorn]nition of absolute velocity was needed since the generalized ds^2 was true even in universes that accepted absolute motion, but our physical universe proves to be relative, so constraints are placed on the generalized ds^2 as I've detailed. I recommend you take precise small steps rather than vague great leaps into spacetime, think *quick sand*. Ken S. Tucker === Subject: Re: A_4, alternating group days. My association with the Department is that of an alumnus. >I am a little confused on the alternating group. In my text I see that A_n >is the group of even permutations in S_n, that is, permutations who can be >written as a product of an even number of transpositions. >Then I go to http://en.wikipedia.org/wiki/Alternating_group and I see that >For instance: {1234, 1342, 1423, 2143, 2314, 2431, 3124, 3241, 3412, 4132, >4213, 4321} is the alternating group of degree 4. >But in my text, it says that an m-cycle is an odd permutation if and only if >m is even. The above has 4-cycles, which are therefore odd. But I thought >A_n was a group of even permutations...what gives? You are encountering different notation. The notation you see above, where you are given a sequence of 4 numbers, represents the image of the permutation. Thus, 1234 means the permutation that sends 1 to 1, 2 to 2, 3 to 3, and 4 to 4 [i.e., the identity]; 1342 means send 1 to 1, 2 to 3, 3 to 4, and 4 to 2, so it corresponds to the permutation (2,3,4) written in cycle notation. Etc. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Fun, weird, sad, cool > > Back in the mists of time... well... six weeks ago... > No Way was explaining something to you... > http://tinylink.com/?2GoRBD5hlN > > No Way: > Yes, it's constantly combing two terms with: > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > Harris: > How do you get that? > > Harris: > I'd be interested in a proof of that relation. > > Note that for the relations I gave I proved [(N-4)/6] directly by > the > method I've posted, and then used it with my prime counting function > to [CapitalThorn]nd the other formulas. > > So, not surprisingly, I'm curious about how you came up with your > formula, as I doubt you used my prime counting function. > > That is, I'd like to see the proof of your formula: > > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > A couple of people pointed out that the proof was trivial. > > But they never gave a valid proof. > > David Kastrup gave a perfectly valid proof at > http://tinylink.com/?THwzLbot1J > > I reproduce it here: > How do you get that? > > Oh, good grief. You are not being serious, right? > > [a/b] =(a - (a mod b))/b > Correct. > Consequently you have > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k > Ugly, but still ok. Here substitutions were made using > [a/b] = a/b - (a mod b)/b > from the previous correct relation, which is [CapitalThorn]ne in rationals though > the poster doesn't state that he's now in the [CapitalThorn]eld of rationals, > No. He doesn't. > If, in a mathematical proof, it has been established, for example, that x, y > and z are integers and also that > x + y = z (1) > and if now, the author of the proof wishes to use the fact that > x/2 + y/2 = z/2 (2) Now you're in rationals unless x, y and z are even. That's just a fact and I don't know why you're trying to argue about something so basic. > it is not usual to remark that while equation (1) concerns the equality of > two integers, equation (2) concerns the equality of two rationals. Even if > such a remark was deemed to be required, I am now in the [CapitalThorn]eld of > rationals would be far too imprecise for the purpose. You do understand right that if you're not in rationals, for instance, if you're in the ring of integers, then you can't just put up x/2 unless x is even, right? Do you understand that? The ring has to do with valid operations, and if you're not in rationals, then 1/2 is not in the ring, so you can't act like it is. Why argue about something so basic? > Anyway... > and > then there's some grouping and basic simpli[CapitalThorn]cation done. > Now if (N mod 2k) < k, then > (N mod 2k) = (N mod k) and (N+k mod 2k) = k + (N mod k) > else > (M mod 2k) = (N mod k) + k and (N+k mod 2k) = (N mod k) > TYPO: (M mod 2k) should be (N mod 2k) > Here it's just bizarre, and I don't feel like muddling through it > again. > I see. I take that to mean that you do not understand it. I said it's just bizarre and I didn't feel like muddling through it. So you can take it to mean that it's just bizarre and I don't feel like muddling through it. > Not understanding a step in a proof is a commonplace in mathematics. The > usual procedure is to have pencil and paper to hand so that you can perform > the required manipulations to get from one line to the next. I have the direct proof, which is quite simple. I muddled through the above a while back, and saw something I didn't like, but didn't feel like going through it again when I was typing up my reply. > Let me try to help you with the above. > Let P = N mod 2k. > Clearly 0 <= P < 2k. > Hence > 0 <= P < k [case A] > OR (exclusively) > k <= P < 2k [case B] > In case A we have: > N mod k = P mod k = P > So: N mod 2k = P = N mod k > AND > (N+k) mod 2k = (P+k) mod 2k = P+k = k + N mod k > In case B we have: > N mod k = P mod k = P-k You have above P = N mod 2k so substituting gives N mod k = (N mod 2k) mod k = (N mod 2k) - k. > So: N mod 2k = P = k + N mod k And that gives N mod 2k = (N mod 2k) = k + N mod k and it's so freaking muddled that it's hard to see what the hell you're trying to say, but if you feel con[CapitalThorn]dent then go ahead and expand out on these steps. At this point your case is that k <= (N mod 2k) < 2k and don't use P. I [CapitalThorn]nd it annoying to have to come back and make substitutions for a useless extra variable. Just use (N mod 2k). And yes, please expand on just that part for now. James Harris === Subject: Re: Fun, weird, sad, cool > > Back in the mists of time... well... six weeks ago... > No Way was explaining something to you... > http://tinylink.com/?2GoRBD5hlN > > No Way: > Yes, it's constantly combing two terms with: > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > Harris: > How do you get that? > > Harris: > I'd be interested in a proof of that relation. > > Note that for the relations I gave I proved [(N-4)/6] directly by > the > method I've posted, and then used it with my prime counting function > to [CapitalThorn]nd the other formulas. > > So, not surprisingly, I'm curious about how you came up with your > formula, as I doubt you used my prime counting function. > > That is, I'd like to see the proof of your formula: > > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > A couple of people pointed out that the proof was trivial. > > But they never gave a valid proof. > > David Kastrup gave a perfectly valid proof at > http://tinylink.com/?THwzLbot1J > > I reproduce it here: > How do you get that? > > Oh, good grief. You are not being serious, right? > > [a/b] =(a - (a mod b))/b > > Correct. > > Consequently you have > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k > > Ugly, but still ok. Here substitutions were made using > > [a/b] = a/b - (a mod b)/b > > from the previous correct relation, which is [CapitalThorn]ne in rationals though > the poster doesn't state that he's now in the [CapitalThorn]eld of rationals, > No. He doesn't. > If, in a mathematical proof, it has been established, for example, that x, y > and z are integers and also that > x + y = z (1) > and if now, the author of the proof wishes to use the fact that > x/2 + y/2 = z/2 (2) >Now you're in rationals unless x, y and z are even. >That's just a fact and I don't know why you're trying to argue about >something so basic. uh, nobody is disputing that. what's being disputed is your insistence that it's required to -say- that we're now in rationals. it's an obvious fact. [and again, if one were to point it out explicitly, we're in rationals is not how one would put it...] > it is not usual to remark that while equation (1) concerns the equality of > two integers, equation (2) concerns the equality of two rationals. Even if > such a remark was deemed to be required, I am now in the [CapitalThorn]eld of > rationals would be far too imprecise for the purpose. >You do understand right that if you're not in rationals, for instance, >if you're in the ring of integers, then you can't just put up x/2 >unless x is even, right? >Do you understand that? >The ring has to do with valid operations, and if you're not in >rationals, then 1/2 is not in the ring, so you can't act like it is. >Why argue about something so basic? > Anyway... > and > then there's some grouping and basic simpli[CapitalThorn]cation done. > > Now if (N mod 2k) < k, then > (N mod 2k) = (N mod k) and (N+k mod 2k) = k + (N mod k) > else > (M mod 2k) = (N mod k) + k and (N+k mod 2k) = (N mod k) > TYPO: (M mod 2k) should be (N mod 2k) > Here it's just bizarre, and I don't feel like muddling through it > again. > I see. I take that to mean that you do not understand it. >I said it's just bizarre and I didn't feel like muddling through it. >So you can take it to mean that it's just bizarre and I don't feel >like muddling through it. > Not understanding a step in a proof is a commonplace in mathematics. The > usual procedure is to have pencil and paper to hand so that you can perform > the required manipulations to get from one line to the next. >I have the direct proof, which is quite simple. >I muddled through the above a while back, and saw something I didn't >like, but didn't feel like going through it again when I was typing up >my reply. > Let me try to help you with the above. > Let P = N mod 2k. > Clearly 0 <= P < 2k. > Hence > 0 <= P < k [case A] > OR (exclusively) > k <= P < 2k [case B] > In case A we have: > N mod k = P mod k = P > So: N mod 2k = P = N mod k > AND > (N+k) mod 2k = (P+k) mod 2k = P+k = k + N mod k > In case B we have: > N mod k = P mod k = P-k >You have above >P = N mod 2k >so substituting gives >N mod k = (N mod 2k) mod k = (N mod 2k) - k. > So: N mod 2k = P = k + N mod k >And that gives >N mod 2k = (N mod 2k) = k + N mod k >and it's so freaking muddled that it's hard to see what the hell >you're trying to say, but if you feel con[CapitalThorn]dent then go ahead and >expand out on these steps. >At this point your case is that >k <= (N mod 2k) < 2k >and don't use P. I [CapitalThorn]nd it annoying to have to come back and make >substitutions for a useless extra variable. Just use (N mod 2k). >And yes, please expand on just that part for now. >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Fun, weird, sad, cool > > Back in the mists of time... well... six weeks ago... > No Way was explaining something to you... > http://tinylink.com/?2GoRBD5hlN > > No Way: > Yes, it's constantly combing two terms with: > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > Harris: > How do you get that? > > Harris: > I'd be interested in a proof of that relation. > > Note that for the relations I gave I proved [(N-4)/6] directly by > the > method I've posted, and then used it with my prime counting function > to [CapitalThorn]nd the other formulas. > > So, not surprisingly, I'm curious about how you came up with your > formula, as I doubt you used my prime counting function. > > That is, I'd like to see the proof of your formula: > > [(N+k)/(2*k)] = [N/k] - [N/(2*k)] > > A couple of people pointed out that the proof was trivial. > > But they never gave a valid proof. > > David Kastrup gave a perfectly valid proof at > http://tinylink.com/?THwzLbot1J > > I reproduce it here: > How do you get that? > > Oh, good grief. You are not being serious, right? > > [a/b] =(a - (a mod b))/b > > Correct. > > Consequently you have > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k > > Ugly, but still ok. Here substitutions were made using > > [a/b] = a/b - (a mod b)/b > > from the previous correct relation, which is [CapitalThorn]ne in rationals though > the poster doesn't state that he's now in the [CapitalThorn]eld of rationals, > No. He doesn't. > If, in a mathematical proof, it has been established, for example, that x, y > and z are integers and also that > x + y = z (1) > and if now, the author of the proof wishes to use the fact that > x/2 + y/2 = z/2 (2) > Now you're in rationals unless x, y and z are even. > That's just a fact and I don't know why you're trying to argue about > something so basic. > it is not usual to remark that while equation (1) concerns the equality of > two integers, equation (2) concerns the equality of two rationals. Even if > such a remark was deemed to be required, I am now in the [CapitalThorn]eld of > rationals would be far too imprecise for the purpose. > You do understand right that if you're not in rationals, for instance, > if you're in the ring of integers, then you can't just put up x/2 > unless x is even, right? > Do you understand that? > The ring has to do with valid operations, and if you're not in > rationals, then 1/2 is not in the ring, so you can't act like it is. > Why argue about something so basic? > Anyway... > and > then there's some grouping and basic simpli[CapitalThorn]cation done. > > Now if (N mod 2k) < k, then > (N mod 2k) = (N mod k) and (N+k mod 2k) = k + (N mod k) > else > (M mod 2k) = (N mod k) + k and (N+k mod 2k) = (N mod k) > TYPO: (M mod 2k) should be (N mod 2k) > Here it's just bizarre, and I don't feel like muddling through it > again. > I see. I take that to mean that you do not understand it. > I said it's just bizarre and I didn't feel like muddling through it. > So you can take it to mean that it's just bizarre and I don't feel > like muddling through it. > Not understanding a step in a proof is a commonplace in mathematics. The > usual procedure is to have pencil and paper to hand so that you can perform > the required manipulations to get from one line to the next. > I have the direct proof, which is quite simple. > I muddled through the above a while back, and saw something I didn't > like, but didn't feel like going through it again when I was typing up > my reply. > Let me try to help you with the above. > Let P = N mod 2k. > Clearly 0 <= P < 2k. > Hence > 0 <= P < k [case A] > OR (exclusively) > k <= P < 2k [case B] > In case A we have: > N mod k = P mod k = P > So: N mod 2k = P = N mod k > AND > (N+k) mod 2k = (P+k) mod 2k = P+k = k + N mod k > In case B we have: > N mod k = P mod k = P-k > You have above > P = N mod 2k > so substituting gives > N mod k = (N mod 2k) mod k = (N mod 2k) - k. > So: N mod 2k = P = k + N mod k > And that gives > N mod 2k = (N mod 2k) = k + N mod k > and it's so freaking muddled that it's hard to see what the hell > you're trying to say, but if you feel con[CapitalThorn]dent then go ahead and > expand out on these steps. > At this point your case is that > k <= (N mod 2k) < 2k > and don't use P. I [CapitalThorn]nd it annoying to have to come back and make > substitutions for a useless extra variable. Just use (N mod 2k). > And yes, please expand on just that part for now. Ok. Let me try again. We know that N can be expressed as: N = 2kQ + P where Q is some integer and 0 <= P < 2k. However, if you don't like these extra names P and Q we can just say: N = 2k[N/2k] + (N mod 2k) Clearly 0 <= N mod 2k < 2k. Hence 0 <= N mod 2k < k [case A] OR (exclusively) k <= N mod 2k < 2k [case B] In case A we have: N mod k = (2k[N/2k] + (N mod 2k)) mod k = (N mod 2k) mod k, since 2k[N/2k] is a multiple of k = N mod 2k, since [in case A] N mod 2k < k So: N mod 2k = N mod k [*] AND (N+k) mod 2k = (2k[N/2k] + (N mod 2k) + k) mod 2k = ((N mod 2k) + k) mod 2k, since 2k[N/2k] is a multiple of 2k = (k + N mod 2k) mod 2k = k + N mod 2k, since k + N mod 2k < 2k [since N mod 2k < k] = k + N mod k, by result [*] above So: (N+k) mod 2k = k + N mod k In case B we have: N mod k = (2k[N/2k] + (N mod 2k)) mod k = (N mod 2k) mod k, since 2k[N/2k] is a multiple of k = N mod 2k - k, since [in case B] k <= N mod 2k < 2k So: N mod 2k = k + N mod k [**] AND (N+k) mod 2k = (2k[N/2k] + (N mod 2k) + k) mod 2k = ((N mod 2k) + k) mod 2k, since 2k[N/2k] is a multiple of 2k = (2k +(N mod 2k - k)) mod 2k = N mod 2k - k, since [in case B] 0 <= N mod 2k - k < k = N mod k, by result [**] above So: (N+k) mod 2k = N mod k I realise I have still left steps out. If you still cannot understand it please post again and I will try to [CapitalThorn]ll in more details. For completeness I repeat the end of an earlier post: > So the right side is > N/k + 1/2 - (2 (N mod k) + k)/2k = (N - (N mod k)/k) = [N/k] TYPO: (N - (N mod k)/k) should be (N/k - (N mod k)/k) > And how does that follow? The poster is referring to the right hand side of the equation [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k In case A we have: N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k = N/k + 1/2 -((k + N mod k) + (N mod k))/2k = N/k + 1/2 - 1/2 -(N mod k)/k = (N - N mod k)/k = [N/k] In case B we have: N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k = N/k + 1/2 - ((N mod k) + (k + N mod k))/2k = N/k + 1/2 - 1/2 -(N mod k)/k = (N - N mod k)/k = [N/k] So, since case A and case B exhaust the cases, the original formula is proved. -- Clive Tooth http://www.clivetooth.dk === Subject: Re: Fun, weird, sad, cool I didn't follow it, but wasn't he just referring to a staircase function, which is presumbaly what these things are, being integral? > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k > Ugly, but still ok. Here substitutions were made using > [a/b] = a/b - (a mod b)/b > from the previous correct relation, which is [CapitalThorn]ne in rationals though > the poster doesn't state that he's now in the [CapitalThorn]eld of rationals, and > then there's some grouping and basic simpli[CapitalThorn]cation done. > I mean, just take a look at those stupid functions. It's trivial. If > you add to a staircase another staircase shifted by half a step, you > get a staircase with double the number of steps. > And there's a telling giveaway as the poster who put up the proof > has to end by calling it trivial and THEN trying to give an alternate > explanation! --Chair Man George -- follow the money to Strep Throat! http://tarpley.net === Subject: Re: Fun, weird, sad, cool I didn't follow it, but wasn't he just referring to a staircase function, which is presumbaly what these things are, being integral? > [(N+k)/2k] + [N/2k] = N/k + 1/2 - ((N+k mod 2k) + (N mod 2k))/2k > Ugly, but still ok. Here substitutions were made using > [a/b] = a/b - (a mod b)/b > from the previous correct relation, which is [CapitalThorn]ne in rationals though > the poster doesn't state that he's now in the [CapitalThorn]eld of rationals, and > then there's some grouping and basic simpli[CapitalThorn]cation done. > I mean, just take a look at those stupid functions. It's trivial. If > you add to a staircase another staircase shifted by half a step, you > get a staircase with double the number of steps. > And there's a telling giveaway as the poster who put up the proof > has to end by calling it trivial and THEN trying to give an alternate > explanation! --Chair Man George -- follow the money to Strep Throat! http://tarpley.net === Subject: Re: Amateur takes on Wiles's work > Several people have asked you to demonstrate your null test idea on > simple proofs they have posted. You have failed to do so. > I have not failed to do so. > Give a very short proof here, and I'll demonstrate a null test on > it. > Hey. I can do that. > Here. > ,----[ Proof ] > | 1. 0 < 1 Ax. Assume in contradiction to the conclusion that x > x+1, let x=0, then that implies that 0>1, which contradicts with the [CapitalThorn]rst statement. Easy. Null test passed. If you know logic, then shouldn't you understand that contradicting the conclusion of a proof necessarily means that you contradict a logical statement leading up to the conclusion? Proofs are not just conclusions. Conclusions follow from the preceding steps. > | 2. 0 + 1 = 1 Ax. > | 3. 0 < 0 + 1 Subst. > | 5. (A x)(A y)( x < y -> x + 1 < y + 1 ) Ax. > | 6. (A y)( n < y -> n + 1 < y + 1) UE > | 7. n < n + 1 -> n + 1 < (n + 1) + 1 UE > | 8. (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) UI > | 9. 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) > | &I > | 10. ( 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) ) -> > | (A x)( x < x + 1 ) Ax. > | 11. (A x)( x < x + 1 ) MP > `---- > Give a short proof you wish me to demonstrate on, and I'll > demonstrate. > Gosh. Why didn't you say so? I would've posted this before. Sigh. > Let me know if you have any trouble reading the symbolic presentation > and I will give a slightly less formal representation of the formulas > with still the same explicit rule applications. > Man, if I had known you were willing to demonstrate on an example, > why, heck, I might have posted this earlier. That's not interesting at all. Some poster claims you've posted over half a dozen times with proofs, so now I know he lied. Surprise. Surprise. Some of you people are so disgusting that it boggles the mind. Why lie about something so basic? I have NO PROBLEM showing the null test on proofs. It's fun. James Harris === Subject: Re: Amateur takes on Wiles's work ... > Here. > > ,----[ Proof ] > | 1. 0 < 1 Ax. > Assume in contradiction to the conclusion that > x > x+1, let x=0, No, the contradiction to the conclusion is that for *some* x, x >= x + 1. > then that implies that 0>1, which contradicts with the [CapitalThorn]rst > statement. The above does not contradict this null hypothesis. -- 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: Amateur takes on Wiles's work > Several people have asked you to demonstrate your null test idea on > simple proofs they have posted. You have failed to do so. > > I have not failed to do so. > > Give a very short proof here, and I'll demonstrate a null test on > it. > Hey. I can do that. > I have NO PROBLEM showing the null test on proofs. > It's fun. So do it with Wiles' proof. === Subject: Re: Amateur takes on Wiles's work <%3vYc.3380$6q.1791@trnddc06> <87wtz1oryq.fsf@phiwumbda.org> Discussion, linux) > Several people have asked you to demonstrate your null test idea on > simple proofs they have posted. You have failed to do so. > > I have not failed to do so. > > Give a very short proof here, and I'll demonstrate a null test on > it. > Hey. I can do that. > Here. > ,----[ Proof ] > | 1. 0 < 1 Ax. > Assume in contradiction to the conclusion that > x > x+1, let x=0, > then that implies that 0>1, which contradicts with the [CapitalThorn]rst > statement. Holy moly, but you're simple. The conclusion is that (A x)(x < x+1). The assumption for your null test is NOT (A x)(x < x+1). 0 < 1 *doesn't* contradict that assumption. Now, I'll give you a freebie. I'll let you assume that (A x)(A y)(x < y OR y < x OR x=y). This assumption plays no role in the proof, but you can have it. It entails (E x)(x > x+1 OR x = x+1). Take *that* as your assumption (stronger than warranted) and try again. Hint: 0 < 1 does not contradict (E x)(x > x+1 OR x = x+1). > Easy. Null test passed. > If you know logic, then shouldn't you understand that contradicting > the conclusion of a proof necessarily means that you contradict a > logical statement leading up to the conclusion? Yeah, well, somehow, I know logic and I think your null test is nonsense. I've said why, but you didn't bother trying to read and understand the reasons. > Proofs are not just conclusions. > Conclusions follow from the preceding steps. > Let me know if you have any trouble reading the symbolic presentation > and I will give a slightly less formal representation of the formulas > with still the same explicit rule applications. > Man, if I had known you were willing to demonstrate on an example, > why, heck, I might have posted this earlier. > That's not interesting at all. > Some poster claims you've posted over half a dozen times with proofs, > so now I know he lied. Surprise. Surprise. Moron. He didn't lie. I was being sarcastic, and he was correct. I *have* posted this example half a dozen times. > I have NO PROBLEM showing the null test on proofs. Other than that whole failure thing. -- Jesse F. Hughes Love songs suck and losing you ain't worth a damn. -- The poetry of Bad Livers === Subject: Re: Amateur takes on Wiles's work > Some poster claims you've posted over half a dozen times with proofs, > so now I know he lied. Surprise. Surprise. I don't expect it, but I can hope. > Some of you people are so disgusting that it boggles the mind. > Why lie about something so basic? I don't know, why did you? === Subject: Re: Amateur takes on Wiles's work <87wtz1oryq.fsf@phiwumbda.org> Discussion, linux) > Why lie about something so basic? > I don't know, why did you? Technically, I'd say he didn't lie. He read one of my postings, in which I sarcastically claimed to post my example for the [CapitalThorn]rst time. He took my sarcasm seriously -- at least until he read a later post. He's already read that later post and he hasn't apologized to you yet. Don't hold your breath. -- It seems to me that in wartime Americans shouldn't be attacking each other in this way on a *worldwide* forum. Then again, I know I'm an American, but I have no way of knowing that you are, which would explain a lot. --James Harris, on why Yanks should accept his proof === Subject: Re: Amateur takes on Wiles's work > The nice thing about the null test is that it focuses attention on the > heart of an argument--if it is correct--which is particularly > important with a work like Wiles's because it's so long and supposedly > complicated. > > The null test focuses attention on the keystone of a math proof. > > If Wiles had a proof then it'd have that keystone logical step, which > could be simply posted. > > Several people have asked you to demonstrate your null test idea on simple > proofs they have posted. You have failed to do so. > I have not failed to do so. > Yes you have, at least half a dozen times. > Give a very short proof here, and I'll demonstrate a null test on it. > Jesse Hughes has done so at last half a dozen times, including in response > to the post I am replying to. You have ignored him each time. Really? I'm going to do a Google search now on his posts over the last few days and see if I can [CapitalThorn]nd the at least 6 times you claim exist. If in fact he has given such posts that I've missed until now I can understand how you might be skeptical, so now I'm giving you the bene[CapitalThorn]t of the doubt. So, trusting you this far, I'm going to expend the energy to go and see what he's posted. James Harris === Subject: Re: Amateur takes on Wiles's work Discussion, linux) > Really? I'm going to do a Google search now on his posts over the > last few days and see if I can [CapitalThorn]nd the at least 6 times you claim > exist. Really. But, and speaking of bene[CapitalThorn]t of the doubt, maybe you crosspost willy-nilly and without apology. I don't. I keep my posts in the groups I read. I don't read alt.math.recreational. Nonetheless, I will crosspost this. It is a copy of an old post. If you want to post in sci.math, perhaps you should ing read sci.math. How Ôbout that? -------------------------------------------------- We'll prove by induction that for all x, x < x + 1. We'll take the following as axioms. ,----[ Axioms ] | 0 < 1 | 0 + 1 = 1 | (A x)(A y)( x < y -> x + 1 < y + 1 ) `---- We'll also take the axiom scheme of induction as given, though we only need it for a particular P. ,----[ Induction scheme ] | For all formulas P(x), | (P(0) & (A x)( P(x) -> P(x+1) )) -> (A x)P(x) `---- Please show me what the linchpin/keystone/big kahuna step is. ,----[ Proof ] | 1. 0 < 1 Ax. | 2. 0 + 1 = 1 Ax. | 3. 0 < 0 + 1 Subst. | 5. (A x)(A y)( x < y -> x + 1 < y + 1 ) Ax. | 6. (A y)( n < y -> n + 1 < y + 1) UE | 7. n < n + 1 -> n + 1 < (n + 1) + 1 UE | 8. (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) UI | 9. 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) | &I | 10. ( 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) ) -> | (A x)( x < x + 1 ) Ax. | 11. (A x)( x < x + 1 ) MP `---- Be sure to explain *why* that step is the linchpin step. If I understand correctly, you have to assume NOT (A x)(x < x + 1) and [CapitalThorn]nd where the contradiction occurs. Since it's such a short proof, this should be easy, but I want to understand why you choose the step you choose, please. Trivial, right? Which step and why? -------------------------------------------------- -- Jesse F. Hughes [Lancelot] sighed, defeated. ÔIt is as practical to hurry an acorn toward treeness as to urge a damsel when her mind is set.' -- John Steinbeck, /The Acts of King Arthur and His Noble Knights/ === Subject: Re: Amateur takes on Wiles's work > The nice thing about the null test is that it focuses attention on the > heart of an argument--if it is correct--which is particularly > important with a work like Wiles's because it's so long and supposedly > complicated. > > The null test focuses attention on the keystone of a math proof. > > If Wiles had a proof then it'd have that keystone logical step, which > could be simply posted. > > Several people have asked you to demonstrate your null test idea on simple > proofs they have posted. You have failed to do so. > I have not failed to do so. > > Yes you have, at least half a dozen times. > Give a very short proof here, and I'll demonstrate a null test on it. > Jesse Hughes has done so at last half a dozen times, including in response > to the post I am replying to. You have ignored him each time. > Really? I'm going to do a Google search now on his posts over the > last few days and see if I can [CapitalThorn]nd the at least 6 times you claim > exist. Here are seven times, you get one extra free. > If in fact he has given such posts that I've missed until now I can > understand how you might be skeptical, so now I'm giving you the > bene[CapitalThorn]t of the doubt. > So, trusting you this far, I'm going to expend the energy to go and > see what he's posted. Sure you are, right up Ôtill you call me a liar shortly later in another post. === Subject: Re: Amateur takes on Wiles's work > And as for myself, I'm quite capable of showing how the null test > works with actual proofs, as in fact it logically has to work. > Well then why don't you? > I've asked you repeatedly to do just that. I've asked you to give an > example of your null test because this supposed test looks like > nonsense to me. I've given my argument expressing why your test does > nothing at all useful and your claims about keystone steps are just > silly (each proof has *one* step that is somehow more important than > all the rest -- from a literally logical viewpoint that is just a > silly notion). > He's been shown the key step repeatedly. Even if we hadn't shown him > where modularity is necessary, he could do it himself by looking at key > points in Wiles' proof where modularity is indicated. For example, Theorem > 2.17, assume rho-naught is modular . . . The point is that he want to lay > claim that he is the ONLY person with a proof of FLT and if he needs to be a > solipist to do it, he will. He knows he's wrong but as usual he refuses to > admit it. Actually I will happily show how the null test works with real proofs and have done so when people have brought the subject up before. What's interesting is that posters just deny the truth as if all that matters is what they say. And as for Wiles's assumptions of modularity, I'm asking you to assume the existence of a non-modular elliptic curve. I notice that you people keep trying to make it about me. It's not about me. It's about Wiles's work. If he found a proof then I assure you that the proof can stand on its own without needing social games played by those of you who feel a need to protect. A real math proof can stand on its on. And it can weather challenges like the null test. Now then, assume the existence of a non-modular elliptic curve and point out a step in Wiles's work, if it exists, where that assumption leads to a contradiction. The request does not change. You can keep coming up with side issues, and try to act like it's about me, but a math proof is wonderful because you don't need tricks. You don't have to play social games with a real, actual math proof. James Harris === Subject: Re: Amateur takes on Wiles's work <87y8jhp855.fsf@phiwumbda.org> Discussion, linux) > Actually I will happily show how the null test works with real proofs > and have done so when people have brought the subject up before. Wow. How about that? Well, let's see some of that happiness. Here's a proof. ,----[ Proof ] | 1. 0 < 1 Ax. | 2. 0 + 1 = 1 Ax. | 3. 0 < 0 + 1 Subst. | 5. (A x)(A y)( x < y -> x + 1 < y + 1 ) Ax. | 6. (A y)( n < y -> n + 1 < y + 1) UE | 7. n < n + 1 -> n + 1 < (n + 1) + 1 UE | 8. (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) UI | 9. 