* For the oneness of a number 'n' : if(n is even) n = n/2; >if(n is odd) n = 3*n + 1; Keep doing this until n = 1. The amount of necessary steps for this is >called the oneness of n. m ==== >>Message-id: I am trying to build fractals generated by the principal of the oneness of >>a > number. >>For the definition of the oneness of a number, view the internet, or >>below.* It's more commonly called the Collatz Conjecture or the 3x+1 problem. >>I have been unseccesful in creating a pattern around the oneness although >>I have struck on some intriguing details. These mainly resolve around the >>issue of, if we take the oneness of the oneness of a number, and keep >>doing that, how long does it take 'til we reach 1? Depends on how many binary digits the number has (which is related to the > magnitude of the number) AND ALSO on the pattern of 1s and 0s in the > binary representaion (which is NOT related to the magniyude of the > number). > >>(note that for 5 this never happens) It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 No, you didn't listen well, I take the oneness of the oneness for the recursive oneness looplength. The oneness of 5 is 5. the oneness of 5 is 5, etc.. >>I think I might be able to find a pattern in that since, if I call >>this the recursive oneness loop length (if anybody can think a better >>name, tell me), It's sometimes refered to as the stopping time. Others make up their own > terminology. > >> then, the recursive oneness loop length of a number like >>20000000 is still only 15. I am trying to make a large-scale 2d map of it, How are you plotting this in 2D? 1. Using complex number representation and messing around with that. 2. Using y as a period of x (y = x/WIDTH, so to speak) -- Quaternion ==== >Message-id: I am trying to build fractals generated by the principal of the oneness of >a >> number. >For the definition of the oneness of a number, view the internet, or >below.* >> >> It's more commonly called the Collatz Conjecture or the 3x+1 problem. >> >> >I have been unseccesful in creating a pattern around the oneness although >I have struck on some intriguing details. These mainly resolve around the >issue of, if we take the oneness of the oneness of a number, and keep >doing that, how long does it take 'til we reach 1? >> >> Depends on how many binary digits the number has (which is related to the >> magnitude of the number) AND ALSO on the pattern of 1s and 0s in the >> binary representaion (which is NOT related to the magniyude of the >> number). >> >(note that for 5 this never happens) >> >> It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 No, you didn't listen well, I take the oneness of the oneness for the >recursive oneness looplength. That is a bit hard to follow. >The oneness of 5 is 5. >the oneness of 5 is 5, etc.. Let me see if I got this straight. If S(n) is the stopping time, you are then finding the stopping time of the stopping time? S(n) S(S(n)) S(S(S(n))) etc. until that reaches 1? So if the stopping time of n is n (as with 5) you get a loop? I haven't considered that before, any other loops other than 5? Anything else interesting about doing that? I think I might be able to find a pattern in that since, if I call >this the recursive oneness loop length (if anybody can think a better >name, tell me), >> >> It's sometimes refered to as the stopping time. Others make up their own >> terminology. >> > then, the recursive oneness loop length of a number like >20000000 is still only 15. I am trying to make a large-scale 2d map of it, >> >> How are you plotting this in 2D? 1. Using complex number representation and messing around with that. >2. Using y as a period of x (y = x/WIDTH, so to speak) That doesn't help much. Can you be more specific? -- >Quaternion -- Mensanator Ace of Clubs ==== >>Message-id: Message-id: I am trying to build fractals generated by the principal of the oneness >>of a > number. >>For the definition of the oneness of a number, view the internet, or >>below.* > > It's more commonly called the Collatz Conjecture or the 3x+1 problem. > > >>I have been unseccesful in creating a pattern around the oneness >>although I have struck on some intriguing details. These mainly resolve >>around the issue of, if we take the oneness of the oneness of a number, >>and keep doing that, how long does it take 'til we reach 1? > > Depends on how many binary digits the number has (which is related to > the magnitude of the number) AND ALSO on the pattern of 1s and 0s in the > binary representaion (which is NOT related to the magniyude of the > number). > >>(note that for 5 this never happens) > > It doesn't? 5 -> 16 -> 8 -> 4 -> 2 -> 1 >>No, you didn't listen well, I take the oneness of the oneness for the >>recursive oneness looplength. That is a bit hard to follow. > >>The oneness of 5 is 5. >>the oneness of 5 is 5, etc.. Let me see if I got this straight. If S(n) is the stopping time, you are > then finding the stopping time of the stopping time? S(n) > S(S(n)) > S(S(S(n))) > etc. until that reaches 1? Yes, and the amount of recursive calls needed for that I call the recursive oneness loop length. Let's call this ROLL(n) > So if the stopping time of n is n (as with 5) you get a loop? Yeah, > I haven't considered that before, any other loops other than 5? Anything > else interesting about doing that? 5 is not the only one that loops, any number of which the oneness is 5 will cause a loop. The oneness of 32 is 5. Very few numbers have a oneness of 5. I'm not sure but something tells me the only next one of which ROLL(n) becomes infinite is 2^32. (which causes the trap 2^32->32->5->5->5->... of course) And very few numbers n cause ROLL(n) to become infinite. Again like I said, it takes a long while for ROLL(n) to become larger than 16. This only happens regularily when n exceeds 2000000000 (2 billion) You just gave me a cool idea btw. Suppose you have a number 's' and you call N(s) the amount of numbers for which the oneness is 's'. I'll try that later. There must be, for each 's', at least one even and at least one odd number that reaches them? Could those two systems form two two-dimensional axises? Also, I made another recursive loop that does S(n) S(S(n)) S(S(S(n))) Until this reaches 'n', not one. Let's call this looplength the recursive oneness identity loop length, ROILL. Note that this only happens for 1,2,3,4,5,7, and 16 Also note that in my definition ROLL(1) = ROILL(1) = 6. Since S(1) = 3 S(3) = 7 S(7) = 16 S(16) = 4 S(4) = 2 S(2) = 1 I think this set of numbers might be interesting, in intuitive terms 3 and 7 might be seen as the 'prime' counterparts of the 2-powers 16 and 4, and 2 as the main axis, being as well prime as the smallest power of 2. Well perhaps you are into intuitive stuff and care to think about it. Again note that this is my definition, some definitions would say S(1) = 0. I'm not a believer in that :) >>I think I might be able to find a pattern in that since, if I call >>this the recursive oneness loop length (if anybody can think a better >>name, tell me), > > It's sometimes refered to as the stopping time. Others make up their own > terminology. > >> then, the recursive oneness loop length of a number like >>20000000 is still only 15. I am trying to make a large-scale 2d map of >>it, > > How are you plotting this in 2D? >>1. Using complex number representation and messing around with that. >>2. Using y as a period of x (y = x/WIDTH, so to speak) That doesn't help much. Can you be more specific? That's because called brainstorm coding and it's a bit hard to express with written language. I did say what is necessary : I've tried to visualize any possible patterns in the one-dimensional series, let's leave it at that because I've tried many things. As for the complex numbers, I tried dividing by 2+2i when the complex and real parts are even, and multiplying with 3+3i and adding 1+1i when they are odd. I'm still experimenting with that.. -- Quaternion ==== Does two have more oneness than one? ==== > Does two have more oneness than one? In my defintion, the oneness of one is 3. 1->4->2->1 takes three iterations It could be that some definitions say that the oneness of 1 is 0, since it starts at the stopping time. It doesn't matter that much, and I prefer it this way, where S(1)=3. And thus larger than S(2) -- Quaternion ==== > Does two have more oneness than one? In my defintion, the oneness of one is 3. 1->4->2->1 > takes three iterations It could be that some definitions say that the oneness of 1 is 0, since it > starts at the stopping time. It doesn't matter that much, and I prefer it > this way, where S(1)=3. And thus larger than S(2) -- > Quaternion If you use an index to represent the collatz-transformations, like C_0 = 1 C_1 = 1*2^1 C_2 = 1*2^2 C_a = 1*2^a C_2,0 = (C_2 -1)/3 C_a,0 = (C_a -1)/3 C_a,b = C_a,0 * 2^b with the periodic property that C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which reflects the loop 1-4-2-1....) and so on, then this indicates in a very natural way the oneness and stopping time of a number: it's just the sum of all indexes plus the number of them: C_4,0 = 5 Stopping-time = (4+0) + (1) = 5 C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7 Because of the loop at 1 it is also C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping time is not exactly unique. With the additional restriction of allowing only the smallest number of indexes the definition is unique then. For 1 the stopping time is then also unique, namely 0, since the representation with the shortest index is C_0 = 1 Gottfried Helms ==== >> >> >> Does two have more oneness than one? >> >> In my defintion, the oneness of one is 3. >> >> 1->4->2->1 >> takes three iterations >> >> It could be that some definitions say that the oneness of 1 is 0, since >> it starts at the stopping time. It doesn't matter that much, and I prefer >> it this way, where S(1)=3. And thus larger than S(2) >> >> -- >> Quaternion If you use an index to represent the collatz-transformations, > like C_0 = 1 > C_1 = 1*2^1 > C_2 = 1*2^2 > C_a = 1*2^a C_2,0 = (C_2 -1)/3 > C_a,0 = (C_a -1)/3 > C_a,b = C_a,0 * 2^b with the periodic property that > > C_0 = C_2,0 = C_2,2,0 = C_2,2,2,2,0 = C_2,...2,0 = 1 (which reflects the loop 1-4-2-1....) and so on, then this indicates in a very natural way the oneness and > stopping time of a number: it's just the sum of all indexes plus the > number of them: > C_4,0 = 5 Stopping-time = (4+0) + (1) = 5 > C_4,1,0 = 3 Stopping-time = (4+1+0) + (2) = 7 Because of the loop at 1 it is also C_4,0 = C_2,4,0 = C_2,2,....,2,4,0 = 5 so the stopping time is not exactly unique. With the additional > restriction of allowing only the smallest number of indexes the > definition is unique then. For 1 the stopping time is then also unique, > namely 0, since the representation with the shortest index is C_0 = 1 Damn let me take that in tonight, it's quite complex for me to understand what you mean to represent with 'C_a1,a2,a3..,an' right now and how it can fight the logic that if you follow the iteration scheme you get a bijection between n and the amount of iterations that are needed to transform it to 1.. -- Quaternion ==== > Damn let me take that in tonight, :-)) So I wish, that the nightmares won't catch you... Gottfried (P.S. I had them already, just some weeks ago :-) ==== Does two have more oneness than one? > That seems vague, and depends on context. Is this ONENESS part of some higher aspect of Mathematics beyond the typical undergraduate study of Calculus? At a simpler level, it seems that you could treat any single group of a given quantity as ONE WHOLE; so a grouped unit is one whole set. This allows us to find a fraction of any number as well as find the fraction of 1. G C ==== > It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: Consider (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 where even by inspection you can see that the constant terms are > separated out, so that you have 1(1)(22) = 22, the constant term of > the polynomial. I'll add that at x=0, a_1(0) = a_2(0) = b_3(0) = 0. Notice, (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) where again you see that the constant terms match as now you have > 7(7)(22) = 1078, which is again the constant term of the polynomial. If 22 does not have 7 as a factor, the former factorization is the > *only* allowed way for 49 to divide through. (For more detail, like what the a's are, see http://mathforprofit.blogspot.com/ where more is explained.) I have a result that shows a problem with a definition that > mathematicians have used for over a hundred years, And, what, exactly, is the problem? > and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. Some posters, not professional mathematicians from what I've gathered, > are at least trying to stand and fight, but because what I have is a > math proof, their claims are necessarily irrational. I need you to stand up for the truth, here and now. Let's chase these mathematician cowards down, and make them face the > music. After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. Let's take 'em out. > James Harris > http://mathforprofit.blogspot.com/ ==== Let's take 'em out. James Harris Take 'em out? What do you mean by that? Are you back to your previous > threat of calling out the military or the CIA and the FBI, or are you > advocating the formation of a private militia? > I think James wants his vast silent audience to treat mathematicians to > dinner and a movie. I suspect that there's an ulterior motive. Physics students are a cool crowd and I know because I was once a physics undergrad, and I'd get into all kinds of fascinating discussions, as physics is that kind of field. When you deal with crackpots and cranks, which mathematicians are now, who have been backed into a logical corner, you go ahead and deliver the coup de grace of the irrefutable facts. That is, you take 'em out. Now then, the crackpots may still believe their nonsense, but civilized society is once again protected by the vigillance of active and highly intelligent minds, as the bulk of society can move on with the truth. I've isolated mathematicians by a logical argument, having made a major discovery--that's what physics people do--in what they probably think of as their area alone, and caught them trying to dodge the result and its implications. Remember mathematicians have claimed that pure math was for the benefit of humanity, but I've shown them trying to hide one of the most astounding results in math history (one of three, mind you). So it's time for the physics people to look at the facts, stand for logic over social needs, and take 'em out. James Harris <8Barb.9828$y95.4857@nwrdny01.gnilink.net> <3c65f87.0311081454.195d813a@posting.google.com> ==== > Remember mathematicians have claimed that pure math was for the > benefit of humanity, but I've shown them trying to hide one of the > most astounding results in math history (one of three, mind you). Now a lot of people think that James is an egotist with no sense of proportion. I hope that this quote lays that accusation to rest. As we can see here, James exhibits an appropriate sense of humility by allowing (charitably, I'm sure) that his result is only one of *three* of the greatest results in math history. Don't leave us hanging here, James. What are the other two? -- Jesse Hughes If you really think there's a bug you should report a bug. Maybe you're not using it properly... It turns out Luddites don't know how to use software properly, so you should look into that. -- Bill Gates ==== >Message-id: <87smkxylda.fsf@phiwumbda.org > Remember mathematicians have claimed that pure math was for the >> benefit of humanity, but I've shown them trying to hide one of the >> most astounding results in math history (one of three, mind you). Now a lot of people think that James is an egotist with no sense of >proportion. I hope that this quote lays that accusation to rest. As >we can see