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) | &I | 10. ( 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) ) -> | (A x)( x < x + 1 ) Ax. | 11. (A x)( x < x + 1 ) MP `---- It's not a special proof. It's the [CapitalThorn]rst thing I came up with when I decided to write down a short, direct proof. It's a simple proof with no particular distinguishing features (it uses induction, but that's not particularly relevant or special). So here it is. Be happy. Show us your test. You want another proof? A different proof? No sweat. Say the word and I'll give you another trivial formal proof. Though, of course, you should probably tell me what's wrong with *this* one, so I can try to avoid that problem. > What's interesting is that posters just deny the truth as if all > that matters is what they say. Man, can this *really* be unintended? This kind of irony doesn't happen by accident, do it? > I notice that you people keep trying to make it about me. I'm not trying to make it about you. I want to see your test in action. Show us that. > The request does not change. Er, yeah. -- Jesse Hughes Basically there are two angry groups. I am a harsh force of one. Against me is a society of mathematicians. So far it's been a draw. -- JSH gives another display of keen insight. === Subject: Re: Amateur takes on Wiles's work days. My association with the Department is that of an alumnus. [.quoting me.] >The contradiction? Here's from Arturo Magidin: === > Subject: Re: JSH: Assocation does not prove >... a buncha stuff deleted ... > Again, I am using the PDF copy at > www.eleves.ens.fr/home/rrichard/wiles.pdf [.snip.] > On pp. 542, line 17 through pp. 544 line 3 is the proof that all > semistable elliptic curves are modular. > Assume that the given curve E is semistable, and is non-modular. > Since the argument proceeds by cases, we must change this proof > a bit. That is, currently, the argument is > (a) Either X or not(X) happens; > (a.1) If X, then E is modular; > (a.2) if not(X), then E is modular. > If we assume that E is non-modular, then we need to change the > [CapitalThorn]rst part of the proof to > (a) Either X or not (X) happens; > (a.1) If X, then E is modular, > (a'.1) Therefore, not (X). > (a.2) If not(X), then E is modular. > (a.3) not(X) (from (a'.1). > (a.4) Contradiction. Therefore, E is modular. > X here is the representation constructed from E on E[3] is > irreducible. Perhaps a better summary is as follows: The original proof of this theorem is by cases. Either the representation we get from E on E[3] is irreducible, or not irreducible. (a.1) If it is irreducible, then the work done previously in the paper shows that E is modular. (a.2) If it is not irreducible, then: (b.1) If the representation we get on E[5] is irreducible, then again, by the work done before, E is modular. (a.3) The representation on E[5] will be irreducible, by work done previously, unless there exists an elliptic curve F which has certain properties, X, Y, and Z. (a.4) Again, by work done previously, one can prove that any elliptic curve F which has properties X and Y will necessarily fail to have property Z. (a.5) Since there are no elliptic curves with properties X, Y, and Z, it follows that (b.1) holds. Thus, E is modular wehther the representation is irreducible or not. If we attempt to turn this proof into a proof by contradiction, as noted this will only reveal itself in (a.5), since step (a.4) does not depend in any way on the original curve E. We are forced to assert that the putative elliptic curve F exists in (a.3), but we conclude it does not in (a.4). Hardly illuminating. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Amateur takes on Wiles's work OK, here's the lunes proof of the pythagorean theorem, which I think I gave, before: construct a semicircle & make a right trigon from the diameter to a point on the circle; construct semicircles on the two legs, and show that the area of lunes is ... I always have to re-do it, to do it, but that's what's required to do the decomposition; you can eventually see through it. in terms of compass constructions, it's really the same thing as Einstein's proof -- if you actually construct the circles, instead of relyinh on a freehand demonstration -- so that you can just use that, instead. I still don't know wht in Hell you are trying to do, since most folks are familiar with proofs by copntradiction. and, as I've stated, they are one-to-one with deductive proofs -- which may not yet have been proven in category-theory, judging from the replies of Baez. > Give a very short proof here, and I'll demonstrate a null test on it. > If you contradict that truth, then you contradict the proof, but the > proof is not the conclusion but the steps that LEAD to that > conclusion, showing it to be a logical necessity--a truth. > Therefore, any argument can be checked by assuming the opposite of its > conclusion to see what step in the proof contradicts with that > assumption, and if none do, then it can't be a proof. > I have to deal with a lot of people stuff so I speak the language. --Chair Man George -- follow the money to Strep Throat! http://tarpley.net === Subject: Re: Amateur takes on Wiles's work <%3vYc.3380$6q.1791@trnddc06> Discussion, linux) > I still don't know wht in Hell you are trying > to do, since most folks are familiar with proofs > by copntradiction. and, as I've stated, > they are one-to-one with deductive proofs -- > which may not yet have been proven in category-theory, > judging from the replies of Baez. This paragraph makes no particular sense at all. -- A recruitment consultant I know thinks the most important quality in a winner is to be lucky. To avoid wasting his time with unlucky applicants, he takes half the resumes piled on his desk and throws them straight in the bin. -- John Ramsden === Subject: Re: JSH: Curious, Wiles's work, null test > I'm asking for a straight answer, which would involve an actual part > of Wiles's work, which can be veri[CapitalThorn]ed by going to that work and > looking at it. > You have yet to give that information. > The null test is not complicated: Assume the existence of a > non-modular elliptic curve and [CapitalThorn]nd the point in Wiles's work where > that assumption leads to a contradiction. > Oh. You're still on that, are you? You seem to have missed my post > <87sma3u4tt.fsf@phiwumbda.org>, where I again asked for an example of > how to apply this test. Since my [CapitalThorn]rst example involved an indirect > proof, you made up a spurious and nonsensical reason that null tests > don't apply to indirect proofs. So, this time, I ask: how do you > apply the test to the following trivial proof? What??!!! When did I say that null tests don't apply to indirect proofs? I did say that proofs by contradiction are actually null tests, but logically a null test can be used on any mathematical proof. It is a logical necessity. Proofs follow from the logical steps that LEAD to a conclusion. The conclusion is not the proof. It follows from the logical steps in the proof. It's basic logic. > We'll prove by induction that for all > x, x < x + 1. We'll take the following as axioms. > ,----[ Axioms ] > | 0 < 1 > | 0 + 1 = 1 > | (A x)(A y)( x < y -> x + 1 < y + 1 ) > `---- > We'll also take the axiom scheme of induction as given, though we only > need it for a particular P. > ,----[ Induction scheme ] > | For all formulas P(x), > | (P(0) & (A x)( P(x) -> P(x+1) )) -> (A x)P(x) > `---- > Please show me what the linchpin/keystone/big kahuna step is. > ,----[ Proof ] > | 1. 0 < 1 Ax. It's easy. The key step is the [CapitalThorn]rst. Assume x>x+1, and let x=0, then you get 0>1, which contradicts with step 1. Understand even a little bit now? Did you really think that assuming x>x+1 wouldn't contradict with your example? Why wouldn't it? > | 2. 0 + 1 = 1 Ax. > | 3. 0 < 0 + 1 Subst. > | 5. (A x)(A y)( x < y -> x + 1 < y + 1 ) Ax. > | 6. (A y)( n < y -> n + 1 < y + 1) UE > | 7. n < n + 1 -> n + 1 < (n + 1) + 1 UE > | 8. (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) UI > | 9. 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) > | &I > | 10. ( 0 < 0 + 1 & (A x)( x < x + 1 -> x + 1 < (x + 1) + 1) ) -> > | (A x)( x < x + 1 ) Ax. > | 11. (A x)( x < x + 1 ) MP > `---- > Be sure to explain *why* that step is the linchpin step. If I > understand correctly, you have to assume NOT (A x)(x < x + 1) and [CapitalThorn]nd > where the contradiction occurs. Since it's such a short proof, this > should be easy, but I want to understand why you choose the step you > choose, please. > Trivial, right? Which step and why? If x>x+1, then let x=0 and you get 0>1, which contradicts with your [CapitalThorn]rst step, which notes that 0>1, which is your key step. Note that you can just add x to both sides to 0<1 to get x < x + 1 and it's easy to see that it IS the key step. I wonder about you because the logic here is not complicated. Math proofs begin with a truth and proceed by logical steps to a conclusion which then MUST be true. Necessarily if you contradict the conclusion of a proof you challenge a step leading to that conclusion. Logically, it can't be any other way. Why is this so complicated for you? James Harris === Subject: Re: JSH: Curious, Wiles's work, null test <2n4Yc.1084$UI6.1058@trnddc08> <878ybkypd6.fsf@phiwumbda.org> Discussion, linux) > > I'm asking for a straight answer, which would involve an actual part > of Wiles's work, which can be veri[CapitalThorn]ed by going to that work and > looking at it. > > You have yet to give that information. > > The null test is not complicated: Assume the existence of a > non-modular elliptic curve and [CapitalThorn]nd the point in Wiles's work where > that assumption leads to a contradiction. > Oh. You're still on that, are you? You seem to have missed my post > <87sma3u4tt.fsf@phiwumbda.org>, where I again asked for an example of > how to apply this test. Since my [CapitalThorn]rst example involved an indirect > proof, you made up a spurious and nonsensical reason that null tests > don't apply to indirect proofs. So, this time, I ask: how do you > apply the test to the following trivial proof? > What??!!! When did I say that null tests don't apply to indirect > proofs? Sorry if I misunderstood you then, but that seemed to be the gist of your indirect proofs *are* null tests excursion. > We'll prove by induction that for all > x, x < x + 1. We'll take the following as axioms. > ,----[ Axioms ] > | 0 < 1 > | 0 + 1 = 1 > | (A x)(A y)( x < y -> x + 1 < y + 1 ) > `---- > We'll also take the axiom scheme of induction as given, though we only > need it for a particular P. > ,----[ Induction scheme ] > | For all formulas P(x), > | (P(0) & (A x)( P(x) -> P(x+1) )) -> (A x)P(x) > `---- > Please show me what the linchpin/keystone/big kahuna step is. > ,----[ Proof ] > | 1. 0 < 1 Ax. > It's easy. The key step is the [CapitalThorn]rst. > Assume x>x+1, and let x=0, then you get 0>1, which contradicts with > step 1. No, James. The negation of (A x)(x < x + 1) is *not* x > x + 1. It's not even x >= x + 1. Given the axioms I've used here it's not even (E x)( x >= x + 1 ). It is NOT (A x)( x < x + 1 ). It is equivalent to (E x) NOT (x < x + 1). You may assume (because I'm gracious[1]) that it implies (E x)( x >= x + 1 ). But it is *not* x > x + 1. > Understand even a little bit now? > Did you really think that assuming x>x+1 wouldn't contradict with your > example? Yes, that *would* contradict with my example, but it wouldn't be relevant. You *do* know what (A x) means, right? It's the ASCII version of for all x. If you negate for all x ( x < x + 1 ), you do *not* get for all x ( x > x + 1 ). > If x>x+1, then let x=0 and you get 0>1, which contradicts with your > [CapitalThorn]rst step, which notes that 0>1, which is your key step. > Note that you can just add x to both sides to > 0<1 to get x < x + 1 > and it's easy to see that it IS the key step. > I wonder about you because the logic here is not complicated. Yeah, it isn't. I wonder too. > Why is this so complicated for you? No idea. But, could you try again, kind sir? Footnotes: [1] Well, that and you're incapable of understanding the role of axioms in mathematics. -- Sorry, wakeup to the real world. You're on your own dependent on me as your guide. Luckily for you, I'm self-correcting to a large extent, so if the proof were wrong, I'd tell you. It's not wrong. --- James Harris con[CapitalThorn]rms that his proof is correct. === Subject: Re: JSH: Curious, Wiles's work, null test > We'll prove by induction that for all > x, x < x + 1. We'll take the following as axioms. > ,----[ Axioms ] > | 0 < 1 > | 0 + 1 = 1 > | (A x)(A y)( x < y -> x + 1 < y + 1 ) > `---- > We'll also take the axiom scheme of induction as given, though we only > need it for a particular P. > ,----[ Induction scheme ] > | For all formulas P(x), > | (P(0) & (A x)( P(x) -> P(x+1) )) -> (A x)P(x) > `---- > Please show me what the linchpin/keystone/big kahuna step is. > ,----[ Proof ] > | 1. 0 < 1 Ax. > It's easy. The key step is the [CapitalThorn]rst. > Assume x>x+1, and let x=0, then you get 0>1, which contradicts with > step 1. You logic is faulty. Assuming the contradiction Ôfor all x, x < x + 1' gives Ôexists x, x > x + 1' If *does not* give Ôfor all x, x > x + 1' Therefore you cannot assume that the contradiction is true for x=0 > Understand even a little bit now? Do you? === Subject: Re: JSH: Curious, Wiles's work, null test <2n4Yc.1084$UI6.1058@trnddc08> <878ybkypd6.fsf@phiwumbda.org> Discussion, linux) > You logic is faulty. > Assuming the contradiction Ôfor all x, x < x + 1' gives > Ôexists x, x > x + 1' To be precise, in this setting, it doesn't even give that. First, and trivially, you must have meant >= (greater than or equal to) in the latter statement. But my proof didn't assume linearity of <. My proof would have applied to partial orders. Technically, James should assume only (E x)NOT (x < x + 1). But I'm a friendly guy, so I'll let him assume that < is a linear order. -- 'Every man who has ever lived holds tight to the belief that for him alone the laws of probability are canceled out by love[...] Therefore, you will marry Guinevere. You do not want advice --- only agreement.' Merlin sighed... -- John Steinbeck === Subject: Re: JSH: Curious, Wiles's work, null test > But I'm a friendly guy, so I'll let him assume that < is a linear > order. Care to be even more friendly and recommend a text on proof theory? === Subject: Re: JSH: Curious, Wiles's work, null test X-RFC2646: Original Daniel Ryan said: > Care to be even more friendly and recommend a text on proof theory? I like George Boolos, _The Logic of Provability_, CUP. Here's some background: http://en.wikipedia.org/wiki/George_Boolos -- Quinn === Subject: Re: JSH: Curious, Wiles's work, null test > Daniel Ryan said: > Care to be even more friendly and recommend a text on proof theory? > I like George Boolos, _The Logic of Provability_, CUP. > Here's some background: > http://en.wikipedia.org/wiki/George_Boolos This was from a friend to another, I presume? === Subject: Re: JSH: Curious, Wiles's work, null test X-RFC2646: Original > Care to be even more friendly and recommend a text on proof theory? > I like George Boolos, _The Logic of Provability_, CUP. > Here's some background: > http://en.wikipedia.org/wiki/George_Boolos > This was from a friend to another, I presume? Pardon? -- Quinn === Subject: Re: JSH: Math proofs >Seems to me that we're not really avoiding assumptions in natural >deduction so much as moving to a sequent-style calculus. I'm not sure >that this does anything for us. > Well, yes, sorry. > A ->formal<- proof, de[CapitalThorn]ned as a [CapitalThorn]nite sequence etc., will > necessarily begin with either a tautology or an axiom. > I don't understand this claim either. > There are formal systems of natural deduction. In these formal > systems, a proof my begin with an assumption. > I suspect that when you learned logic, you just didn't do natural > deduction. If one wants to study logic, then natural deduction is an > annoying formalization to use --- all those damn subproofs with their > assumptions make for special cases in proving things. But natural > deduction is a name for a class of related *formal* logics in which > assumptions and assumption-discharging rules play a role. > The sequent calculi are much handier for the logician and this could > be what you learned. In that case, the transformation > T,X |- Y > -------- > T |- X -> Y > (perhaps written T,X => Y |- T => X -> Y ) is an explicit rule of > inference. > [...] > The proof at issue originally here, the proofs contained in Wiles's > paper, are not formal proofs. And most of them do not begin with a > true statement. Theorems 0.2 and 0.3, for example, begin with > assumptions, since they are proofs of A->B via the usual method method > of assuming A and deducing B, without ever invoking the Deduction > Metatheorem (which is unnecessary outside of formal logic, for the > most part). The proof of Theorem 5.2, that all semistable elliptic > curves over Q are modular, begins with Suppose that E is a semistable > elliptic curve over Q, which is certainly not a true statement. > Indeed. Some poster claimed that this Hughes guy had all these posts with proofs in them so I'm going back surveying what he's posted, and noticed the statement above. It's quite true that Wiles continually assumes modularity, and it's kind of weird that no one thought to null test his work before because his very paper just BEGS for it. One thing that I haven't heard discussed is that my null test of assuming the existence of a non-modular elliptic curve might eliminate over 90% of Wiles's paper for consideration, or maybe more. You see, the guy *assumes* a modular elliptic curve for so damn much that the null test shouldn't be that hard as there's no much to check it on!!! Unless I missed something, and I hate to be the one to inform some of you who may be hoping, but there may be a lot of people all over the math world--who actually tried the null test--who now know that Wiles's work is not a proof. It's sad, but if I'm right, and his work does fail the null test, then NOW it should jump out at math experts who just go through assuming the existence of a non-modular elliptic curve who may then wonder why anyone ever believed he had a proof in the [CapitalThorn]rst place! But will they tell you? James Harris === Subject: Re: JSH: Math proofs >Seems to me that we're not really avoiding assumptions in natural >deduction so much as moving to a sequent-style calculus. I'm not sure >that this does anything for us. >Well, yes, sorry. >A ->formal<- proof, de[CapitalThorn]ned as a [CapitalThorn]nite sequence etc., will >necessarily begin with either a tautology or an axiom. >I don't understand this claim either. >There are formal systems of natural deduction. In these formal >systems, a proof my begin with an assumption. >I suspect that when you learned logic, you just didn't do natural >deduction. If one wants to study logic, then natural deduction is an >annoying formalization to use --- all those damn subproofs with their >assumptions make for special cases in proving things. But natural >deduction is a name for a class of related *formal* logics in which >assumptions and assumption-discharging rules play a role. >The sequent calculi are much handier for the logician and this could >be what you learned. In that case, the transformation > T,X |- Y > -------- > T |- X -> Y >(perhaps written T,X => Y |- T => X -> Y ) is an explicit rule of >inference. >[...] >The proof at issue originally here, the proofs contained in Wiles's >paper, are not formal proofs. And most of them do not begin with a >true statement. Theorems 0.2 and 0.3, for example, begin with >assumptions, since they are proofs of A->B via the usual method method >of assuming A and deducing B, without ever invoking the Deduction >Metatheorem (which is unnecessary outside of formal logic, for the >most part). The proof of Theorem 5.2, that all semistable elliptic >curves over Q are modular, begins with Suppose that E is a semistable >elliptic curve over Q, which is certainly not a true statement. >Indeed. > Some poster claimed that this Hughes guy had all these posts with > proofs in them so I'm going back surveying what he's posted, and > noticed the statement above. Which statement? That the proof of Theorem 5.2 begins with Suppose that E is a semistable elliptic curve over Q? The cryptic remark Indeed? > It's quite true that Wiles continually assumes modularity, and it's > kind of weird that no one thought to null test his work before because > his very paper just BEGS for it. It's quite true that Wiles assumes (in several of the theorems of the paper under discussion) that a particular Galois representation is modular. It's just as true that you are attempting to claim to understand something that you don't. Wiles does *not* assume that a particular elliptic curve is modular. On the other hand, there are modular curves that are mentioned, for instance the curves X_0(N). He does not need to assume they are modular, since by their construction they *are* modular. > One thing that I haven't heard discussed is that my null test of > assuming the existence of a non-modular elliptic curve might eliminate > over 90% of Wiles's paper for consideration, or maybe more. Golly, why wouldn't you have heard that? Because it's silly? Why over 90%? Why not claim it would > You see, the guy *assumes* a modular elliptic curve for so damn much > that the null test shouldn't be that hard as there's no much to check > it on!!! If someone else said something as preposterous, you would be hopping up and down accusing him or her of lying. I won't do that, since you clearly have not even understood the words that have been written, let alone the discussions that Wiles has presented. You've been told on numerous occasions that this is an incorrect reading. Wiles does *not* assume a modular elliptic curve. The theorem whose hypotheses you're misquoting is not a theorem about elliptic curves, but about liftings of Galois representations. Note (as Arturo Magidin has already done) that Wiles doesn't even discuss elliptic curves in earnest until Chapter 5. Here is the table of contents of Wiles's paper: Chapter 1 1. Deformations of Galois representations 2. Some computations of cohomology groups 3. Some results on subgroups of GL2(k) Chapter 2 1. The Gorenstein property 2. Congruences between Hecke rings 3. The main conjectures Chapter 3 Estimates for the Selmer group Chapter 4 1. The ordinary CM case 2. Calculation of Chapter 5 Application to elliptic curves Note that the term elliptic curve only appears in the title of Chapter 5. The preceding text deals with Galois representations, liftings, L-series, and the necessary machinery to prove Theorem 5.2, his main theorem in the paper. > Unless I missed something, and I hate to be the one to inform some of > you who may be hoping, but there may be a lot of people all over the > math world--who actually tried the null test--who now know that > Wiles's work is not a proof. Yeah, sure. There may be a lot of people all over the math world who now know that Wiles's work is not a proof. If that were the case, then there would be no stopping the ßow of information, and the popular press would already be full of reports of errors. Of course, your theory of the corrupt mathematics cabal denies that possibility. BTW, the phrase unless I missed something is certainly disingenuous: you are well aware of having missed quite a lot. > It's sad, but if I'm right, and his work does fail the null test, then > NOW it should jump out at math experts who just go through assuming > the existence of a non-modular elliptic curve who may then wonder why > anyone ever believed he had a proof in the [CapitalThorn]rst place! You couldn't [CapitalThorn]nd the smoking gun by looking for his use of the fallacious Cum Hoc, Ergo Propter Hoc form of argument, couldn't [CapitalThorn]nd any reference to the magical 4 parameters that you were so sure would doom his arguments, and got into deep poop when you attempted to penetrate the text. Since you must always be correct, you have seized on this current idee [CapitalThorn]xee of yours, the null test, and since you can't do your own grunt work, you rant and rave that someone *else* must do it for you. > But will they tell you? Try to answer this question without an appeal to the assumption of base motives. Not everyone has to be dragged kicking and screaming to a point of making a retraction. The [CapitalThorn]rst paper apparently *did* have an error, and Wiles wasted no time in accepting that realization and eliminating the error. The [CapitalThorn]rst reports of a proof of FLT were followed *quickly* by reports of a hole in the logic, and that was in the popular press. What rational basis is there to assume that the discovery of a subsequent error would have been hidden from the public? > James Harris Dale. === Subject: Re: JSH: Math proofs > > > > > >Seems to me that we're not really avoiding assumptions in natural >deduction so much as moving to a sequent-style calculus. I'm not sure >that this does anything for us. > >Well, yes, sorry. > >A ->formal<- proof, de[CapitalThorn]ned as a [CapitalThorn]nite sequence etc., will >necessarily begin with either a tautology or an axiom. > >I don't understand this claim either. > >There are formal systems of natural deduction. In these formal >systems, a proof my begin with an assumption. > >I suspect that when you learned logic, you just didn't do natural >deduction. If one wants to study logic, then natural deduction is an >annoying formalization to use --- all those damn subproofs with their >assumptions make for special cases in proving things. But natural >deduction is a name for a class of related *formal* logics in which >assumptions and assumption-discharging rules play a role. > >The sequent calculi are much handier for the logician and this could >be what you learned. In that case, the transformation > > T,X |- Y > -------- > T |- X -> Y > >(perhaps written T,X => Y |- T => X -> Y ) is an explicit rule of >inference. > > >[...] > > >The proof at issue originally here, the proofs contained in Wiles's >paper, are not formal proofs. And most of them do not begin with a >true statement. Theorems 0.2 and 0.3, for example, begin with >assumptions, since they are proofs of A->B via the usual method method >of assuming A and deducing B, without ever invoking the Deduction >Metatheorem (which is unnecessary outside of formal logic, for the >most part). The proof of Theorem 5.2, that all semistable elliptic >curves over Q are modular, begins with Suppose that E is a semistable >elliptic curve over Q, which is certainly not a true statement. > >Indeed. > > > > Some poster claimed that this Hughes guy had all these posts with > proofs in them so I'm going back surveying what he's posted, and > noticed the statement above. > > Which statement? That the proof of Theorem 5.2 begins with Suppose that > E is a semistable elliptic curve over Q? The cryptic remark Indeed? Actually I'm talking about his statement that the proofs do not begin with a true statement. My thought at the time was the he's questioning my statements about math proofs, to whit, a math proof begins with a truth and proceeds by logical steps to a conclusion which then must be true. > It's quite true that Wiles continually assumes modularity, and it's > kind of weird that no one thought to null test his work before because > his very paper just BEGS for it. > > It's quite true that Wiles assumes (in several of the theorems of > the paper under discussion) that a particular Galois representation > is modular. It's just as true that you are attempting to claim to > understand something that you don't. Wiles does *not* assume that > a particular elliptic curve is modular. On the other hand, there are > modular curves that are mentioned, for instance the curves X_0(N). > He does not need to assume they are modular, since by their construction > they *are* modular. The null test is simple: assume the existence of a non-modular elliptic curve and [CapitalThorn]nd some point in Wiles's work where that assumption leads to a contradiction. Math proofs begin with a truth and proceed by logical steps to a conclusion which then must be true. What I call a null test is to assume the opposite of the conclusion of an argument and see if that assumption leads to a contradiction with some logical step in that argument. If it does not, then the argument is not a proof. Proofs by contradiction are actually null tests. All math proofs necessarily logically connect truths. James Harris === Subject: Re: JSH: Math proofs <87vfeze6cf.fsf@phiwumbda.org> <871xhmeßs.fsf@phiwumbda.org> <59U0d.14646$QJ3.263@newssvr21.news.prodigy.com> Discussion, linux) > One thing that I haven't heard discussed is that my null test of > assuming the existence of a non-modular elliptic curve might eliminate > over 90% of Wiles's paper for consideration, or maybe more. > Golly, why wouldn't you have heard that? > Because it's silly? Why over 90%? Why not claim it would *Two* line? Ha! It reduces it to a *single* line: the keystone step. Don't underestimate the power of the NULL! -- Jesse Hughes She moaned, in pain and pleasure, as, in a confused whirlwind, she glimpsed an image of Saint Sebastian riddled with arrows, cruci[CapitalThorn]ed and impaled. --Mario Vargas Llosa on category theory === Subject: Re: 3D Geometry vector problem > However, now I've got it working, I can't get it to fail! (as in alex's > comment: it's easy to choose setups whereby there is no answer). > Which isn't a problem, but given that there are solutions around the > cone, how can there fail to be a solution where V2.y = V1.y? > Is there a nasty bug waiting under certain conditions? Don't worry about my thing about Ôit can be made to fail', I think I was misunderstanding the problem. I understand the problem you're trying to solve now. Your idea - that as far as the point is concerned, the sphere's outline will look like a circle - is correct, and this is key to an Ôintuitive' So, let P be the points position, and S be the centre of the sphere. The sphere has radius R. V1 is the vector from P to S: V1 = S - P Now, as Silvain noted, if you have a vector that is at a right angle to V1, call it Vr, it's a simple case to work out V2 as the sum of V1 and V2 multipled by the sphere radius: V2 = V1 + R * Vr [1] So the only remaing problem is [CapitalThorn]nding a vector Vr that is at a right angle to V1. You can choose your own as you see [CapitalThorn]t, but if you want an automatic way to make Vr, read on... A guaranteed way to generate a valid value of Vr is to take the cross product of V1 and any another vector, L, such that V1 and L are *not* colinear (i.e. do not represent the same direction). How to generate L based on V1? One way (which I have just made up but it looks ok to me) is to de[CapitalThorn]ne it as: if V1 = (Vx, Vy, Vz), then L = (Vy, -Vz, -Vx) (I made this by doing a 90 degree X-Y rotation on V1, followed by a 90 degree Y-Z rotation, which will never produce the same vector as you started with, unless it is (0, 0, 0), which is not under consideration in this problem - i.e. I assume V1 will never be (0, 0, 0), as it makes the problem have no solution anyway.) So, the equation to de[CapitalThorn]ne Vr (with a cross product) is: Vr = (Vx, Vy, Vz) X (Vy, -Vz, -Vx) = (-Vy*Vx + Vz^2, Vx^2 + Vz*Vy, - Vx*Vz - Vy^2) So, plug this value of Vr into equation [1] noted earlier, and bobs yer uncle. Or your aunty, after too many whiskies, wearing a whig and putting on a man's voice. alex === Subject: Re: 3D Geometry vector problem > Now, as Silvain noted, if you have a vector that is at a right angle to > V1, call it Vr, it's a simple case to work out V2 as the sum of V1 and > V2 multipled by the sphere radius: > V2 = V1 + R * Vr [1] A mistake: this only works if Vr is normalised (has modulus of 1). So that should have been written: V2 = V1 + R * norm(Vr) alex === Subject: Re: Null test, Wiles's work > If you wish to push the point, give an algebraic proof, and I'll point > out to you where your claim will draw a contradiction *before* the > conclusion. > I [CapitalThorn]nd the following proof easier to present, so would you please > apply the test to it? (I've presented this in another post, but I > repeat it here since you offered.) Hell yeah!!! It's quite fun to do null tests!!! I did mention that a while back, and for me it's also relaxing. > ,---- > | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. Ok, going to assume that the square root of 2 is rational. > | > | Proof: Assume that sqrt(2) is rational. Let x and y be given such > | that > | > | (1) sqrt(2) = x / y > | (2) gcd(x,y) = 1 > | > | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 > | divides x, so let z be given such that x = 2 z. Not stated but clearly assumed is that x and y are integers. Maybe the poster feels that use of gcd means that x and y are integers. Here's where sloppiness can be painful, as this poster at the end claims to not see a contradiction with assuming that the square root of 2 is irrational, but notice he doesn't declare a ring for x and y. But look further down and you see him trying to use x and y being integers with something about divisibility by 2, so he's sloppy. Ok, back to the null test. Notice that x is now shown to be even. > | > | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. And now y is shown to be even, which contradicts with 2. Notice that a proof by contradiction is just a null test. Not surprisingly, here you can do a proof by contradiction by assuming that sqrt(2) is rational, which actually is usually how it's taught! So it's kind of funny that the poster chose this example. I have posted on this one before, so yes, posters also were wrong in claiming that I haven't shown null tests before. > | Therefore y^2 is divisible by 2 and hence y is also divisible by 2. > | > | But we have thus proven that both x and y are divisible by 2, > | contradicting (2). > `---- > Now let's see how the null test works. It's like this, right? > We assume (contrary to the above theorem) that the square root of 2 is > rational and then we see which step in the proof is invalid, right? > Okay, so I'm assuming... Golly. Every step still seems to follow from > previous steps or previously proven theorems. Uh oh. But you see, that's not logically possible, and in fact, what makes your post kind of funny is the fact that the *classical* proof that sqrt(2) is irrational is a proof by contradiction--a null test--which works by assuming that sqrt(2) is rational!!! What is your math training? > Help me out here. You might need to go to school and take some math classes...um, hey that's what a lot of people have told me!!! Who knows, maybe that may be the only way to settle this crap. It's not like it'd be that hard for me to go back and get a math degree. I can penetrate the math social group and then [CapitalThorn]nish the takedown from inside. James Harris === Subject: Re: Null test, Wiles's work >[...] >Who knows, maybe that may be the only way to settle this crap. It's >not like it'd be that hard for me to go back and get a math degree. >I can penetrate the math social group and then [CapitalThorn]nish the takedown >from inside. you should do that. probably you need to enroll under an assumed name, of course, since the powers that be have already decided that james harris shall not be admitted to the math social group. >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Null test, Wiles's work <87k6vjvce7.fsf@phiwumbda.org> Discussion, linux) > If you wish to push the point, give an algebraic proof, and I'll point > out to you where your claim will draw a contradiction *before* the > conclusion. > I [CapitalThorn]nd the following proof easier to present, so would you please > apply the test to it? (I've presented this in another post, but I > repeat it here since you offered.) > Hell yeah!!! It's quite fun to do null tests!!! > I did mention that a while back, and for me it's also relaxing. > ,---- > | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. > Ok, going to assume that the square root of 2 is rational. > | > | Proof: Assume that sqrt(2) is rational. Let x and y be given such > | that > | > | (1) sqrt(2) = x / y > | (2) gcd(x,y) = 1 > | > | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 > | divides x, so let z be given such that x = 2 z. > Not stated but clearly assumed is that x and y are integers. > Maybe the poster feels that use of gcd means that x and y are > integers. > Here's where sloppiness can be painful, as this poster at the end > claims to not see a contradiction with assuming that the square root > of 2 is irrational, but notice he doesn't declare a ring for x and y. > But look further down and you see him trying to use x and y being > integers with something about divisibility by 2, so he's sloppy. > Ok, back to the null test. Notice that x is now shown to be even. > | > | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. > And now y is shown to be even, which contradicts with 2. > Notice that a proof by contradiction is just a null test. Notice that this is just bull. A null test, according to James Harris: Take a proof, assume the negation of the conclusion, look for [CapitalThorn]rst line which contradicts assumption. (1) There was no existing proof to which you applied the null test. (2) The assumption sqrt(2) is rational does not contradict any of the lines in the proof, save the conclusion. In fact *every line* from the proof follows *from* the assumption sqrt(2) is rational, save the conclusion. What is your de[CapitalThorn]nition of contradict? How does the assumption that the square root of two is rational *contradict* that y is even? > Okay, so I'm assuming... Golly. Every step still seems to follow from > previous steps or previously proven theorems. Uh oh. > But you see, that's not logically possible, and in fact, what makes > your post kind of funny is the fact that the *classical* proof that > sqrt(2) is irrational is a proof by contradiction--a null test--which > works by assuming that sqrt(2) is rational!!! Every step *was* indeed a valid step. There was no invalid step. Each step follows from either the assumption, previous theorems or rule of inference applied to prior steps. At the time of this post, you were going on about how an assumption of the negation would make some step *invalid*. Now you say it would *contradict* some step. The two explanations are not synonymous. > You might need to go to school and take some math classes...um, hey > that's what a lot of people have told me!!! > Who knows, maybe that may be the only way to settle this crap. It's > not like it'd be that hard for me to go back and get a math degree. > I can penetrate the math social group and then [CapitalThorn]nish the takedown > from inside. Yeah, do that. Finish the takedown from the inside, would you? -- Jesse F. Hughes Wiles made somewhere around half a million dollars U.S. that I heard about, and I know he didn't take major endorsements. --JSH on the rewards of proving Fermat's last theorem. === Subject: Re: Null test, Wiles's work > > If you wish to push the point, give an algebraic proof, and I'll point > out to you where your claim will draw a contradiction *before* the > conclusion. > > I [CapitalThorn]nd the following proof easier to present, so would you please > apply the test to it? (I've presented this in another post, but I > repeat it here since you offered.) > Hell yeah!!! It's quite fun to do null tests!!! > I did mention that a while back, and for me it's also relaxing. > > ,---- > | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. > Ok, going to assume that the square root of 2 is rational. > | > | Proof: Assume that sqrt(2) is rational. Let x and y be given such > | that > | > | (1) sqrt(2) = x / y > | (2) gcd(x,y) = 1 > | > | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 > | divides x, so let z be given such that x = 2 z. > Not stated but clearly assumed is that x and y are integers. > Maybe the poster feels that use of gcd means that x and y are > integers. > Here's where sloppiness can be painful, as this poster at the end > claims to not see a contradiction with assuming that the square root > of 2 is irrational, but notice he doesn't declare a ring for x and y. > But look further down and you see him trying to use x and y being > integers with something about divisibility by 2, so he's sloppy. > Ok, back to the null test. Notice that x is now shown to be even. > | > | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. > And now y is shown to be even, which contradicts with 2. > Notice that a proof by contradiction is just a null test. > Notice that this is just bull. Nope. REPEATEDLY I've said that a null test of a proof is to assume the opposite of its conclusion and [CapitalThorn]nd a logical step in the proof where you get a contradiction. Mathematicians might have immediately recognized that to be a proof by contradiction!!! Now then, if you never did [CapitalThorn]gure that out until now, despite my coming back over a week ago in a post to note that proofs by contradiction are null tests, then it's not reason to curse mathematics itself! So you didn't get it. Get over it. For those who still don't quite get it, my assertion is that any proof that's not a proof by contradiction can be turned into one by assuming the opposite of the conclusion, and tracing through to a logical step where that gives a contradiction! That is, I'm saying you can get a proof by contradiction from a proof that's not one. Basic. > A null test, according to James Harris: Take a proof, assume the > negation of the conclusion, look for [CapitalThorn]rst line which contradicts > assumption. > (1) There was no existing proof to which you applied the null test. > (2) The assumption sqrt(2) is rational does not contradict any of the > lines in the proof, save the conclusion. > In fact *every line* from the proof follows *from* the assumption > sqrt(2) is rational, save the conclusion. > What is your de[CapitalThorn]nition of contradict? How does the assumption that > the square root of two is rational *contradict* that y is even? Now it turns out that this Jesse F. Hughes clearly did NOT get it, and now shows why he didn't. Notice in his statement he seems to think that a proof IS the conclusion. But the conclusion *FOLLOWS* from the logical steps that precede it. What's worse here is that he doesn't seem to have taken many math classes as the the classical proof that sqrt(2) is irrational is to assume that it is rational, which is a proof by contradiction. That is, the classical proof that square root of 2 is irrational is a null test. > Okay, so I'm assuming... Golly. Every step still seems to follow from > previous steps or previously proven theorems. Uh oh. > But you see, that's not logically possible, and in fact, what makes > your post kind of funny is the fact that the *classical* proof that > sqrt(2) is irrational is a proof by contradiction--a null test--which > works by assuming that sqrt(2) is rational!!! > Every step *was* indeed a valid step. There was no invalid step. > Each step follows from either the assumption, previous theorems or > rule of inference applied to prior steps. A math proof begins with a truth and proceeds by logical steps to a conclusion which then MUST BE TRUE. Necessarily every step in a math proof is a valid one. Some people seem to get confused on this issue because they hear proof of this or proof of that when there are CLAIMS of proof. But if someone says they have proof that you robbed a bank, and then it turns out that they're wrong, and you didn't rob the bank, did they have proof? No. Then you'd say they never had proof, but were just wrong, but in mathematics people routinely will instead say something like a ßawed proof. Can you have ßawed proof that you robbed a bank, when the supposed proof is just bogus? Nope. It's just a place where the math world has not been appropriately diligent in enforcing logical usage. The criminal justice system has, so why haven't math people? My guess is laziness. > At the time of this post, you were going on about how an assumption > of the negation would make some step *invalid*. Now you say it would > *contradict* some step. The two explanations are not synonymous. Now you're back to lying. Go back, look at the record, and pay attention to what I've said from the beginning. That should be easy for you as you seem to like to quote me, so no excuses!!! Go back to what I said and pay attention this time. And think a lot about logic functionally versus as some abstraction you can get a hold of from reading a book! Logic is quite powerful when you understand it, and it's not just some social trophy to put out to say, hey I know logic! It's for real. James Harris === Subject: Re: Null test, Wiles's work <87k6vjvce7.fsf@phiwumbda.org> <87oekcm1fb.fsf@phiwumbda.org> Discussion, linux) > > If you wish to push the point, give an algebraic proof, and I'll point > out to you where your claim will draw a contradiction *before* the > conclusion. > > I [CapitalThorn]nd the following proof easier to present, so would you please > apply the test to it? (I've presented this in another post, but I > repeat it here since you offered.) > > Hell yeah!!! It's quite fun to do null tests!!! > > I did mention that a while back, and for me it's also relaxing. > > ,---- > | Theorem: The square root of 2 (hereafter sqrt(2)) is irrational. > > Ok, going to assume that the square root of 2 is rational. > > | > | Proof: Assume that sqrt(2) is rational. Let x and y be given such > | that > | > | (1) sqrt(2) = x / y > | (2) gcd(x,y) = 1 > | > | Then 2 = x^2 / y^2 so x^2 = 2 y^2. Now, if 2 divides x^2, then 2 > | divides x, so let z be given such that x = 2 z. > > Not stated but clearly assumed is that x and y are integers. > > Maybe the poster feels that use of gcd means that x and y are > integers. > > Here's where sloppiness can be painful, as this poster at the end > claims to not see a contradiction with assuming that the square root > of 2 is irrational, but notice he doesn't declare a ring for x and y. > > But look further down and you see him trying to use x and y being > integers with something about divisibility by 2, so he's sloppy. > > Ok, back to the null test. Notice that x is now shown to be even. > > | > | Then (2 z)^2 = 2 y^2, so 4 z^2 = 2 y^2 and hence 2 z^2 = y^2. > > And now y is shown to be even, which contradicts with 2. > > Notice that a proof by contradiction is just a null test. > Notice that this is just bull. > Nope. REPEATEDLY I've said that a null test of a proof is to assume > the opposite of its conclusion and [CapitalThorn]nd a logical step in the proof > where you get a contradiction. Yes. I agree. You have said that. And proof by contradiction is nothing like that. Your hallowed null test *begins with a proof* and applies some method to it. Proof by contradiction doesn't start with an extant proof. > Mathematicians might have immediately recognized that to be a proof by > contradiction!!! > Now then, if you never did [CapitalThorn]gure that out until now, despite my > coming back over a week ago in a post to note that proofs by > contradiction are null tests, then it's not reason to curse > mathematics itself! Er, I'll keep that in mind. I'll try to refrain from cursing mathematics itself, but it will be tough. Kind of a habit, you know, but I'll give it my best. > A null test, according to James Harris: Take a proof, assume the > negation of the conclusion, look for [CapitalThorn]rst line which contradicts > assumption. > (1) There was no existing proof to which you applied the null test. > (2) The assumption sqrt(2) is rational does not contradict any of the > lines in the proof, save the conclusion. > In fact *every line* from the proof follows *from* the assumption > sqrt(2) is rational, save the conclusion. > What is your de[CapitalThorn]nition of contradict? How does the assumption that > the square root of two is rational *contradict* that y is even? > Now it turns out that this Jesse F. Hughes clearly did NOT get it, and > now shows why he didn't. > Notice in his statement he seems to think that a proof IS the > conclusion. No, tain't so. Nothing I said above suggests that I confuse a proof with its conclusion. They taught me better than that at DeVry Business School and Logic Academy. > Every step *was* indeed a valid step. There was no invalid step. > Each step follows from either the assumption, previous theorems or > rule of inference applied to prior steps. > A math proof begins with a truth and proceeds by logical steps to a > conclusion which then MUST BE TRUE. James, the proof that square root of two is irrational does not begin with a truth (which I assume means an axiom). It starts with *an assumption*. Didn't you notice? [snip silliness about bank robberies] Damn you, Mother Mathematics! Damn you to hell! Oh. Sorry. Slipped out. > At the time of this post, you were going on about how an assumption > of the negation would make some step *invalid*. Now you say it would > *contradict* some step. The two explanations are not synonymous. > Now you're back to lying. > Go back, look at the record, and pay attention to what I've said from > the beginning. > That should be easy for you as you seem to like to quote me, so no > excuses!!! Here it is (emphasis added): | If I'm right, you can trace through Wiles's entire work--dropping | the assumption he puts forward so much himself about a certain | something being modular--and see that you don't *invalidate* the | conclusion, which should be a bit weird given some of the the other | works he cites. `---- What does invalidate mean? What would it mean to invalidate the conclusion? Mother Mathematics is a harlot! Damn. Sorry. Happened again. Nasty habit, that. -- Jesse F. Hughes Leaving things always seems to [CapitalThorn]x me, Running seems to ease my worried mind. -- Bad Livers, Honey, I've Found a Brand New Way === Subject: Re: Positive and Negative numbers > > Both proofs may be presented more clearly as follows. > > Recall inverses are unique: if b has two inverses a,A > > then a = a + (b + A) = (a + b) + A = A. Using this: > > (1) a and -(-a) are inverses for -a so are equal > > (2) -a + -b and -(a+b) are inverses for a+b so are equal. > > This is a special case of the Law of Signs, which > has an analogous proof -- see my many prior posts: > > You said that these are special cases of ab = (-a)(-b). > I don't agree with this. I'll get to this in detail later below. > I mean (2) is viewable as a special case of the Law of Signs (in Rings). > I don't see how. The Law of Signs: -x -y = x y when specialized for x = -1 is -y = -1 y [-] for y = a+b is -(a + b) = -1 (a + b) = -1 a + -1 b via Distributive Law = -a + -b via [-], y = a, b Hence, indeed, (2) is a special case of the Law of Signs, via special cases of Distributive Law and (1): -(-1) = 1. Anyway that was a digression from the main point of my post, which was to show how Uniqueness Theorems provide a uni[CapitalThorn]ed framework for the results under discussion. I am happy to see from your reply below that I did have better success sharing this much more signi[CapitalThorn]cant mathematical morsel. > Hopefully this message has helped convince you of the power > and utility of various uniqueness theorems in mathematics. > Yes, I'm convinced. Splendid. Signing off, --Bill Dubuque -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Re: Positive and Negative numbers > The Law of Signs: -x -y = x y when specialized > for x = -1 is -y = -1 y [-] > > for y = a+b is -(a + b) = -1 (a + b) > = -1 a + -1 b via Distributive Law > = -a + -b via [-], y = a, b > Hence, indeed, (2) is a special case of the Law of Signs, > via special cases of Distributive Law and (1): -(-1) = 1. Of course. The relevant substitutions were right there, but I still didn't see them. (I'm now doing what they do in that V8 commercial.) > Anyway that was a digression from the main point of my post, > which was to show how Uniqueness Theorems provide a uni[CapitalThorn]ed > framework for the results under discussion. I am happy to > see from your reply below that I did have better success > sharing this much more signi[CapitalThorn]cant mathematical morsel. > Hopefully this message has helped convince you of the power > and utility of various uniqueness theorems in mathematics. > > Yes, I'm convinced. > Splendid. Signing off, amplifying them. I'll be using them. Paul -- submissions: post to k12.ed.math or e-mail to k12math@k12groups.org private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Subject: Gamma function - Graphical Analysis In order to test a (precise and very fast) library of transcendental functions, together with a graphical interface for real-time analysis, selected functions are being made available to the public for critique. Currently you may enjoy the marvels of the: Gamma Function - Gamma(x) The computer program runs in MS DOS or Windows on standard x86 machines with an FPU and SVGA (VESA standard). Detailed speci[CapitalThorn]cations and GUI description are available at: www.iging.com/Transcendental. Your constructive feedback is very much appreciated. Dan Baruth x86igingcom === Subject: Fractan by John Conway Hi all, I'm searching for a paper by John Conway called: Fractan : A Logical Computing Language As far as I know it is only in two places, the book Open Problems in Communication and the Eureka magazine, the cambridge students magazine afaik. Well, I don't have access to any of these however, I'd like to read the paper. Does anybody know if it's anywhere online available to download? -- Paulo J. Matos : pocm [_at_] mega . ist . utl . pt Instituto Superior Tecnico - Lisbon Computer and Software Eng. - A.I. - > http://mega.ist.utl.pt/~pocm --- -> God had a deadline... === Subject: Re: Fractan by John Conway >I'm searching for a paper by John Conway called: >Fractan : A Logical Computing Language >As far as I know it is only in two places, the book Open >Problems in Communication and the Eureka magazine, the cambridge > students magazine afaik. >Well, I don't have access to any of these however, I'd like to >read the paper. Does anybody know if it's anywhere online >available to download? The language is called FRACTRAN and there is a link and references on Mathworld at . John Conway gave a seminar on FRACTRAN at Princeton while I was a grad student there. Generally, it takes a lot of code to write even simple programs, and even the short programs usually take a long time to run, but it is neat to have a list of rational numbers as code. A Simple Universal Programming Language for Arithmetic. See However, the book in which it appears, Open Problems in Communication and Computation is out of print. You can try to order it used from amazon.com: Good luck! Rob Johnson take out the trash before replying === Subject: Sources for FRACTRAN (was: Fractan by John Conway) > I'm searching for a paper by John Conway called: > Fractan : A Logical Computing Language Watch the spelling! http://mathworld.wolfram.com/FRACTRAN.html Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Sources for FRACTRAN >I'm searching for a paper by John Conway called: >Fractan : A Logical Computing Language > Watch the spelling! Which spelling? http://www.cs.ualberta.ca/events/dls/2000-2001/Conway.php > http://mathworld.wolfram.com/FRACTRAN.html I couldn't [CapitalThorn]nd a link to the paper on that page. > Rainer Rosenthal > r.rosenthal@web.de Thx anyway. -- Paulo J. Matos : pocm [_at_] mega . ist . utl . pt Instituto Superior Tecnico - Lisbon Computer and Software Eng. - A.I. - > http://mega.ist.utl.pt/~pocm --- -> God had a deadline... === Subject: Re: Sources for FRACTRAN Paulo Jorge O. C. Matos > Which spelling? > http://www.cs.ualberta.ca/events/dls/2000-2001/Conway.php Searching this site gives: fractan = 0 hits fractran = 2 hits Rainer Rosenthal r.rosenthal@web.de === Subject: Re: Question about [CapitalThorn]eld > If F={0,1,a,b} is a [CapitalThorn]eld of order 4, then > How can I determine a+b , a*b ? > I know a+b=/=a because b=/=0 > also, a+b=/=b since a=/=0. > But I can not determine wheather a+b = 0 or 1. > I know a*b=/=a, a*b=/=b because b=/=1, a=/=1. > But I can not judge wheather a*b = 0 or 1 > Would you please help me ? For one thing, the non-zero elements of a [CapitalThorn]eld must form a multiplicative group. and in this case, a group with 3 elements. That should straighten out your multiplication tables with a little thught. Then use the distributive property to straighen out your addition. === Subject: Re: Question about [CapitalThorn]eld >If F={0,1,a,b} is a [CapitalThorn]eld of order 4, then >How can I determine a+b , a*b ? >I know a+b=/=a because b=/=0 >also, a+b=/=b since a=/=0. >But I can not determine wheather a+b = 0 or 1. Since a+a = 0, then a+b cannot also equal 0. So... (Or you could construct the [CapitalThorn]eld: it is isomorphic to F_2[x]/(x^2+x+1), where F_2 is the [CapitalThorn]eld of 2 elements. >I know a*b=/=a, a*b=/=b because b=/=1, a=/=1. >But I can not judge wheather a*b = 0 or 1 If it is a [CapitalThorn]eld, then surely a*b=0 would imply a=0 or b=0. So... -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: ßoor(xn)/n > Hi Ruben, > Ruben schrieb im Newsbeitrag > Somewhere, I came across a post where someone was trying to prove that > the sequence ßoor(xn)/n is strictly increasing if x is a non-integer. > I can't [CapitalThorn]nd the post now, but someone then replied - it's going to be > hard to prove because it's not true! And it's not - consider x=1/2 > No I'm modest--the sequence de[CapitalThorn]ned by a_n:=ßoor(nx)/n needn't be > strictly monotonous increasing. > But - I wanted to ask - does such an x exist such that ßoor(xn)/n is > a strictly increasing sequence (ie for n=1,2,3...)? > Two things I noticed - x would have to be irrational since if it is > p/q then n=q+1 will be a smaller term than n=q. Secondly x > 1 since > ßoor(xn) will have to be strictly increasing. > Take x=sqrt(2). Then a_1=1, a_2=1, a_3=4/3, a_4=5/4; thus a_3>a_4. Even the > values of ßoor(nx), n=1,2,3 are not strictly monotonous increasing. > I'm afraid that strict monotony can't be achieved since the ßoor function > has the property that for any r in [a,a+1) (i.e the interval [a,a+1] without > a+1) ßoor(r)=ßoor(a). > Thomas > Now, can anyone prove or disprove that such an x exists? > Ruben What the hell are you talking about? of course ßoor(n * sqrt(2)) is strictly increasing. Show me a counterexample to this. But ßoor(n * sqrt(2))/n is not - i agree with this. But my question was DOES THERE EXIST an x with this property - you've eliminated root two - so that still leaves - every other number. Ruben === Subject: Re: ßoor(xn)/n > But my question was DOES THERE EXIST an x with this property - you've > eliminated root two - so that still leaves - every other number. The answer is no. De[CapitalThorn]ne frac(x) = x - ßoor(x). Note that ßoor(nx)/n = (nx - frac(nx))/n = x - frac(nx)/n By Kronecker's Theorem (http://mathworld.wolfram.com/ KroneckersApproximationTheorem.html) we can choose arbitrarily large n such that frac(nx) < 1 - frac(x). We would then have frac((n+1)x) = frac(nx) + frac(x). The difference between successive terms in the sequence would be: ßoor((n+1)x)/(n+1) - ßoor(nx)/n = frac(nx)/n - frac((n+1)x)/(n+1) = ((n+1)frac(nx) - nfrac((n+1)x))/(n(n+1)) = (nfrac(nx) + frac(nx) - n(frac(nx) + frac(x)))/(n(n+1)) = (frac(nx) - nfrac(x))/(n(n+1)) If frac(x) >= 1/2, choose n such that frac(nx) < 1 - frac(x) and n >= 2. We have frac(nx) < 1 and nfrac(x) >= n/2 >= 1, so frac(nx) - nfrac(x) is negative. If 0 <= frac(x) < 1/2, choose n such that frac(xn) < 1 - frac(x) and n > 1/frac(x). We have frac(nx) < 1 and nfrac(x) > 1, so frac(nx) - nfrac(x) is negative again. In either case, ßoor((n+1)x)/(n+1) - ßoor(nx)/n < 0. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: ßoor(xn)/n > But my question was DOES THERE EXIST an x with this property - you've > eliminated root two - so that still leaves - every other number. > The answer is no. > De[CapitalThorn]ne frac(x) = x - ßoor(x). > Note that ßoor(nx)/n = (nx - frac(nx))/n = x - frac(nx)/n > By Kronecker's Theorem > (http://mathworld.wolfram.com/ KroneckersApproximationTheorem.html) we > can choose arbitrarily large n such that frac(nx) < 1 - frac(x). We > would then have frac((n+1)x) = frac(nx) + frac(x). The difference > between successive terms in the sequence would be: > ßoor((n+1)x)/(n+1) - ßoor(nx)/n > = frac(nx)/n - frac((n+1)x)/(n+1) > = ((n+1)frac(nx) - nfrac((n+1)x))/(n(n+1)) > = (nfrac(nx) + frac(nx) - n(frac(nx) + frac(x)))/(n(n+1)) > = (frac(nx) - nfrac(x))/(n(n+1)) > If frac(x) >= 1/2, choose n such that frac(nx) < 1 - frac(x) and n >= 2. > We have frac(nx) < 1 and nfrac(x) >= n/2 >= 1, so frac(nx) - nfrac(x) is > negative. > If 0 <= frac(x) < 1/2, choose n such that frac(xn) < 1 - frac(x) and n > > 1/frac(x). > We have frac(nx) < 1 and nfrac(x) > 1, so frac(nx) - nfrac(x) is > negative again. > In either case, ßoor((n+1)x)/(n+1) - ßoor(nx)/n < 0. theorem does prove it. It's interesting that it turned out not to be a trivial question - it pretty much relies on Kronecker's theorem - which, I think may not be straight forward to prove (it has to do with rational aproximations which I don't know a lot about). Although I'd be happy for you to prove me wrong on this point if you can show that {nx} is dense in the unit interval easily. It's one of those things which seems like it pretty much has to be right - although I don't know enough to prove it, I don't think. Ruben === Subject: Re: ßoor(xn)/n > theorem does prove it. And if x is a non-integral rational, you can simply use its denominator as n. > It's interesting that it turned out not to be a trivial question - it > pretty much relies on Kronecker's theorem - which, I think may not > be straight forward to prove (it has to do with rational aproximations > which I don't know a lot about). Although I'd be happy for you to > prove me wrong on this point if you can show that {nx} is dense in the > unit interval easily. It's one of those things which seems like it > pretty much has to be right - although I don't know enough to prove > it, I don't think. For the bene[CapitalThorn]t of anyone reading this who hasn't checked the Mathworld site, Kronecker's Theorem states that for a given irrational number x, for any fraction f in the unit interval and any positive epsilon, we can [CapitalThorn]nd a positive integer k such that abs(frac(kx)-f) < epsilon. There is actually a straightforward proof using the pigeonhole principle. What follows is not completely formal: We need to [CapitalThorn]nd a k such that f-epsilon < frac(kx) < f+epsilon . Let us choose positive integers p and q such that f-epsilon < p/q < (p+1)/q < f+epsilon. If nothing else, we can choose q > 1/epsilon and then set p = ßoor(q*(f-epsilon)) + 1. Now consider the q equal pieces of the unit interval and the q fractions frac(x), frac(2x), ..., frac(qx). If one of those fractions lands in the p+1'th piece, we are done. Otherwise, the pigeonhole principle puts two of the fractions in the same piece. If the two fractions are frac(ax) and frac(bx) with b>a, consider d = b-a. If frac(bx) > frac(ax), the difference gives us 0 < frac(dx) < 1/q. If frac(bx) < frac(ax), the difference gives us 1 - 1/q < frac(dx) < 1. (If frac(bx) = frac(ax), someone slipped us a rational x with d as the denominator.) In the case frac(dx) < 1/q, choose an integer m such that m > 1/frac(dx). Then consider the sequence frac(dx), 2frac(dx), ..., mfrac(dx). It will have representatives in each of the q pieces of the unit interval, since each step either stays in the same piece or goes to the next one (and it eventually goes out the far end of the interval). One of those, call it n, is in the targeted piece. Then nd is the desired k. In the case frac(dx) > 1 - 1/q, choose m such that m > 1/(1-frac(dx)). The rest of the case is much the same. -- Daniel W. Johnson panoptes@iquest.net http://members.iquest.net/~panoptes/ 039 53 36 N / 086 11 55 W === Subject: Re: ßoor(xn)/n > But my question was DOES THERE EXIST an x with this property - you've > eliminated root two - so that still leaves - every other number. > The answer is no. > De[CapitalThorn]ne frac(x) = x - ßoor(x). > Note that ßoor(nx)/n = (nx - frac(nx))/n = x - frac(nx)/n > By Kronecker's Theorem > (http://mathworld.wolfram.com/ KroneckersApproximationTheorem.html) we > can choose arbitrarily large n such that frac(nx) < 1 - frac(x). We > would then have frac((n+1)x) = frac(nx) + frac(x). Alternatively, we can use Hurwitz's Irrational Number Theorem that for any irrational x, there exist (in[CapitalThorn]nitely many) rationals p/q ((p,q)=1) s.t. |x - p/q| < 1/{sqrt(5) q^2) http://mathworld.wolfram.com/ HurwitzsIrrationalNumberTheorem.html Then Ruben's argument for the rational case of x can be applied in a straightforward manner. # Note that when x is rational, then we have x-p/q = 0 for # some p, q. p/q - 1/{sqrt(5)q^2} < x < p/q + 1/{sqrt(5) q^2} Then take q, q+1 for n. I believe the rest can be worked out. Actually, we can also start from the original, more coarse evaluation by Dirichlet: |x - p/q| < 1/q^2 But then we have to take care that (p,q) ful[CapitalThorn]lls: p = aq + r where 0 < r < q-1 < q. - Yuzuru Hiraga === Subject: Re: JSH: Simplify, right? >I learned a while back that math people easily lie about mathematics, >and can get away with it. >Still I had a choice, give up or come back at them. >I decided not to give up, so I simplify. >I give simpler relations, and talk about subjects easier to understand >like counting prime numbers. >Sure, posters come back and say the same things, but it's about time. >Some people pay attention and learn that math people lie a lot. Whom? >Some math people pay attention and leave the [CapitalThorn]eld. Whom? >Over time the impact builds as people lose faith. C'mon, James - you keep appealing to these unseen Ôpeople' - everybody who follows sci.math can read y'know - who are these Ôpeople'? Do they communicate by e-mail? Post their responses (sans details of course). -- Min I blame the jelly === Subject: Re: Statics Word Problems > A horizontal bar of negliglible weight hangs from two cables, one at > each end. When a 624-N force is applied vertically downward from a > point 185 cm from the left end of the bar, the right cable is seen to > have a tension of 341 N. [CapitalThorn]nd the length of the bar. If your textbook gives a compressed treatment of the subject, [CapitalThorn]nd an introductory engineering book on statics and read the general statements of the basic principles (do not just look at examples). The solution the problem aboves goes like this (view with [CapitalThorn]xed font to see the ascii diagrams): ^ ^ | | |P |Q | L | ------------------------ ------------------------ d | |F v d = 1.85 meters L = unknown length of the bar Q = 341 N (newtons) cable force F = 624 N applied force Equilibrium of moments about the left endpoint of the bar: Q L - F d = 0 from which the length of the bar is L = F d/Q = 624*1.85/341 = 3 meters > A uniform horizontal beam is 19.80 ft long and weighs 1360 lb. It is > supported at either end. A vertical load of 13,510 lb is applied to > the beam 8.450 ft from the left end. Find the reaction at each end of > the beam ^ ^ | | |P |Q | L | ------------------------ ------------------------ d | | |F |W v | v L = 19.80 ft W = 1360 lb the weight acting at the midpoint F = 13510 lb applied force d = 8.45 ft P and Q unknown cable tensions To [CapitalThorn]nd Q, equilibrium of moments about the left end: Q L - F d - W (L/2) = 0 which gives Q = (F d)/L + W/2 = 6445 lb P can now be found from force equilibrium: P + Q - F - W = 0 P = F + W - Q = 8425 lb In general, you need to draw a free body diagram from the problem statement, showing dimensions and forces. Then apply conditions of equilibrium, either that the sum of forces is zero or that the sum of the moments about a point is zero. Because these are basic principles, you need to fully understand them by reading the developments given in a textbook in order to be able to apply them correctly.Good luck. === Subject: Re: Is there a possible mathematical function for this? > I need a mathematical parameter to quantify direction with the > following qualities. > 1) It must be continuous at all points on the unit circle so a polar > representation won't work since it is discontinous at 2pi. > 2) It must be one parameter so x,y cartesian coordinates won't work. > 3) All points must be uniquely de[CapitalThorn]ned so slope won't work since pi/4 > and 5pi/4 are the same. > 4) it must be relatively computationally undemanding since I will be > calculating it a lot. > It doesn't need to have a purely linear correspondance since I will be > feeding it into a neural network and the system can learn to adapt to > non-linear functions, but the closer it is to a linear function the > better. > I suspect that there is some topology based proof that makes this sort > of function impossible, but I [CapitalThorn]gured I would ask since I need such a > function and I only have a hunch that it is impossible. Ok, I'm new to newsgroups, I admit - I thought I posted a reply to this yesterday and it hasn't shown up! So, here goes a second time... What you want is a continuous map from the real number line (or some portion of it) to the unit circle (since each point on the circle corresponds to a unique direction). Would this be right? Once you had the coordinates of this point, you could work out an angle using inverse trig functions if you needed to. This is possible using the following construction: imagine the unit circle centred at the origin, then imagine the line y = -1. A given real x0 corresponds to a point on this line (x0,-1). Now imagine a ray starting at the top of the circle (0,1) passing through the line at (x0,-1). This will intersect the circle in two places - at (0,1) trivially and at another point which depends uniquely on the value of x0. You can solve for the coordinates of this point, as I have just done, and you get: x1 = 4*x0 / (4 + x0)^2 y1 = (4 - x0^2) / (4 + x0^2) to get the point (0,1) on the circle, you need to allow x0 be be in[CapitalThorn]nity. Then you have a bijection between the unit circle and the real number line plus the point at in[CapitalThorn]nity. So the catch here is that you must include the point at in[CapitalThorn]nity to [CapitalThorn]ll up the discontinuity at (0,1). It seems clear to me, that one could not obtain a continuous mapping from a [CapitalThorn]nite line segment to the unit circle. This is because you will necessarily have discontinuities at the end points of the line. Similarly you could not obtain a mapping from a ray to the unit circle. I think that this inadvertantly proves that there is no continuous map from a line segment or ray to the complete real number line. Am I right in this? I think that is interesting. Out of interest, what is your neural network supposed to be doing? (I know very little about neural networks) Ruben === Subject: Re: Is there a possible mathematical function for this? Oh, sorry the damn thing did post twice - and, after all, I realised that my suggestion probably doesn't solve your problem anyway - and I got a bit confused at the end saying that there was no mapping from a [CapitalThorn]nite line segment to the real number line. In fact, it is on the contrary - the construction I gave tells us a mapping from [0..2pi) onto the real number line! Ruben === Subject: Re: point of inßection AND extreme point > Just a hunch, I think there isn't. For a minimum, f'(x)=0 and > f''(x)>0; for a maximum, f'(x)=0 and f''(x) < 0. For a point of > inßection, f''(x)=0, but then you have neither a maximum nor a > minimum. >That's not strictly true. f(x) = x^4 has a minimum at x=0, but f''(0) >= 0. True, but (0,0) is not a point of inßection, since f''(0) is positive on both sides of it. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: point of inßection AND extreme point days. My association with the Department is that of an alumnus. >I was wondering if there is a function f(x) such that at x=a we have both a >max point (or min) and a point of inßection? >I think not. A point of inßection is where the second derivative >changes sign. But at a maximum he [CapitalThorn]rst derivative changes sign >(positive to negative), which means that the second derivative is >negative on both sides of that point. >I'm certain your condition can't be ful[CapitalThorn]lled for a continuous >function. Maybe some really oddball function does what you want, >with strategically placed discontinuities, but I can't think of one. Take f(x) = -x^2 for x<0 and -sqrt(x) for x>=0 This function is continuous everywhere, though not differentiable at 0. At x=0, a critical point, the derivative is positive on the left, and negative on the right of 0, so this is a relative maximum (in fact, an absolute maximum). Note, however, that it is false that the second derivative is negative on both sides of the point: the second derivative is negative on the left of 0, put positive on the right. Your error was to conclude that the second derivative would be negative on both sides of the point, presumably because you imagined the [CapitalThorn]rst derivative had to be decreasing on a neighborhood of the point in order to go from positive to negative. But this is not the case if the [CapitalThorn]rst derivative is not continuous at the point (or does not exist). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: point of inßection AND extreme point days. My association with the Department is that of an alumnus. >I was wondering if there is a function f(x) such that at x=a we have both a >max point (or min) and a point of inßection? If yes, can you list some. If >not can we prove this? Depends on your exact de[CapitalThorn]nition of point of inßection. But here's an easy one: f(x) = x^2 if x<0, and f(x) = sqrt(x) if x>= 0. This function has an absolute minimum at 0, the second derivative is positive on x<0 (so it is concave up), and on x>0 the second derivative is -(1/4)x^{-3/2} hence negative, so it is concave down. -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: The real numbers, and general comments >Perhaps I should try to explain again my argument. (Provided that you >haven't just decided to run away from me.) > >I generally don't post much on the weekends. > > I apologise, I should have thought of that. Note that I tend to post more on the weekends, as I have time. >R is normally conceptualised as an in[CapitalThorn]nite geometric line. I say that >what we call a number is a point on this line, and so many points as >we can locate can be called numbers. But, it is not a valid inference >from this to state that the line is made up of numbers! > >No, it's an axiom that each of the points *corresponds* to a distinct >number. There is no inference, just an axiom... an assumption. > > > Ihis Ôaxiom' is actually a rede[CapitalThorn]nition of Ônumber'. This is justi[CapitalThorn]ed > based on the completeness axiom, isn't it? > > By de[CapitalThorn]nition no computable sequence can have a non-computable limit, > and no computable set can have a non-computable supremum, so what do > we really need completeness for? > Dealing with the points on the real line that are non-computable. Circular argument, isn't it? You are justifying the Ôaxiom' that gives us non-computable numbers by using the concept of non-computable numbers. Andrew Usher === Subject: Re: The real numbers, and general comments > Now, in homogeneous coordinates, > it's obvious that the rationals are isomorphous to a dense set on the > line at in[CapitalThorn]nity (its rational points). > Isomorphous? Is this a misprunt for amorphous? No. Andrew Usher === Subject: Re: The real numbers, and general comments >But that *is* a [CapitalThorn]nite de[CapitalThorn]nition. I de[CapitalThorn]ned them, I didn't tell you >how to enumerate them, since that is impossible. Also, I didn't know >that you believe quantifying over all subsets of an in[CapitalThorn]nite set is >illegitimate. If that is the case, then you are working with a >completely different set theory from ZFC, or even ZF. > > In another thread (ÔUncountable sets in CZF?') we are discussing the > Loewenheim-Skolen paradox, which shows that ZFC is incomplete. Given > that, why should ZFC have any preferred status? > Only because it's the one I'm most familiar with. What are the axioms > of CZF? More to the point, how is it different from ZFC? I don't know what CZF is. That thread was started by Ross Finlayson, not myself. I was asking if the incompleness of ZFC should cause us to reject it as the true set theory. Is a complete set theory possible, in the sense that no additional axioms can falsify any of its theorems? >There's a simple problem with that: Let f be the following function: >f(x) = 0 when x is irrational, f(x) = 1 when x is rational. > >This is not piecewise continuous. (Note the restriction above.) > >Good point, however not all functions in real analysis are piecewise >continuous. > > Can you give an example of a useful function that isn't? (I don't > count the delta Ôfunction', as that isn't even a valid function.) > Useful for what? The above is useful for demonstrating that a function > can have countably in[CapitalThorn]nitely many non-zero values, yet have an integral > = 0. Pathology for the sake of pathology is not what I consider Ôuseful'. Interesting, yes, but not useful. > Though I'd rather de[CapitalThorn]ne continuity to be uniform continuity over an > interval, following the naive Ôcan be drawn without lifting the > pencil'. An open interval can be de[CapitalThorn]ned as the union of a sequence of > closed intervals, hence we can still say that > > f(x) = 1/x > > is continous on its entire domain, and similarly all algebraic > functions. > Fair enough. Do you agree with my general point here, that analysis can be done over any countable dense set, and for all piecewise continuous functions, will give the same results? >If R is not just numbers, what are the other things? > > Uncomputable sequences would be a good description. > It seems to me that it would be easier to use a new de[CapitalThorn]nition of R in > which you restrict yourself to computable numbers. Of course, that > would result in R being countable. I'm not sure if there's a good, short de[CapitalThorn]nition of the computables that doesn't mention any speci[CapitalThorn]c computing machine. My de[CapitalThorn]nition is the union of all de[CapitalThorn]nable archimedean ordered [CapitalThorn]elds. In this case, it can only be gotten as the sup of an in[CapitalThorn]nite sequence: Q < F1 < F2 < F3 < ... to in[CapitalThorn]nity = S where the Fn are archimedean ordered [CapitalThorn]elds (Not all such sequences will give all of S, and any that does will be uncomputable.). > I don't see a problem with calling an uncomputable sequence a number, > however. You can't calculate with it. > I hope you aren't just being obtuse here; the de[CapitalThorn]nition of parallel > of course is changed in the obvious fashion. I said that no actual > theorems of either geometry are disproved, and that is clearly true. > That is because each geometry is consistent. That does not mean they > have the same results. If you are changing de[CapitalThorn]nitions to accomodate > the axioms (which is common), then you can still state different results > by replacing parallel in one system with the de[CapitalThorn]nition from the the > other system. > In one system, there are lines that do not intersect (Euclidean). In > the other, every pair of lines intersects (projective). These are still > different results. Given that the two geometries are consistent with each other, what is the problem with unifying them as one? Andrew Usher === Subject: Wiki as A Time Saver for Researchers Wiki as A Time Saver for Researchers The intention of this message is to draw researchers attention to Wiki as a potentially ef[CapitalThorn]cient new means for scienti[CapitalThorn]c/technical communication/education/collaboration. The author believes the current approach (papers) alone is not very ef[CapitalThorn]cient for idea exchange among researchers. Assume a learner new to a [CapitalThorn]eld has already read some classic textbooks for systematic knowledge acquisition, and already had the framework of the [CapitalThorn]eld on his mind, and is going to specialize on a speci[CapitalThorn]c problem. At this point, he has to search for and read existing papers on this topic, which can be an extremely painful hunting and digestion progress. Whether if he just wants to get informed of as many previous efforts on the problem as possible, or if he believes he has got an original idea and wants to ensure the originality, he probably has to undertake an extensive search with Google/CiteSeer for all papers available online with a seemingly relevant title. He has to scan through a screenful of downloaded papers. Even after doing this, there can still be new papers found later to be relevant. Papers are essentially individual units of information shattered over the Internet, and the current approach to [CapitalThorn]nd them -- search engines connect researchers to them via a keyword combination. Not to mention that we can't be sure the wanted papers all contain such a keyword phrase, the prohibitively enormous amount of search results can always bury critical information. On the other hand, web sites that have a well organized collection of papers on a topic, as an information gateway alternative to search engines, are easy to get out of date and miss the latest useful discoveries for a topic. The idea of using Wiki came to my mind last night. It can be useful in both scenarios below: (1) A newcomer to a [CapitalThorn]eld who wants to investigate all existing efforts on a speci[CapitalThorn]c task; (2) Experienced researchers who want to effortlessly keep track of new ideas/solutions to a speci[CapitalThorn]c task (or any speci[CapitalThorn]c task in a [CapitalThorn]eld). So what is Wiki? Wiki is an easy way to collaboratively edit online documentation via a Web interface. A live example is Wikipedia (http://www.wikipedia.org), which is an online encyclopedia on general knowledge. How can Wiki be used in scienti[CapitalThorn]c communication? Usually, what a researcher discovers is a new idea, or an enhancement to an existing idea. He can contribute this new idea to a Wiki page where all historical efforts for a task are documented in an organized manner. Readers interested in a speci[CapitalThorn]c task can directly go down to that context and get all the relevant details about previous efforts. This is like a precisely targeted advertising model which immediately connects scienti[CapitalThorn]c authors and readers of the same speci[CapitalThorn]c research interest. It also helps a new idea to quickly propagate to researchers who set a news alert to capture all new efforts on a speci[CapitalThorn]c problem. How speci[CapitalThorn]c can a problem be de[CapitalThorn]ned? It's unlimited and up to your needs. It can be far more speci[CapitalThorn]c than what categories are de[CapitalThorn]ned in Yahoo Directory or Dmoz Directory. If the Wiki way of scienti[CapitalThorn]c communication becomes popular, researchers can save countless hours from search, and put more time on problem solving. To generalize my initiative, the Wiki way is useful not only in sci/tech communication, but also in any general domains where the current state of the art of search engines can't return relevant and comprehensive results. Yao Ziyuan http://www.babelcode.org yao@babelcode.org === Subject: Re: Uncountable sets in CZF? > This is no doubt because of our intuition that sets ought to have a > largest element > our intuition? Speak for yourself, mate! > - if N has no [CapitalThorn]nite upper bound, Ôobviously' it must > contain elements that are not [CapitalThorn]nite. > obviously that is a non-sequitur. Obviously I don't believe it, I was asking if Ross does. Duh. Andrew Usher === Subject: Re: Uncountable sets in CZF? > Why do we insist upon only [CapitalThorn]rst order logic > That's a very good question. Perhaps you could answer it, then. > I see, your argument is that we can formalise Ôuncountability' in ZFC. > But clearly, this doesn't have external value, given what you have > said. > No, it doesn't. So what? Therefore, it is not a consistent de[CapitalThorn]nition of Ôuncountable', which I think we agree, does have a good de[CapitalThorn]nition in English. >Set-domain models of ZFC are only interesting in so far as they allow >one to establish the consistency of certain theories (independence of >certain assumptions from others). They are no realistic models anyway, >as the Ôreal' universe of sets has a proper class for its domain.) > So you would prefer to use ZFC without a model? What does that mean? > The universe of sets is envisioned as a model, but having a domain of > discourse that is too large to be a set. For such models, [CapitalThorn]rst order > sentences also make sense, as well as the inference rules of FOL. Oh, I get it. Any model of ZFC can be a set in a larger model, but there must be some Ôultimate model' that is not. Unfortunately ZFC isn't capable of establishing such a model. Is this right? Andrew Usher === Subject: Re: Uncountable sets in CZF? >The universe of sets is envisioned as a model, but having a domain of >discourse that is too large to be a set. For such models, [CapitalThorn]rst order >sentences also make sense, as well as the inference rules of FOL. > Oh, I get it. Any model of ZFC can be a set in a larger model, but > there must be some Ôultimate model' that is not. Unfortunately ZFC > isn't capable of establishing such a model. Is this right? Why should there be an ultimate model. As soon as you have one, I can and will imagine the category of all sets, and may even want to talk about the category of categories, functor categories etc. There are large number of existing papers that use such things. How are you going to accommodate them? Nath Rao === Subject: Re: Easiest problem on the Earth... (but not for me) > Show, that ideal of Z[x] generated by > (two) [b:6600c5e757]members > 2, > x^4 + x^2 + 2 > is maximal ideal.[/b:6600c5e757] > ---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** > ---------------------------------------------------------- > http://www.usenet.com I don't think so. First off, you can factor out 2. That is, Z[x]/(2,p] (p being any polynomial) is isomorphic to (Z/(2))/(p). Now mod 2, p is just x^4 + x^2 = x^2(x^2 + 1). Since neither x^2 nor x^2 + 1 is a multiple of p, the ideal is not prime and therefore not maximal. As a matter of fact, x^2 + 1 = (x + !)^2, so the quotient mod the ideal is the product of four copies of Z/(2). === Subject: Re: Easiest problem on the Earth... (but not for me) Please use descriptive subject lines, instead of whining about how stupid you are. That becomes progressively more and more unattractive. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups X-RFC2646: Original > sci.math: >Herc's decision to disrupt newsgroups: > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? Someone wondered wether or not Herc was a troll. I cleared it up for him. HTH === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups >Herc's decision to disrupt newsgroups: > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? fock its a kook fest here in math and logic fess up, save me valuable search time when I go exterminating Herc === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups X2b5S?7at*2R/5vY{L[AI_LKLHh.E >Herc's decision to disrupt newsgroups: > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? > fock its a kook fest here in math and logic > fess up, save me valuable search time when I go exterminating Is this a k0oKsuit threat, Hercie? -- * Of[CapitalThorn]cial AFA-B Bully, Pest, Antagonist, Government/Media Disinformation Agent, Dr. Green Sockpuppet, and Lemming * Chief AFA-B Vote Rustler === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups >Herc's decision to disrupt newsgroups: > > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? > fock its a kook fest here in math and logic > fess up, save me valuable search time when I go exterminating > Is this a k0oKsuit threat, Hercie? I'm not giving you your main couse already, I know you all like to get everyone participating in your mass sadistic gang bashing then get the quotes prepared for your bravado displays of public service. Why don't you go to alt.astrology and bash some people for kicks if you're hungry? Herc === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups double secret probation because: >Herc's decision to disrupt newsgroups: > > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? > > fock its a kook fest here in math and logic > > > fess up, save me valuable search time when I go exterminating > Is this a k0oKsuit threat, Hercie? >I'm not giving you your main couse already, I know you all like to get everyone >participating in your mass sadistic gang bashing then get the quotes prepared for your >bravado displays of public service. >Why don't you go to alt.astrology and bash some people for kicks if you're hungry? >Herc Look out Carl, the k0ok will [CapitalThorn]nd out your real name and out you. He Where do these k0oks buy their egos? It sure as hell is not self endowed. ÔRatz === Subject: Re: Herc, rat poison, and deliberate decisions to disrupt newsgroups X2b5S?7at*2R/5vY{L[AI_LKLHh.E > double secret probation because: > >Herc's decision to disrupt newsgroups: > > So if you catch a cold in one place, you [CapitalThorn]nd it necessary to go to > another place and start sneezing on everyone? > > fock its a kook fest here in math and logic > > > fess up, save me valuable search time when I go exterminating > > Is this a k0oKsuit threat, Hercie? >I'm not giving you your main couse already, I know you all like to get >everyone >participating in your mass sadistic gang bashing then get the quotes >prepared for your >bravado displays of public service. >Why don't you go to alt.astrology and bash some people for kicks if you're >hungry? >Herc > Look out Carl, the k0ok will [CapitalThorn]nd out your real name and out you. He > Where do these k0oks buy their egos? It sure as hell is not self > endowed. It might be time for Guido to take a nice little vacation Down Under. -- * Of[CapitalThorn]cial AFA-B Bully, Pest, Antagonist, Government/Media Disinformation Agent, Dr. Green Sockpuppet, and Lemming * Chief AFA-B Vote Rustler === Subject: Re: Arxiv paper supposedly gives roots of any polynomial? ... > Actually I did not look at that aspect, and now that you mention it I am > wondering. I am even wondering about why a school would like to bring it > to the world in the way it is done. (No, actually I am not wondering, it > is about money.) > Money? > Can you explain? It is actually quite simple. Government money is distributed to schools, but the distribution depends on quite a few things. As the ultimate deciders on the distribution are actually civil servants, it is quite simple. Get a big name in the press about what your students do accomplish, and there you are. (Yes, I am being quite a bit cynical.) What it shows to me is complete incompetence by the people involved at that school and a bit of greed. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Journals Call for Papers Journals Call for Papers Do you have research results, scienti[CapitalThorn]c opinions or review papers to be published? The new international academic journal Nature and Science and The Journal of American Science are inviting you to publish your papers. Please visit http://www.sciencepub.net and send your manuscript to sciencepub@gmail.com or editor@sciencepub.net. === Subject: Re: Journals Call for Papers > Journals Call for Papers > Do you have research results, scienti[CapitalThorn]c opinions or review papers to > be published? The new international academic journal Nature and > Science and The Journal of American Science are inviting you to > publish your papers. Please visit http://www.sciencepub.net and send > your manuscript to sciencepub@gmail.com or editor@sciencepub.net. Can nonAmericans do American Science? Can Americans do unAmerican Science? -- 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: Call for Papers Call for Papers Do you have research results, scienti[CapitalThorn]c opinions or review papers to be published? The new international academic journal Nature and Science and The Journal of American Science are inviting you to publish your papers. Please visit http://www.sciencepub.net and send your manuscript to sciencepub@gmail.com or editor@sciencepub.net. Papers in all the [CapitalThorn]elds are accepted: including natural science and social science. === Subject: Journals Call for Papers Journals Call for Papers Do you have research results, scienti[CapitalThorn]c opinions or review papers to be published? The new international academic journal Nature and Science and The Journal of American Science are inviting you to publish your papers. Please visit http://www.sciencepub.net and send your manuscript to sciencepub@gmail.com or editor@sciencepub.net. Papers in all the [CapitalThorn]elds are accepted: including natural science and social science. === Subject: Re: easy algebra problem [ but not for me :( ] by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8C2leK29826; Find the greatest common divisor d=gcd(2,x^4 + x^2 + 2) by Euclide's algorithm. If d is any integer it follows M==<1>=Z[x] joccis >Show, that ideal of Z[x] generated by >(two) [b:e46927f1d5]members >x^4 + x^2 + 2 >is maximal ideal.[/b:e46927f1d5] >---------------------------------------------------------- > ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY ** >---------------------------------------------------------- > http://www.usenet.com === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Beckwith said: > I am a mathematical physics PhD. > Kolowski replied: > I frankly doubt this. Your writing is semi-literate. > Take a gander at: > http://www.arxiv.org/[CapitalThorn]nd/math-ph/1/au:+Beckwith_A/0/1/0/all/0 /1 > This merely demostrates that physicists need not be very literate. I bet there are even a few people with math PhDs that are also nuts! === Subject: Re: How long would it take a computer to completely solve chess? > Does anyone really care? If computers were able to solve the game, we > would simply go to Fischer Random chess where it would be in[CapitalThorn]nitely > more dif[CapitalThorn]cult to solve. I doubt it. Even if computers were to solve chess, there'd remain the much more complicated problem of mining that gargantuan database for human- usable patterns. My guess is, that would take far longer than coming up with the database in the [CapitalThorn]rst place! === Subject: Re: How long would it take a computer to completely solve chess? > richard miller ha scritto nel > If the media or Scientists ever announce, Ôit will completely > revolutionize > this or that', you know it is doomed. My favourite bet, Fusion, > currently promised as ÔUnlimitless, CLEAN energy at very low cost'. - > kiss of death IMHO. Fission was ÔUnlimitless energy at very low > cost'. It costs an absolute packet. Fusion, I bet the reactors cost > an absolute packet too, > the > reactor walls will be humming, it won't be cheap that's for sure. By > 2000 > we > were supposed to have robots doing all the housework, no one would > work, space stations the size of the Enterprise, moon bases, mars > bases, cars > that > run on nowt, you name it... > You're so right, the only revolution in the last ten years has been > Internet. Not really the Internet itself, which has been around a lot longer than ten years, but certainly the WWW. However, the WWW came to develop its own set of magni[CapitalThorn]cent promises which turned out empty, much to stockholder chagrin. > Luigi Caselli === Subject: Re: How long would it take a computer to completely solve chess? > David Bandel ha scritto nel messaggio > there are more board positions in the tree than atoms in the universe. > assuming reasonable pruning occured.. i imagine a computer > calculuating a trillion positions per second (and i'm talking about > the kind of pruning fritz products are known for where the branching > factor is about 2.8 to 3.5) this could probably get nowhere near > approaching an optimal solution.. though if the win in chess is forced > in less than 40 moves.. it's likely the computer would play perfectly > or near perfectly when given days per move.. because 10^12 pos/sec > would get about 30-ply analyzed in 3 minutes > > > You are looking at 3.15^80 = 10^40 positions > 10^12 pos/s = 3*10^19 pos/year > so it would take 300000000000000000000 years! > > Many people expressed the strategy to wait for a faster computers. > I would not :-) > > Theo > no offense but you obviously know notihng abuot the chess game tree. > Maybe Theo doesn't know anything about the chess tree, but in any case you > have to spent 10^n years to solve chess, where n is so high that you need > the lifetime of a lot of universes... and the Theo number is not so wrong... > Luigi Caselli again i disagree and i think you suffer from the same misconception about chess that he does. chess will certainly be solved before the year 2100.. there are too many directions the solution is being approached from for this not to be the case. if you look into computer chess and endgame tablebases.. and the depth of opening research.. and the awesome pruning capabvility of the top chess engines.. and the brute processing strength computers are gaining (yes they are approaching some physical limits.. but in the laboratory.. NOT at home yet) if you look at all this and put it together.. you'll have to agree the game is nearing tractibility === Subject: Re: How long would it take a computer to completely solve chess? >again i disagree and i think you suffer from the same misconception >about chess that he does. chess will certainly be solved before the >year 2100.. there are too many directions the solution is being >approached from for this not to be the case. if you look into >computer chess and endgame tablebases.. and the depth of opening >research.. and the awesome pruning capabvility of the top chess >engines.. and the brute processing strength computers are gaining (yes >they are approaching some physical limits.. but in the laboratory.. >NOT at home yet) if you look at all this and put it together.. you'll >have to agree the game is nearing tractibility But on the other hand if one has a basic understanding of numbers, a good calculator or spreadsheet, and is willing to take a bit of time to actually think about it and do some actual calculations, they will understand that pretty well all you say above is stuff and nonsense. Ed === Subject: Re: How long would it take a computer to completely solve chess? Ed Seedhouse says... >again i disagree and i think you suffer from the same misconception >about chess that he does. chess will certainly be solved before the >year 2100.. there are too many directions the solution is being >approached from for this not to be the case. if you look into >computer chess and endgame tablebases.. and the depth of opening >research.. and the awesome pruning capabvility of the top chess >engines.. and the brute processing strength computers are gaining (yes >they are approaching some physical limits.. but in the laboratory.. >NOT at home yet) if you look at all this and put it together.. you'll >have to agree the game is nearing tractibility >But on the other hand if one has a basic understanding of numbers, a >good calculator or spreadsheet, and is willing to take a bit of time >to actually think about it and do some actual calculations, ...all of which are based on the assumptions that all computers will forever be non neumann machines and that the only possible algorithm for solving chess is a brute-force search. >they will understand that pretty well all you say above is stuff >and nonsense. Or perhaps they will question your unexamined assumptions. === Subject: Re: How long would it take a computer to completely solve chess? >But on the other hand if one has a basic understanding of numbers, a >good calculator or spreadsheet, and is willing to take a bit of time >to actually think about it and do some actual calculations, >...all of which are based on the assumptions that all computers will >forever be non neumann machines It is of course based on no such thing, which merely shows Mr. Macon's fundamental innumeracy. >and that the only possible algorithm for solving chess is a brute-force search. It is always possible that someone will come up with an algorithm for generating the perfect chess move without any lookahead. *But* in more than half a century of consideration by programmers and mathemeticians no one has done so. Moreover there is no evidence that it can or will happen soon or ever. To rely on this as evidence that chess can or soon will be solved is to rely on a chimera. If it's so easy why doesn't Mr. Macon show us this algorithm and become rich? Ed === Subject: Re: How long would it take a computer to completely solve chess? Ed Seedhouse says... >But on the other hand if one has a basic understanding of numbers, a >good calculator or spreadsheet, and is willing to take a bit of time >to actually think about it and do some actual calculations, >...all of which are based on the assumptions that all computers will >forever be Von Neumann machines >It is of course based on no such thing, which merely shows Mr. Macon's >fundamental innumeracy. Get out your good calculator or spreadsheet, do your calculations with the assumption that a working Quantum computer capable of solving chess will be invented in less time than it would take for a conventional computer to do the job, and post the numbers. Actual calculations are far more convincing than childish namecalling. >and that the only possible algorithm for solving chess is a brute- >force search. >It is always possible that someone will come up with an algorithm for >generating the perfect chess move without any lookahead. *But* in >more than half a century of consideration by programmers and >mathemeticians[sic] no one has done so. Moreover there is no >evidence that it can or will happen soon or ever. How long have programmers and mathemeticians been doing serious work on programming computers so far? 60-60 years? That's not what I call a convincing argument that they won't be able to do it in less time than the many thousands of years that it would take to solve chess with a brute force search using today's computers and algorithms. >To rely on this as evidence that chess can or soon will be solved >is to rely on a chimera. You are confusing me with the fellow who said it would be solved soon. I am the fellow who is saying that your claim of being able to calculate the time it will take to solve chess is a big stinking pile of manure. There are too many unknowns such as time to invent a quantum computer (50 years? 