here, James exhibits an appropriate sense of humility by >allowing (charitably, I'm sure) that his result is only one of *three* >of the greatest results in math history. Don't leave us hanging here, James. What are the other two? Didn't he solve The Riemann Hypothesis with his prime counting function? -- >Jesse Hughes >If you really think there's a bug you should report a bug. Maybe >you're not using it properly... It turns out Luddites don't know how >to use software properly, so you should look into that. -- Bill Gates -- Mensanator Ace of Clubs ==== It turns out that I can isolate the current dispute easily enough by > focusing on the factorizations: Consider [snipped] Crank Information > http://www.crank.net/harris.html > http://www.crank.net/usenet.html > http://www.google.com/search?q=harris+site%3Awww.crank.net > http://www.google.com/search?q=%22james+harris%22+site%3Ausers.pandora.be Readers should please check out *all* the links Sam the Worm listed. As for the math, notice that with (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 no other factorization works as long as 7 is not a factor of 22. You have only demonstrated anything for the case x=0, not 'x' in general. -- 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 ==== I hope you're not speaking poorly about my Venus lizard folk, Cathars and all. This may be getting some folks a wee bit off topic from group focus, or that of my intended agenda upon Venus life (lizard folk and all), but according to some fairly recent feedback, I've learned a thing or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby recedes form Earth at 38 mm/year. As worthy feedback provided from: Ami Silberman (silber@mitre.org) The mechanisms for the lunar recession have been well understood for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer of energy is essentially a transfer of angular momentum, which is a conserved quantity.) The historical (over geological eras) rate of recession has varied due to varying amounts of tidal friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.htm ==== Unfortunately Scott, this sting/ruse is way so much bigger than even you can imagine. I've posted this folowing in at least a half dozen topics that I believe are related to obtaining the truth. This may be getting some folks a wee bit off topic from group focus, or that of my intended agenda upon Venus life (lizard folk and all), but according to some fairly recent feedback, I've learned a thing or two about our nasty moon, as a place that I believe we've long needed to establish a lunar space elevator (LSE) in order to be efficiently getting ourselves off to visiting the wizard of Oz at Venus L2, as well as for reaching out to those thoroughly irradiated to death snowman/snowwoman situated on Mars: http://guthvenus.tripod.com/gv-cm-ccm-01.htm Here's a little typical feedback of supposed facts from: Jay Windley (webmaster@clavius.org) outside the solar system, and our sun is the primary defense against lower energy and thus their secondary effects in the ambient are minimal. My thoughts: http://guthvenus.tripod.com/gv-moon-radiation.htm Unfortunately, I may have incorrectly utilized the terminology of cosmic rays, though fortunately, I never specified upon any specific spectrum of high-energy cosmic rays, just pointing out that our sun is certainly capable of tossing out its fair share of far worse things than visible photons plus IR worth of BTUs and of those nasty UVs. Obviously a supernovae is worth a thousand fold in terms of being nasty, thereby from the far off generated galactic influx must offer a measurable degree of such events, and of the secondary radiation given off by all that infamous clumping lunar dirt should become a fairly darn good as well as unobstructed indicator of whatever cosmic/galactic influx. Seems fairly odd that shuch measurements aren't common place, as where's the justification for not otherwise providing this level of information? The assertions or premise offered by the likes of Jay Windley, that of our moon not only lacking an atmosphere but also without a Van Allen buffer zone is not such a bad thing if you're out and about on the lunar surface, seems somewhat risky if not downright lethal. I might have come into that understanding if we're referring to an earthshine illuminated lunar surface, but I'm not going so far if that's of any fully solar illuminated environment while wearing a mostly synthetic moon suit because, we're not talking about avoiding a 270 nm UV sun burn. Sorry about all my make-due reverse engineering logic, or lack thereof. I was simply trying to establish upon the amount of solar radiation that becomes hard X-Ray class. So exactly how much is it on a typical lunar day, or how about on a good day as well as on a truly Do we suppose that happens to include the likes of the last couple of weeks of horrific solar flak? Seems there should be some specific knowledge (excluding Apollo) of what's what pertaining to the solar illuminated surface, as opposed to the absolute lunar nighttime environment and, of something specific pertaining to whatever earthshine contributes. This task of obtaining knowledge is somewhat like my getting a grasp upon the applied energy (thrust or torque) involved in accelerating something the size and mass of the moon, so that it accelerates and thereby recedes form Earth at 38 mm/year. As worthy feedback provided from: Ami Silberman (silber@mitre.org) The mechanisms for the lunar recession have been well understood for decades. In a nutshell, tides cause friction between the oceans and the ocean floors, which transfers energy from the solid part of the earth to the oceans. One of the effects of this friction is that the tidal bulge is off-center, and is located eastward of the moon. (So the high tide actually occurs when the moon is west of overhead.) The result of the tidal bulge being off center is that there is a torgue effect placed on the moon, and this in turn transfers energy from the earth to the moon. The earth's spin rate slows, the moon is speeded in its orbit and therefor moves further away from the earth. (This transfer of energy is essentially a transfer of angular momentum, which is a conserved quantity.) The historical (over geological eras) rate of recession has varied due to varying amounts of tidal friction due to shallower or deeper oceans, and the positions of the continents. For the benefit of all my loyal critics, I've conceded that there's a darn good chance that the likes of Tim Thompson has more than a few valid points as to his version of what's what. This following page is just another example of my learning from the pros, of accepting other input, which may even including the likes of what Ami Silberman just presented, that I'd not be calling flak, as there actually seems to be some considerable worth to at least Tim's version of the lunar recession, if I don't say so myself. http://guthvenus.tripod.com/earth-moon-energy.htm ==== I posted an example earlier today to try and help those of you who've been so thoroughly confused by posters like Arturo Magidin, Nora Baron and Dik Winter to see how it all works, and it turns out that if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you would have a problem realizing that only two of the c's can have 7 as a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit fighting me on my work. My guess is that polynomial irreducible over Q, but I haven't checked it since I recently deleted off my math software because I had to reinstall Windows XP. Now then, that example should cause certain posters to post retractions, or they can try and run. Some of you might want to get legal advice. James Harris http://mathforprofit.blogspot.com/ ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you > would have a problem realizing that only two of the c's can have 7 as > a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let > x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit > fighting me on my work. My guess is that polynomial irreducible over Q, but I haven't checked > it since I recently deleted off my math software because I had to > reinstall Windows XP. Now then, that example should cause certain posters to post > retractions, or they can try and run. Some of you might want to get legal advice. This example is interesting for several reasons. One, it was shown almost immediately by more than one poster that there is no factorization with integer coefficients of the form that Harris claimed. In spite of that for a while he did not retract his claim that c_1 etc. *should* be algebraic integers. Two, he never acknowledged his error. Three, he started a new thread elsewhere on the same topic. Four, he claimed in that thread that c_1, etc. - which he himself described as algebraic *numbers* - could not be written as the quotient of two coprime algebraic integers. The latter contradicts an old and very deep theorem of Dedekind. Altogether, a fiasco. Confusion, and one error piled on top of another. But what is really ironic in this is that Harris *himself* proved that there is no factorization of the polynomial above with integer coefficients - and he used the identical method that we have used many times to show that his claims regarding the factorization of the famous P(x) [i.e., that two of the a's must be divisible by 7 or 5 or f or whatever] are false. So he evidently now understands and accepts that. But he does not react to it logically. He continues to say, I believe, that the a's *should* be divisible by f, with never a real explanation of *should*. He clings to a clearly-incorrect argument based on constant terms. He ignores the fact that if his central argument were correct, mathematics would be inconsistent. He concludes I think that the a's divided by f are objects, however with no clear method of telling what is an object and what is not, and no body of theorems regarding objects such as one has for algebraic integers. Why should anyone care if you prove something about objects, if no one can tell what an object is ? Actually I think the meaning of *should* is pretty clear. The a's *should* be divisible by f because, if they are not, his entire proof collapses. Thus the true translation of *should* is: *I desperately want it to be true!*, because if it's not, 8 years of work, torment, hope, and dreams of fame and riches go down the drain! Best summed up as: backwards logic, and pathos. Plus he is now running away from all substantive objections. He responds to superficial issues on notation and deletes substantive parts of posts where the real problems with his logic are confronted. He is starting new threads at a frantically high rate, interspersed with rants, whines, paranoid delusional claims, even legal threats, posted to several different news groups simultaneously: the actions of a frightened and desperate person, it would appear. Nora B. > James Harris > http://mathforprofit.blogspot.com/ ==== ... > Four, he > claimed in that thread that c_1, etc. - which he himself > described as algebraic *numbers* - could not be written as > the quotient of two coprime algebraic integers. The latter > contradicts an old and very deep theorem of Dedekind. The deepness is not that an algebraic number can be written as the quotient of two algebraic integer, but that they also can be chosen to be made coprime. That it is possible is because the algebraic integers form a gcd ring. That is you can define some (sensible) form of gcd. I do not know whether this was already known by Dedekind. Arturo will probably know (he has already posted about it). -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >... > Four, he > claimed in that thread that c_1, etc. - which he himself > described as algebraic *numbers* - could not be written as > the quotient of two coprime algebraic integers. The latter > contradicts an old and very deep theorem of Dedekind. The deepness is not that an algebraic number can be written as the >quotient of two algebraic integer, but that they also can be chosen >to be made coprime. That it is possible is because the algebraic >integers form a gcd ring. That is you can define some (sensible) >form of gcd. I do not know whether this was already known by >Dedekind. Arturo will probably know (he has already posted about it). In fact, stronger than just a gcd domain: you can write the gcd as a linear combination of the two numbers (i.e., a Bezout domain). Yes, this was known by Dedekind and by Kummer. In Dedekind's _Theory of Algebraic Numbers_, for example (available from Cambridge Mathematical Library, translated and introduced by John A deeper investigation will enable us to see that two nonzero [algebraic] integers a,b have a ->greatest common divisor<- [emphasis in the original], which ca be put in the form aa' + bb' where a' and b' are [algebraic] integers. This important theorem is not at all easy to prove with the help of the principles developed so far, but we shall later (section 30) be able to derive it simply from the theory of ideals. In section 30, after proving the finiteness of the class number, and that any ideal A raised to the class number is principal (about two pages), he proves the result [I am changing, as I did above, some greek letters into latin letter]: Any two algebraic integers, a,b have a common divisor d which can be expressed in the form d= aa'+ bb', where a' and b' are likewise algebraic integers. Proof. We assume that the two numbers a,b are nonzero, otherwise the theorem is evident. Then it is easy to see that there is a field F of finite degree including both a and b. Let R be the domain of integers of this field, and let h be the number of classes of ideals. Now let Ra = AD, Rb = BD, D^h = Rd_1 where D is the greatest common divisor of the ideals Ra and Rb, and d_1 is in R. Since a^h and b^h are divisible by D^h, we can write a^h = a_1*d_1, b^h = b_1*d_1, Ra_1 = A^h, Rb_1 = B^h where a_1,b_1 are likewise in R. Also, since A and B are relatively prime ideals, D will be the greatest common divisor of Ra_1 and Rb_1 and, since the number 1 is in R, there will be two numbers a_2 and b_2 in R satisfying the condition a_1*a_2 + b_1*b_2 = 1, or a^h*a_2 + b^h*b_2 = d_1. If we now put d_1 = d^h, then the integer d will be a common divisor of a and b, since a^h and b^h are divisible by d_1, and hence, since h>=1, we can put a_2*a^{h-1} = a'*d^{h-1}, b_2b^{h-1} = b'*d^{h-1}, where a', b' are integers satisfying the condition aa'+bb' = d. QED If at least one of the two numbers a,b is nonzero, then the number d, and any of its associates, deserves the name ->greatest<- common divisor of a,b [emphasis in the original]. If d is a unit, then a,b may be called ->relatively prime<- [emphasis in the original] and two such numbers enjoy the characteristic property that any number m divisible by a and b is also divisible by ab. This because the equations m=aa''=bb'', and 1=aa'+bb' imply m=ab(a'b''+b'a''), and the converse is equally valid, since a,b are both nonzero. That's the end of Dedekind's book. By the way, I highly recommend Dedekind's book (and Stillwell's introduction). It is not only a very good introduction to Alg. Number Theory, but it is amazing how modern it is for something written over 100 years ago. Modulo a few precisions (Dedekind calls something modules when we would call them Z-modules, and so on), it could work as a textbook in a class today. ====================================================================== 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 ==== welcome to Platform 3 and 1/7, the short proof of the last theorem of Fermat; Usenet is just to small to contain it! > Some of you might want to get legal advice. > But what is really ironic in this is that Harris *himself* > proved that there is no factorization of the polynomial > above with integer coefficients - and he used the identical > method that we have used many times to show that his claims > regarding the factorization of the famous P(x) [i.e., that > two of the a's must be divisible by 7 or 5 or f or whatever] are > false. So he evidently now understands and accepts that. > Actually I think the meaning of *should* is pretty > clear. The a's *should* be divisible by f because, if they > are not, his entire proof collapses. Thus the true > translation of *should* is: *I desperately want it to > be true!