10,000 years? Never?) and whether an algorithm that has eludes programmers for 50 years will continue to elude them for hundreds or even thousands of years. >If it's so easy why doesn't Mr. Macon show us this algorithm and >become rich? I never said it was easy. I am challenging the assumption that it will never be done that underlies your claim to be able to calculate how long it will take to solve chess. === Subject: Re: How long would it take a computer to completely solve chess? > > ...all of which are based on the assumptions that all computers will > forever be non neumann machines > von Neumann (after John von Neumann) Shedar === Subject: Re: How long would it take a computer to completely solve chess? Luigi Caselli says... >Maybe Theo doesn't know anything about the chess tree, but in any case you >have to spent 10^n years to solve chess, where n is so high that you need >the lifetime of a lot of universes... and the Theo number is not so wrong... Nope. You only have to spent T+N time to solve chess, where T is the amount of time it takes to make a working quantum computer that can solve chess (maybe a few years, maybe never) and N is the time it will take to run the program (likely between a few minutes and a few days). === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) > >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- > > if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... > > so, there is a positive integer p such that n^p >= m. > **** > I don't follow you here....if a^n = a, and a^n^2 = a, and a^n^3 = a, > ...doesn't a^n^p = a, and so doesn't n^p = n? if not, why not, and/or why > can we then call n^p > less than or equal to m ? I understand the problem solution, if you > can > say that n^p >= m, the whole thing makes sense, but I miss you on that > step. > (I know I'm jumping in on someone else's problem, but I hate to not > understand something that is *close* to within my grasp : ) > -karinne > **** > so, a^n^p = 0 ....contradiction. > > thus a = 0. > > > oh......i'm sorry. my mistatke. > there is a condition which n>1. sorry. > but, there does not exist concrete condition about m, n. > namely, m and n is integer ? positive ? negative ?...i don't know.. > thank you very much for your advice. I once asked a question like this here, thinking that m was [CapitalThorn]xed. What the problem was really saying was, there is some m in Z such that a^m = 0. I think that is what is meant here. Isn't this what is called a nilpotent element? === Subject: Re: algebra...; > >suppose that >x^n = x for all x in R (R is ring) > > > >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- > > > > if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... > > so, there is a positive integer p such that n^p >= m. > > **** > I don't follow you here....if a^n = a, and a^n^2 = a, and a^n^3 = > a, > ...doesn't a^n^p = a, and so doesn't n^p = n? if not, why not, > and/or why > can we then call n^p > less than or equal to m ? I understand the problem solution, if > you > can > say that n^p >= m, the whole thing makes sense, but I miss you on > that > step. > (I know I'm jumping in on someone else's problem, but I hate to not > understand something that is *close* to within my grasp : ) > -karinne > **** > > so, a^n^p = 0 ....contradiction. > > thus a = 0. > > > oh......i'm sorry. my mistatke. > there is a condition which n>1. sorry. > but, there does not exist concrete condition about m, n. > namely, m and n is integer ? positive ? negative ?...i don't know.. > thank you very much for your advice. > I once asked a question like this here, thinking that m was [CapitalThorn]xed. > What the problem was really saying was, there is some m in Z such that > a^m = 0. I think that is what is meant here. Isn't this what is called > a nilpotent element? ****** Ok, this is beyond my Ôtraining' and into stuff I just play with at this point, trying to get it. Messing about with this, (and I don't know what a nilpotent element is), if in this R, a^n = a, then n*ln a = ln a, and then (I think?) n^2*ln a = n*ln a , etc etc, which would tell me that for this R, n^2 = n^3 = n^p, but then n>1, so it all falls apart in my head. Scratching my head, what m could give a^m = 0? I don't know. but, if I think about it conceptually, and say, ok, n>1, and if there is no de[CapitalThorn]nition or condition given for m, where do we get to n^p = m? if a^n = a, how is a^(anything)= 0? Is there a condition for m? -karinne ****** === Subject: Re: algebra...; > >suppose that >x^n = x for all x in R (R is ring) > > > >show that >if a in R is such that a^m =0, then a=0 > >-------------------------------------------------- > > > > if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... > > so, there is a positive integer p such that n^p >= m. > > **** > I don't follow you here....if a^n = a, and a^n^2 = a, and a^n^3 = > a, > ...doesn't a^n^p = a, and so doesn't n^p = n? if not, why not, > and/or why > can we then call n^p > less than or equal to m ? I understand the problem solution, if > you > can > say that n^p >= m, the whole thing makes sense, but I miss you on > that > step. > (I know I'm jumping in on someone else's problem, but I hate to not > understand something that is *close* to within my grasp : ) > -karinne > **** > > so, a^n^p = 0 ....contradiction. > > thus a = 0. > > > > oh......i'm sorry. my mistatke. > > there is a condition which n>1. sorry. > > but, there does not exist concrete condition about m, n. > namely, m and n is integer ? positive ? negative ?...i don't know.. > > thank you very much for your advice. > I once asked a question like this here, thinking that m was [CapitalThorn]xed. > What the problem was really saying was, there is some m in Z such that > a^m = 0. I think that is what is meant here. Isn't this what is called > a nilpotent element? > ****** > Ok, this is beyond my Ôtraining' and into stuff I just play with at this > point, trying to get it. > Messing about with this, (and I don't know what a nilpotent element is), if > in this R, > a^n = a, then n*ln a = ln a, and then (I think?) n^2*ln a = n*ln a , etc Can one take the log of an element in a ring which is arbitrary apart from x^n = x for all x in R (R is ring)? > etc, which would tell me that for this R, n^2 = n^3 = n^p, but then n>1, so > it all falls apart in my head. > Scratching my head, what m could give a^m = 0? I don't know. but, if I > think about it conceptually, and say, ok, n>1, and if there is no de[CapitalThorn]nition > or condition given for m, where do we get to n^p = m? if a^n = a, how is > a^(anything)= 0? > Is there a condition for m? > -karinne > ****** === Subject: Re: A simple question about integers > Hello Eray, > Eray Ozkural exa schrieb im Newsbeitrag > Hello there, > > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > > In Z, are there integers with an in[CapitalThorn]nite number of digits? > Suppose there were only numbers with [CapitalThorn]nitely many digits (say N). (The base > in which you represent won't matter; but to make it clear, take decimal > digits). Let x be a number with N digits (we can admit leading zeroes here). > Then the absolute value of all these numbers is less than or equal to > 9.10` 0 +... +9.10^{N-1} > this means that there is a smallest negative / greatest positive integer; > this is false. > To put it in other words: If x is a number with N digits (including leading > zeroes), then say 10x+1 has N+1 digits, so there cannot be a greatest number > of decimal digits. > Thomas > > > -- > Eray Ozkural > idiot. that doesn't prove there are any integers in Z with in[CapitalThorn]nite > digits. all it proves is there is no [CapitalThorn]nite limit to the amount of > digits of integers in Z. this kind of misleading bull is what > plagues these newgroups I thought in[CapitalThorn]nity was just short hand for no [CapitalThorn]nite limit, or bigger than any [CapitalThorn]xed n in Z. === Subject: Re: A simple question about integers > idiot. that doesn't prove there are any integers in Z with in[CapitalThorn]nite > digits. all it proves is there is no [CapitalThorn]nite limit to the amount of > digits of integers in Z. this kind of misleading bull is what > plagues these newgroups > I thought in[CapitalThorn]nity was just short hand for no [CapitalThorn]nite limit, or > bigger than any [CapitalThorn]xed n in Z. So which n are you claiming has in[CapitalThorn]nitely many digits? -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: I do not have any more time for you cowards, but James is more a man than any of you Your little group of censorious little cretins ßunked out in my estimation. James has been intemperate and hot headed. What is NOT missing is his heart and his basic sense of being a MAN, rather than a spineless whelp like many of you who worship orthodoxy without understanding it. I have had my disagreements with James on certain issues, but he is more a MAN in the sense of what counts than any of you whelps. And with that, I go back to my work. I despise you for picking on him . I LOATHE cowards, and that is what most of you are. Your group ßunked out in my estimation. While you are at it, grow some balls if you can and attack some REAL basic problems in mathematics, rather than act like censorious little cretins who only know how to critique, rather than create something new. Andrew Beckwith, PhD === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you >Your little group of censorious little cretins ßunked out in my >estimation. James has been intemperate and hot headed. What is NOT >missing is his heart and his basic sense of being a MAN, rather than a >spineless whelp like many of you who worship orthodoxy without >understanding it. >I have had my disagreements with James on certain issues, but he is >more a MAN in the sense of what counts than any of you whelps. And >with that, I go back to my work. >I despise you for picking on him . I LOATHE cowards, and that is what >most of you are. >Your group ßunked out in my estimation. While you are at it, grow >some balls if you can and attack some REAL basic problems in >mathematics, rather than act like censorious little cretins who only >know how to critique, rather than create something new. giggle. you really think we care what you think? >Andrew Beckwith, PhD ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > Your little group of censorious little cretins ßunked out in my > estimation. James has been intemperate and hot headed. What is NOT > missing is his heart and his basic sense of being a MAN, rather than a > spineless whelp like many of you who worship orthodoxy without > understanding it. You have either never read any post of James Harris, or you are just the same kind of scum as he is. > I have had my disagreements with James on certain issues, but he is > more a MAN in the sense of what counts than any of you whelps. And > with that, I go back to my work. Please could you explain this MAN thing to me. As far as I know, a MAN is a kind of lorry, as described at www.man-nutzfahrzeuge.de and I cannot see any similarity to James Harris. If you mean to say that he is not a woman, you may be right, and he may be more of a man than a small percentage of the posters here (I wonder if he is more of a man than Nora Baron or not). If you are a feminist, who thinks that all men are pondscum, then maybe he is more of a man than most here. > I despise you for picking on him . I LOATHE cowards, and that is what > most of you are. > Your group ßunked out in my estimation. While you are at it, grow > some balls if you can and attack some REAL basic problems in > mathematics, rather than act like censorious little cretins who only > know how to critique, rather than create something new. Asshole. === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you === >Subject: I do not have any more time for you cowards, but James is more a man >than any of you >Your little group of censorious little cretins ßunked out in my >estimation. James has been intemperate and hot headed. What is NOT >missing is his heart and his basic sense of being a MAN, rather than a >spineless whelp like many of you who worship orthodoxy without >understanding it. >I have had my disagreements with James on certain issues, but he is >more a MAN in the sense of what counts than any of you whelps. And >with that, I go back to my work. >I despise you for picking on him . I LOATHE cowards, and that is what >most of you are. >Your group ßunked out in my estimation. While you are at it, grow >some balls if you can and attack some REAL basic problems in >mathematics, rather than act like censorious little cretins who only >know how to critique, rather than create something new. >Andrew Beckwith, PhD Well, it's not like you were contributing anything to this forum. -- Mensanator Ace of Clubs === Subject: Re: The real numbers, and general comments by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8C4G7304718; > > In another thread (ÔUncountable sets in CZF?') we are discussing the > Loewenheim-Skolen paradox, which shows that ZFC is incomplete. ... >... That thread was started by Ross Finlayson, >not myself. Here is a minor point of attribution: Uncountable sets in CZF? was started by Agamemnon. Ross F. === Subject: Re: map between torus > Suppose f: T^2-->T^2 is a map between torus which induces the identity > on the fundamental group pi_1(T^2). How can I see that f is homotopic to > identity? Consider the induced map on the universal cover f':R^2-->R^2. Since the map on pi_1 is the identity, f'(m,n)=(m,n) for all pairs (m,n) in Z x Z. Then try to build a homotopy F:R^2 x I --> R^2, with F_0=f' and F_1=identity in such a way that induces a homotopy on T^2 x I --> T^2. -Ron === Subject: Re: Fibonacci numbers, Lucas numbers and Huffman codes > De[CapitalThorn]nition 1.6. A binary tree is called elongated > if at least one of any two sibling nodes is terminal. > An elongated binary tree is called left-sided ------------------------------------------------------------- -------------- > if the left node in each pair of sibling nodes is terminal. Must be: if the _right_ (!) node in each pair of sibling nodes is terminal. ------------------------------------------------------------- -------------- -- Alex Vinokur http://mathforum.org/library/view/10978.html http://sourceforge.net/users/alexvn === Subject: Re: Comments on the proof of cyclic Z*_p^k by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8C4v2F07387; >I have been looking some more at Z*_p^k from a different >point of view. (As I am not a mathematician, its an elementary >POV). Each to his own taste, of course. The bit I gave about exp, log, and the p-adics actually feels simpler to me than messing around with the binomial theorem; the opposite may be true for you. I have no quarrel with that. >One can show from the binomial thm. that >(1 + p)^(p^b) mod p^k > 1 if b < k-1, and >(1 + p)^(p^(k-1)) mod p^k = 1, so o(1 + p) = p^(k-1), >and so ker f is a cyclic subgroup of Z*_p^k. Not meaning to cavil, but I would avoid statements of the form m mod n > k because there is no order structure on Z_n compatible with the ring structure. You meant of course (1 + p)^(p^b) mod p^k =/= 1 if ... >Finding the elements of Z*_p^k which form a cyclic subgroup of >order p-1 is a bit more dif[CapitalThorn]cult--I haven't done this yet. >But Z*_p^k/ker f =~ Z*_p which is cyclic, and since ker f and >Z*_p are cyclic, (p-1,p) = 1 and Z*_p / ker f = (1), I >think it follows that Z*_p^k is cyclic. >(I was thinking of the thm. that if A,B are normal subgroups of G >with G = AB (g = ab with a in A anad b in B, for all g in G), and >if A / B = (1), then G = A x B. Then if (|A|,|B|) = 1, G is cyclic (?) Yes, if A and B are normal *sub*groups of G and their orders are relatively prime, then G ~ A x B; if in addition A and B are cyclic then so is G. But what we have here is a subgroup A (ker f) and the *quotient* B = Z_{p^k}*/ker f, which is an entirely different story. Unless of course we could view B as a subgroup of Z_{p^k}* (by *splitting* the surjection Z^{p^k}* --> B, i.e., giving an injection B --> Z_{p^k}* whose composite with the surjection is the identity on B). But this brings us back to the problem of [CapitalThorn]nding a suitable cyclic subgroup of order p-1 in Z_{p^k}*! On the other hand, there is a theorem that any [CapitalThorn]nite abelian group is isomorphic to a direct sum of [CapitalThorn]nite cyclic groups of prime power orders, and this very quickly leads to a proof of the thing you want. Although that's like cracking a nut with a sledgehammer -- the argument using short exact sequences is a somewhat less brutal tool, IMO. Todd === Subject: Graduate school question I was wondering if anyone could give me some feedback as to how graduate programs look at students taking a year off between receiving their bachelors and applying to a PhD program. Speci[CapitalThorn]cally would taking a year off hurt an otherwise strong application? === Subject: Re: Graduate school question X-RFC2646: Original > I was wondering if anyone could give me some feedback as to how > graduate programs look at students taking a year off between receiving > their bachelors and applying to a PhD program. Speci[CapitalThorn]cally would > taking a year off hurt an otherwise strong application? I think that you could reasonably expect schools to express an interest in what you did during the year away. At a competitive school, what you did during that year could very well be a deciding factor one way or the other. === Subject: Help Solving FT Problems I'm trying to get a refresher (developing software for sometime..) on Fourier Transforms and while perusing a text, I'm having dif[CapitalThorn]culty determining two fourier transform problems and seek assistance with a solution. 1. Obtain Fourier Transform F(w) for: f(t) = exp (-alpha*t)U(t) where alpha > 0 and U(t) is the heaviside step function de[CapitalThorn]ned to be U(t) = 1 t >= 0 0 otherwise 2. Given F(w) = Integral (-in[CapitalThorn]nity, in[CapitalThorn]nity) f(t) exp (-jwt) dt By repeated differentiation and induction, obtain the fourier transform pair (-jt)^n f(t) <--> D^n F(w) / d w^n === Subject: Re: JSH circa 1996 by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i8C2ldc29804; >Interesting ...James Harris, the early years: >QUOTE: >I went ahead and did a text version for those who wouldn't download a >Write [CapitalThorn]le and for all of those folks at institutions who don't have >Windows. They'll have to be the [CapitalThorn]nal judges anyway. >------------------------------------------------------------ >Introduction. >Fermat's Last Theorem has long been a magnet to the amateur and >professional mathematician alike because of its seeming simplicity; >yet, >extraordinary dif[CapitalThorn]culty. Although there is a proof by Andrew Wiles, >think it is understandable that the problem still would incite >curiosity. I would also assume that a simpler solution would also be >interest. >Note: The following proof makes extensive use of Fermat's Little >Theorem which isn't usually stated. I also make use of accepted >results >which have came up in my previous posts on sci.math without going over >them in detail again. >1. Statement of the Problem: Fermat's Last Theorem >Given x,y,z, relatively prime, n odd prime >no solution exists for the equation x^n + y^n = z^n >2. Proof for Cases where x,y or z are divisible by n. >Let x=af, y=bg, z=ch means that > x+y=h^n or n^{n-1}h^n, z-x=g^n or > n^{n-1}g^n, and z-y=f^n or n^{n-1}f^n. *** Some important value is missing here once division by n should be introduced as: x+y = h^n or n^(nu-1) h^n ; z-x = g^n or n^(nu-1) g^n ; z-y = f^n or n^(nu-1) f^n ; where u is some natural number but u >=2 . For u=1 it were some possibility for to use Eisenstein criterion, but the true developments of FLT can not show u=1 . I think, that this and following arguments used already to be corrected by James Harris shortly after 1996... Anybody like to belive to 1996 Year could do it. ( I myself used to be more correct to these developments shortly before 1996 calling them as TAB developments and showing a lot of general formulas.) Nowadays I have some important step once using PQR input... Roman B. Binder 91035 Jerusalem POBox 3962 Israel === Subject: Re: Raatikainen's critique of Chaitin > Not true. Godel's results are that for any [CapitalThorn]nite axiom system that > has enough power to produce theorems about natural numbers, there must > exist a statement that is not provable in the axiom system, but > nevertheless true if we were to assume that the axiom system is > consistent. Correction. I should have said any FAS that is as powerful as PA or more powerful than PA. > Chaitin's Theorem (about Omega, which is really his big result) shows > that only a [CapitalThorn]nite number of bits of Omega are possible to determine; > the only way to determine more bits is to call them axioms. Since the > bits can be represented as exponential diophantine equations (1 if > there are in[CapitalThorn]nitely many solutions, 0 if only [CapitalThorn]nitely many > solutions), we see that there are some statements in number theory > that are unknowable. Much stronger than Godel's result, as Godel's > theorem gives a way to get around the incompleteness, by assuming that > the axiom system is consistent. Chaitin's theorem shows that there is > no way to get around it, since assuming that the axiom system is > consistent only improves the information complexity of the axiom > system by a [CapitalThorn]nite amount, so you can still only know a [CapitalThorn]nite number > of bits of Omega. I take this back that Chaitin's theorem shows that there is no way to get around this. One could conceivably get around it by adding lots of axioms to the FAS. But the important question (which is really what I was getting at) is for suf[CapitalThorn]ciently large N, can one conceive of enough plausible axioms with an information complexity greater than N? > Therefore, mathematics is as Chaitin calls it, quasi-empirical, just > like physics, in which theorems can only be proven by statistics and > observations; they are never certain, always subject to being possibly > refuted. Mathematicians don't like to hear this, since most of them > almost religiously believe that mathematics is absolute certainty. But > in reality, the absolute certainty is only for the trivial theorems > that they prove, as Chaitin showed. The majority of mathematical facts > are simply unattainable through deductive reasoning; only > inductive/empirical methods work (and not so well). > > Wrong, as a matter of fact. Did Godel show that randomness had > something to do with incompleteness? Did he show the existence of a > maximally unknowable number? These are not the only things, btw.... > Many of these statements are actually philosophically very > signi[CapitalThorn]cant, because they have something to do with epistemology! > > Bollocks the lot of it. All these discoveries of Chaitin's > are simply repackagings of the well-established notion of completeness. > Nope, only parts of Chaitin's discoveries are repackagings of Godel's > results, i.e., using the Berry paradox instead of the liar paradox. > The main part about Omega is not. > > It was the work of Godel, Turing and Church which > shook the foundations of mathematics in the 1930s not Chaitin many > decades later. > All of them did. The only difference is that because Chaitin's results > are so much stronger and are so shocking and go against the grain of > how most mathematicians think, most mathematicians don't want to > believe them; therefore, you get reactions that [CapitalThorn]t into the 4 > categories that I mentioned when I started this thread. I should add > that most mathematicians don't even understand Chaitin's results, > mainly because they don't want to understand them. > > This thread started with some good observations about the weakness of > Raatikainen's critique and then degenerated into a mudball about how > bad and naughty Chaitin is. If you are saying such a thing, give a > proper argument. He does not claim to shake the foundations of > mathematics, but he thinks he has shown that incompleteness should be > taken more seriously. > > Perhaps you would be better served by learning to read. I was replying > to Feinstein not Chaitin. shake up the foundations of mathematics > is Feinstein's phrase. > > Why do you object to that? > > Have you stopped beating your wife? > Chaitin did shake up the foundations of mathematics. The only problem > is mathematicians don't like his result, so they lose out and remain > ignorant. (This is my opinion.) > Craig === Subject: Re: Raatikainen's critique of Chaitin > Correction. I should have said any FAS that is as powerful as PA or > more powerful than PA. No, you should have removed the unnecessary assumption that the system has only a [CapitalThorn]nite number of axioms. As for the amount of arithmetic that the system needs to encompass, Robinson arithmetic will suf[CapitalThorn]ce. PA is overkill. But these are technicalities, and invoking such technicalities is itself overkill in the present context. We should instead all babble heartily, drinking mug upon mug of ale, thumping the table as we sing the hearty and barely intelligible songs of our respective countries. === Subject: Re: Raatikainen's critique of Chaitin > > Question the basic dogmas prevalent in modern science, for instance > evolution or the big bang or whether certain famous conjectures like > the Riemann Hypothesis or the Collatz Conjecture can even be proved > deductively, > What dogma? I am aware of no one who insists that either > of these problems *must* be decidable, say within ZFC. > Another pathetic straw man attack. > and suddenly you are the heretic. State arguments > informally (as Chaitin did when explaining his theories) and suddenly > you are a heretic. Everything must be spelled perfectly and presented > perfectly within the standard conventions or else it is wrong. If your > conclusions do not match the dogma of the times, then they are wrong. > If the readers don't understand your ideas, then the ideas are wrong. > Your paper on the Collatz is wrong, Mr Feinstein. http://arxiv.org/abs/math.GM/0312309 > If you state that your conclusion is that some theorems are true for > no reason (like Chaitin's conclusion about his work), the Pavlovian > response is not to try to understand what was meant by his comment in > the context of his arguments, > So when readers like Raatikainen do just that, you attack them > for so doing. > but rather the assertion that because > Chaitin did not de[CapitalThorn]ne true for no reason, he is babbling nonsense. > There we are. He is making a hyperbolic assertion, and not de[CapitalThorn]ning > his terms. > Metaphoric and informal statements are blasphemy in today's science. > Scientists (like those who posted in this thread) believe that it is > their duty to misinterpret informal statements, play dumb, and claim > I don't know what your talking about. > My what? > Or You didn't de[CapitalThorn]ne what an > algorithm is. Or You didn't de[CapitalThorn]ne what Ôis' is. > I am no scientist, but to de[CapitalThorn]ne your terms is part of the burden > of proof, which you so notably shirk in your paper on the Collatz > conjecture, Mr Feinstein. > See http://www.amasci.com/weird/wclose.html > Why do you favour credulousness over careful scepticism, Mr Feinstein? Craig === Subject: Feinstein's paper on the Collatz problem was Re: Raatikainen's critique of Chaitin > Your paper on the Collatz is wrong, Mr Feinstein. > http://arxiv.org/abs/math.GM/0312309 This paper is nonsense. I refer to version 12 (sic). The statement of theorem 2 mentions a sequence of integers modulo 2 as being initially unknown. What does that mean? Which sequences of integers are initially unknown? The statement also speaks of something requiring at least m computations? What does that mean? The author seems to have a private theory that decomposes an algorithmic operation into a number of discrete calculations. But he doesn't tell us what is a calculation in his theory. Now consider the alleged proof of this theorem. Feinstein states that in order to determine the value of T^(m)(n) it is necessarly to determine the values of (n, T^(1)(n),...,T^(m-1)(n)) mod 2. This isn't so. A smarter algorithm that Feinstein's would look at the power of 2 dividing a T^(j)(n) and take that out immediately (easy on a binary computer) and not even compute some of the subsequent T^(k)(n). In general then, it is not necessary to compute what Feinstein says we have to compute. Of course, there may be smarter algorithms which perform faster. Feinstein argues that because his (the crudest possible algorithm) takes m steps then all algorithms take m computations. This is a non-sequitur. The proof of Theorem 3 is nonsense too. I note that Mr Feinstein has already withdrawn three papers from the arXiv. One wonders what's so different about this one; that he hasn't withdrawn it too. > Why do you favour credulousness over careful scepticism, Mr Feinstein? Again, why? -- 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: Feinstein's paper on the Collatz problem was Re: Raatikainen's critique of Chaitin > > > Your paper on the Collatz is wrong, Mr Feinstein. > > http://arxiv.org/abs/math.GM/0312309 > This paper is nonsense. > I refer to version 12 (sic). > The statement of theorem 2 mentions a sequence of integers > modulo 2 as being initially unknown. What does that mean? > Which sequences of integers are initially unknown? > The statement also speaks of something requiring at least m > computations? What does that mean? The author seems to have > a private theory that decomposes an algorithmic operation into > a number of discrete calculations. But he doesn't tell us what > is a calculation in his theory. > Now consider the alleged proof of this theorem. Feinstein > states that in order to determine the value of T^(m)(n) it is > necessarly to determine the values of (n, T^(1)(n),...,T^(m-1)(n)) > mod 2. This isn't so. A smarter algorithm that Feinstein's > would look at the power of 2 dividing a T^(j)(n) and take that out > immediately (easy on a binary computer) and not even compute > some of the subsequent T^(k)(n). In general then, it is not necessary > to compute what Feinstein says we have to compute. Of course, there > may be smarter algorithms which perform faster. Feinstein > argues that because his (the crudest possible algorithm) takes > m steps then all algorithms take m computations. This is > a non-sequitur. > The proof of Theorem 3 is nonsense too. > I note that Mr Feinstein has already withdrawn three papers from the arXiv. > One wonders what's so different about this one; that he hasn't withdrawn > it too. > > Why do you favour credulousness over careful scepticism, Mr Feinstein? > Again, why? http://arxiv.org/abs/math.GM/0312309 publicity. Unfortunately, I don't have time to argue with you about it. I am of the opinion that even though I did not de[CapitalThorn]ne computations or initially unknown, an intelligent reader should be able to overcome these dif[CapitalThorn]culties by looking at these terms within the context of my arguments. Craig === Subject: Re: Feinstein's paper on the Collatz problem was Re: Raatikainen's critique of Chaitin > I am of the opinion that even > though I did not de[CapitalThorn]ne computations or initially unknown, an > intelligent reader should be able to overcome these dif[CapitalThorn]culties by > looking at these terms within the context of my arguments. More precisely, by gulping down a very large number of mugs of ale! === Subject: Re: Raatikainen's critique of Chaitin > Craig, > you make some valid points about unhealthy aspects of the modern > scienti[CapitalThorn]c enterprise. Many other people feel the same way. However, > you haven't proposed any alternative to improve what you don't like. > The only suggestion you have made so far is that mathematics should > be more like physics; that would not solve any of the problems > listed above, and it could make the situation in mathematics > publishing worse (depending on exactly what it would mean, which > you did not de[CapitalThorn]ne). See http://amasci.com/freenrg/newidea1.html> I don't have any solutions. My suggestion (which came from Chaitin) that mathematics be thought of more like physics has nothing to do with the unhealthy aspects of the modern scienti[CapitalThorn]c enterprise. > Don't romanticize the science during Renaissance, or any other time in > history. Heretics were not encouraged in their heresies, and some were > burned at stake. Scientists had to be very careful not to upset their > political masters. Dissent in science, as in other [CapitalThorn]elds, is more accepted > today than in any previous century. You are right that NSF will not > give you a grant for research that does not follow the methodology > established for a given [CapitalThorn]eld; but you still have more freedom to > pursue your ideas than someone in a similar situation in the past. That is de[CapitalThorn]nitely true. > Seems you have some ideas that you would like to have accepted by > mainstream mathematicians. In mathematics, new ideas are accepted when they > (a) help solve known open problems, or > (b) simplify existing theories, or > (c) unify seemingly unrelated areas of mathematics, or > (d) lead to new nontrivial problems. > One can also add > (e) translate some area outside of mathematics into mathematical language, > which is really applied math. Typically mathematicians are interested in (e) > if it also means (d). > If your ideas don't contribute any of (a) to (d), you will have hard time > getting mathematicians interested. > Then of course there is everyday mathematics research, which doesn't do > any of the above but it ßeshes out existing theories in routine fashion. > Not exciting perhaps, but necessary and useful nevertheless. > So if you have new revolutionary ideas that you want accepted by > mathematicians, you will have to do (a) or (b) or (c) or (d). If you don't > care about being accepted, then everything is much easier --- you don't > have to worry about any of the above. Craig === Subject: Re: Sin Cos Tan, why not Sin Sec Tan? > Narasimham G.L. schrieb im Newsbeitrag > The word is derived from Sanskrit. > Am more interested in maths and less in its history or claims relating > to historical precedences. :)... > Then don't make claims about historical precedences. > Phil My responses were based on post number 27 James Dolan about name trigonometry and post number 40 Thomas Heye about Greek connection. In post number 41 I have mentioned more about etymology, not history. So that this thread discusion be of some use to the OP Cassandra, further historical talk may please be avoided. === Subject: Re: Sin Cos Tan, why not Sin Sec Tan? top-posting imposes a bit of need for including referents, maybe less conversational, instead of just via context of the inundation of prior threading, which is supposed to be readily available, anyway, herein the googlag archipelagoog. mainly, it saves a lot of scrolling, as cooly paleological as that might be for you monastery mathguys! yes, the Sanskrit mathguys could have been speaking of trigonometry, when teh Gree mathguys were writing it on impermanent stuff ... one of my favorite books is _The Myth of Invariance_. > Up until now I thought top posting was annoying. And after now too. --Chair Man George -- follow the money to Strep Throat! http://tarpley.net === Subject: Re: Cyclic (Z_p^k)^* rom: Van Jacques === Subject: Re: Cyclic (Z_p^k)^* >I have gone over your posts, and am convinced. Good good. I've revised my rendition of Beachy's approach to directly construct a generator of (Z_p^k)^* from a generator of Z_p^* This is approach is more direct than the approach you presented (in next section) and now that I've converted Beachy's approach into a construction of a generator, I see no advantage to the other approach, Beachy's does the same and is much more direct about it. For example, Beachy's construts a generator for (Z_p^2)^*, while the other approach hems and haws about there being an unspeci[CapitalThorn]ed generator to be found in a group of up to four integers which is then the generator for the rest of the (Z_p^k)^*'s for k > 2. prime p ==> Z_p^* cyclic. Previous theorem. 2 <= p, 1 <= k, a = b (mod p^k) ==> a^p = b^p (mod p^(k+1)) some n with a = b + n.p^k a^p = b^p + p.b^(p-1) n.p^k + ... = b^p (mod p^(k+1)) odd prime p, 2 <= k, [1 + p] in (Z_p^k)^* ==> o(1 + p) = p^(k-1) (1 + p)^(p^(k-2)) = 1 + p^(k-1) (mod p^k) (1 + p)^p^0 = 1 + p^1 (mod p^2) (base step) (1 + p)^p^(j-1) = (1 + p^(j-1))^p (induction step) = 1 + p.p^(j-1) + p(p-1)/2 * p^2(j-1) + ... = 1 + p^j (mod p^(j+1)). 3(j-1) = j + 2j-3 >= j + 1 (1 + p)^p^(k-1) = 1 (mod p^k). 1 + p = 1 (mod p) & induction (1 + p)^p^(k-2) /= 1 (mod p^k). Otherwise p^(k-1) = 0 (mod p^k) odd prime p, 2 <= k ==> (Z_p^k)^* cyclic let g generate Z_p; g^(p-1) = 1 (mod p); let b = g^p^(k-1) b^(p-1) = 1 (mod p^k); o(b) | p-1; b^o(b) = 1 (mod p^k) g^o(b) = b^o(b) = 1 (mod p); p-1 | o(b); o(b) = p-1 o(1 + p) = p^(k-1); p^(k-1), p-1 coprime o(g(1 + p)) = o(g).o(1 + p) = phi(p^k) -- other approach prime p, g generates (Z_p^m)^* ==> o(g) in (Z_p^(m+1))^* = phi(p^m) or phi(p^(m+1)) g^o(g) = 1 (mod p^(m+1)); g^o(g) = 1 (mod p^m); phi(p^m) | o(g) o(g) | phi(p^(m+1)); o(g)/phi(p^m) | p; o(g)/phi(p^m) = 1 or p prime p, 2 <= m, g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod p^(m+2)) let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops prime p, g generates Z_p^* ==> g or g + p generates (Z_p^2)^* if (g + p)^(p-1) = 1 = g^(p-1) (mod p^2): 1 = g^(p-1) + (p-1) g^(p-2) p + ... = 1 - p.g^(p-2) (mod p^2) p.g^(p-2) = 0 (mod p^2); p = p.g^(p-1) = 0 (mod p^2), whoops odd prime p ==> some g generator (Z_p^2)^* with g^p(p-1) /= 1 (mod p^3) some d generates Z_p^*; d or d + p generates (Z_p^2)^* d + 2p generates Z_p^*; d + 2p or d + 3p generates (Z_p^2)^* some a in { 1,2,3 }, g with g, g + ap generate (Z_p^2)^* (g + ap)^p(p-1) = (g^p + p.ap.g^(p-1) + p(p-1)/2 * (ap)^2 g^(p-2) + ... )^(p-1) = (g^p + a^2 p^2 g^(p-1))^(p-1) = g^p(p-1) + (p-1)a^2 p^2 g^(p-1)^2 + ... = g^p(p-1) + (p-1)a^2 p^2 g^(p-1)^2 = g^p(p-1) - a^2 p^2 g^(p-1)^2 (mod p^3) if g^p(p-1) = 1 = (g + ap)^p(p-1) (mod p^3): 0 = a^2 p^2 g^(p-1)^2 (mod p^3) 0 = a^2 g^(p-1)^2 = a^2 (mod p); p | a; p = 3 2 generates (Z_3^2)^* = { 1,2, 4,5, 7,8 }; 2^6 = 10 (mod 3^2) ---- === Subject: Re: strange 1 > 1. Log z = ln r + i@ (r>0 , -Pi < @ < Pi) > 2. Log z = ln r + i@ (r>0, -Pi < @ <= Pi) > which of case is right ?? > is it no problem ? Yes, it is so simple, even 3. Log z = ln r + i@ (r>0, -Pi <= @ < Pi) says the same thing, because on the Argand diagram @ undergoes one full rotation whichever way, @= +Pi or -Pi are same directions, viz., negative y-axis. === Subject: Re: strange 1 > The pricipal branch is a complete plane, so > 2. Log z = ln r + i@ (r>0, -Pi < @ <= Pi > is true. > Alternative: > Log z = ln r + i@ (r>0, 0 <= @ < 2*Pi. > Go to Advanced Groups search in Google-Groups > enter log z (or whatever You are on , residue,...) > and sci.math.. > If You get too much results, re[CapitalThorn]ne the search by > entering a name of Your vib (=vibration, someone, > You can understand), lets say KPledger. > In this case here Title branch of log z might help. > Have a nice weekend. > Hero > P.S. Don't forget: > i is no loger imaginary. but, i have the book of James Ward Brown Ruel V Churchill. there is -Pi < @ < Pi. but i have the other book. there is -Pi < @ <= Pi. um.....i am confusing. === Subject: Re: strange 1 Take another function, let's say cosinus: It's de[CapitalThorn]ned for every real number r, for every r You can compute or by geometrical means one can [CapitalThorn]nd a f(r)=cos r. Now, Inverse function - that's a holy wish, nearly never comes true. Either it's not invers, or not a function (for one s You can [CapitalThorn]nd several r's with f(r1)=s and f(r2)=s) or there are numbers s with: there is no r with f(r)=s. Now a mathematical plane, every point z has it's coordinates z=(a,b) and the function e gives to every point z one f(z)=e^z (e to the power of z)=e^((a,b))=e^(a+i*b) e^((a,0))=e^(a) is a positive number, e^((0,b))=e^(i*b)=(e^(b) )^(i), this is a direction, to be more precise,it has unit 1 length and has the direction b as it can also be expressed by e^(i*b)= (cos b, sin b) So, for every z one can [CapitalThorn]nd it's length and it's direction. The invers, the log, for a length and a direction [CapitalThorn]nd a point z (with reference to the origin) with e^z = length *(cos @, sin @) There is the zero, which gives troubles, so exclude it in Your de[CapitalThorn]nition. And the @'s , the angels ? You can look at angels as rotations, as movements - or as difference of direction. In the case of rotations, You can do several rounds, repeatedly looking into the same direction, so one chooses only a principal branch or one rotation (to keep it to a function) Or as difference of direction, one can have every angle from zero to nearly 2*pi (but 2*pi itself is the same diretion as zero). (A day makes 24 hours, but there's no 24 o'clock or ten minutes past 24 o'clock - it' zero hours ten minutes) Puh, [CapitalThorn]nally You start with minus pi (but not the direction itself), and crossing the zero coming to plus pi (but not the point itself) then You exclude the negative numbers, the left part of the x-axis. Is there any reason, why one should do it ? So, either include + pi or minus pi. That is my opinion, and You have an opinion of one book and an opinion of another book - which one is right, You can [CapitalThorn]nd out Yourself. Have fun Hero === Subject: Re: A simple question about integers >Hello Eray, >Eray Ozkural exa schrieb im Newsbeitrag > Hello there, > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > In Z, are there integers with an in[CapitalThorn]nite number of digits? >Suppose there were only numbers with [CapitalThorn]nitely many digits ok. > (say N). no, you can't say N here. that's assuming that the number of digits is -bounded-, which saying much more than that it's a;ways [CapitalThorn]nite. ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: A simple question about integers >Suppose there were only numbers with [CapitalThorn]nitely many digits > (say N). > no, you can't say N here. that's assuming that ... Well he can, because he said Suppose .... What you wanted to express is: if you do so, dear TH, then you will get an answer to a question, which wasn't posed. > the number of digits is -bounded-, which is saying much > more than that it's always [CapitalThorn]nite. Quite so. But you can express a true statement in a wrong way too :-)) Rainer Rosenthal r.rosenthal@web.de === Subject: Re: A simple question about integers >Suppose there were only numbers with [CapitalThorn]nitely many digits > (say N). > no, you can't say N here. that's assuming that ... >Well he can, because he said Suppose .... >What you wanted to express is: if you do so, dear TH, then >you will get an answer to a question, which wasn't posed. i did -not- say that he couldn't say what you say i said he > the number of digits is -bounded-, which is saying much > more than that it's always [CapitalThorn]nite. >Quite so. But you can express a true statement in a wrong >way too :-)) >Rainer Rosenthal >r.rosenthal@web.de ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: A simple question about integers > i did -not- say that he couldn't say what you > say i said he couldn't say! Ah well - *now* everything's becoming more clear :-)) Sorry for that. I really tried now as you suggested. But - hmmm... I did express what I thought should be expressed expressis verbis to express my impression of your expressions. Which may be wrong (the impression). One thing is really awkward: I saw OP in a lengthy philosophical discussion. Don't they need for all and there exists there? I agree with Robin Chapmans word juvenile usage and estimate the OP's age < 20. So there is good hope for him learning all the necessary basic stuff. I was pleased to hear him being in doubt about his intuition. Most of the time people stubbornly stick to this idea of an in[CapitalThorn]nite number as a must-be in Z. These people came to de.sci.mathematik as well and they will appear again - for sure :-(( Rainer Rosenthal r.rosenthal@web.de === Subject: Re: A simple question about integers > Hello there, > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > In Z, are there integers with an in[CapitalThorn]nite number of digits? This question shows that you are mired in crass formalism, unable to distinguish between mathematical objects and their representations. Also talking about an in[CapitalThorn]nite number of things rather than in[CapitalThorn]nitely many is a juvenile usage. Let k be an integer with k >= 2. The positive integer n has at most n digits in base k. The proof is an easy induction on n. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9 Francis Wheen, _How Mumbo-Jumbo Conquered the World_ === Subject: Re: A simple question about integers > Hello there, > > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > > In Z, are there integers with an in[CapitalThorn]nite number of digits? > No. Absolutely not. > There is no bound on the number of digits they may have, but > integers must always have a [CapitalThorn]nite number of digits. > The distinction between Ôpotentially in[CapitalThorn]nite' and Ôunbounded > but always [CapitalThorn]nite' often causes confusion, but is essentially > not a dif[CapitalThorn]cult concept. So, it's like the case with the ID of a TM that is Ôunbounded but always [CapitalThorn]nite'. Is there a canonical reference for this distinction which I defended vigorously in the past? I ask this because in simple intuitive terms I was unable to accomplish as such. I am afraid there is something wrong with my intuition. -- Eray Ozkural === Subject: Re: A simple question about integers > Hello there, > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > In Z, are there integers with an in[CapitalThorn]nite number of digits? > -- > Eray Ozkural No! But there is always another integer with more digits than any one you point to. === Subject: Re: A simple question about integers > Hello Eray, > Eray Ozkural exa schrieb im Newsbeitrag > Hello there, > I was trying to explain some basic ideas in number theory to a friend > in plain language, and I had a dif[CapitalThorn]culty with a concept. > In Z, are there integers with an in[CapitalThorn]nite number of digits? > Suppose there were only numbers with [CapitalThorn]nitely many digits (say N). (The base > in which you represent won't matter; but to make it clear, take decimal > digits). Let x be a number with N digits (we can admit leading zeroes here). > Then the absolute value of all these numbers is less than or equal to > 9.10` 0 +... +9.10^{N-1} > this means that there is a smallest negative / greatest positive integer; > this is false. > To put it in other words: If x is a number with N digits (including leading > zeroes), then say 10x+1 has N+1 digits, so there cannot be a greatest number > of decimal digits. > Thomas > -- > Eray Ozkural idiot. that doesn't prove there are any integers in Z with in[CapitalThorn]nite digits. all it proves is there is no [CapitalThorn]nite limit to the amount of digits of integers in Z. this kind of misleading bull is what plagues these newgroups === Subject: Convergence speed The last graphic in the section Finding Roots Of Equations Numerically With Bucky Numbers at: http://users.adelphia.net/~cnelson9/Links/index_lnk_11.html which is one of the sections of Synergetics Coordinates Applications at: http://users.adelphia.net/~cnelson9/ is followed by a question about the speed of the convergence of a general root [CapitalThorn]nding method. The guesses for the third iteration on, could be found with the vector equation method shown in the section Solving Matrix Problems Using Bucky Numbers at: http://users.adelphia.net/~cnelson9/Links/index_lnk_10.html I don't think Newton's method can be beat, I've tried and tried, but if anyone could beat Newton it would be Bucky Fuller. How does the speed of convergence compare for polynomials in general? What is the traditional name for this kind of root [CapitalThorn]nding method? Cliff Nelson Dry your tears, there's more fun for your ears, Forward Into The Past 2 PM to 5 PM, Sundays, California time, at: http://www.kspc.org/ Don't be a square or a blockhead; see: http://users.adelphia.net/~cnelson9/ === Subject: Re: NEWSFLASH: Closed form solution of the quintic & above > I'm VERY skeptical. > http://science.slashdot.org > http://arxiv.org/pdf/math.CA/0408264 > No problem! Supply a polynomial whose exact solution is not amenable > to the proposed method. Put up or shut up. > A relevant criticism would be the time needed to calculate the > solution vs. the degree of the polynomial. x^2-1=0 As Robin Chapman has been pointing out in a different thread the method requires that the solution x of xf(x)=s can be expressed in a power series and therefore f(x) must not vanish in x=0. The author does not mention this condition. Whether the method has any practical value at all has not been a consideration of the author. Probably not, because the degree of the polynomials involved to calculate de coef[CapitalThorn]cients of the ODE increase exponentially with the degree of the original polynomial. -- Jos van Kan === Subject: Re: Cyclic (Z_p^k)^* >I have gone over your posts, and am convinced. > Good good. I've revised my rendition of Beachy's approach to > directly construct a generator of (Z_p^k)^* from a generator of Z_p^* > This is approach is more direct than the approach you presented > (in next section) and now that I've converted Beachy's approach into a > construction of a generator, I see no advantage to the other approach, > Beachy's does the same and is much more direct about it. For example, > Beachy's construts a generator for (Z_p^2)^*, while the other approach > hems and haws about there being an unspeci[CapitalThorn]ed generator to be found > in a group of up to four integers which is then the generator for > the rest of the (Z_p^k)^*'s for k > 2. I assume that you are speaking of some printed notes, not available online, as I haven't been able to [CapitalThorn]nd anything like this online. > prime p ==> Z_p^* cyclic. Previous theorem. > 2 <= p, 1 <= k, a = b (mod p^k) ==> a^p = b^p (mod p^(k+1)) > some n with a = b + n.p^k > a^p = b^p + p.b^(p-1) n.p^k + ... = b^p (mod p^(k+1)) > odd prime p, 2 <= k, [1 + p] in (Z_p^k)^* ==> o(1 + p) = p^(k-1) > (1 + p)^(p^(k-2)) = 1 + p^(k-1) (mod p^k) > (1 + p)^p^0 = 1 + p^1 (mod p^2) (base step) > (1 + p)^p^(j-1) = (1 + p^(j-1))^p (induction step) This is only true for the 1st term in the binomial expansions, and fails for the 2nd term. e.g., p = 3, j-1 = 2 gives 4^9 /= 10^3 > = 1 + p.p^(j-1) + p(p-1)/2 * p^2(j-1) + ... > = 1 + p^j (mod p^(j+1)). 3(j-1) = j + 2j-3 >= j + 1 Perhaps you mean mod p^(j+1). That must be it. > (1 + p)^p^(k-1) = 1 (mod p^k). 1 + p = 1 (mod p) & induction > (1 + p)^p^(k-2) /= 1 (mod p^k). Otherwise p^(k-1) = 0 (mod p^k) Yes. > odd prime p, 2 <= k ==> (Z_p^k)^* cyclic > let g generate Z_p; g^(p-1) = 1 (mod p); let b = g^p^(k-1) > b^(p-1) = 1 (mod p^k); o(b) | p-1; b^o(b) = 1 (mod p^k) > g^o(b) = b^o(b) = 1 (mod p); p-1 | o(b); o(b) = p-1 I don't see why g^o(b) = b^o(b) mod p, unless you are using b = g^p^(k-1) = g mod p (from g^p = g mod p). Must be this. OK. > o(1 + p) = p^(k-1); p^(k-1), p-1 coprime > o(g(1 + p)) = o(g).o(1 + p) = phi(p^k) Yes. Very clear. I like this better than the other method. > -- other approach This is related to the problem from Wilkins that I posted. > prime p, g generates (Z_p^m)^* > ==> o(g) in (Z_p^(m+1))^* = phi(p^m) or phi(p^(m+1)) > g^o(g) = 1 (mod p^(m+1)); g^o(g) = 1 (mod p^m); phi(p^m) | o(g) > o(g) | phi(p^(m+1)); o(g)/phi(p^m) | p; o(g)/phi(p^m) = 1 or p > prime p, 2 <= m, g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod > p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops This isn't clear to me. Van === Subject: Re: Cyclic (Z_p^k)^* > I assume that you are speaking of some printed notes, not available > online, as I haven't been able to [CapitalThorn]nd anything like this online. Beachy is on line and was quoted in one of these related threads. What I'm presenting is my work guided by his outline. > prime p ==> Z_p^* cyclic. Previous theorem. > 2 <= p, 1 <= k, a = b (mod p^k) ==> a^p = b^p (mod p^(k+1)) > some n with a = b + n.p^k > a^p = b^p + p.b^(p-1) n.p^k + ... = b^p (mod p^(k+1)) > odd prime p, 2 <= k, [1 + p] in (Z_p^k)^* ==> o(1 + p) = p^(k-1) > (1 + p)^(p^(k-2)) = 1 + p^(k-1) (mod p^k) > (1 + p)^p^0 = 1 + p^1 (mod p^2) (base step) > (1 + p)^p^(j-1) = (1 + p^(j-1))^p (induction step) > = 1 + p.p^(j-1) + p(p-1)/2 * p^2(j-1) + ... > = 1 + p^j (mod p^(j+1)). 3(j-1) = j + 2j-3 >= j + 1 > Perhaps you mean mod p^(j+1). That must be it. Yup, a = b = c = d (mod n) is all in Z_n > (1 + p)^p^(k-1) = 1 (mod p^k). 1 + p = 1 (mod p) & induction > (1 + p)^p^(k-2) /= 1 (mod p^k). Otherwise p^(k-1) = 0 (mod p^k) > odd prime p, 2 <= k ==> (Z_p^k)^* cyclic > let g generate Z_p; g^(p-1) = 1 (mod p); let b = g^p^(k-1) > b^(p-1) = 1 (mod p^k); o(b) | p-1; b^o(b) = 1 (mod p^k) > g^o(b) = b^o(b) = 1 (mod p); p-1 | o(b); o(b) = p-1 > I don't see why g^o(b) = b^o(b) mod p, unless you are using > b = g^p^(k-1) = g mod p (from g^p = g mod p). Must be this. OK. > o(1 + p) = p^(k-1); p^(k-1), p-1 coprime > o(g(1 + p)) = o(g).o(1 + p) = phi(p^k) > Yes. Very clear. I like this better than the other method. > -- other approach > prime p, g generates (Z_p^m)^* > ==> o(g) in (Z_p^(m+1))^* = phi(p^m) or phi(p^(m+1)) > g^o(g) = 1 (mod p^(m+1)); g^o(g) = 1 (mod p^m); phi(p^m) | o(g) > o(g) | phi(p^(m+1)); o(g)/phi(p^m) | p; o(g)/phi(p^m) = 1 or p > prime p, 2 <= m, g generates (Z_p^m)^*, g^phi(p^m) /= 1 (mod p^(m+1)) > ==> g generates (Z_p^(m+1))^*, g^phi(p^(m+1)) /= 1 (mod > p^(m+2)) > let b = g^phi(p^m); if b^p = g^phi(p^(m+1)) = 1 (mod p^(m+2)): > (b - 1)(b^(p-1) +..+ 1) = b^p - 1 = 0 (mod p^(m+2)) > p^m | b-1; not p^(m+1) | b-1; p^2 | b^(p-1) +..+ 1 > b = 1 (mod p^m); b = 1 (mod p^2); b^j = 1 (mod p^2) > 0 = b^(p-1) +..+ 1 = p (mod p^2); p^2 | p, whoops > This isn't clear to me. Proof by contradiction of g^... /= 1 (mod p^...). === Subject: Math is such a BORING science ! Numbers numbers numbers, really B O R I N G ! Real men study science like chemistry, where you can blow yourself up our dissolve yourself with acid. Math is for wieners and pussies only. === Subject: Re: Math is such a BORING science ! > Numbers numbers numbers, really B O R I N G ! > Real men study science like chemistry, where you can blow yourself up > our dissolve yourself with acid. > Math is for wieners and pussies only. Numbers? I think you're confusing math with accounting. :p === Subject: Re: Math is such a BORING science ! > Numbers numbers numbers, really B O R I N G ! > Real men study science like chemistry, where you can blow yourself up > our dissolve yourself with acid. > Math is for wieners and pussies only. math is philosophy, not science.. but regarding your last statement.. it sounds like the perfect discipline for you! === Subject: Re: Math is such a BORING science ! >Numbers numbers numbers, really B O R I N G ! >Real men study science like chemistry, where you can blow yourself up >our dissolve yourself with acid. >Math is for wieners and pussies only. And yet you hang around here... -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Re: Math is such a BORING science ! >Math is for wieners and pussies only. >And yet you hang around here... ... and Matthew P. Wiener no longer does. -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril === Subject: Re: Math is such a BORING science ! > Numbers numbers numbers, really B O R I N G ! > Real men study science like chemistry, where you can blow yourself up > our dissolve yourself with acid. > Math is for wieners and pussies only. Math can help blow things up. Check out http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/ Ulam.html for a brief biography of Stanislaw Marcin Ulam. He was a pretty good mathematician (though not as well known as many others). He devised the Monte Carlo method. Anyway, this person, trained as a topologist under the mentoring of Banach, went on the create the hydrogen bomb. He solved the problem on how to start fusion which was necessary for the loud boom. He also devised nuclear pulse propulsion, which NASA uses to power space craft. I think with a hydrogen bomb you could blow up more than yourself. Talk about boring, dissolving yourself in acid. Who cares? I guess the janitor that has to clean up a pile of mess that was once your pathetic life. Nuclear war, well, there wouldn't be a janitor left to clean up any mess. See, math can be used to blow things up. - Tim -- Timothy M. Brauch NSF Fellow Department of Mathematics University of Louisville email is: news (dot) post (at) tbrauch (dot) com === Subject: Re: Math is such a BORING science ! >Numbers numbers numbers, really B O R I N G ! >Real men study science like chemistry, where you can blow >yourself up our dissolve yourself with acid. >Math is for wieners and pussies only. But as James Harris has demonstrated, real men can be exploded and dissolved in acid even when they're only doing mathematics. -- Angus Rodgers (angus_prune@ eats spam; reply to angusrod@) Contains mild peril === Subject: Re: Comments on the proof of cyclic Z*_p^k >I have been looking some more at Z*_p^k from a different >point of view. (As I am not a mathematician, its an elementary >POV). > Each to his own taste, of course. The bit I gave about exp, > log, and the p-adics actually feels simpler to me than messing > around with the binomial theorem; the opposite may be true for > you. I have no quarrel with that. It gives me something to learn though. >One can show from the binomial thm. that >(1 + p)^(p^b) mod p^k > 1 if b < k-1, and >(1 + p)^(p^(k-1)) mod p^k = 1, so o(1 + p) = p^(k-1), >and so ker f is a cyclic subgroup of Z*_p^k. > Not meaning to cavil, but I would avoid statements of the > form m mod n > k because there is no order structure on > Z_n compatible with the ring structure. You meant of > course (1 + p)^(p^b) mod p^k =/= 1 if ... You are quite right, of course. One should not use statements about order when there is no order. Doing math means being careful about things like this, right? >Finding the elements of Z*_p^k which form a cyclic subgroup of >order p-1 is a bit more dif[CapitalThorn]cult--I haven't done this yet. >But Z*_p^k/ker f =~ Z*_p which is cyclic, and since ker f and >Z*_p are cyclic, (p-1,p) = 1 and Z*_p / ker f = (1), I >think it follows that Z*_p^k is cyclic. >(I was thinking of the thm. that if A,B are normal subgroups of G >with G = AB (g = ab with a in A anad b in B, for all g in G), and >if A / B = (1), then G = A x B. Then if (|A|,|B|) = 1, G is cyclic (?) > Yes, if A and B are normal *sub*groups of G and their orders > are relatively prime, then G ~ A x B; if in addition A and B > are cyclic then so is G. But what we have here is a subgroup > A (ker f) and the *quotient* B = Z_{p^k}*/ker f, which is an > entirely different story. Unless of course we could view B > as a subgroup of Z_{p^k}* (by *splitting* the surjection > Z^{p^k}* --> B, i.e., giving an injection B --> Z_{p^k}* whose > composite with the surjection is the identity on B). But this > brings us back to the problem of [CapitalThorn]nding a suitable cyclic > subgroup of order p-1 in Z_{p^k}*! Yes. > On the other hand, there is a theorem that any [CapitalThorn]nite abelian > group is isomorphic to a direct sum of [CapitalThorn]nite cyclic groups of > prime power orders, and this very quickly leads to a proof of > the thing you want. I am missing something here. Although that's like cracking a nut with > a sledgehammer -- the argument using short exact sequences is > a somewhat less brutal tool, IMO. > Todd === Subject: Re: I do not have any more time for you cowards, but James is more a man than any of you > Your little group of censorious little cretins ßunked out in my > estimation. James has been intemperate and hot headed. What is NOT > missing is his heart and his basic sense of being a MAN, rather than a > spineless whelp like many of you who worship orthodoxy without > understanding it. > I have had my disagreements with James on certain issues, but he is > more a MAN in the sense of what counts than any of you whelps. And > with that, I go back to my work. > I despise you for picking on him . I LOATHE cowards, and that is what > most of you are. Yes. It takes a lot of courage to type nasty things on your keyboard, locked safely away in your house. > Your group ßunked out in my estimation. While you are at it, grow > some balls if you can and attack some REAL basic problems in > mathematics, rather than act like censorious little cretins who only > know how to critique, rather than create something new. > Andrew Beckwith, PhD There are a few people here who have at least tried to be of some help to people who come here with questions. I'm not clear on what you are doing here, except calling people names. I fear most of us will be forgotten completely in a couple hundred years or so. It would be nice to imagine that there is a Gauss or Galileo here, but I fear its not so, we just have to do our best with our meager gifts. === Subject: Re: Amateur math, neat relation > I'm going to give a derivation again, and talk about what I face with > today's mathematicians so you can see how they operate and how they > make certain that amateur mathematicians are cut out. > The simplest way to understand what I did is to consider the count of > odd composites up to and including N that are divisible by 3, as > 3C <= N > where C is the count of all composites, so C <= N/3, but every odd > natural besides one can be written as 2k+1, where k is a natural, so > you can replace C with 2k+1, to get > 3(2k+1) <= N, > and solving you get > k <= (N-3)/6, and using ßoor(x) = [x], you have > k = [(N-3)/6] as the count of composites. > When I did my original research over two years ago, knowing that N was > to be even, I only cared about odds so I used the equivalent of > 3(2k+1) <= N-1 > which gives k = [(N-4)/6], which does work as long as N is even. > Here the brackets mean to just take the integer part, like [1.3333...] > = 1. > Now, notice that instead of 3 I can use any odd prime, so I have > p(2k+1)<=N, so k = [(N-p)/2p] > is the count of *odd* composites up to and including N that have p as > a factor, where p is an odd prime. > BUT, you can also get that count by using [N/p] - [N/2p] - 1, so I > have as a fundamental relationship that > [N/p] - [N/2p] - 1 = [(N-p)/2p] > and in general > [N/j] - [N/2j] - 1 = [(N-j)/2j] > where j is a natural greater than 1. > You can of course tweak it a bit so that you have > j(2k-1)<=N, so k = [(N+j)/2j] > and then the formula becomes > [N/j] - [N/2j] = [(N+j)/2j] > which was actually guessed by an anonymous poster a while back when I > was talking about my tidbits. > Some poster on sci.math made a post like that was a big deal, but in > mathematics guessing is one thing, proving is another. > I've seen a lot of talk about these simple relations already, as > if...as if mathematicians knew them before my work, and the > discussions I started. I think the story is getting more interesting as someone pointed out that you can get my relation using the trivial relation [x] + [x + 1/2] = [2x] in reals with x = N/2j, but how would you know to try? Check this out. It's also trivially true (still in reals) that [x] + [x + 1/k] = [2x], with k>1, and that means I can get an even more general relation, with x = N/2kj, where N, k and j are naturals, as I substitute to get [N/2kj] + [(N+2j)/2kj] = [N/kj] so that's [(N+2j)/2kj] = [N/kj] - [N/2kj] which is terribly neat!!! Ok that was mainly just for fun, returning to the subject at hand, let's look at [N/p] - [N/2p] - 1 = [(N-p)/2p] once again. I think what's interesting here is that mathematicians don't seem to have [CapitalThorn]gured out that they could compress counting composites in this way! So now I actually still need a citation!!! After all [x] + [x + 1/2] = [2x] IS trivial, but it takes a deeper understanding to [CapitalThorn]gure out that you can use that to get [N/p] - [N/2p] - 1 = [(N-p)/2p] which gives you that [(N-p)/2p] is the count of odd *composites* up to and including N that have p as a factor! That is, math is full of tools. Some people can use basic tools to give surprising and interesting answers but just because the tools are simple, does it make the work useless? Understand? Do any of you understand that just because you can derive something deep from simple tools that doesn't mean that the derived result is worthless? It's one thing to have a simple way to derive a relation. It's another to even know the relation exists to be derived! How about this? Go [CapitalThorn]nd a reference that talks about counting composites, like to count primes, using compressed forms in any math book out there in all of the world. Go see for yourself that simple does not necessarily mean easy. Or never learn. I think some of you think that math must mean really very complicated and so dif[CapitalThorn]cult to understand that few people in the world can get it so it must be good. That's not true. Math is a lot more than showing off to other people. It's far deeper. And if any of you love mathematics itself, versus needing to show off on sci.math, then you might listen. James Harris === Subject: Re: Amateur math, neat relation > So now I actually still need a citation!!! A professor once told me, If you need to know something, [CapitalThorn]nd it out. If you don't know how to [CapitalThorn]nd it out, [CapitalThorn]nd out how to [CapitalThorn]nd out! That's what science requires from a researcher. -- 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: Amateur math, neat relation > It's also trivially true (still in reals) that > [x] + [x + 1/k] = [2x], with k>1, It is also trivially false in many cases. Take x = 2/3, k = 6. -- 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: Amateur math, neat relation > It's also trivially true (still in reals) that > > [x] + [x + 1/k] = [2x], with k>1, > It is also trivially false in many cases. Take x = 2/3, k = 6. Aha! Finally, proof that mathematics is inconsistent. Jim Burns === Subject: Re: Amateur math, neat relation > [...] >I think the story is getting more interesting as someone pointed out >that you can get my relation using the trivial relation >[x] + [x + 1/2] = [2x] >in reals with x = N/2j, but how would you know to try? uh, right. nobody could possibly [CapitalThorn]gure out any of this trivia without you to guide them. >Check this out. It's also trivially true (still in reals) that >[x] + [x + 1/k] = [2x], with k>1, right. how do you -prove- that? [no help from the audience, please.] >[...] >I think what's interesting here is that mathematicians don't seem to >have [CapitalThorn]gured out that they could compress counting composites in this >way! i wouldn't know about that, but i'm -certain- that they have not [CapitalThorn]gured out that [x] + [x + 1/k] = [2x], with k>1. when you show us the proof of -that- you're going to be famous. >So now I actually still need a citation!!! >After all [x] + [x + 1/2] = [2x] IS trivial, but it takes a deeper >understanding to [CapitalThorn]gure out that you can use that to get >[N/p] - [N/2p] - 1 = [(N-p)/2p] >which gives you that [(N-p)/2p] is the count of odd *composites* up to >and including N that have p as a factor! >That is, math is full of tools. Some people can use basic tools to >give surprising and interesting answers but just because the tools are >simple, does it make the work useless? >Understand? Do any of you understand that just because you can derive >something deep from simple tools that doesn't mean that the derived >result is worthless? >It's one thing to have a simple way to derive a relation. It's >another to even know the relation exists to be derived! >How about this? Go [CapitalThorn]nd a reference that talks about counting >composites, like to count primes, using compressed forms in any math >book out there in all of the world. >Go see for yourself that simple does not necessarily mean easy. >Or never learn. I think some of you think that math must mean really >very complicated and so dif[CapitalThorn]cult to understand that few people in the >world can get it so it must be good. >That's not true. Math is a lot more than showing off to other people. >It's far deeper. And if any of you love mathematics itself, versus >needing to show off on sci.math, then you might listen. >James Harris ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Amateur math, neat relation > [N/j] - [N/2j] = [(N+j)/2j] > > Therefore > > [N/j] = [(N+j)/2j] + [N/2j] > > Next substituting [N/j]=N/j- (N mod j)/j we have > > N/j - (N mod j)/j = (N+j)/(2j)- ((N+j) mod 2j)/(2j) + N/(2j)- (N mod > 2j)/(2j) > > which modulo distribution law allow to expand to > > N/j - (N mod j)/j = N/j + 1/2 - (N mod 2j)/(2j) - (j mod 2j)/(2j) + N/(2j)- > (N mod 2j)/(2j) > Sorry but in my sense you made a small typo (?) error here since: > N/j + 1/2 != (N+j)/(2j) > Check it by yourself : for N = 3 and j = 2 we have 3/2 + 1/2 != (3+2)/4 > which could be reduced to > > (N mod j)/j = 1/2 - (N mod 2j)/j - (j mod 2j)/(2j) + N/(2j) > You forgot the minus before (N mod j)/j I think > Clearly (j mod 2j)/(2j)=1/2, so we have > > (N mod j)/j = (N mod 2j)/j + N/(2j) > You forgot the minus again before (N mod 2j)/j > or just > > N mod j = N mod 2j +1/2 > Why ? I received (omitting the previous errors) N mod j = N mod 2j + N/2 > which is absurd since > > integer != integer + 1/2 > > > Where is the error? > Are you aware of some relation between N and j that I'm not ? > I believe that that proof is false ... am I wrong ? N mod j = N mod 2j + N/2 is still absurd with the suitable selection of N. The bigger problem is distributive law, which is in fact: a+b mod c = ((a mod c) + (b mod c)) mod c Then Harris identity reduces to: (N mod 2j + j) mod 2j = j + 2(N mod j) - N mod 2j which I called Harris-Harakiri identity. (As I see no folowups in the thread, I'm beginning to suspect that the joke was not really that good). === Subject: Re: Amateur math, neat relation > [N/j] - [N/2j] - 1 = [(N-j)/2j] > where j is a natural greater than 1, in a math textbook. Claim 1: Let x be any positive real number and let [x] denote ßoor(x). Then [x] + [x + 1/2] = [2x] Proof: If the fractional part of x (i.e. x mod 1) is less than 1/2, and [x] = n, then [x + 1/2] = n and [2x] = 2n. If the fractional part of x is greater than or equal to 1/2, and [x] = n, then [x + 1/2] = n + 1 and [2x] = 2n + 1. Claim 2: For integer N, j, j > 1, [N/j] - [N/2j] - 1 = [(N-j)/2j] Proof: Let x = N/(2j). Then by Claim 1: [N/(2j)] + [N/(2j) + 1/2] = [N/j] so [N/j] - [N/(2j)] = [N/(2j) + 1/2] = [(N + j)/(2j)] = [(N - j)/(2j) + 1] = 1 + [(N - j)/(2j)] Knuth, The Art of Computer Programming, Volume 1, page 43, Exercise 38 has the more general formula [x] + [x + 1/n] + ... + [x + (n - 1)/n] = [n x] Knuth cites Charles Hermite, Acta Mathematica, Volume 5, page 315, (1884) as the original source for this formula. === Subject: Re: Amateur math, neat relation > > [N/j] - [N/2j] - 1 = [(N-j)/2j] > > where j is a natural greater