*, because if it's not, 8 years of work, torment, > hope, and dreams of fame and riches go down the drain! > Best summed up as: backwards logic, and pathos. > http://mathforprofit.blogspot.com/ --ils duces d'Enron! http://larouchepub.com/ ==== > Consider > > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) > > where the c's *should* be algebraic integers. Why should they be algebraic integers? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. Why should they be? Please provide a proof. -- 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 ==== In sci.math, C. Bond <3FAE78C0.71CE871C@ix.netcom.com>: > >> I posted an example earlier today to try and help those of you who've >> been so thoroughly confused by posters like Arturo Magidin, Nora >> Baron and Dik Winter to see how it all works, and it turns out that >> if you were smart enough you should have noticed something. >> Consider >> (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = >> 49(x^3 + 5x^2 + 3x + 2) >> where the c's *should* be algebraic integers. Why should they be? Please provide a proof. If he does provide a valid proof I'll be very surprised; I've already proven that they're not. :-) (How I prove that my proof is valid is not entirely clear but I think I can at least explain most of my steps. :-) ) [.sigsnip] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. No, you're completely off base on this. There is no factorization of the form you propose where c1, c2, and c3 are algebraic integers. This is easy to see. First, note that the roots of [1] R(x) = x^3 + 5*x^2 + 3*x + 2 are all algebraic integers - say, a, b, and c - and their product is -2: a*b*c = -2. Since R(x) is irreducible, a, b, and c are non-units. Now you can write your original polynomial as 49*(x - a)*(x - b)*(x - c), and then distribute the 49 among the factors however you want - for example, as (7*x - 7*a)*(7*x - 7*b)*(x - c), but then note that neither 7*a nor 7*b can equal 7: remember, a, b, and c are nonunits. It's easy to show no other distribution works either. Bottom line: there is no factorization that looks like what you have above. This is a dead end. > I think few of you > would have a problem realizing that only two of the c's can have 7 as > a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let > x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit > fighting me on my work. > Obviously the roots cannot be algebraic integers. This is just another way of proving what I showed above, that is, there is no factorization of the form you started with. And so what??? > My guess is that polynomial irreducible over Q, but I haven't checked > it since I recently deleted off my math software because I had to > reinstall Windows XP. > It's trivial to check. The only possible roots are 2, -2, 1, and -1. None work. > Now then, that example should cause certain posters to post > retractions, or they can try and run. > Ridiculous. You blundered on this one. Best admit it and move on. > Some of you might want to get legal advice. > You're going to sue us for pointing out your mistake? Nora B. James Harris > http://mathforprofit.blogspot.com/ ==== >[...] Now then, that example should cause certain posters to post >retractions, or they can try and run. Some of you might want to get legal advice. Oops. Sounding like a raving lunatic _again_. You know some people are starting to wonder about that... >James Harris >http://mathforprofit.blogspot.com/ David C. Ullrich ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you > would have a problem realizing that only two of the c's can have 7 as > a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let > x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit > fighting me on my work. Now what exactly is that supposed to show? Looks like you are completely incapable of writing down one coherent thought. You may want to check out http://www.mentalhealth.com/rx/p23-pe07.html which won't explain the maths, but it will explain the real problem you are having. ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. Why? ==== In sci.math, James Harris <3c65f87.0311081551.261b9a0b@posting.google.com>: > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. Sorry, they're not. If P(x) = 49*(x^3 + 5*x^2 + 3*x + 2) = 49 * (x - x_1) * (x - x_2) * (x - x_3) for some x_i, then for some permutation of these x_i, c_1 = -7/x_1 c_2 = -7/x_2 c_3 = -2/x_3 (note that x_1 * x_2 * x_3 = -2 so c_1 * c_2 * c_3 = 49, as required) If one substitutes x = -7/c, one gets c^3 * P(-7/c)/49 = 2*c^3 - 21*c^2 + 245*c - 343 which is irreducible. Sorry James. If one substitutes x = -2/c, c_3 satisfies c^3*P(-2/c)/98 = c^3 - 3*c^2 + 10*c - 4 so we get one out of three. [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >I posted an example earlier today to try and help those of you who've >been so thoroughly confused by posters like Arturo Magidin, Nora >Baron and Dik Winter to see how it all works, and it turns out that >if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you >would have a problem realizing that only two of the c's can have 7 as >a factor with that example!!! Let A denote the ring of all algebraic integers and let r1, r2, r3 denote the zeros of f(x) = x^3 + 5x^2 + 3x + 2. Since f is monic and its coefficients are rational integers, its zeros are in A, but none of the zeros is a unit in A since f is irreducible over Z and its constant coefficient is not a unit. The r's are not all coprime to 2 in A since 2 divides r1 * r2 * r3 = -2 in A. If exactly two of the r's, say r1 and r2, were coprime to 2 in A then 2 would divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in A and each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in A. If exactly one of the r's, say r1, was coprime to 2 in A then 2 would divide r2 * r3 in A, so (r2 * r3) / 2 = -1 / r1 would be in A and r1 would be a unit in A. Hence, in the factorization f(x) = (x - r1)(x - r2)(x - r3), none of the r's is coprime to 2 in A. The polynomial 49 * f(x) is not primitive and there is an infinity of ways in which its content may be distributed over its factors. If c1, c2, c3 are any algebraic integers whose product is 49 then there is a factorization of 49 * f(x) in A[x] as 49 * f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3), and, since it has the same zeros as f, every factorization of 49 * f(x) in A[x] has this form. Since none of the r's is coprime to 2 in A, none of the c * r's is coprime to 2 in A. In particular, none of them can be equal to -7. The supposed example cannot exist. I can see no way out. You will have this problem in any ring in which 2 and 7 are coprime. Now then, you have as a zero of the factorization x = -7/c_1, so let >x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit >fighting me on my work. My guess is that polynomial irreducible over Q, but I haven't checked >it since I recently deleted off my math software because I had to >reinstall Windows XP. Now then, that example should cause certain posters to post >retractions, or they can try and run. Some of you might want to get legal advice. >James Harris >http://mathforprofit.blogspot.com/ John Roberts-Jones ==== >I posted an example earlier today to try and help those of you who've >been so thoroughly confused by posters like Arturo Magidin, Nora >Baron and Dik Winter to see how it all works, and it turns out that >if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you >would have a problem realizing that only two of the c's can have 7 as >a factor with that example!!! Let A denote the ring of all algebraic integers and let r1, r2, r3 > denote the zeros of f(x) = x^3 + 5x^2 + 3x + 2. > > Since f is monic and its coefficients are rational integers, its > zeros are in A, but none of the zeros is a unit in A since f is > irreducible over Z and its constant coefficient is not a unit. The r's are not all coprime to 2 in A since 2 divides > r1 * r2 * r3 = -2 in A. If exactly two of the r's, say r1 and r2, were coprime to 2 in A > then 2 would divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in > A and each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in A. If exactly one of the r's, say r1, was coprime to 2 in A then 2 > would divide r2 * r3 in A, so (r2 * r3) / 2 = -1 / r1 would be in A > and r1 would be a unit in A. Hence, in the factorization f(x) = (x - r1)(x - r2)(x - r3), none > of the r's is coprime to 2 in A. The polynomial 49 * f(x) is not primitive and there is an infinity > of ways in which its content may be distributed over its factors. > If c1, c2, c3 are any algebraic integers whose product is 49 then > there is a factorization of 49 * f(x) in A[x] as > 49 * f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3), > and, since it has the same zeros as f, every factorization of > 49 * f(x) in A[x] has this form. Since none of the r's is coprime to 2 in A, none of the c * r's is > coprime to 2 in A. In particular, none of them can be equal to -7. > The supposed example cannot exist. Now that's funny. You must mean that the c's can't be algebraic integers as, of course, they can be algebraic numbers, right? I mean, you are NOT trying to say that the c's just can't exist at all, right? > I can see no way out. You will have this problem in any ring in > which 2 and 7 are coprime. So I guess you're saying that if 2 and 7 are coprime in a ring then the c's, can't be in that ring. You're wrong, and it's easy to prove. I'll use a separate thread though, for my own reasons. James Harris ==== >> >>I posted an example earlier today to try and help those of you who've >>been so thoroughly confused by posters like Arturo Magidin, Nora >>Baron and Dik Winter to see how it all works, and it turns out that >>if you were smart enough you should have noticed something. >>Consider >>(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = >> >> 49(x^3 + 5x^2 + 3x + 2) >>where the c's *should* be algebraic integers. I think few of you >>would have a problem realizing that only two of the c's can have 7 as >>a factor with that example!!! >> >> Let A denote the ring of all algebraic integers and let r1, r2, r3 >> denote the zeros of f(x) = x^3 + 5x^2 + 3x + 2. >> >> Since f is monic and its coefficients are rational integers, its >> zeros are in A, but none of the zeros is a unit in A since f is >> irreducible over Z and its constant coefficient is not a unit. >> >> The r's are not all coprime to 2 in A since 2 divides >> r1 * r2 * r3 = -2 in A. >> >> If exactly two of the r's, say r1 and r2, were coprime to 2 in A >> then 2 would divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in >> A and each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in A. >> >> If exactly one of the r's, say r1, was coprime to 2 in A then 2 >> would divide r2 * r3 in A, so (r2 * r3) / 2 = -1 / r1 would be in A >> and r1 would be a unit in A. >> >> Hence, in the factorization f(x) = (x - r1)(x - r2)(x - r3), none >> of the r's is coprime to 2 in A. >> >> The polynomial 49 * f(x) is not primitive and there is an infinity >> of ways in which its content may be distributed over its factors. >> If c1, c2, c3 are any algebraic integers whose product is 49 then >> there is a factorization of 49 * f(x) in A[x] as >> 49 * f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3), >> and, since it has the same zeros as f, every factorization of >> 49 * f(x) in A[x] has this form. >> >> Since none of the r's is coprime to 2 in A, none of the c * r's is >> coprime to 2 in A. In particular, none of them can be equal to -7. >> The supposed example cannot exist. Now that's funny. You must mean that the c's can't be algebraic >integers as, of course, they can be algebraic numbers, right? I mean, you are NOT trying to say that the c's just can't exist at >all, right? We seem to be agreed that there is no factorization in the given form of 49 * f(x) in A[x]. If we instead work in the ring of polynomials over the algebraic numbers, then of course 49 * f(x) = (c1 * x + 7)(c2 * x + 7)(c3 * x + 2), where c1 = -7 / r1, c2 = -7 / r2 and c3 = -2 / r3 = r1 * r2. Only c3 is an algebraic integer. I hope you are not suggesting that c1 and c2 should be algebraic integers on the grounds that they may be written as the ratio of two algebraic integers. > >> I can see no way out. You will have this problem in any ring in >> which 2 and 7 are coprime. So I guess you're saying that if 2 and 7 are coprime in a ring then >the c's, can't be in that ring. Yes, if 2 and 7 are not units. > You're wrong, and it's easy to prove. I'll use a separate thread though, for my own reasons. I look forward to seeing it. >James Harris John Roberts-Jones ==== I posted an example earlier today to try and help those of you who've >been so thoroughly confused by posters like Arturo Magidin, Nora >Baron and Dik Winter to see how it all works, and it turns out that >if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you >would have a problem realizing that only two of the c's can have 7 as >a factor with that example!!! Let A denote the ring of all algebraic integers and let r1, r2, r3 > denote the zeros of f(x) = x^3 + 5x^2 + 3x + 2. Since f is monic and its coefficients are rational integers, its > zeros are in A, but none of the zeros is a unit in A since f is > irreducible over Z and its constant coefficient is not a unit. The r's are not all coprime to 2 in A since 2 divides > r1 * r2 * r3 = -2 in A. If exactly two of the r's, say r1 and r2, were coprime to 2 in A > then 2 would divide r3 in A, so r3 / 2 = -1 / (r1 * r2) would be in > A and each of -r1 * r2, r1 * r2, r1 and r2 would be a unit in A. If exactly one of the r's, say r1, was coprime to 2 in A then 2 > would divide r2 * r3 in A, so (r2 * r3) / 2 = -1 / r1 would be in A > and r1 would be a unit in A. Hence, in the factorization f(x) = (x - r1)(x - r2)(x - r3), none > of the r's is coprime to 2 in A. The polynomial 49 * f(x) is not primitive and there is an infinity > of ways in which its content may be distributed over its factors. > If c1, c2, c3 are any algebraic integers whose product is 49 then > there is a factorization of 49 * f(x) in A[x] as > 49 * f(x) = (c1 * x - c1 * r1)(c2 * x - c2 * r2)(c3 * x - c3 * r3), > and, since it has the same zeros as f, every factorization of > 49 * f(x) in A[x] has this form. Since none of the r's is coprime to 2 in A, none of the c * r's is > coprime to 2 in A. In particular, none of them can be equal to -7. > The supposed example cannot exist. Now that's funny. You must mean that the c's can't be algebraic > integers as, of course, they can be algebraic numbers, right? I mean, you are NOT trying to say that the c's just can't exist at > all, right? I can see no way out. You will have this problem in any ring in > which 2 and 7 are coprime. So I guess you're saying that if 2 and 7 are coprime in a ring then > the c's, can't be in that ring. You're wrong, and it's easy to prove. I'll use a separate thread though, for my own reasons. > Can't get out of the argument again, eh? MAYBE if you took a few years and learned some math, you'd see that we're not lying like you think we are. I'd think if you did this 7 years ago, you could almost have a Ph.D by now. But as I said before, you obviously don't care about the math, you just want the fame and money. I do have a challenge for you; go an entire thread without any childish insults. David Moran James Harris ==== I can see no way out. You will have this problem in any ring in >which 2 and 7 are coprime. I should have said ... in which 2 and 7 are coprime non-units. John Roberts-Jones ==== My guess is that polynomial irreducible over Q, but I haven't checked > it since I recently deleted off my math software because I had to > reinstall Windows XP. Now then, that example should cause certain posters to post > retractions, or they can try and run. Some of you might want to get legal advice. > James Harris > http://mathforprofit.blogspot.com/ psychopathology: narcissism ---------------------------------------------------------------------------- ---- A pattern of traits and behaviours which signify infatuation and obsession with one's self to the exclusion of all others and the egotistic and ruthless pursuit of one's gratification, dominance and ambition. ==== > >>My guess is that polynomial irreducible over Q, but I haven't checked >>it since I recently deleted off my math software because I had to >>reinstall Windows XP. >>Now then, that example should cause certain posters to post >>retractions, or they can try and run. >>Some of you might want to get legal advice. >>James Harris >>http://mathforprofit.blogspot.com/ > psychopathology: narcissism ---------------------------------------------------------------------------- > ---- A pattern of traits and behaviours which signify infatuation and obsession > with one's self to the exclusion of all others and the egotistic and > ruthless pursuit of one's gratification, dominance and ambition. I think you've nailed it. Gib ==== Dear Flip, http://malignantselflove.tripod.com/npdglance.html There is a lot more here: http://malignantselflove.tripod.com/ Take care. Sam Vaknin ==== Dear Flip, > http://malignantselflove.tripod.com/npdglance.html There is a lot more here: http://malignantselflove.tripod.com/ Take care. Sam Vaknin you have excellent writing skills. I wonder if you could answer a couple of questions regarding JSH's NPD? 1. Is it bad to feed the troll? 2. Does feeding the troll only serve to increase his high from all of the attention, particularly when the comments are against his point of view? 3. How can he be helped? 4. What can we do to subdue the troll? ==== > I wonder if you could answer a couple of questions regarding JSH's NPD? 1. Is it bad to feed the troll? 2. Does feeding the troll only serve to increase his high from all of the > attention, particularly when the comments are against his point of view? 3. How can he be helped? 4. What can we do to subdue the troll? > You may take a look at http://www.mentalhealth.com/dis/p20-pe07.html which discusses diagnosis and treatment. Treatment on sci.math is just plain impossible. Ignoring him completely is probably the best we could do for his mental health. To be able to treat him, you would have to be a person of the highest possible authority (and if you read his opinion of Andrew Wiles, you will see that this person will be hard to find), and that person would have to not contradict him (which would only make him feel threatened and strengthen his illness), but gently lead him to a realistic outlook on the world. Forget it, you won't find anyone who could do that. ==== I will not go into any specifics in my response. Please do not forget that only a qualified mental health diagnostician - can determine whether someone suffers from NPD and this, following lengthy tests and personal interviews. Still, these may be of interest: http://malignantselflove.tripod.com/journal67.html http://malignantselflove.tripod.com/faq4.html Take care. Sam ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you > would have a problem realizing that only two of the c's can have 7 as > a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let > x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit > fighting me on my work. My guess is that polynomial irreducible over Q, but I haven't checked > it since I recently deleted off my math software because I had to > reinstall Windows XP. Now then, that example should cause certain posters to post > retractions, or they can try and run. Some of you might want to get legal advice. > James Harris > http://mathforprofit.blogspot.com/ James, I thought you were going back to lancing blisters and emptying bedpans. What happened? Lurch ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. Consider (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) where the c's *should* be algebraic integers. I think few of you > would have a problem realizing that only two of the c's can have 7 as > a factor with that example!!! Now then, you have as a zero of the factorization x = -7/c_1, so let > x= -7/c_1, so you have 49(-7^3/c_1^3 + 5(49)c_1^2 - 21/c_1 + 2) = 0 which is 2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0. Please tell me that example is simple enough that now you'll quit > fighting me on my work. My guess is that polynomial irreducible over Q, but I haven't checked > it since I recently deleted off my math software because I had to > reinstall Windows XP. Now then, that example should cause certain posters to post > retractions, or they can try and run. Some of you might want to get legal advice. > Legal advice? For what? Maybe if you took some math courses, you'd have an idea of what everyone is talking about and that none of us are lying. David Moran James Harris > http://mathforprofit.blogspot.com/ ==== > I posted an example earlier today to try and help those of you who've > been so thoroughly confused by posters like Arturo Magidin, Nora > Baron and Dik Winter to see how it all works, and it turns out that > if you were smart enough you should have noticed something. > Hey James. Nora has the patience of an absolute saint. You should be honored that she spends so much time on her thoughtful reasoned replies to you. But no. You'd rather avoid answering questions, avoid any response to the points brought up, and avoid avoiding calling people dog shit. You are pitiful. ==== those of us who try to killfile him. As you've no doubt noticed though Robin, his posts (to sci.math anyway) do have an Achilles heel - a four-digit year at the end of the title. If your newsreader supports wildcard titles in kill rules, all one needs is a filter on titles of the form /(19|20)dd$/ (in Perl wildcard syntax). --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen ==== >Indeed, algebraic topology seems to be an extremely powerful tool: Yes! >S^2 is simply connected, Yes! >so every continuous image of it also is No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe or something to get an interval, and then join the ends together to get a circle, which is not simply connected. Sort of at the basis of the theory of covering spaces is the observation that the continuous image of a simply-connected space is not necessarily simply connected. dave ==== > >>so every continuous image of it also is simply connected. > No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe > or something to get an interval, and then join the ends together > to get a circle, which is not simply connected. What do you mean with roll it up like a crepe? And joining the ends together is no continuous operation!? Sort of at the basis of the theory of covering spaces is the > observation that the continuous image of a simply-connected space is > not necessarily simply connected. > Yes, I see that I was wrong. But given a simply connected top. space X and a top. space Y together with an epimorphism f:X->Y, can we say that Y is simply connected? My idea was that any continuous curve s:[0,1]->Y can be pulled back to a continuous curve r:[0,1]->X such that f(r)=s so that we can prove that Y has to be simply connected. Can we turn this idea into a proof? According to your post, the answer seems to be no. So can you give an example where this pull-back does not work? In the present case, we would have X=S^2, Y=R^2{(0,0)} and f any continuous map X->Y. Of course it need not be an epimorphism, so the conclusion was wrong even if the proposition mentioned above was correct. I'll think about it further, but this is it for now. Tobias -- ==== > so every continuous image of it also is simply connected. >> No! Take a balloon (S^2), deflate it to a disk, roll it up like a crepe >> or something to get an interval, and then join the ends together >> to get a circle, which is not simply connected. > What do you mean with roll it up like a crepe? Sorry, too many math for poets lectures. That really just means to take a projection to an interval. Everything to this point is just the projection pi : R^3 --> R^1, restricted to S^2. (That's continuous.) >And joining the ends together is no continuous operation!? Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ). Perhaps you're thinking of continuous maps _with continuous inverses_, which this isn't. If that's what you mean by continuous image, then the right phrase is homeomorphic image, and then yes, the homeomorphic image of a simply-connected space is also simply-connected. >> Sort of at the basis of the theory of covering spaces is the >> observation that the continuous image of a simply-connected space is >> not necessarily simply connected. >Yes, I see that I was wrong. But given a simply connected top. space X and a > top. space Y together with an epimorphism f:X->Y, can we say that Y is > simply connected? No, that is still exactly the setting for covering spaces. The usual example is the same f I gave above, where X is the real line and Y is the circle. > My idea was that any continuous curve s:[0,1]->Y can be > pulled back to a continuous curve r:[0,1]->X such that f(r)=s That's called the path-lifting property; it's a consequence of f being a covering map. (It's also possessed by other maps, such as fibrations.) > so that we can prove that Y has to be simply connected. No! First of all, the fact that you can lift paths does not mean that you can lift loops; that is, if s is a loop in Y, then it's also a path, and so it lifts to a path r if f is a covering (say); but this lifted path r need not be a loop, i.e., there's no reason you must have r(0)=r(1). If you can't see this with formulas, draw a picture of a helix (a spring) X, casting a circular shadow. The shadow, Y is a circle and the projection is the map f. Draw a loop around Y, and then lift it back to X and you'll see you don't get a loop. Also you want to be a little careful : the fact that you can lift paths does not by itself guarantee that you can lift other maps such as homotopies (maps from [0,1]x[0,1] into Y). If you _are_ in the context of covering spaces, there is a pretty characterization of what can be lifted: a map r : Z --> Y induces a homomorphism r* : pi_1(Z) --> pi_1(Y), from which you can show that a _necessary_ condition for liftability is that the subgroup r*(pi_1(Z)) must be contained in the image f*(pi_1(X)). If f is a covering, then this condition is also sufficient. In particular, any map from a contractible space Z into a space Y will lift to a map from Z into any cover X of Y. I know you never said you interested only in covering maps, but these things put your confusion into stark relief. You might want to think about some of the classical examples of these. Start with any simply-connected space X and a discrete group G of self-homeomorphisms of X and then let Y be the quotient X/G . You could take (X,G) to be (R, Z), (R^2, Z^2), (S^2, Z/2Z), (complex half-plane, modular group), (S^3, binary icosahedral group), etc. dave ==== > Sorry, too many math for poets lectures. That really just means to take > a projection to an interval. Everything to this point is just the > projection pi : R^3 --> R^1, restricted to S^2. (That's continuous.) Sure it is; it's the map f : t |--> ( cos(2pi t), sin(2pi t) ). > Ah, now your example is clear. Nice! > Perhaps you're thinking of continuous maps _with continuous inverses_, > which this isn't. If that's what you mean by continuous image, then > the right phrase is homeomorphic image, and then yes, the > homeomorphic image of a simply-connected space is also simply-connected. > >> Yes, I see that I was wrong. But given a simply connected top. space X >> and a top. space Y together with an epimorphism f:X->Y, can we say that Y >> is simply connected? > In the meantime, I could figure this out. A morphism (here: a continuous map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such that fg=id_Y. Using this definition, and any closed curve s:[0,1]->Y, we can easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1). Because X is simply connected, there is continuous c:[0,1]^2->X with c(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives us fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simply connected. But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to: for every S subset Y we have g-(f-(S))=S (where the - sign denotes pre-images; this is just set-theoretic). Because of continuity reasons, a necessary condition is S open <=> f-(S) open (*) For surjective f with the property (*), you can take any set-theoretic g:Y->X with fg=id_Y to prove the other direction of the following theorem: f is an epimorphism in the category of topological spaces, iff it is surjective and has property (*). This is equivalent to that f be a quotient map. So being an epimorphism is pretty special, so I agree with you that continuous images of simply-connected spaces need not be simply-connected. Tobias -- ==== >> Yes, I see that I was wrong. But given a simply connected top. space X >> and a top. space Y together with an epimorphism f:X->Y, can we say that Y >> is simply connected? In the meantime, I could figure this out. A morphism (here: a continuous > map) f:X->Y is called an epimorphism iff there is a morphism g:Y->X such > that fg=id_Y. No - that's _not_ the usual (category-theoretic) definition of epimorphism. Instead, f: X --> Y is an epimorphism iff whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f, it follows that g_1 = g_2. (Your notion might be called a split epimorphism ... ) > Using this definition, and any closed curve s:[0,1]->Y, we > can easily lift it to gs:[0,1]->X and still it is closed: gs(0)=gs(1). > Because X is simply connected, there is continuous c:[0,1]^2->X with > c(0,t)=gs(t) and c(1,t)=x for some x in X. Applying f gives us > fc:[0,1]^2->Y with fc(0,t)=fgs(t)=s(t) and fc(1,t)=f(x). So Y is simply > connected. In the context of covering spaces (where Dave Rusin was explaininf stuff), the only covering maps f: X --> Y with X simply connected and f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_ be simply connected in this (fairly uninteresting) case. > But what does it mean for f to be an epimorphism? fg=id_Y is equivalent to: > for every S subset Y we have g-(f-(S))=S (where the - sign denotes > pre-images; this is just set-theoretic). Because of continuity reasons, a > necessary condition is S open <=> f-(S) open (*) For surjective f with the property (*), you can take any set-theoretic > g:Y->X with fg=id_Y to prove the other direction of the following theorem: f is an epimorphism in the category of topological spaces, iff it is > surjective and has property (*). > This is equivalent to that f be a quotient map. ... Yep ... > So being an epimorphism is > pretty special, so I agree with you that continuous images of > simply-connected spaces need not be simply-connected. Tobias ==== No - that's _not_ the usual (category-theoretic) definition > of epimorphism. Instead, f: X --> Y is an epimorphism iff > whenever g_1,g_2 : Y --> Z are morphisms such that g_1 o f =g_2 o f, > it follows that g_1 = g_2. (Your notion might be called a split epimorphism ... ) > Sorry for this one, the only reference I have for category theory is Lang's Algebra, and he defines epimorphism and monomorphism only for abelian categories. My notion is much stronger; it implies the other, but if we take a set X and let X_1 be X with the discrete topology, X_2 with the indiscrete topology, and define f:X_1->X_2 as the identity, then f is continuous. It is an epimorphism, but does not have a right inverse. In the context of covering spaces (where Dave Rusin was explaininf > stuff), the only covering maps f: X --> Y with X simply connected and > f an epimorphism in your sense are _homeomorphisms_ ... so Y _would_ > be simply connected in this (fairly uninteresting) case. > Maybe, but since I am not familiar with covering maps, I did not consider this case. -- Tobias Fritz Student of Mathematics and Physics University of Heidelberg, Germany ==== >Does anyone know if there is any potential one-way function >which is associative other than modular exponentiation? How is this associative? Is (a^b)^c = a^(b^c)? I think not! (And therefor I am not?) >That is, given a function f(x,y) (which is easy to compute), >- given z and x, it is difficult to find y such that z = f(x,y) >- f(f(x, y1), y2) = f(f(x, y2), y1) That doesn't look like the associative law, either. Setting the words aside, it sort of looks like what you want occurs in any group: given a group element x and an integer n we compute the power x^n in the group; certainly (x^n)^m = (x^m)^n, which is what you appear to request in your last line. And yes, it's often difficult to take an element in a cyclic group and decide what power of the generator it is. Your initial example is of this type, where the group is (Z/kZ)^* , the multiplicative group mod k (where k is any fixed integer). Another example commonly used for one-way functions is to use the group of points in an elliptic curve mod p. dave ==== As most here are already aware (but just in case some aren't), > if > you shift the digits of a number to the left or right one time then it's > like multiplying or dividing the number by its radix. So if you perform a > binary left shift on say: 00011011 (27 base 10) you get: 00110110 (54 base 10 or 27 * 2) So you can multiply a number by any power of two very quickly (for a > computer, register shifts are much faster than mult/div. instructions) by > shifting. What about numbers that are not powers of two? They can be done > by distributing the operation across numbers that are powers of two and sum > to the number you're trying to multiply by. Say you want to multiply by > 800. 