than 1, in a math textbook. > Claim 1: Let x be any positive real number and let > [x] denote ßoor(x). Then > [x] + [x + 1/2] = [2x] Yup. And an important point is demonstrated here which I'll make at the end. > Proof: If the fractional part of x (i.e. x mod 1) is less > than 1/2, and [x] = n, then [x + 1/2] = n and [2x] = 2n. > If the fractional part of x is greater than or equal to 1/2, > and [x] = n, then [x + 1/2] = n + 1 and [2x] = 2n + 1. > Claim 2: For integer N, j, j > 1, > [N/j] - [N/2j] - 1 = [(N-j)/2j] > Proof: Let x = N/(2j). Then by Claim 1: > [N/(2j)] + [N/(2j) + 1/2] = [N/j] > so > [N/j] - [N/(2j)] = [N/(2j) + 1/2] = [(N + j)/(2j)] > > = [(N - j)/(2j) + 1] > = 1 + [(N - j)/(2j)] > Knuth, The Art of Computer Programming, Volume 1, page 43, > Exercise 38 has the more general formula > [x] + [x + 1/n] + ... + [x + (n - 1)/n] = [n x] > Knuth cites Charles Hermite, Acta Mathematica, Volume 5, > page 315, (1884) as the original source for this formula. Neat! In retrospect that IS quite obvious. I wonder if anyone has used it in counting primes... In any event, remember a while back I said that I didn't think my relation was new, and asked for a citation. So some posters come back talking about how trivial it is, and basically annoying me. So what if it's simple? I wanted a citation. Now I get one, and it's [CapitalThorn]ne. I don't know why some people have to make this an antagonistic and hateful process. If I put something out, and you're not interested, ignore it. If you think my posts aren't of interest, ignore them. If you [CapitalThorn]nd something of interest then why not try to deliver more facts than hostility? It'd be easier all around. James Harris === Subject: Re: Amateur math, neat relation > In any event, remember a while back I said that I didn't think my > relation was new, and asked for a citation. > So some posters come back talking about how trivial it is, and > basically annoying me. So what if it's simple? I wanted a citation. > Now I get one, and it's [CapitalThorn]ne. > I don't know why some people have to make this an antagonistic and > hateful process. The reason is simple. You persist in making outrageous statements and claims, then demand that others do the Ôresearch' that you should have done. Shame on you. That's decidedly unscienti[CapitalThorn]c. Do your research and homework *before* you start ponti[CapitalThorn]cating. -- 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: Amateur math, neat relation >[...] > Knuth, The Art of Computer Programming, Volume 1, page 43, > Exercise 38 has the more general formula > [x] + [x + 1/n] + ... + [x + (n - 1)/n] = [n x] > Knuth cites Charles Hermite, Acta Mathematica, Volume 5, > page 315, (1884) as the original source for this formula. >Neat! In retrospect that IS quite obvious. >I wonder if anyone has used it in counting primes... >In any event, remember a while back I said that I didn't think my >relation was new, and asked for a citation. a citation for an utter triviality. -i- just proved that 9767654654 + 90867531496 = 100635186150. i'm certain that this -is- new! anyone who says otherwise should give a citation. >So some posters come back talking about how trivial it is, and >basically annoying me. So what if it's simple? I wanted a citation. >Now I get one, and it's [CapitalThorn]ne. >I don't know why some people have to make this an antagonistic and >hateful process. >If I put something out, and you're not interested, ignore it. >If you think my posts aren't of interest, ignore them. >If you [CapitalThorn]nd something of interest then why not try to deliver more >facts than hostility? maybe it has something to do with the way you behave? let's look at some things you said in the post at the start of this thread: >I'm going to give a derivation again, and talk about what I face with >today's mathematicians so you can see how they operate and how they >make certain that amateur mathematicians are cut out. >[...] >I've seen a lot of talk about these simple relations already, as >if...as if mathematicians knew them before my work, and the >discussions I started. >But that's how today's mathematicians operate. >They don't care about the math, only about what's said about the math. >They are all about prestige and holding on to their social position >and the math be damned. >Go out, look in number theory texts or other math books and [CapitalThorn]nd that >relation. >I hope you do [CapitalThorn]nd it, as if this is new then mathematicians are worse >than pathetic, because if the people of sci.math who are still trying >to push the idea that my work is useless or old never learned that >relation but STILL are trying to downgrade it, then there's just no >doubt that they hate mathematics. >Math is not a social game. People who truly love mathematics don't >stop to [CapitalThorn]rst decide if they like someone before they accept a result. >That's the beauty of mathematics, but there are these people who call >themselves mathematicians who other people call mathematicians who get >paid, and they probably [CapitalThorn]gure that if they don't play these social >games then maybe they can't get their wife that car, or maybe they >can't put their kid through school. >But they are willing to sacri[CapitalThorn]ce you for their needs. [...] >Today's mathematicians have to hate mathematics because mathematics >doesn't look out for them. It doesn't pay attention to their needs. >It doesn't worry about their bills. >Mathematics just doesn't care, so they don't care about mathematics. >They may SAY they care, but that's just to get money and prestige. [...] >When you don't [CapitalThorn]nd it, understand the people you're dealing with, no >matter what label they have on themselves. >They may call themselves mathematicians, but they hate mathematics. see, those things you said are not nice. ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: Re: Amateur math, neat relation >Don't believe me? Then [CapitalThorn]nd >[N/j] - [N/2j] - 1 = [(N-j)/2j] >where j is a natural greater than 1, in a math textbook. > You probably won't [CapitalThorn]nd it in a textbook, because it's (a) not very > interesting and (b) not very dif[CapitalThorn]cult. There are thousands of > relationships like this which are true but not interesting enough to > be shown for their own sake. > You might [CapitalThorn]nd it used in a proof, and if you did it quite likely > would be considered obvious enough not to prove. > -- Richard Well, one thing I've noticed time and time again from sci.math'ers is a knee-jerk response. It's animalistic to me in its simplicity and motivation. First sci.math'ers need to attack what I present, and make the effort to put up a post to try and tell others to ignore me. But let's consider the facts here. There are any number of rather simple results all over mathematics which have people's names attached to them, when to us today they're so trivial that it seems rather odd, but that's how mathematics works. Discoverers get credit. If my result is out there then show it. But don't claim that it's supposedly so trivial when my proof of it came late as I waited to see if any of you could manage one, and you couldn't. But now you say it's trivial. Animalistic reßex response. I'll give you a test. Prove the relation that you call trivial. More than likely you'll be stuck with my proof, but if you're not just reacting like an animal as I think you are, then assuredly you can give your own proof of what you call trivial. James Harris === Subject: Re: Amateur math, neat relation >Well, one thing I've noticed time and time again from sci.math'ers is >a knee-jerk response. It's animalistic to me in its simplicity and >motivation. >First sci.math'ers need to attack what I present, and make the effort >to put up a post to try and tell others to ignore me. >But let's consider the facts here. >There are any number of rather simple results all over mathematics >which have people's names attached to them, when to us today they're >so trivial that it seems rather odd, but that's how mathematics works. >Discoverers get credit. If my result is out there then show it. But >don't claim that it's supposedly so trivial when my proof of it came >late as I waited to see if any of you could manage one, and you >couldn't. >But now you say it's trivial. Animalistic reßex response. >I'll give you a test. Prove the relation that you call trivial. > Comforting to [CapitalThorn]nd that there still is some permanence nowadays, and on > usenet out of all places ;-) > > [ snip ] > > Don't believe me? Then [CapitalThorn]nd > > [N/j] - [N/2j] - 1 = [(N-j)/2j] > > where j is a natural greater than 1, in a math textbook. > ...which, after trivial manipulation (x = N/2j) , amounts to > [2x] = [x] + [x+1/2] > I, too, wonder why math textbooks wouldn't carry it. That centuries-old more > general [Nx] identity should be copyright-free by now. > Liviu > [ snip ballistic tangent ] > So alternatively the relation can be written as > ßoor(N/j) - ßoor(N/2j) - 1 = ßoor((N-j)/2j) > and I've shown the derivation in my original post. Which, again, is a trivial case (n = 2) of an old and known identity [nx] = [x] + [x+1/n] + ... + [x+(n-1)/n] See for example problem A1 at http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ ßoor-sums or Knuth's TAOCP Vol 1 section 1.2.4 exercise 38. Hardly groundbreaking _and_ in the textbooks, if only as a token curiosity. Hope this answers your original question. > [ more snip ] > James Harris Liviu Does this work? (seeing as I cut Ôn' pasted - I should make clear that this is Liviu's work, not mine) -- Min I blame the jelly === Subject: Re: Amateur math, neat relation Originator: richard@cogsci.ed.ac.uk (Richard Tobin) >I'll give you a test. Prove the relation that you call trivial. I gave an informal proof in a posting yesterday, . -- Richard === Subject: Re: Amateur math, neat relation Discussion, linux) > I'll give you a test. Prove the relation that you call trivial. > More than likely you'll be stuck with my proof, but if you're not just > reacting like an animal as I think you are, then assuredly you can > give your own proof of what you call trivial. Oh, I get it. Here's one. A & B & C -> C. Maybe that one's in a logic text somewhere, but if so, just increase the number of conjuncts in the antecedent. Eventually, you will reach a formula that has never been published and also that formula is provable and also I have a proof and if you say it's trivial, then I bet your proof is the same as mine so I'm a discoverer. Wow. Whew. Emulating the master leaves me breathless. -- Jesse Hughes She moaned, in pain and pleasure, as, in a confused whirlwind, she glimpsed an image of Saint Sebastian riddled with arrows, cruci[CapitalThorn]ed and impaled. --Mario Vargas Llosa on category theory === Subject: Re: Amateur math, neat relation Discussion, linux) > There are any number of rather simple results all over mathematics > which have people's names attached to them, when to us today they're > so trivial that it seems rather odd, but that's how mathematics works. Example? -- Jesse F. Hughes [I]t's the damndest thing. There's something wrong with every last one of you, and I *never* thought that was a possibility. But now I feel it's the only reasonable conclusion. --JSH sees some sorta light === Subject: Re: Amateur math, neat relation > Comforting to [CapitalThorn]nd that there still is some permanence nowadays, and on > usenet out of all places ;-) > > [ snip ] > > Don't believe me? Then [CapitalThorn]nd > > [N/j] - [N/2j] - 1 = [(N-j)/2j] > > where j is a natural greater than 1, in a math textbook. > ...which, after trivial manipulation (x = N/2j) , amounts to > [2x] = [x] + [x+1/2] > I, too, wonder why math textbooks wouldn't carry it. That centuries-old more > general [Nx] identity should be copyright-free by now. > Liviu > [ snip ballistic tangent ] > So alternatively the relation can be written as > ßoor(N/j) - ßoor(N/2j) - 1 = ßoor((N-j)/2j) > and I've shown the derivation in my original post. Which, again, is a trivial case (n = 2) of an old and known identity [nx] = [x] + [x+1/n] + ... + [x+(n-1)/n] See for example problem A1 at http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/ ßoor-sums or Knuth's TAOCP Vol 1 section 1.2.4 exercise 38. Hardly groundbreaking _and_ in the textbooks, if only as a token curiosity. Hope this answers your original question. > [ more snip ] > James Harris Liviu === Subject: Sphere Construction Hi everybody. Could someone please tell me how can I construct a sphere based on a system of coordinates? === Subject: Re: Sphere Construction > Could someone please tell me how can I construct a sphere based on a > system of coordinates? A sphere of radius r centered at (0,0,0) is all the points (x,y,z) with x^2 + y^2 + z^2 = r^2. If you want the sphere to be solid like a ball, then the points with x^2 + y^2 + z^2 <= r^2. === Subject: JSH: Proof by contradiction So then what I call a null test is more traditionally called a proof by contradiction, but I wasn't trying to be tricky. First off, it was after a bit that I thought about proofs by contradiction, and it was after I talked about a null test, as I resolved what might seem problematic in a de[CapitalThorn]nition I've given for a while: A math proof begins with a truth and proceeds by logical steps to a conclusion which then must be true. But a proof by contradiction begins with an assumption that is false, so what gives? Well I call a proof by contradiction a null test, where you hold the negative of the conclusion in mind to test a proof. If you have a proof then your assumption will be false, will contradict with a step in that proof, and you'll also have what's called a proof by contradiction. But notice I ended up with some posters after me as usual. And that's in a way a proof by contradiction as I've long noticed that there's a group effect on sci.math where people get away with just disagreeing with me. Like math is nothing to many of you, so it doesn't matter if the math I present is correct or interesting, because you have this group effect which you obey quite diligently. So recently I've talked more about areas not so speci[CapitalThorn]c to my own work, like math proofs in general and Wiles's work. Ok, Wiles's work, so my null test is essentially reversing his argument to get a proof by contradiction--if it exists. I can logically prove that *any* math proof can be so reversed, and in fact, for some simple reasons, proofs by contradiction are SHORTER than the full proof. Logically, they must be shorter. So I ask to assume the existence of a non-modular elliptic curve and trace through his argument. The more mathematically astute of you would have realized from the request that I was asking for you to try to reverse his work to a proof by contradiction. The rest of you seemed to just listen to the group effect, and posters who criticized. Now I glanced over Wiles's paper, and it just begs for a null test as he begins assuming a modular elliptic curve. Assuming a non-modular elliptic curve yanks out quite a bit of his paper! But that doesn't mean he doesn't have a proof, but if what remains can't contradict the assumption, then he doesn't. Now some of you I hope like mathematics because it is not a social pheonomenom. In America today we have President Bush, who many love, who hated President Clinton--why? They say because he lied to them. In fashion, you have one style one day, and yet another the next, that's to die for, and stars are here today and gone tomorrow. Social stuff abounds. Now yes, you may not like me, you may hate me, or just not want to think about me, but if you're a mathematician, a real one, then you know it's not about me. Math is not just a social exercise. It's not about committee decisions, and it's not about what makes you feel good. Math is hard not necessarily because of dif[CapitalThorn]culty in the math itself, but because it is so unforgiving, and I guess the way most would see it, uncaring. There are too many people with the label of mathematician in the world today which probably has a lot to do with why the [CapitalThorn]eld is corrupted. Mathematics is too hard of a discipline for it to be otherwise. There just aren't that many people dedicated to such hard truths in such an area in this world. There just aren't that many people who can handle the truth of mathematics. But so many have the label, but they don't like the hardness of math, so they make up things. So yeah, you are paying your bills. You keep the labels, and the prestige that goes with them, and maybe many of you don't even believe in math anymore, so you don't think you lost anything. But you see, you're not REALLY mathematicians. For mathematicians, the truth IS the thing. Not the prestige. Not the labels of professor or mathematician. Necessarily there are very few real mathematicians in the world. James Harris http://mathforpro[CapitalThorn]t.blogspot.com/ === Subject: Re: Proof by contradiction OK, let's test your claim. Give a direct proof of the irrationality of the square root of two. > So then what I call a null test is more traditionally called a proof > by contradiction, but I wasn't trying to be tricky. > First off, it was after a bit that I thought about proofs by > contradiction, and it was after I talked about a null test, as I > resolved what might seem problematic in a de[CapitalThorn]nition I've given for a > while: > A math proof begins with a truth and proceeds by logical steps to a > conclusion which then must be true. > But a proof by contradiction begins with an assumption that is false, > so what gives? Well I call a proof by contradiction a null test, > where you hold the negative of the conclusion in mind to test a proof. > If you have a proof then your assumption will be false, will > contradict with a step in that proof, and you'll also have what's > called a proof by contradiction. > But notice I ended up with some posters after me as usual. And that's > in a way a proof by contradiction as I've long noticed that there's a > group effect on sci.math where people get away with just disagreeing > with me. > Like math is nothing to many of you, so it doesn't matter if the math > I present is correct or interesting, because you have this group > effect which you obey quite diligently. > So recently I've talked more about areas not so speci[CapitalThorn]c to my own > work, like math proofs in general and Wiles's work. > Ok, Wiles's work, so my null test is essentially reversing his > argument to get a proof by contradiction--if it exists. > I can logically prove that *any* math proof can be so reversed, and in > fact, for some simple reasons, proofs by contradiction are SHORTER > than the full proof. > Logically, they must be shorter. > So I ask to assume the existence of a non-modular elliptic curve and > trace through his argument. > The more mathematically astute of you would have realized from the > request that I was asking for you to try to reverse his work to a > proof by contradiction. > The rest of you seemed to just listen to the group effect, and posters > who criticized. > Now I glanced over Wiles's paper, and it just begs for a null test as > he begins assuming a modular elliptic curve. Assuming a non-modular > elliptic curve yanks out quite a bit of his paper! > But that doesn't mean he doesn't have a proof, but if what remains > can't contradict the assumption, then he doesn't. > Now some of you I hope like mathematics because it is not a social > pheonomenom. > In America today we have President Bush, who many love, who hated > President Clinton--why? They say because he lied to them. > In fashion, you have one style one day, and yet another the next, > that's to die for, and stars are here today and gone tomorrow. > Social stuff abounds. > Now yes, you may not like me, you may hate me, or just not want to > think about me, but if you're a mathematician, a real one, then you > know it's not about me. > Math is not just a social exercise. It's not about committee > decisions, and it's not about what makes you feel good. > Math is hard not necessarily because of dif[CapitalThorn]culty in the math itself, > but because it is so unforgiving, and I guess the way most would see > it, uncaring. > There are too many people with the label of mathematician in the world > today which probably has a lot to do with why the [CapitalThorn]eld is corrupted. > Mathematics is too hard of a discipline for it to be otherwise. There > just aren't that many people dedicated to such hard truths in such an > area in this world. There just aren't that many people who can handle > the truth of mathematics. > But so many have the label, but they don't like the hardness of math, > so they make up things. > So yeah, you are paying your bills. You keep the labels, and the > prestige that goes with them, and maybe many of you don't even believe > in math anymore, so you don't think you lost anything. > But you see, you're not REALLY mathematicians. > For mathematicians, the truth IS the thing. > Not the prestige. Not the labels of professor or mathematician. > Necessarily there are very few real mathematicians in the world. > James Harris > http://mathforpro[CapitalThorn]t.blogspot.com/ === Subject: Re: Proof by contradiction > OK, let's test your claim. Give a direct proof of the irrationality of the > square root of two. Euclid's proof is direct. === Subject: Re: JSH: Proof by contradiction >So then what I call a null test is more traditionally called a proof >by contradiction, but I wasn't trying to be tricky. >First off, it was after a bit that I thought about proofs by >contradiction, and it was after I talked about a null test, as I >resolved what might seem problematic in a de[CapitalThorn]nition I've given for a >while: uh, the word is Ôwrong', not Ôproblematic'. >A math proof begins with a truth and proceeds by logical steps to a >conclusion which then must be true. >But a proof by contradiction begins with an assumption that is false, >so what gives? what gives is that what you said was not true. nobody's surprised by this but you. [nobody is suprised by the fact that this had to be pointed out several times before you got it, either. nor by the fact that now that you've [CapitalThorn]nally realized it's false you're trying to explain why it's true.] > Well I call a proof by contradiction a null test, >where you hold the negative of the conclusion in mind to test a proof. quiz: how many legs does a dog have if you call the tail a leg? ************************ David C. Ullrich sorry about the inelegant formatting - typing one-handed for a few weeks... === Subject: probability...~ hello....doctor~ suppose that X_1, X_2 is stochastically independent variables, and P(a < X_1 < b) = 2/3 P(c < X_2 < d) = 5/8 and event A = a < X_1 < b , -00 < X_2 < 00 event B = -00 I wonder whether anyone here would be able to help me with a question > on Dirac delta functions. Suppose (x,y) are two dimensional cartesian > co-ordinates. If I write > delta(x) * delta(y) > this is a two dimensional delta function at (0,0), and I understand > that > delta(x) * delta(x) > is unde[CapitalThorn]ned (although my knowledge of the theory of distributions is > somewhat limited). I'm interested in products of the form > delta(x) * delta(alpha * x + beta * y) > is this well de[CapitalThorn]ned? If so, what is its value? I'm also interested > in the more general case, > prod_{i=1}^{N} delta( alpha_i * x + beta_i * y ) Let D(x,y) be any product of delta functions you are interested in. Taking a very low brow approach to distribution theory, simply try to evaluate the integral Int D(x,y) f(x,y) dx dy for an arbitrary continuous function f(x,y). If you get in[CapitalThorn]nity for the integral, your product D(x,y) is not well de[CapitalThorn]ned. Igor === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > I most certainly do exist. As it is, I am asking you to view the > situation as it is viewed by others. Show us proof (like for instance a post from a real adress... hint : rwill9955@hotmail.com is not very convincing for such a high-level veteran you seem to be) > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > Van === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > I most certainly do exist. As it is, I am asking you to view the > situation as it is viewed by others. > Show us proof (like for instance a post from a real adress... hint : > rwill9955@hotmail.com is not very convincing for such a high-level > veteran you seem to be) Then it seemed that there was a real A. W. Beckwith, with a PhD in physics . So what ? 1) it is not necessarily andy 2) UnaBomber was a mathematician 3) Math ideas are judged for their worth, not any other way, and *certainly* not because a PhD (in physics) says something 4) JSH is loathable, and to try to defend him just lowers one position by association 5) Threats by this PhD are worth a good laugh, but nothing more. > LOL! Do these people really exist, or is it just JSH posting under > another name. I have a hard time believing he really has > got someone to come over here and stick up for him. > Van === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Gentlemen, > This is Dr. Andrew Beckwith. I am spaking to you after having had > people from another forum tell me that James Harris's patriotism has > been questioned. > I urged James not to react to baiting. Now, after I have said this, I > am going to respond to you in this forum. > I am also a fellow veteran. A disabled veteran who is legally nearly > stone deaf . Many of my ilk have looked in abject horror at the > presidential elections where both Kerry and Bush have thrown > accusations at each other about who is unpatriotic and who has the > mojo with respect to the military. > This is despicable, and I urge you, in this forum , not to harass > another military veterans mainly, James Harris. > Those who have served and who have suffered for it know the > uncalculatable pain of permanent severance of health and of life of > those who have been in harms way. Those who have served also respect > the rights of veterans in the United States to entertain their own > points of view and to question authority. > You should respect James Harris's right to question authority and to > entertain unpopular points of view. This is what we should allow ALL > Americans. > If you honor our own democratic heritage, you should value that > dispensation and to never, NEVER penalize a person , especially a > MILITARY person, for having their own view point. > I am a mathematical physics PhD. I may return in due course to this > forum to discuss professional issues with you all in this discussion > group. A pre requisite for me returning though will be a cessation of > attacks upon James and at least a modicum of respect for his veterans > status. Especially on the part of sunshine patriots who have never had > the gumption to be potentially or actually in the line of [CapitalThorn]re. > Andrew Beckwith, PhD. James has to give respect to earn respect, idiot. Dave === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. Discussion, linux) > I may return in due course to this forum to discuss professional > issues with you all in this discussion group. Uh oh. > A pre requisite for me returning though will be a cessation of > attacks upon James and at least a modicum of respect for his > veterans status. Whew. Threat averted. -- Jesse F. Hughes Usenet is demonstrably dangerous. It needs to be regulated. --James S. Harris, voice of reason and moderation === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Gentlemen, > This is Dr. Andrew Beckwith. I am spaking to you after having had > people from another forum tell me that James Harris's patriotism has > been questioned. > I urged James not to react to baiting. Now, after I have said this, I > am going to respond to you in this forum. > I am also a fellow veteran. A disabled veteran who is legally nearly > stone deaf . Many of my ilk have looked in abject horror at the > presidential elections where both Kerry and Bush have thrown > accusations at each other about who is unpatriotic and who has the > mojo with respect to the military. > This is despicable, and I urge you, in this forum , not to harass > another military veterans mainly, James Harris. > Those who have served and who have suffered for it know the > uncalculatable pain of permanent severance of health and of life of > those who have been in harms way. Those who have served also respect > the rights of veterans in the United States to entertain their own > points of view and to question authority. > You should respect James Harris's right to question authority and to > entertain unpopular points of view. This is what we should allow ALL > Americans. > If you honor our own democratic heritage, you should value that > dispensation and to never, NEVER penalize a person , especially a > MILITARY person, for having their own view point. > I am a mathematical physics PhD. I may return in due course to this > forum to discuss professional issues with you all in this discussion > group. A pre requisite for me returning though will be a cessation of > attacks upon James and at least a modicum of respect for his veterans > status. Especially on the part of sunshine patriots who have never had > the gumption to be potentially or actually in the line of [CapitalThorn]re. > Andrew Beckwith, PhD. You see me now, a veteran Of a thousand psychic wars I've been living on the edge so long Where the winds of limbo roar And I'm young enough to look at And far too old to see All the scars are on the inside I'm not sure that there's anything left of me ... You ask me why I'm weary Why I can't speak to you You blame me for my silence Say it's time I changed and grew But the war's still going on, dear And there's no end that I know And I can't say if we're ever I can't say if we're ever gonna be free ... Don't let these shakes go on It's time we had a break from it Send me to the rear Where the tides of madness swell And been sliding into hell Oh, please don't let shakes go on Don't let these shakes go on Don't let these shakes go on Excerpts from Veteran of the Psychic Wars by B.85C === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Gentlemen, > This is Dr. Andrew Beckwith. I am spaking to you after having had speaking > people from another forum tell me that James Harris's patriotism has > been questioned. Harris has threatened the use of *military* force against people here who have proved that his mathematics is incorrect. He has threatened to expose his opponents to the FBI, the CIA, and the NSA. He has threatened legal action, again, against people here. For maligning or libelling him or for calling him unpatriotic? No! For having dared to post valid counterexamples to his mathematical proofs! He has tried to silence at least one of his critics by contacting his employer and charging him with racism (for having *considered* making a racist remark, not for actually making it). In my view these are potentially abuses of the freedoms we are guaranteed by the Constitution. I have no idea who has been questioning his patriotism in this forum or any other, but I would say that if such questioning has occurred, there may be a rational basis for it. > I urged James not to react to baiting. Now, after I have said this, I > am going to respond to you in this forum. > I am also a fellow veteran. A disabled veteran who is legally nearly > stone deaf . Many of my ilk have looked in abject horror at the > presidential elections where both Kerry and Bush have thrown > accusations at each other about who is unpatriotic and who has the > mojo with respect to the military. > This is despicable, and I urge you, in this forum , not to harass > another military veterans mainly, James Harris. veteran namely > Those who have served and who have suffered for it know the > uncalculatable pain of permanent severance of health and of life of incalculable > those who have been in harms way. Those who have served also respect harm's > the rights of veterans in the United States to entertain their own > points of view and to question authority. Of course. But are you saying veterans have more such rights than the rest of us? > You should respect James Harris's right to question authority and to > entertain unpopular points of view. This is what we should allow ALL > Americans. Agreed. The point has been made many times that James Harris himself is perhaps the only person here here who has actually taken action to suppress other people's freedom of expression, by explicitly contacting his opponent's employer *and* the Attorney General of the state in which his opponent lived. > If you honor our own democratic heritage, you should value that > dispensation and to never, NEVER penalize a person , especially a > MILITARY person, for having their own view point. Veterans deserve respect. They do not deserve obsequious worship or absolute freedom from criticism for their words or actions. No one does. > I am a mathematical physics PhD. I frankly doubt this. Your writing is semi-literate. > I may return in due course to this > forum to discuss professional issues with you all in this discussion > group. A pre requisite for me returning though will be a cessation of prerequisite > attacks upon James and at least a modicum of respect for his veterans veteran's > status. Especially on the part of sunshine patriots who have never had > the gumption to be potentially or actually in the line of [CapitalThorn]re. Was Harris ever in the line of [CapitalThorn]re? He has been quite reluctant to answer certain questions regarding his military service. Andrzej > Andrew Beckwith, PhD. === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Gentlemen, > I am a mathematical physics PhD. > I frankly doubt this. Your writing is semi-literate. non sequitur. === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. X-RFC2646: Original Beckwith said: > I am a mathematical physics PhD. Kolowski replied: > I frankly doubt this. Your writing is semi-literate. Take a gander at: http://www.arxiv.org/[CapitalThorn]nd/math-ph/1/au:+Beckwith_A/0/1/0/all/0 /1 Kolowski, I utterly do not *get* how you draw that conclusion about Beckwith. Why are you willing to jump into an ad hominem so quickly? Find reference 3 here: http://www.arxiv.org/ftp/math-ph/papers/0407/0407016.pdf Read it. Reference 3. Jesus H. Christ, people! Maybe Dr. Beckwith is a professional mathematician from that other list who has now come public with his support for James Harris. Holy Frick! What *is* it with you all? Do you