800 is not a power of two, but 32, 256, and 512 are (32+256+512=800). > So: n*800 = (n*32)+(n*256)+(n*512) = (n<<5)+(n<<8)+(n<<9) The same is true for division. Binary right shifting is the same as > dividing by two. I can't seem to work out the algebra though. Can someone > tell me how (or is it possible) to do division by a general integer using > shifts? > -- Best wishes, > Allen You could try not working binary but convert to the divisor as radix instead, and then reconvert to binary. Does this help? ==== >> Ok, I've probably asked this question before, but I've forgot the answer... >> What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? >> I guess asking for the integral of these functions would be foolish? >x^x = e^(x*log(x)) >Now use the chain and product rules. Another method, more general, is to differentiate with respect to each x as if all the others are constant. So the derivative of x^a is a*x^(a-1); setting a=x, x^x, and the derivative of b^x is b^x*log(b), and this yields x^x*log(x). Add the two. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== > What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? >x^x = e^(x*log(x)) >Now use the chain and product rules. Another method, more general, is to differentiate with > respect to each x as if all the others are constant. So the derivative of x^a is a*x^(a-1); setting a=x, x^x, > and the derivative of b^x is b^x*log(b), and this yields > x^x*log(x). Add the two. > Hm, the chain rule for functions of two variables. df(u,v)/dx = f_u(u,v) du/dx + f_v(u,v) dv/dx Thus df(x,x)/dx = f_x1(x,x) + f_x2(x,x) where _x1 is with respect to the first x. ==== >Ok, I've probably asked this question before, but I've forgot the answer... >What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? Logarithmic differentiation. Doug ==== >Ok, I've probably asked this question before, but I've forgot the answer... >What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? In general, (u^v)' = v u^{v-1} u' + u^v ln(u) v', wherever it's defined. The answer can be derived from this. ==== >Ok, I've probably asked this question before, but I've forgot the answer... >What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? Think u^v = exp(v*ln(u)), with standard rules for derivatives. ==== > Ok, I've probably asked this question before, but I've forgot the answer... > What's the derivate of f(x)=x^x and f(x)=x^(x^(x-1)) ? > I guess asking for the integral of these functions would be foolish? Use logarithmicdrivative: f(x) = x^x ===> Ln(f(x)) = x*Ln(x) ===> f'(x)/f(x) = 1*Ln(x) + x *(1/x) = 1 + Ln(x) ===> f'(x) = x^x(1 + Ln(x)) For the other, aply that twice. -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com ==== Some people may think that this test is irrelevant for admissions committees for pure math, but there were some interesting questions that were asked on the exam (it was held today and I took it)... For instance, 1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what? 2. Let f(x) = x^2 + Bx + C, let B and C be independent random variables uniformly distributed on [0,1]. What is the probability that the roots of f(x) are distinct and real? There were other interesting ones as well, but I can't remember them...These are pretty easy if you're not under pressure to do them in less than average 3 minutes each (or maybe you think 30 seconds under pressure is sufficient...but give undergraduates some slack :-) ) It's good to see that the test requires more than memorization and some thinking...needless to say I got the 1st one wrong and the 2nd one right... In any case, I don't think I did as well as I would have liked, and hope that doesn't impact my chances at some of the top schools in representation theory and algebra in general (princeton, harvard, berkeley), but I have heard that this test isn't looked at very much, it's more recommendations and extra stuff/research...only I have heard that if you do poorly it raises a red flag (but I'm confident I don't have to worry about that) Just a post if anyone's interested, Mike ==== [GRE questions] >1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what? >2. Let f(x) = x^2 + Bx + C, let B and C be independent random variables >uniformly distributed on [0,1]. What is the probability that the roots of >f(x) are distinct and real? After 10 seconds of thought: 2:1 odds against. I first learned to count at the poker table, at the age of 1 or so, going A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. :) ==== > [GRE questions] >1. Hom(Z/2Z x Z/2Z, S_3) is isomorphic to what? >2. Let f(x) = x^2 + Bx + C, let B and C be independent random variables >uniformly distributed on [0,1]. What is the probability that the roots of >f(x) are distinct and real? After 10 seconds of thought: 2:1 odds against. > That seems to be incorrect. Explanation? We look for the probability that B^2/4 > C...We can immediately put an upper bound on the probability at 1/4 since if B = 1, then C must be less than or equal to 1/4... > I first learned to count at the poker table, at the age of 1 or so, going > A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K. :) ==== > >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. > > Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? I think that when I order soup in France (in English), I would get soup. As the French have the word soupe which is some form of soup. E.g. bouillabaisse is a soupe. But in that case it will be the main course. A better example would be ordering cold water in Italy. Do not be surprised when you get hot water. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >Message-id: And believe me, if you don't speak any French, the >waiters in France won't bring you soup. > Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? I think that when I order soup in France (in English), I would get soup. >As the French have the word soupe which is some form of soup. E.g. >bouillabaisse is a soupe. But in that case it will be the main >course. A better example would be ordering cold water in Italy. Do not be >surprised when you get hot water. I ordered two lattes at an AutoGrill in Italy, and got exactly (and only) what I ordered...two cups of milk. Marvin Sebourn osugeography@aol.com >dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, >+31205924131 >home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >And believe me, if you don't speak any French, the >waiters in France won't bring you soup. Have you considered the possibility that the reason waiters don't > bring you soup might *not* be that you don't speak French? This is unthinkable if you have ever been to Quebec or France. The only way I could get a cup of coffee in Quebec City was for my wife to order it for me. :-) Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net ==== |> Have you considered the possibility that the reason waiters don't |> bring you soup might *not* be that you don't speak French? | |This is unthinkable if you have ever been to Quebec or France. The only |way I could get a cup of coffee in Quebec City was for my wife to order |it for me. :-) Since this is a thread about social problems.... ;-) I think it varies depending on the part of the city. I went on a trip to Quebec city with friends, mostly to the old city (inside the walls once set up to defend the city), which gets quite a bit of tourism. We were there to ring bells in the cathedral there. Some of us knew a bit of French, and tried using it, but the waiters and waitresses would always use English with us. I think sometimes the first person to order was someone who didn't know any French, and simply ordered in English, and the net result was much the same. The Vancouver Sun once ran a piece on language. They arranged for two people to ask where the nearest laundromat was, one asking in French in Vancouver, and the other asking in English in Quebec. They got a much higher rate of success in Quebec. ObMath anecdote. In Littlewood's Miscellany, he describes a lecture given by someone from a region where the 'u' sounds in Duke and duck were swapped over. He made repeated references to Fuchs in his talk, but pronounced the other way.... Afterward someone pointed this out, and he said yes, he'd realized this was not the usual pronunciation, but he said it like that because Ladies were present. Keith Ramsay ==== if f is differentialbe on (0, infinite) and lim [f(x) +f'(x)] = L (x->infinite) show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) ------------------------------------ it seems to be trivial. but i no touch.....oops........ help.....me......please...... ==== if f is differentiable on (0, infinite) and lim [f(x) + f'(x)] = L (x->infinite) show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) Note: this classic problem is well-known due to the fact that it appeared in Hardy's classic calculus textbook with a solution more awkward than the elegant L'Hopital approach. For further details and references on the L'Hospital-based approach see my prior post [1], and [2]. -Bill Dubuque [1] http://google.com/groups?threadm=y8zognpn1wa.fsf%40berne.ai.mit.edu [2] Martin D. Landau; William R. Jones. A Hardy Old Problem. Math. Magazine 56 (1983) 230-232. ==== 0 = L - L = lim f+f' - lim f = lim f' However when f e^x -> k; then f = fe^x e^-x -> 0 f e^-x = f e^x e^-2x -> 0 L = lim f+f' - lim 2f = lim f'-f 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L 0 = lim f+f' - lim f = lim f' So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f' k = lim(x->oo) f = k + lim f'; lim f' = 0 If lim f' doesn't exist, then for n >= 2 f_n(x) = sin(x^n)/x -> 0 (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation ---- ==== >> >> if f is differentiable on (0, infinite) >> >> and lim [f(x) + f'(x)] = L (x->infinite) >> >> show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) http://google.com/groups?threadm=y8zognpn1wa.fsf%40berne.ai.mit.edu >| >| f e^x (f + f') e^x >| lim f + f' = L => lim f = lim ----- = lim ------------ = L >| x->oo x->oo x->oo e^x x->oo e^x Provided f e^x -> +-oo, in which case > 0 = L - L = lim f+f' - lim f = lim f' No proviso is needed. This form of L'Hopital's rule needs only that the denominator is infinite, not also the numerator [1][2]. So there's no need to treat this case specially as you do below. -Bill Dubuque [1] A. E. Taylor, L'Hospital's Rule Amer. Math. Monthly, Vol. 59, No. 1 (Jan., 1952), pp. 20-24. http://links.jstor.org/sici?sici=0002-9890(195201)59:3%3C20%3E [2] A. M. Ostrowski, Note on the Bernoulli-L'Hospital Rule Amer. Math. Monthly, Vol. 83, No. 4 (Apr., 1976), pp. 239-242. http://links.jstor.org/sici?sici=0002-9890(197604)83:3%3C239%3E > However when f e^x -> k; then > f = fe^x e^-x -> 0 > f e^-x = f e^x e^-2x -> 0 L = lim f+f' - lim 2f = lim f'-f 0 = lim f = lim f e^-x / e^-x = (f' - f)e^-x / -e^-x = lim f-f' = -L > 0 = lim f+f' - lim f = lim f' So if lim f = k and lim f' exists, then lim(x->oo) f+f' = k + lim f' > k = lim(x->oo) f = k + lim f'; lim f' = 0 If lim f' doesn't exist, then for n >= 2 > f_n(x) = sin(x^n)/x -> 0 > (f_n)'(x) = nx^(n-1) cos (x^n)/x - (sin x^n)/x^2 -> oscillation ==== > if f is differentialbe on (0, infinite) and lim [f(x) +f'(x)] = L (x->infinite) show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) > Let's suppose L=0 (substitute f by f-L). We must prove that if f(x)+f'(x)->0 (x->inf), then f(x)->0 (x->inf) f(x)+f'(x)=o(1) (x->inf) multiply by e^(x) : e^(x)[f(x)+f'(x)]=o(e(x)) (x->inf) integrate from 0 to t : (the derivative of e^(x)f(x) is e^(x)[f(x)+f'(x)]) e^(t)f(t)-f(0)=o(e(t)) (t->inf) Divide by e^t : f(t)=o(1) which means that f(t)->0 (t->inf) (excuse my english) ==== > if f is differentialbe on (0, infinite) and lim [f(x) +f'(x)] = L (x->infinite) show that lim f(x) = L (x->infinite) and lim f'(x) = 0 (x->infinite) Well. Perhaps L'Hopital's Rule has some use, after all. --Ron Bruck ==== X,Y is topology space. function f : X -> Y any B <= Y (symbol <= is inclusion relation), cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy (symbol cls = closure) show that f is continuous. -------------------------------- um......i think...... i use that any closed A <= Y, f^(-1) (A) is closed set. <=> f : conti i would show that f^(-1) cls(B) is closed set. cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B)) thus f^(-1) cls(B) is closed. it is right?? please......advice....for me....thank you ==== >i would show that f^(-1) cls(B) is closed set. >cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B)) >thus f^(-1) cls(B) is closed. >it is right?? Almost. You forgot to start with f^-1(cl B) subset cl f^-1(cl B) hypothesis and closed B ==> f^-1(B) closed f^-1(B) subset cl f^-1(B) subset f^-1(cl B) = f^-1(B) f^-1(B) = cl f^-1(B) Extend exercise unto theorem f continuous iff for all A, f(cl A) subset cl f(A) iff for all A, cl f^-1(A) subset f^-1(cl A) iff for all A, bd f^-1(A) subset f^-1(bd A) ---- ==== bok3np$ocd$1@news.hananet.net... > X,Y is topology space. function f : X -> Y any B <= Y (symbol <= is inclusion relation), cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy (symbol cls = closure) show that f is continuous. > Take B a closed subset of Y; you have cl(f^(-1)(B)) C f^(-1)(cl(B)) = f^(-1)(B). Hence cl(f^(-1)(B)) = f^(-1)(B), and f^(-1)(B) is closed. -- Julien Santini, Etudiant en licence au CMI de Ch.89teau-Gombert, FRANCE. ==== Julien Santini grava .88 la saucisse et au marteau: > Julien Santini, > Etudiant en licence au CMI de Ch.89teau-Gombert, > FRANCE. Est-ce vraiment n.8ecessaire? :) -- Nicolas ==== Nicolas Le Roux scribbled the following: > Julien Santini grava .88 la saucisse et au marteau: >> Julien Santini, >> Etudiant en licence au CMI de Ch.89teau-Gombert, >> FRANCE. > Est-ce vraiment n.8ecessaire? :) Que est-ce que est vraiment n.8ecessaire? -- /-- Joona Palaste (palaste@cc.helsinki.fi) ------------- Finland -------- -- http://www.helsinki.fi/~palaste --------------------- rules! --------/ Ice cream sales somehow cause drownings: both happen in summer. - Antti Voipio & Arto Wikla ==== > X,Y is topology space. function f : X -> Y any B <= Y (symbol <= is inclusion relation), cls{f^(-1) (B)} <= f^(-1) cls(B) satisfy (symbol cls = closure) show that f is continuous. -------------------------------- um......i think...... i use that any closed A <= Y, f^(-1) (A) is closed set. <=> f : conti i would show that f^(-1) cls(B) is closed set. cls{f^(-1) cls(B)} <= f^(-1) cls(cls(B)) = f^(-1) (cls(B)) thus f^(-1) cls(B) is closed. it is right?? No. You said that you wanted to proved that a certain condition implies that f is continuous, but what you did prove was the reverse implication. Jose Carlos Santos ==== Can the direct isogeny be found between these two elliptic curves? It is known that they are in a set of 8 isogenous curves, but what is the exact transformation that takes x <--> X, i.e. points on (1) to points on (2)? (1) y^2 = (x + 366)*(x^2 - 366*x + 2625489) (2) Y^2 = (X - 1551)*(X^2 + 1551*X + 497214) Both curves have conductor 210. More generally, how might one find the isogeny between y^2 = x^3+ax+b and Y^2 = X^3+AX+B, when the curves have the same conductor? (a,b,A,B are known) ==== > There's more to my work than just arguing on Usenet, so I'd like to > point out that my paper Advanced Polynomial Factorization is slated > to be published: > The Mega Foundation is an organization of high IQ people, and I'm glad > to be associated with them. To learn further about the organization > you can use Google, or see: Why a group like the Mega Foundation? > http://www.ultrahiq.org/Mega/WhyMega.htm > Mr. Harris, to me, this seems equivalent to the copyright you had on the web page you created for your famed Proof of FLT. Now, you are once again hell-bent on crying, cursing, cheating and trying to do anything to propagate your flawed lies. You also have some sort of greatness disorder that makes you think you are above every person you come in contact with. IIRC, this is a Narcissistic Personality Disorder. Here is excerpt from the web site: http://www.suite101.com/welcome.cfm/npd Narcissists attract abuse. Haughty, exploitative, demanding, insensitive, and quarrelsome - they tend to draw opprobrium and provoke anger and even hatred. Sorely lacking in interpersonal skills, devoid of empathy, and steeped in irksome grandiose fantasies - they invariably fail to mitigate the irritation and revolt that they induce in others. Here is a clinical definition: http://www.behavenet.com/capsules/disorders/narcissisticpd.htm *** Diagnostic criteria for 301.81 Narcissistic Personality Disorder (cautionary statement) A pervasive pattern of grandiosity (in fantasy or behavior), need for admiration, and lack of empathy, beginning by early adulthood and present in a variety of contexts, as indicated by five (or more) of the following: (1) has a grandiose sense of self-importance (e.g., exaggerates achievements and talents, expects to be recognized as superior without commensurate achievements) (2) is preoccupied with fantasies of unlimited success, power, brilliance, beauty, or ideal love (3) believes that he or she is special and unique and can only be understood by, or should associate with, other special or high-status people (or institutions) (4) requires excessive admiration (5) has a sense of entitlement, i.e., unreasonable expectations of especially favorable treatment or automatic compliance with his or her expectations (6) is interpersonally exploitative, i.e., takes advantage of others to achieve his or her own ends (7) lacks empathy: is unwilling to recognize or identify with the feelings and needs of others (8) is often envious of others or believes that others are envious of him or her (9) shows arrogant, haughty behaviors or attitudes Reprinted with permission from the Diagnostic and Statistical Manual of Mental Disorders, fourth Edition. Copyright 1994 American Psychiatric Association *** Perhaps instead of spewing your useless nonsense, you should study this disorder, go see a doctor and then drug and drink yourself to death. Think about, eh, let's wack you! ==== ... > > 1. I define: > > w1(x) = gcd(5 a1(x) + 7, 49) > > > > Ok, w_1(x) = 7, which works. > > Where do you see w1(x) = 7? That can only be true when you *know* that > (5 a1(x) + 7) is divisible by 7 in the algebraic integers, otherwise the > greatest common divisor can *not* be 7. > > Which is the problem with the ring of algebraic integers. Nope, there is *not yet* a problem. I have defined w1(x) *as I defined it*. You are immediately seeing that that w1(x) equals 7, which is blatantly false as the definition of gcd *does not warrant that*. > > At this moment the only thing we know about w1(0) = 7. For some > values of x w1(x) can even be larger than 7. > > That's nonsense, easily proven by the factorization. That is *not* nonsense. > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 Yup, this is the polynomial. Note that when x = 7y + 5 it can be divided by additional factors 7. So when x is of that form, it is *not possible* that the gcd's deliver 7, 7 and 1. For instance, when x = 5 the value is 2^3 * 7 * 13 * 23 * 2213. See that 7 in the factors? > where no other factorization works as long as 7 is not a factor of 22. > > That's because I've isolated the factors of 22, which is obvious by > inspection. > > So you think that (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 makes a > difference Dik Winter? Not as long as 7 is not a factor of 22, it > doesn't. *Why* do you think that 22/w3(x) must *also* be an algebraic integer? That is just plain stupid. You only need that (5 b3(x)/w3(x) + 22/w3(x)) is an algebraic integer. > Understand Dik Winter? You do understand factors, right? Yes, but I seriously doubt your understanding. > So you know where I got (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 from, > right? Yes, that is true with my factorisation too. > Now adding in more variables, like sticking in w's may look like a > good way to hide the truth Dik Winter, but it's childish. > > Besides, I'm bored now that I've shown that math society is so stupid > as to try and fight such a basic result. Now the only question is how > to start yanking funds so that they can be redirected to real > research. You would not know a basic result when you saw one. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== > ... > > Pray answer the following quesions: > > Ok. > > 1. I define: > > w1(x) = gcd(5 a1(x) + 7, 49) > > Ok, w_1(x) = 7, which works. Where do you see w1(x) = 7? That can only be true when you *know* that > (5 a1(x) + 7) is divisible by 7 in the algebraic integers, otherwise the > greatest common divisor can *not* be 7. Which is the problem with the ring of algebraic integers. At this moment the only thing we know about w1(0) = 7. For some > values of x w1(x) can even be larger than 7. That's nonsense, easily proven by the factorization. (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 b_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where no other factorization works as long as 7 is not a factor of 22. That factorization has only been demonstrated for the case x = 0. It has not been proven for 'x' generally. Since you *know* this, your presentation is deceptive and dishonest. -- 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 ==== ... > Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing. > > After all the factorization is > > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, so > doing that gives > > (5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22 > > and notice that necessarily, if you multiply everything out and > simplify, you have > > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 > > which leaves you shit out of luck Dik Winter. Yes, the product of those three terms is 22. What is the problem? You have (7/7)(7/7)(22/1) = 22. Looks pretty similar to me. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== ... > Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing. > After all the factorization is > (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 22) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > where you want to divide through by w_1(x) w_2(x) w_3(x) = 49, so > doing that gives > (5 a_1(x)/w_1(x) + 7/w_1(x))(5 a_2(x)/w_2(x) + 7/w_2(x))(5 > b_3(x)/w_3(x) + 22/w_3(x)) = 300125 x^3 - 18375 x^2 - 360 x + 22 > > and notice that necessarily, if you multiply everything out and > simplify, you have > > (7/w_1(x))(7/w_2(x))(22/w_3(x)) = 22 > > which leaves you shit out of luck Dik Winter. > > Why are you so hateful toward people who make mistakes? Everyone makes > mistakes, at least he admits when he is wrong. By the way, your use of > swearing makes you look very immature. The strangest about this is that there is no mistake. The product indeed comes to 22. When you do the multiplication James suggest you come at 22, just what was needed. So I really do not understand what James is doing here. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== >Oh yeah, easy mistake, but yours Dik Winter is MUCH more embarrassing. Much more embarrassing than starting a new thread every time you can't talk your way out of a corner? Doug X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== >1. Why is there a cross product in the definition of torque T = r x >F ? Because you're doing Physics in 3 dimensions, you can map skew-symmetric tensors into vectors. For more than 3 dimensions, you have to explicitly treat the torque as a skew symmetric tensor or a 2-form. >2. How is torque a tensor? Because the definition of cross product makes it transform like one. If you don't want to treat it as a tensor or a 2-form, than you need to treat it as a vector and pay close attention to your weights. Either way, you need to be careful to distinguish covariant indices from contravariant. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but I'm not quite seeing > how that matches the A' = Q^{-1} A Q form. As you say, under the change of orthonormal basis, the vectors a and b > go to a' = Qa and b' = Qb. In component form, a_i = (sum over m) Q_(im) a_m and b_i = (sum over n) Q_(in) b_n I've used different summation indices in order that the substitutions > below makes sense. Because both the new and old bases are orthonormal, Q is an orthogonal > matrix, i.e., Q^T = Q^(-1), i.e if M = Q^(-1), M_(ij) = Q_(ji). Thus, > det Q = +/- 1. If in addition, right-handed orthonormal bases are > taken to right-handed orthonormal bases, then det Q = 1, and Q is called > a special orthogonal matrix. The set of all special orthogonal matrices > forms the group SO(3), where the notation is fairly self-explanatory - > S for special (det +1), O for orthogonal (inverse = transpose), 3 for > 3 by 3. Define matrix A by A_(ij) = a_i b_j - a_j b_i. Under a transformation, A'_ij = a'_i b'_j - a'_j b'_i = (sum over m,n) [Q_(im) a_m Q_(jn) b_n - Q_(jn) a_n Q_(im) b_m] = (sum over m,n) [Q_(im) a_m b_n Q_(jn) - Q_(im) a_n b_m Q_(jn)] = (sum over m,n) [Q_(im) (a_m b_n - a_n b_m) Q_jn] = (sum over m,n) [Q_(im) A_(mn) Q^(-1)_nj] A = Q A Q^(-1) This is a little different than A = Q A Q^(-1) because, the way Feynman > defines things, a'_i = (sum over n) Q^(-1)_(in) a_n. It's a good > exercise to show this. sure what you were meaning to show here, if you will note the last 4 lines of your posting. that is inconsistent with how I specified the change of basis, it should have been A' = Q A Q^{-1} and that is what you have. I also note that Feynman's matrix object T_ij = x_i F_j - x_j F_i does indeed transform as T' = Q T Q^{-1}, as you have shown. I am still mystified as to what real value there is in taking a nice axial vector object like Torque and making the anti-symmetric matrix out of it... <2202379a.0310270854.d34b089@posting.google.com> <2202379a.0310281028.2884dbfa@posting.google.com> <09iupvkca3n0ur4e7bcvroeviu9isfbbit@4ax.com> <2202379a.0310310905.20a78472@posting.google.com> <77seqvotf0ul8f8nmds64ilpesolg1dhhm@4ax.com> <2202379a.0311050918.1ff64462@posting.google.com> <3fa9ddbf$4$fuzhry+tra$mr2ice@news.patriot.net> <2202379a.0311061125.782cb723@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 11:25 AM, dynamics@vianet.on.ca (Ken S. Tucker) said: >But the spectra itself is formed by photons, No. The term continuous spectra is not the same as[1] the usage of the term spectrum in optics. An observable in QM is a Hermitian operator, and such an operator has a characteristic that is known in the literature as its spectrum. Read any good book on Functional Analysis for a explanation. Or read Principles of Quantum Mechanics (P.A.M. Dirac) for an explanation from a Physics perspective. [1] But the spectrum in optics is related to the spectrum of the Hamiltonian. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >>But the spectra itself is formed by photons, >No. The term continuous spectra is not the same as[1] the usage of >the term spectrum in optics. An observable in QM is a Hermitian >operator, and such an operator has a characteristic that is known in >the literature as its spectrum. Actually, the term spectrum is generic to linear algebras, and is not specific to operator algebras. If A is an algebra over a field F, the spectrum of an element a of A would be spec(a) = { f in F: (a-f) has no two-sided inverse in A }. <3fad7a93$1$fuzhry+tra$mr2ice@news.patriot.net> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 12:11 AM, whopkins@alpha2.csd.uwm.edu (Mark) said: >Actually, the term spectrum is generic to linear algebras, and is >not specific to operator algebras. Correct, although I don't know whether the more general meaning is ever used in QM. I'm certainly not aware of the field[1] being other than R, C or H. [1] If you use the definition that requires commutativity, then omit H. -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== >But the spectra itself is formed by photons, >Ken >>No. The term continuous spectra is not the same as[1] the usage of >>the term spectrum in optics. An observable in QM is a Hermitian >>operator, and such an operator has a characteristic that is known in >>the literature as its spectrum. >>Seymour >Actually, the term spectrum is generic to linear algebras, and is not >specific to operator algebras. If A is an algebra over a field F, the >spectrum of an element a of A would be >spec(a) = { f in F: (a-f) has no two-sided inverse in A }. >Mark This is amusing, see this thread has been cross posted to sci-math and sci-physics, so there are different meanings to words. Anyway I'm the guy who is talking about physics, but I'll try to make it interesting for the higher class posters in sci-math, who may disagree. An invariant has no units, specifically invariant mass or energy. Two invariant masses can be equated on a balance without resort to units. I argue the invariant mass/energy can only change by small but finite amounts, hence discontinuously. Let E be the invariant then dE =0 as it has no continuous function and derivative, although E =/=constant. Transactions are quantized to the nearest penny, thus d(money) =0. For mathematical masochists, one might suggest a Fourier series to form a digital pulse (penny), however, the wavelengths required exceed the size of the penny to form the square pulse, ((SUM i, (n=1+2*i) 1/n sin n*w*t)). So a wave mechanical solution is an approximation.but is DOA mathematically. Returning to physics, I maintain we cannot infinitely divide the two masses on the balance, nor can we infinitesmally vary either mass. Therefore dE =0 is a Law of Physics and describes how invariant energy varies. This might look like a peculiar law but compare this to the GR geodesical equation of motion, given by the absolute derivative DU^u/ds =0. (Absolute acceleration does not exist) This equation is very successful. Now 4-momentum is E*U^u so a quantum- mechanical generally relativistic geodesic