think Beckwith and Jackson have never written a paper before? What *is* it with you, I ask again? Good God. -- Quinn === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. > Beckwith said: > I am a mathematical physics PhD. > Kolowski replied: > I frankly doubt this. Your writing is semi-literate. > Take a gander at: > http://www.arxiv.org/[CapitalThorn]nd/math-ph/1/au:+Beckwith_A/0/1/0/all/0 /1 This merely demostrates that physicists need not be very literate. === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. Discussion, linux) > Harris has threatened the use of *military* force against people > here who have proved that his mathematics is incorrect. He has > threatened to expose his opponents to the FBI, the CIA, and the > NSA. He has threatened legal action, again, against people here. > For maligning or libelling him or for calling him unpatriotic? > No! For having dared to post valid counterexamples to his > mathematical proofs! He has tried to silence at least one of > his critics by contacting his employer and charging him with > racism (for having *considered* making a racist remark, not for > actually making it). In my view these are potentially abuses of > the freedoms we are guaranteed by the Constitution. I have no > idea who has been questioning his patriotism in this forum or any > other, but I would say that if such questioning has occurred, > there may be a rational basis for it. Aw, don't take either of these guys so seriously. Andrew Beckwith is just another garden variety troll (but JSH is one of a kind). -- Jesse F. Hughes What you call reasonable is suspect since you've proven yourself to be an enemy of mathematics. -- James S. Harris defends the cause. === Subject: Re: Harassment of James Harris: A necessary rebuke to those who mocked his veterans status. === >Subject: Re: Harassment of James Harris: A necessary rebuke to those who >mocked his veterans status. >I read your group posts. All of them as of the last month. As I >expected, your group does not have the entire picture . >James wanted unconditional backing, and I do not exactly work that >way. As it is, the discussions as to the goings on went to the >ultranet, which I am a part of and I felt obligated to respond to the >upset engendered. === >Subject: Harassment of James Harris: A necessary rebuke to those who >mocked >his veterans status. > >Gentlemen, > Who, me? >What should I call you. Hey you ? I was merely wondering if this diatribe was directed at sci.math readers in general or me speci[CapitalThorn]cally. > >This is Dr. Andrew Beckwith. I am spaking to you after having had >people from another forum tell me that James Harris's patriotism has >been questioned. > And did you make any effort to verify this, or are you just responding > to hearsay? >Not quite. I read the entire business. I spent an afternoon reading >this groups goings on, at the expense of my duties at Fermi >laboratory. Which I cannot do very often. > >I urged James not to react to baiting. > Other people have urged James to behave properly to no avail. > Why did you think he'd listen to you? >I regard him as having been booted in the balls one too many times. >That makes people hyper sensitive. That is not a fault in itself. Unless one has provoked the ball booting, in which case it is a fault. Or did you not bother to read those posts in your review. >Now, after I have said this, I am going to respond to you in this forum. > In other words, he won't listen to you either. So you plan to solve > the problem by dealing with the symptoms rather than the cause. > Have you ever heard the phrase putting a band-aid on a brain tumor? >The [CapitalThorn]rst step in responding to upset is to get people to lower their >voices. It's too late for that. The way to deal with a troll is to shout back just as loud. >Someone ADULT enough has to make the [CapitalThorn]rst step. Sorry, there are no adults here, just veterans and critic trolls. >That datum appears to be beyond the ability of sci.math to parse. We parse it just [CapitalThorn]ne, thank you. It's just that your datum is wrong. > >I am also a fellow veteran. A disabled veteran who is legally nearly >stone deaf . Many of my ilk have looked in abject horror at the >presidential elections where both Kerry and Bush have thrown >accusations at each other about who is unpatriotic and who has the >mojo with respect to the military. > Patriotic? I thought the issue was who served versus who got his daddy > to pull strings and get him into the National Guard. The issue has > nothing to do with patriotism, it's about who's the coward. >We know that Bush had strings pulled. That is obvious. But it, the >dispute, went way beyond that. >As you know, James is extremely controversial > >This is despicable, and I urge you, in this forum , not to harass >another military veterans mainly, James Harris. > Do you consider Bush a veteran? I consider him a spoiled rich-kid > draft dodger. As for Harris, I have no information on his service. >I consider Bush to be a fraud. Pass the salt. > >Those who have served and who have suffered for it know the >uncalculatable pain of permanent severance of health and of life of >those who have been in harms way. > Maybe you know. Did Harris suffer from his service? >Let me say that from what I learned that he was not treated with >respect. Really? Does he give a reason? Discrimination? It wouldn't be because he's an asshole, would it? There was a rumor going around that Harris is currently unemployed. Does he have a job? Could you e-mail the name and address of his employer? >Those who have served also respect the rights of veterans >in the United States to entertain their own points of view >and to question authority. > And to spew anti-American rhetoric while we are at war? >The test of patriotism is to encourage debate, and to not brand anti >American opinions which differ from the norm. >Nothing Harris said was treasonable. The statements were hot headed, >but that is something some super patriots ought to learn how to deal >with. If you're going to make hot-headed statements, then you can't demand respect for being a veteran. If Harris wants the respect he's due as a veteran, then maybe he should learn to think before he shoots his mouth off. Everybody makes hot-headed statements. It's the cold, calculated statements, such as harassing someone's employer in an attempt to get them [CapitalThorn]red, that are the real issue. Try observing the forest instead of focussing on a single tree. > >You should respect James Harris's right to question authority and to >entertain unpopular points of view. This is what we should allow ALL >Americans. > >If you honor our own democratic heritage, you should value that >dispensation and to never, NEVER penalize a person , especially a >MILITARY person, for having their own view point. > Here's a little something you are probably not aware of. It's the > Oath of Of[CapitalThorn]ce of the President of the United States: > I do solemnly swear (or af[CapitalThorn]rm) that I will faithfully execute the of[CapitalThorn]ce > President of the United States, and will to the best of my ability, >preserve, > protect and defend the Constitution of the United States. > Note that the president doesn't swear to protect the country or its > people. He swears to PROTECT THE CONSTITUTION. You do > know that the President is the Commander In Chief of the > Armed Forces? For ANY MILITARY person to advocate the > abrogation of the Constitution is TREASON! >I know that better than you. > >I am a mathematical physics PhD. I may return in due course to this >forum to discuss professional issues with you all in this discussion >group. A pre requisite for me returning though will be a cessation of >attacks upon James > Keep dreaming. When Harris learns to behave the forum will > follow suit. >You are playing a game of chickens which is not going to earn anyone >any credit. Credit? What credit? Do you know who I am? How can I cash in this credit I'll get by being civil to Harris? >and at least a modicum of respect for his veterans status. > Sorry. Veterans who turn traitor don't get no respect. When Harris > shows sympathy for the future veterans who are suffering in Iraq > instead of sympathy for terrorists, then maybe he'll get some > respect. >Let me get this through your rock head. Harris did not utter >treasonous statements. They were not well thought out, but that is not >treason. Ok, if you say so. There's still that matter of terrorist sympathies though. >For your information, I was a counter intelligence of[CapitalThorn]cer and I have >dealt with security threats. Jesus H. Christ! We're doomed. >Harris is not a threat. He is being >extremely hot headed. >Especially on the part of sunshine patriots who have never had >the gumption to be potentially or actually in the line of [CapitalThorn]re. > Look pal, I didn't have a rich daddy to get me into the National > Guard. I avoided Vietnam the hard way: I won the draft lottery. >I am severely disabled from military duty. Ditto. The whine fest stops >here. Right now, I have had maybe about 15 years sliced off my life >due to service injuries. > >Andrew Beckwith, PhD. -- Mensanator Ace of Clubs === Subject: Q: integrating rational functions -which formula to remember. There is one complicated formula we should know to be able to integrate rational functions. That is, the recursion formula for I_n(x) = integral[ dx / (x^2+a^2)^n ] ,where n is an integer >= 2. Another version is the recursion formula for J_n(x) = integral[ dx/(a*x^2+b*x+c)^n] ,where b^2 is not equal to 4*a*c. The formula is from the page: [ http://www.sosmath.com/calculus/integration/rational/ Tablefraction/Tablefrac tion.html ] I_n formula is a little bit easier to remember than J_n formula. But I_n formula requires you to complete the square, while J_n formula does not. So the question arises: which formula should one remember? Or which formula do you prefer? === Subject: Spherical curves very confused Let f be a unit-speed curve with kappa > 0, torsion not equal to 0. If f lies on a sphere of center c and radius r, show that f(t) - c = -rho N - rho' sigma B where N is the principal normal, B is binormal, rho is 1/kappa, sigma = 1/torsion I don't see how to do this. Let's say I dot product the left hand side with the left hand side and dot product the right hand side with the right hand side (it's ok since LHS = RHS) Then the left side is ||f(t)-c||^2 and the right hand side is ||-rho N||^2 + ||rho Ô sigma B||^2 - (2 rho N) dot product (rho Ô sigma B) (since v-w dot v-w = ||v||^2 + ||w||^2 - 2 v dot w) But N has norm 1, B has norm 1, and N dot B = 0 since they are perpindicular. So we're left with ||f(t)-c||^2 = ||rho||^2 + ||rho' sigma||^2 But for any spherical curve, it is true that kappa is greater than or equal to 1/r where r is the radius of the sphere. Thus ||rho||^2 = 1/kappa^2 is less than or equal to r^2 But ||f(t) - c||^2 = r^2 since f(t) lies on sphere centered at c of radius r. So how can we have this? rho' would have to equal 0, and as well, the curve would have to lie on a circle. What's going on? Isaac === Subject: Re: These are the folks from the.... > These are the folks from the profession that always > criticizes Bush for his use of the English language..... > .......ahahahahaha.........AHAHAHAHA......... === > Subject: FW: Olympic Commentators GOOFS! > Here are the top nine comments made by NBC sports commentators > during the Summer Olympics that they would like to take back: > 1. Weightlifting commentator: This is Gregoriava from Bulgaria. I saw her > snatch this morning during her warm up and it was amazing. > 2. Dressage commentator: This is really a lovely horse and I speak from > personal experience since I once mounted her mother. > 3. Paul Hamm, Gymnast: I owe a lot to my parents, especially my mother and > father. > 4. Boxing Analyst: Sure there have been injuries, and even some deaths in > boxing, but none of them really that serious. > 5. Softball announcer: If history repeats itself, I should think we can > expect the same thing again. > 6. Basketball analyst: He dribbles a lot and the opposition doesn't like > it. In fact you can see it all over their faces. > 7. At the rowing medal ceremony: Ah, isn't that nice, the wife of the IOC > president is hugging the cox of the British crew. > 8. Soccer commentator: Julian Dicks is everywhere. It's like they've got > eleven Dicks on the [CapitalThorn]eld. > 9. Tennis commentator: One of the reasons Andy is playing so well is that, > before the [CapitalThorn]nal round, his wife takes out his balls and kisses them... Oh > my God, what have I just said? http://members.lycos.co.uk/cabbagesofdoom/colemanballs.htm -- Dirk The Consensus:- The political party for the new millenium http://www.theconsensus.org === Subject: Beatty sequences A Beatty sequence is de[CapitalThorn]ned as ßoor(nx) for n=1,2,3... for a given irrational number x. There is a theorem (Beatty's theorem?) which states that if a and b are irrational numbers and (1/a) + (1/b) = 1 then the beatty sequences of a and b together contain all the positive integers with no repetition. That is, [a], [2a], [3a]... [b], [2b], [3b]... contains all the integers with no repetitions. (See http://mathworld.wolfram.com/BeattySequence.html) Two questions: [CapitalThorn]rst - can anyone give a relatively simple proof of this. Second, and more dif[CapitalThorn]cult I suppose - does anyone know if it is possible to make a kind of generalisation of this theorem to something involving a, b, and c? Is it possible to [CapitalThorn]nd three irrational numbers a, b, c, such that their Beatty sequences contains all the integers with no repetitions? Ruben === Subject: Re: Is Stephen Wolfram (mathematica) delusional? > I don't know if delusional is quite the right word. Obsessed with > himself, narcissistic, etc. ... The question is about behavior. BTW,how is inspired creativity distinguished from delusional activities? What is the dividing line? === Subject: non noetherian ls, is a non-noetherian ring, together with a euclidean function, principal? singau === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping > I say this because if I ask Dik Winter or Andrew Wiles > or most every other mathematician, ask them why they cannot sit down and > state that the formulation of 4 Color Mapping is wrong and imprecise > because it assumes the borderline is no color at all when we all know > the borderline is a Black color and thus 5 colors are involved. What is the color of the borderlines in: ? How would you colour the four regions with less than four colours? > So can Dik or Andrew thence sit down and state that the FLT is > also i-m-p-r-e-c-i-s-e because [CapitalThorn]nite-integers are ill-de[CapitalThorn]ned. So you say... > NaturalNumbers are de[CapitalThorn]ned by the Peano axioms but those axioms have a > Successor Axiom which is a series of endless adding of 1. P-adics are > de[CapitalThorn]ned similarly as a series of endless adding of 1. So there is no > distinction in the Peano Axioms themselves that the NaturalNumbers are > Adics and not some so called [CapitalThorn]nite-integer. Eh? The [CapitalThorn]nite integers satisfy some nice axioms, like when x.y = 0 either x or y is 0. This is not true in your p-adics. Moreover, you never explain how the Peano axioms, when starting with 0, ever can come up with an in[CapitalThorn]nite integer, so I fail to see how they can ever de[CapitalThorn]ne the p-adics. Further, your p-adics are ill-de[CapitalThorn]ned. You see them as encompassing all possible adics, however you never came to de[CapitalThorn]ne the addition or multiplication of an arbitrary 2-adic with an arbitrary 3-adic. > And the reason most every mathematician cannot sit down and really talk > about the foundational de[CapitalThorn]nitions and whether the Mapping or FLT are > precisely stated or illogically stated, is because like most > mathematicians who are paid money, hand to mouth, where money, career > and fame and book writing fortunes are more important to them than is > the simply truth of the matter in discussion. Eh? I did not know I earned money with my mathematical ramblings. Where can I pick up that money? > Time is on my side. First there was only me who saw that FLT is false > and that Natural Numbers are the In[CapitalThorn]nite Integers and that > [CapitalThorn]nite-integers are a illusion or mirage. Bear in mind that the in[CapitalThorn]nite integers are *not* your p-adics. Check for instance Conway's numbers... -- 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: the imprecision of 4 color mapping and why it should be 2 color mapping and why FLT is also imprecise and thus false Re: E.E.E. also claims he disproved or neutralized Wiles proof! > if I ask Dik Winter or Andrew Wiles > or most every other mathematician, ask them why they cannot sit down and > state that the formulation of 4 Color Mapping is wrong and imprecise > because it assumes the borderline is no color at all when we all know > the borderline is a Black color and thus 5 colors are involved.... > Black borderlines in mapping is ill-de[CapitalThorn]ned. It is a 2 color mapping > easily proven by the Jordan Curve theorem. Is it really that hard for you to conceive of a map with no borderlines drawn? Look halfway down this page: http://www.maps2anywhere.com/Maps/World_map_folded.htm === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping and why FLT is also imprecise and thus false Re: E.E.E. also claims he disproved or neutralized Wiles proof! > Is it really that hard for you to conceive of a map with no borderlines drawn? Part of the trouble of Mapping in mathematics is the constant switching back and forth between the practicality of drawing and the theoretical precsion of math. This switching is one of the reasons it took 150 years or more to get a handle on this conjecture. In answer to the question from Harvard above as to whether is it really that hard for you..... My answer: is it really that hard for Harvard to conceive that every sheet of paper ever written on in the history of humanity is a 2 Color Mapping. Every post to the Internet is a 2 Color Mapping. The letter M is a borderline only with no interior. The letters o, p, q, a have interiors and can be considered the insides of a country. A checkerboard is a 2 color mapping because the borderlines and white interiors is all that is needed. So, once mathematicians inspect Mapping and inspect their understanding of how borderlines are integrally related to interiors, then 4 color mapping disappears in the junkpile of mathematics history as a ill-de[CapitalThorn]ned and ill-formulated problem. The proof of all Mappings rests squarely on the Jordan Curve Theorem and a bright High School student can prove it. One can say that mathematicians for the past 150 years have been more psychologists and hippies about color on maps than they have been Gaussian stern de[CapitalThorn]nition and formulators of precision. > Right the difference would be: > Borderline (of [CapitalThorn]nite thickness > 0) > / > ||#### > ||#### > ||#### > ||#### > Ô' > vs. > #### > #### > #### > #### > (no Borderline) > Or if we consider a model in one dimension: > The interval [-1,0[ may be colored red, > the interval [0,1] may be colored green. > There is no border line (point) between > these two intervals. > It's just: > +++++++~~~~~~~ > red green > F. You cannot have disjoint open sets for maps. If you have Interiors, then you necessarily have Borderlines. The Jordan Curve Theorem would be false if you had a map of Interiors but no borderlines at all. As I said in the beginning of this post that every sheet of paper ever written on has borderlines and interiors and most of those papers are just borderlines with no interior. Something that the Appel and Haken alleged proof fails to deal with that you can have a Map that is mostly borderlines and few if any interiors. Which should be a Counterexample that the Appel & Haken proof is false. post of a global Mapping where no country interior exists but where borderlines mostly exist except for the land surrounding the Is and the Ss. So what does Appel and Haken say about such a Map where there is only one land interior but disjoint and disconnected borderlines of Is and Ss. So you see, all maps are 2 Colorable and the 4 Color Mapping problem begun around 1850 where many famous mathematicians never could get a handle on the problem because they never injected Precision of de[CapitalThorn]ning what the problem really was. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping and why FLT is also imprecise and thus false Re: E.E.E. also claims he disproved or neutralized Wiles proof! > > It is sad that mathematicians today act more like politicians rather > than scientists. I say this because if I ask Dik Winter or Andrew Wiles > or most every other mathematician, ask them why they cannot sit down and > state that the formulation of 4 Color Mapping is wrong and imprecise > because it assumes the borderline is no color at all when we all know > the borderline is a Black color and thus 5 colors are involved. > I'm sure I've done this before for you, but here we go again. Below, I > preceed the terms I am de[CapitalThorn]ning with em Well, [CapitalThorn]rst of all, can you admit that all maps that need to show all countries require 2 and only 2 colors for the black borderlines and for white interiors. Can you understand and admit that. This should have been your [CapitalThorn]rst beginnings and anyone else who wanders into studying 4 Color Mapping. You see, all the information needed about countries on planar maps suf[CapitalThorn]ces with just black and white. And then of course 4 colors are superßuous. > Firstly however, note that the following setup is equivalent to the > rather more convenient setup of planar graphs that you would [CapitalThorn]nd in > the litterature. If you looked... Also a jigsaw puzzle is a good analogy where the borderlines are the cut between any two pieces. So in a jigsaw puzzle we match the 4 Color Mapping in that it too is really a 2 Color Mapping. > Defn. > 1. Let n be a natural number. Let {X_i:i disjoint open connected two dimentional subsets of R^2. Suppose that > R^2 union of {X_i:i (n,{X_i,i (intuativela, a map...) > 2. Given a map (n,{X_i,i adjacent to X_j iff there exists some Y_1,Y_2 connected one > dimentional subsets of R^2 with the following properties: > Firstly, for each k in {1,2} > a) Y_k intersects both X_i and X_j, and > b) |Y_k (X_i union X_j)|=1 > Also Y_k (X_i union X_j) differs from Y_2 (X_i union X_j) > (intuatively, two areas of the map are adjacent iff they share a > boundry of more than one point) > 3. Given a map (n,{X_i,i colouring of (n,{X_i,i > {1,2,...,m} with the following property: > For each i f(X_j). > (we have to colour adjacent maps differently) > Theorem [Appel and Haken 1977] > Given any map (n,{X_i,i {1,2,3,4}. You neglected Borderlines. What you have missing above are the points that consist of borderlines. Points that are interiors are reckoned with, but points that are borderlines you fail to address. Your failure geometrically is that you must think (as per your above) that interior points come close to borderline points and that visually those borderlines will not be observable and that the viewer can somehow estimate where the borderline is from the difference in color of one interior to another interior. So the gapping hole in your above is a total absence of reckoning with borderline points. And that is the same failing of Appel and Haken. A true mathematical statement of the Map Coloring Problem is All planar maps are 2 colorable and the proof is simple in using the Jordan Curve theorem. So Tim, getting back to my [CapitalThorn]rst question to you. Can you admit that all maps are 100 percent presentable of information if you color the Borderlines in black and color the interiors in white. Can you Tim think of a map in which you need more than 2 colors of black and white because more information is needed to be conveyed? Of course not. Why do you skip borderlines? Why do you ignore them? Do you think that coloring the interiors allows you to completely ignore borderlines? Tim, can you provide the exact quote by Gauss when he said words to the effect that whenever a huge problem arises in mathematics it is usually because of a lack of understanding of the primal de[CapitalThorn]nitions being used. In the case of Color Mapping it is the borderlines that are utterly ignored and that is why the Appel Haken alleged computer proof is a fake and why 4 Color Mapping has never been proved and never will be because it is ßawed in its formulation. The proper formulation is that all planar Maps are 2 Colorable. Colorable of the borderlines as blue and the interiors as off-white is that binary digits can cover all other number systems. We need only 2 bits of 0 and 1 to cover all number systems and we need only 2 Colors to cover all planar maps. If you cannot admit that blue and white suf[CapitalThorn]ce to cover all the information ever needed in any Map, then please forego responding because I do not want to waste time on you. Archimedes Plutonium www.archimedesplutonium.com www.iw.net/~a_plutonium whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies === Subject: Re: the imprecision of 4 color mapping and why it should be 2 color mapping and why FLT is also imprecise and thus false Re: E.E.E. also claims he disproved or neutralized Wiles proof! You are interested in a different question to that studied for one hundred and [CapitalThorn]fty odd years under the title four colour problem. That is [CapitalThorn]ne, of course, you can be interested in whatever you like. As you point out, however, the solution to your question is trivial. Why rant and rave about other people studying more involved problems? === Subject: Vector Spaces Suppose i have a vectorspace V which is spannend by three polynomials p0,p1 and p2, such that V=span{p0,p1,p2}. Using a basis transformation, i generate three other polynomials which span the same basis, such that V=span{q0,q1,q2}=span{p0,p1,p2}. Now my question is, is the following statement true? W=span{1/q0,1/q1,1/q2}=span{1/p0,1/p1,1/p2} i don't think it is, but i don't know how to proove it. can someone point this out for me? (sorry if i didn't explain it well enough, since i'm not a mathematician) paul === Subject: Re: Vector Spaces > Suppose i have a vectorspace V which is spannend by three polynomials > p0,p1 and p2, such that V=span{p0,p1,p2}. Using a basis transformation, i > generate three other polynomials which span the same basis, such that > V=span{q0,q1,q2}=span{p0,p1,p2}. > Now my question is, is the following statement true? > W=span{1/q0,1/q1,1/q2}=span{1/p0,1/p1,1/p2} > i don't think it is, but i don't know how to proove it. > can someone point this out for me? > (sorry if i didn't explain it well enough, since i'm not a mathematician) > paul To disprove that assertion, you need only provide a counterexample. For example, span{1,x,x^2} = span{1,x-1,x^2} but 1/(x-1) is not in the span of {1, 1/x, 1/x^2}. === Subject: Re: algebra...; > If x^n = x for all x in ring R > and a^m = 0 then a = 0 HINT: a^k = a for k = n^i unbounded but a^k = 0 for k > m --Bill Dubuque === Subject: Re: algebra...; === Subject: algebra...; >suppose that >x^n = x for all x in R (R is ring) Require n > 1. n = 1 does nothiing. >show that >if a in R is such that a^m =0, then a=0 Require m in N. a^0 = 0 not possible. a^-1, a^-m not de[CapitalThorn]ned unless a unit. If a is unit, then a /= 0. Now some q,r in N with m = qn + r, 0 <= r < n a = a^m = a^qn a^r = a^q a^r = a^(q + r) notice that as 0 < m, 0 < q + r Repeat as necessary to obtain some k with a^k = 0, 0 < k < n What's next? Inquiring minds want to know. ---- === Subject: Re: algebra...; > suppose that > x^n = x for all x in R (R is ring) Suppose n = 1. > show that > if a in R is such that a^m =0, then a=0 === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- >um......i can't do it. >help me, please~ >thank you very much. Hint: What if you did x^n again, and perhaps again and again? Perhaps doing it some number of times that would relate what you are given and what you need? I'm [CapitalThorn]rmly convinced a good hint is better than having someone just blurt out the answer. === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- >um......i can't do it. >help me, please~ >thank you very much. > Hint: What if you did x^n again, and perhaps again and again? > Perhaps doing it some number of times that would relate what you > are given and what you need? > I'm [CapitalThorn]rmly convinced a good hint is better than having someone > just blurt out the answer. um....thank you very much. easy. if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... so, there is a positive integer p such that n^p >= m. so, a^n^p = 0 ....contradiction. thus a = 0. === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- >um......i can't do it. >help me, please~ >thank you very much. > Hint: What if you did x^n again, and perhaps again and again? > Perhaps doing it some number of times that would relate what you > are given and what you need? > I'm [CapitalThorn]rmly convinced a good hint is better than having someone > just blurt out the answer. >um....thank you very much. easy. >if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... >so, there is a positive integer p such that n^p >= m. >so, a^n^p = 0 ....contradiction. >thus a = 0. Now a few questions, just to get you to think about this. have given you a particular speci[CapitalThorn]c n and m, rather than letting you choose what n and m might be. But perhaps I misunderstood. If all you knew was that x^n=x and a^m=0 (for all x and a) but for [CapitalThorn]xed constants n and m that they provided you, can you still really use that >= that you used there, if you were beeing really careful and really precise? Do you need to be a little more careful about your argument? Could there be examples where your argument could have problems? And your if a =/= 0, a^n = a statement, isn't this also true if a == 0? Does it matter? Does it mean you want to state this differently? Or not? All this is practice and training, for what is likely to come soon. === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- > if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... > so, there is a positive integer p such that n^p >= m. **** I don't follow you here....if a^n = a, and a^n^2 = a, and a^n^3 = a, ...doesn't a^n^p = a, and so doesn't n^p = n? if not, why not, and/or why can we then call n^p less than or equal to m ? I understand the problem solution, if you can say that n^p >= m, the whole thing makes sense, but I miss you on that step. (I know I'm jumping in on someone else's problem, but I hate to not understand something that is *close* to within my grasp : ) -karinne **** > so, a^n^p = 0 ....contradiction. > thus a = 0. === Subject: Re: algebra...; >suppose that >x^n = x for all x in R (R is ring) > >show that >if a in R is such that a^m =0, then a=0 >-------------------------------------------------- > if a =/= 0, a^n = a and a^n^2 = a and a^n^3 = a ...... > so, there is a positive integer p such that n^p >= m. > **** > I don't follow you here....if a^n = a, and a^n^2 = a, and a^n^3 = a, > ...doesn't a^n^p = a, and so doesn't n^p = n? if not, why not, and/or why > can we then call n^p > less than or equal to m ? I understand the problem solution, if you can > say that n^p >= m, the whole thing makes sense, but I miss you on that step. > (I know I'm jumping in on someone else's problem, but I hate to not > understand something that is *close* to within my grasp : ) > -karinne > **** > so, a^n^p = 0 ....contradiction. > thus a = 0. oh......i'm sorry. my mistatke. there is a condition which n>1. sorry. but, there does not exist concrete condition about m, n. namely, m and n is integer ? positive ? negative ?...i don't know.. thank you very much for your advice. === Subject: ALL THE MATEHMATICS YOU MISSED by Garrity Someone (can't remember who) suggested this book as a survey of undergraduate mathematics. Whoever it was, thank you. I ordered it from interlibrary loan, and I'm [CapitalThorn]nding it fascinating. (I can't remember in which of the two groups the suggestion appeared. I've set followups to alt.math.undergrad only.) -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com You want an intelligent conversation? Do what I do: talk to yourself. It's the only way. -- /Torch Song Trilogy/ === Subject: Spherical curves very confused In Message-ID > Let f be a unit-speed curve with kappa > 0, torsion not equal to 0. > If f lies on a sphere of center c and radius r, show that > f(t) - c = -rho N - rho' sigma B > where N is the principal normal, B is binormal, rho is 1/kappa, sigma = > 1/torsion Prime denotes differentiation with respect to s (arc length). Unit speed curve means velocity has magnitude 1, so that v = df/dt = df/ds ds/dt = df/ds = T where T = unit tangent vector. Since the curve is on a sphere with center c || f(t) - c ||^2 = R^2 where R = radius of the sphere. This relation (f(t) - c) . (f(t) - c) = R^2 differentiated with respect to s gives f'(t) . (f(t) - c) = 0 But f'(t) = T = unit tangent vector, so this means that the vector f - c may have nonzero components along the N (principal normal) and B (binormal) directions only. Hence the vector f - c may be written as f(t) - c = a N + b B (1) where a, b are unknown and possibly time-dependent, hence s-dependent, coef[CapitalThorn]cients. We calculate df/ds from (1) as f'(t) = a' N + a N' + b' B + b B' (2) N' and B' are known from Frenet's equations as N' = - kappa T + sigma B B' = - sigma N Equation (2) may be rewritten as T = f'(t) = a' N - a kappa T + a sigma B + b' B - b sigma N The T component of this vector equality gives 1 = -a kappa so that a = -1/kappa = -rho. The N component gives 0 = a' - b sigma b = a'/sigma = -rho'/sigma (1) now becomes f(t) - c = a N + b B = -rho N - rho'/sigma B === Subject: Re: Are logarithms still useful? > [...] > Why was this at all needed to produce tables of > trigonometric functions, especially when the > argument was in degrees, minutes, seconds? > Er, Herman? You're rather wasting your time here. ZZBunker says > nothing coherent at all, but merely gives the ßeeting and false > impression of coherence from time to time. > You cannot peel away the obscure bits and get at the nugget of insight > underneath. It's obscure bits all the way down. I've wondered if ZZBunker is a bot, possibly Bloxy release 2.0. -- http://hertzlinger.blogspot.com