is, DE^u/ds =0 . (absolute force does not exist) This equation rules out any force like f^u (or f_u). So it sets all the components of Lorentz force to zero to describe how a charge moves geodesically. That's a powerful constraint, and *appears* to be verified so far. One might call DE^u =0 a unified field theory, but I think it's only a requirement. The metric that satisfies this equation is the field, and I suspect the unified field equation will turn out to be G_uv=kT_uv. Einstein has suggested (in his later years) using anti-symmetrical metrics to go from G_uv = kT_uv to DE^u=0, to unify. Ken S. Tucker <2202379a.0311052323.77555cbc@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft ==== at 11:23 PM, dynamics@vianet.on.ca (Ken S. Tucker) said: >As I understand, the indices forming the relatively >defined tensor K are meaningless after the definition, >but I have some uncertainty. I'm not sure what you're trying to say, but change to polar coordinates and observe what happens. >The relative tensor K_uv is actually a 4th rank >(counting weight) like the Riemann-Christoffel >Curvature tensor. The weight has nothing to do with the rank. >It is very evident the Kronecker delta (d^u_v) >serves as an orthogonal metric, No it doesn't. >relativity of K_uv ? -- Shmuel (Seymour J.) Metz, SysProg and JOAT not reply to spamtrap@library.lspace.org ==== Mr. Harris, Psychopathology ------------------------ Narcissism: A pattern of traits and behaviors which signify infatuation and obsession with one's self to the exclusion of all others and the egotistic and ruthless pursuit of one's gratification, dominance and ambition. Perhaps research in to your delusional fits of grandeur would be a more appropriate vocation for you? Still pissing out your research? ==== >> Similarly, Cramer's proof of the Levy-Cramer theorem that >> the sum of two independent random variables is not normal >> unless they both are is probably the only easy proof, but >> it likewise obscures the ideas. Any time characteristic >> functions are used to prove a probability theorem, the >> concepts are not even present. >Do you happen to know of a probabilistic proof of the Cramr-Lvy >theorem? There are entropy proofs. Any published? -- A. ==== > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2) with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 as long as you're in a ring where 7 is not a factor of 22. I want to emphasize that point as notice there's only *one* way to > divide through by 49 if 7 is not a factor of 22. Usually you can *see* the other factors of 7, but I want you to > abstract, and generalize. Please pay careful attention to that example. You may see people who reply claiming that the word polynomial has > some significance, as if it's a mystical thing which refutes basic > logic, so if something isn't polynomial it no longer behaves > logically. All functions that I know of, all of them behave logically when examined from the view point of calculus. EXAMPLE:Describe the graph of f(x)=sin(x) without graphing. Well, I know that a derivative measures change and I also know that when the instantaneous change is 0, there is some sort of a critical point (determined by the 2nd derivative). Let's say we want to describe the graph of f(x) in words. Begin by finding f'(x). f'(x)=cos(x) We know that a function has a critical point at a point where the derivative is 0. cos(x)=0 at x=pi/2 and 3pi/2 Ok, we know where the critical points are, but we don't know what happens at those points. So take the 2nd derivative. f''(x)=-sin(x) When x=pi/2, f''(x)=-1. When x=3pi/2 f''(x)=1 Therefore I can now analyze my results. When x=pi/2, the 2nd derivative is negative, which means it's decreasing so pi/2 must be a maximum. When x=3pi/2, the 2nd derivative is positive so 3pi/2 must be a minimum. In fact, if I graph this, my predictions are correct. My point is that all elementary functions that I know of behave logically if you examine it from a calculus viewpoint, which may not be a bad idea for you to do. David Moran Now then, in my advanced factorization work, I just use functions of x > that are a lot more complicated than f_1(x) = c_1 x, and unfortunately > there are people who can use an unfamiliar leap in complexity to > confuse others. Some of you have learned various advanced math topics, now imagine if > in your classrooms there were some hecklers who continually hollered > out at your teacher, or otherwise disrupted the class? What if when there were difficult concepts those hecklers would try to > confuse everyone as they sought to discredit the mathematics? If you find that hard to imagine, imagine me in your class with you > questioning the professor and calling him names. How much would you have learned? I need those of you interested in mathematics to focus on the basics, > so that you can understand the advanced. > James Harris > http://mathforprofit.blogspot.com/ ==== You may see people who reply claiming that the word polynomial has > some significance, as if it's a mystical thing which refutes basic > logic, so if something isn't polynomial it no longer behaves > logically. > No both polynomial and non-polynomial functions behave logically. It is just that polynomials have some properties that are not true in general. Let P(x) by a polynomial divisible by p, and let g_1(x) and g_2(x) be polynomials such that P(x) = g_1(x) * g_2(x) Then p divides at least one of g_1(x) and g_2(x). If g_1(x) and g_2(x) are not polynomials, this is no longer true. Counterexample. Let P(x) = 15-3x. Then P(x) is divisible by 3. Let g_1(x)=4-sqrt(1+3x) and g_2(x)=4+sqrt(1+3x). Then P(x) = g1(x)*g2(x) list a few values P( 0) = 15, g_1( 0) = 3, g_2( 0) = 5 P( 1) = 12, g_1( 1) = 2, g_2( 1) = 6 P( 5) = 0, g_1( 5) = 0. g_2( 5) = 8 P( 8) = -9, g_1( 8) = -1. g_2( 8) = 9 P(16) = -33, g_1(16) = -3. g_2(16) = 11 Note how the factor of 3 switches back and forth between g_1 and g_2 depending on the value of x. Thus neither g_1(x) nor g_2(x) is divisible by 3 for all x. In particular the fact that g_1(0) is divisible by 3 does not mean that g_1(1) is divisible by 3. - William Hughes ==== > If you saw > > (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) > > with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. How do you know that this is possible with algebraic integer c1 to c3? I do not think this is a valid factorisation in the algebraic integers. Could you provide the monic polynomial with integer coefficients of which the c's are the roots? More to the point, x^3 + 5x^3 + 3x + 2 factors as (x - r1)(x - r2)(x - r3) in the algebraic integers. Where the r's are the roots of the polynomial. We can multiply the factors by 7/r1, 7/r2 and 2/r3 to get your factorisation. So we get: (7/r1 x + 7)(7/r2 x + 7)(2/r3 x + 2). So c1 = 7/r1, c2 = 7/r2, c3 = 2/r3. Of these only c3 is an algebraic integer (r3 is a divisor of 2). Neither 7/r1, nor 7/r2 are algebraic integers. Unless you claim that r1 and r2 are units (they are divisors of 2). But they are not, because the constant term of a polynomial (see the correct usage here, James) must be 1 for a root to be a unit. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ ==== In sci.math, James Harris <3c65f87.0311080557.3c8f97f1@posting.google.com>: > If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) with the c's algebraic integers, I think few of you would have a > problem realizing that only two of the c's have 7 as a factor. Not clear at all without knowing precisely what the roots are. Since two of the roots are complex in this case I might have a chance of not using trig, but it would take some work. I can tell you that the roots, as reported by Pari/GP, are approximately: -4.424076847035404679966063971 -0.2879615764822976600169680144 - 0.6075770270241740255965589263*I -0.2879615764822976600169680144 + 0.6075770270241740255965589263*I but this doesn't help much as algebraic units are provably dense around (0 + 0i) -- sqrt(n+1) - sqrt(n) is a unit. But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as > (f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 as long as you're in a ring where 7 is not a factor of 22. Only an issue modulo 3 or 5, AFAICT.... :-) I can posit a simpler example. Let's assume for giggles the following equation, similar to yours: (d_1 * x + 7) * (d_2 * x + 7) * (d_3 * x + 2) = 49 * (x^3 + 2) and carry through your logic. I know the roots to this equation. :-) We know that x^3 + 2 = (x + 2^(1/3)) * (x + 2^(1/3) * (-1/2 + sqrt(3) * I/2)) * (x + 2^(1/3) * (-1/2 - sqrt(3) * I/2)) (GP/Pari verifies this nicely, as it turns out). Therefore, one can pick d_1 = 7 / (2^(1/3)), d_2 = 7 / (2^(1/3) * (-1/2 + sqrt(3) * I/2)), d_3 = 2 / (2^(1/3) * (-1/2 - sqrt(3) * I/2)). Since -1/2 + sqrt(3) * I/2 and -1/2 - sqrt(3) * I/2 are both units (as is easily verified by multiplying them together), they factor little in the analysis. I can just as easily posit d_1' = d_2' = 7 / (2^(1/3)) d_3' = 2 / (2^(1/3)) = 2^(2/3) Now d_3' turns out to be an algebraic integer, solving the equation y^3 - 4 = 0. (d_3 solves this equation too.) d_1' ? Well, lessee; the most obvious equation is 2 * y^3 - 343 = 0 which is clearly not going to lead to anything resembling an algebraic integer. Since d_1' requires a cube root no equation of lesser degree will do. But I may need those roots to your equation, and not just approximations, either. If one posits N(x) = x^3 + 5*x^2 + 3*x + 2 then, if one substitutes x = y-5/3, one can rewrite this as: N(y-5/3) = y^3 - 16/3 * x + 169/27 Now substitute y = w + 16/(9*w) (Vi.8fta's substitution, with p = -16/3) and get N(w+16/(9*w)-5/3) = (729*w^6 + 4563*w^3 + 4096)/(729*w^3) Surprise: w^3 = (-4563 ± sqrt(4563^2-4*729*4096)) / (2 * 729) = (-4563 ± sqrt(8877033)) / (2 * 729) , by our old friend, the quadratic formula. Since 8877033 = 3^9*11*41 and 4563 = 3^3 * 13^2, we can divide by 3^3/3^3 (or sqrt(3^6)/3^3), resulting in w^3 = (-169 ± sqrt(12177)) / (2 * 27) or w^3 = (-169 ± 3*sqrt(1353)) / (2 * 27) I'm going to take the + sign; the - sign will yield slightly identical results, as it turns out. Because GP/Pari does odd (but somewhat logical) things with (-1)^3, I'm going to take the sign out now. The results can be written: x1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 + sqrt(3) * I/2) x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 16/(3*((169 - 3*sqrt(1353)) / 2)^(1/3)) * (-1/2 - sqrt(3) * I/2) It's worth noting that (169 - 3*sqrt(1353)) * (169 + 3*sqrt(1353)) = 16384 = 2^14, so one can rewrite the above as x1 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) x2 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 - sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 + sqrt(3) * I/2) x3 = -5/3 - 1/3 * ((169 - 3*sqrt(1353)) / 2)^(1/3) * (-1/2 + sqrt(3) * I/2) - 1/3 * ((169+3*sqrt(1353))/2)^(1/3) * (-1/2 - sqrt(3) * I/2) If you look carefully you'll see that using the alternate solution w^3 = (-169 - 3*sqrt(1353)) / (2 * 27) will merely swap x2 and x3 in the final list. In this form, figuring out whether xi is an algebraic integer is a Lost Cause(tm), although we know the answer is yes since they all solve the equation x^3 + 5*x^2 + 3*x + 2 = 0, as one can verify if one multiplies (x-x1)*(x-x2)*(x-x3) in GP/Pari. However, there's a different route, making the exact roots entirely irrelevant. (Sorry, guys -- but it *was* an interesting solution using Vi.8fta's substitution. :-) ) We are hypothesizing (c_1*x + 7)(c_2*x + 7)( c_3*x + 2) = 49(x^3 + 5 x^2 + 3x + 2). for some algebraic numbers ci. This hypothesis turns out to be quite valid, as it turns out -- but what are the ci? Since we know x^3 + 5x^2 + 3x + 2 = (x-x1)*(x-x2)*(x-x3) we can deduce that c1 = -7/x1, c2 = -7/x2, and c3 = -2/x3, for some rotation of x1, x2, x3. Are these algebraic integers? We note that, if xi solves x^3 + 5 x^2 + 3x + 2 = 0, then 1/xi has to solve 2*z^3 + 3*z^2 + 5*z + 1 = 0, 7/xi has to solve 2*z^3 + 3*7*z^2 + 5*7^2*z + 7^3 = 0, and -7/xi has to solve 2*z^3 - 3*7*z^2 + 5*7^2*z - 7^3 = 0. -7/xi is clearly *not* an algebraic integer (as the above is not a reducible polynomial over Q). Fortunately, -2/xi is: 2*z^3 - 3*2*z^2 + 5*2^2*z - 2^3 = 0 or z^3 - 3*z^2 + 10*z - 4 = 0 Well, 1 out of 3 ain't bad. This counter-proof generalizes quite nicely for most cubic polynomials x^3 + a_2 * x^2 + a_1 * x + a_0 with non-unit a_0 not divisible by 7. I'm afraid you may be out of luck, James. :-) [rest snipped] -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== >If you saw (c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = > > 49(x^3 + 5x^2 + 3x + 2) with the c's algebraic integers, I think few of you would have a >problem realizing that only two of the c's have 7 as a factor. > No - the problem here is that this polynomial does not factor at all in the algebraic integers in the form you have shown. Its factorization is necessarily of the form 49*(x - r1)*(x - r2)*(x - r3), where r1, r2, and r3 are nonunit *algebraic integers* which are the roots of x^3 + 5*x^2 + 3*x + 2 = 0, and are such that -r1*r2*r3 = 2. Knowing this, you can distribute the 49 among the factors in many different ways. However, *none* of those ways will end up having constant terms of 7, 7, and 2. Why? Because 7 and 2 are coprime in the algebraic integers. It is not possible to write 2, which you want as the constant term of the third factor, as a product of a nonunit factor of 7 and a nonunit factor of 2. Andrzej. >But, of course, you're looking at *functions* of x, as you have f_1(x) = c_1 x, f_2(x) = c_2 x, and f_3(x) = c_3 x, so I could also write it as >(f_1(x) + 7)(f_2(x) + 7)( f_3(x) + 2) = 49(x^3 + 5 x^2 + 3x + 2). Notice that dividing both sides by 49 gives (f_1(x)/7 + 1)(f_2(x)/7 + 1)( f_3(x) + 2) = x^3 + 5 x^2 + 3x + 2 as long as you're in a ring where 7 is not a factor of 22. I want to emphasize that point as notice there's only *one* way to >divide through by 49 if 7 is not a factor of 22. Usually you can *see* the other factors of 7, but I want you to >abstract, and generalize. Please pay careful attention to that example. You may see people who reply claiming that the word polynomial has >some significance, as if it's a mystical thing which refutes basic >logic, so if something isn't polynomial it no longer behaves >logically. Now then, in my advanced factorization work, I just use functions of x >that are a lot more complicated than f_1(x) = c_1 x, and unfortunately >there are people who can use an unfamiliar leap in complexity to >confuse others. Some of you have learned various advanced math topics, now imagine if >in your classrooms there were some hecklers who continually hollered >out at your teacher, or otherwise disrupted the class? What if when there were difficult concepts those hecklers would try to >confuse everyone as they sought to discredit the mathematics? If you find that hard to imagine, imagine me in your class with you >questioning the professor and calling him names. How much would you have learned? I need those of you interested in mathematics to focus on the basics, >so that you can understand the advanced. >James Harris >http://mathforprofit.blogspot.com/ ==== Is there a simple proof, or counterexample, to the proposition that if we consider the expansion of the fractional part of some (irrational) number x in some base, and know that it is normal, then it follows that the expansion of 1/x is also normal? I'd appreciate a pointer to any literature in this area. Rob Shaw ==== >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? I'd appreciate a pointer to any literature in this area. > What do you mean by normally distributed digits? The normal distribution is a continuous one. Nor would it make sense to speak of x - floor(x) being normally distributed, since the normal dsitrubtion is supported by the entire real line. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu ==== >Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? I'd appreciate a pointer to any literature in this area. What do you mean by normally distributed digits? The normal > distribution is a continuous one. Nor would it make sense to speak of x - floor(x) being normally > distributed, since the normal dsitrubtion is supported by the entire > real line. A real number is said to be normal (to the base 10) if for all n every string of digits of length n appears with frequency 10^(-n) in its decimal expansion. This has nothing to do with the (Gaussian) normal distribution that you are thinking of. As others have noted, it seems unlikely that x normal implies 1/x normal, but I don't know what's in the literature. Let y be normal to base 2, and let x be the number you get by making believe y is a decimal expansion. Then x isn't normal - it has only the digits 0 and 1 - but I'd expect 1/x to be normal (though I wouldn't be able to prove it). -- ==== Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? I seriously doubt it. Here's an indication of why: Let's talk about base 2. Start with x = .01010101... . Of course x is not normal, but it does contain the right number of 0's and 1's; it's weakly normal in base 2, in a sense. But x = 1/3, so 1/x = three = 11.000... , which is far from normal. I wouldn't be surprised if you could actually construct a counterexample along these lines - you'd need to verify a few things... >I'd appreciate a pointer to any literature in this area. >Rob Shaw David C. Ullrich ==== Is there a simple proof, or counterexample, to the proposition >that if we consider the expansion of the fractional part >of some (irrational) number x in some base, and know that it is normal, >then it follows that the expansion of 1/x is also normal? I seriously doubt it. Here's an indication of why: Let's talk about base 2. Start with x = .01010101... . > Of course x is not normal, but it does contain the right > number of 0's and 1's; it's weakly normal in base 2, in > a sense. But x = 1/3, so 1/x = three = 11.000... , which is > far from normal. You're in the land of rationals. Rationals and normality don't mix. (Take that out of context, man!) Phil -- Unpatched IE vulnerability: window.open search injection Description: cross-domain scripting, cookie/data/identity theft, command execution Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm ==== > >>Is there a simple proof, or counterexample, to the proposition >>that if we consider the expansion of the fractional part >>of some (irrational) number x in some base, and know that it is normal, >>then it follows that the expansion of 1/x is also normal? >> >> I seriously doubt it. Here's an indication of why: >> >> Let's talk about base 2. Start with x = .01010101... . >> Of course x is not normal, but it does contain the right >> number of 0's and 1's; it's weakly normal in base 2, in >> a sense. But x = 1/3, so 1/x = three = 11.000... , which is >> far from normal. You're in the land of rationals. >Rationals and normality don't mix. (Take that out of context, man!) Well, that's amusing enough to make it seem possible that you were just trying to be funny. In case you were also trying to make a serious point: I certainly didn't mean that this was a counterexample, or something that would lead to a counterexample with no new ideas required. But what I had in mind was something _vaguely_ like this: Suppose we could find x_n such that each n-bit sequence in the binary expansion of x_n occurs with the right frequency, but 1/x_n has a terminating binary expansion. Then let x consist of a long stretch of the bits of x_1, followed by a much longer stretch of the bits of x_2, etc. For suitable values of long and longer x will be normal, but it doesn't seem so unlikely that 1/x would turn out to be abnormal. >Phil -- >Unpatched IE vulnerability: window.open search injection >Description: cross-domain scripting, cookie/data/identity > theft, command execution >Exploit: http://safecenter.net/liudieyu/WsFakeSrc/WsFakeSrc-MyPage.htm David C. Ullrich ==== What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) ? Is it a catenary ? If not, what is it? ==== In 3D, y+b=0 is a plane. Right? Are you looking for its reflection, or reflection of a line? > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? ==== > In 3D, y+b=0 is a plane. Right? Are you looking for its reflection, or > reflection of a line? > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? It's probably safe to assume, since this is a line on the floor of the mall, that the equation is y+b = 0 and z = 0. Now that it's been a few days, can I give a hint? The symmetry of the problem -- what coordinate system should you be using? ==== When is the homework due? > What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? ==== > When is the homework due? :) this is my curiosity about curvarture of lines when the virtual image is distorted in reflection. I observed clear images of square lines of floor mosaic in a polished cylindrical flowerpot in Twelve Oak Mall, Detroit. What is commonly known is the optics: image-object distances are connected by standard formula (u-f)(v-f)=f^2 where f=a/2 or focal length, u,v are object & image distances etc. in reflection. I did not do much of image / ray tracing in 3D before . I believed that cos(slope angle) in the x-z projection is inversely proportional to z-coordinate, as it is for a catenary z=A cosh(x/A). I asked this question expecting that there could be elementary geometrical operations with straight lines, cylinders, reflection &c. My imagination surrounded vague thoughts about geodesic deviation, but I right now I would thank in advance for any ensuing pointers and clarifications. ==== What is the reflection in a cylindrical mirror x^2+y^2 = a^2 of a > straight line y+b = 0 as seen from a viewpoint (0,-c,h) ( a < b << c) > ? Is it a catenary ? If not, what is it? What have you done so far? ==== Why don't you guys go to sci.numerology? > a) E N G E L E > 5 14 7 5 12 5 = 48 48+ Dad 21 6 23 48+ Mom 4 11 23 188 Paulette 8 8 59 220/145 902 > Paulette 100 Linda 40 Engele 48 Paulette was born on the 8th day of the 8th month, her name begins with > the 16th (8+8th) letter of the alphabet. Her given initials add together for > 28. Her last two names differ in value by 8 and add together for 88. Her > full name adds to 188. She has 19 letters (8th prime). Her names add to the > 75th, 28th and 33rd non-primes, together for 136, corresponding to Numbers > 19 (8th prime), prettier as the numbers 1 through to 16 (8+8) add together > for 136. * * * * * * * > b) This family stopped to provide stats. 200 Mathew 13 11 66 317/48 > Mathew 70 William 79 Shaw 51 195 Jeanine 7 10 77 280/85 > Jeanine 58 Marie 46 Taylor 91 131 Aidan 7 5 03 127/238 > Aidan Michael 51 Shaw 51 * * * * * * * > c) 97+ Dad 3 3 47 62/303 218 Ambrose 27 3 84 87/279 > Ambrose 73 James 48 Pritchard 97 Ambrose was born on the 27th, the vowels in his given names add to 27. Ambrose was bon in 84 (7 times the 7th non-prime), he was born on the prime). > He has 21 (7+7+7) letters, it's the number of chapters in Bible Book 7. is > Bible Book 59 (the 17th prime). The vowels in his first name add to the 21 > (7+7+7) chapters of Bible Book 7. He has 7 vowels in all, they add to 37 chapter > 7). Dad and Ambrose were born on days of the year adding to 149, there are 149 > verses in Bible Book 48 while his middle name adds to 48. Ambrose was born on day 87 (29+29+29), his first name adds to the 73 > adds to 218 (109+109, or twice the 29th prime). Daryl Shawn Kabatoff > Box 7134 Saskatoon Sask. > S7K 4J1 ==== This guy is totally out of touch with reality. So what would be the harm in simply ignoring him from this day forward and forever, and permitting him to retire in the triumphant knowledge (as far as he's concerned) that he is absolutely and totally correct in everything he's ever said? He would be quite happy, and then you all could have a lot more fun communicating with real mathematicians. You all must know by now that you are wasting your time. I mean the ones of you who are responding in a thoughtful reasoned way to James. The others, who are simply calling him a horse's ass, should join another group, maybe the mutualasskicking.math group. ==== > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? he wouldn't retire. he is a world-class crackpot and troll. have you seen his picture? > He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. ==== > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. irresistable. You yourself have added yet another JSH-related post to this newsgroup. ==== >>This guy is totally out of touch with reality. So what >>would be the harm in simply ignoring him from this >>day forward and forever, and permitting him to retire >>in the triumphant knowledge (as far as he's concerned) >>that he is absolutely and totally correct in everything >>he's ever said? He would be quite happy, and then you >>all could have a lot more fun communicating with real >>mathematicians. You all must know by now that you >>are wasting your time. I mean the ones of you who >>are responding in a thoughtful reasoned way to James. >>The others, who are simply calling him a horse's ass, >>should join another group, maybe the >>mutualasskicking.math group. >> > irresistable. You yourself have added yet another JSH-related post to > this newsgroup. And you another ... and me too. Gib ==== Francis Harrington > This guy is totally out of touch with reality. So what > would be the harm in simply ignoring him from this > day forward and forever, and permitting him to retire > in the triumphant knowledge (as far as he's concerned) > that he is absolutely and totally correct in everything > he's ever said? He would be quite happy, and then you > all could have a lot more fun communicating with real > mathematicians. You all must know by now that you > are wasting your time. I mean the ones of you who > are responding in a thoughtful reasoned way to James. > The others, who are simply calling him a horse's ass, > should join another group, maybe the > mutualasskicking.math group. Talking to Harris _about mathematics_ is a waste of time, yes. After eight years of claiming to have a proof of FLT, he doesn't know an algebraic integer is, or what a ring extension is. Conclusion? He doesn't _care_ whether his statements are true or false. Even if he had no audience at sci.math, would he stop posting? It costs him nothing to cross-post to other rooms from which he would still get his arse kicked. The JSH thing is tedious most of the time, and morbid all the time, but still good for the occasional laugh. Try Google with these keywords: guffaw idiot polynomial LH ==== I'm having a hard time with the following problem: Let f be a continuous function on [1,oo) such that f^4 is Lebesgue integrable. Prove that if a > 4 then |integral of f(x^a)| < oo where the integral is also over [1,oo). Also, give an example where the integral of f(x^4) is infinite. Any hints or input on this would be greatly appreciated. Hugh ==== > I'm having a hard time with the following problem: Let f be a continuous function on [1,oo) such that > f^4 is Lebesgue integrable. Prove that if a > 4 then |integral of f(x^a)| < oo where > the integral is also over [1,oo). Also, give an example where the integral of f(x^4) is infinite. Any hints or input on this would be greatly appreciated. Can you write the integral of f(x^a) differently? Do you know Hoelders inequality? HTH, ==== > I'm having a hard time with the following problem: Let f be a continuous function on [1,oo) such that > f^4 is Lebesgue integrable. Prove that if a > 4 then |integral of f(x^a)| < oo where > the integral is also over [1,oo). Also, give an example where the integral of f(x^4) is infinite. I'll give you a hint for a = 5: consider a change of variable in int_1^infty f(x^5)^4 x^4 dx, then think Holder. --Ron Bruck ==== > William Elliot ha scritto nel messaggio > >Is there a fast way to determine X and Y both integer such that: >X^2 - Y^2 = A >Of course the answer to your question is yes. >>Indeed, you phrased your question incorrectly. >>Factor x^2 - y^2 = (x-y)(x+y) = a >>Thus x-y and x+y are factors of a >>For each factorization of a = nm you have >>x-y = n; x+y = m >>solve for x and y. Beware, >>if the solution turns out not be an integer, discard it. >>Do this for all factorizations of a. >>Collect the solutions you find. >>To make sure you understand the process >>solve x^2 - y^2 = 12 for integers, x,y. >>First list all the ways of factoring 12. >>Let's see what you try. >>What are your answers? >>x^2 - y^2 = 66 has no solution. Why? bacause it has tre factors. Think of the form of the factorization of the left side: x^2 - y^2 = (x - y)(x + y) Now, let A = x - y, and B = x + y. What can you say about the relationship between A and B? What sorts of numbers can occur as A and B? For instance, you can probably see that B - A = 2y. What does that say about A and B (thought of together, not just one at a time)? Having decided what numbers A and B that can appear in this factorization, now look at all the factorizations of 66. What sorts of numbers do you get from those factorizations? If this isn't enough of a hint, you're trying too hard. The answer is really pretty simple. Thanx marco Dale. ==== > Saying that one random variable is > equal to another one means that the value of the first is, with > probability 1, equal to the value of the second. Huh? I think you're understating. Saying that one random variable > is equal to another is stronger than what you're stating -- it means > that *every instance* of the value of the first variable is equal to > the value of the second one. (yes, what I just said implies that > the probability is one -- but probability one does not imply what I > said). Please correct me if I'm mistaken I won't insist on this point. I think it depends on your point of view. E.g. in analysis L^1 functions are not really functions, but rather equivalence classes for equality almost everywhere. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Here is an important equation I am encountering. > >b <= x * ( (1-x)^a ) Inequality, not equation. I accept. Solve for x in terms of a and b. Unlikely to have a closed-form solution in general. > Okay, for my purposes, a is an even integer and b is a ratioanl number less > than one. > I am looking for a solution in the range (0,1). Particularly, I am interested in the smallest value of x which will satisfy > the inequality. If you can't get me the smallest value, can you get me > something as small as possible in a closed form solution. Right now the > smallest I have is the place where x * ( (1-x)^a ) is maximised. ... that place being x_0 = 1/(1+a). I assume b < f(x_0) = a^a/(1+a)^(1+a), where f(x) = x (1-x)^a. It seems that f'''(x) > 0 for 0 < x < 1/(1+a), so if x_1 is any point with 0 <= x_1 <= x_0 with f(x_1) < b, f(x) > f(x_1) + f'(x_1) (x-x_1) + 1/2 f''(x_1) (x - x_1)^2. So if x_1 is not itself a solution, solve the quadratic f(x_1) + f'(x_1) (x - x_1) + 1/2 f''(x_1) (x - x_1)^2 = b; any solution x_2 of this with x_1 < x_2 < x_0 will have f(x_2) > b. In particular, with x_1 = 0 you get x_2 = (1 - sqrt(1-4ab))/(2a) if b<1/(4a). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >but the Greeks started it. > Just about every thing the Greeks did was rediscovered by others > as it was needed in their time. Much of Archemedies work was for War > efforts as they were needed. Sure they have being first name > recognition but played no more role than India and other countries > to thought today. I have heard that Euclid's Elements is the book with more editions > published than any other except the Bible. But wasn't his school in > Alexandria? So perhaps he should be assigned to Egypt (even if he was > Greek by culture and ancestry). And in fact the MacTutor site does assign Euclid to Egypt, although there's no reliable information about his birthplace or ancestry. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >>but the Greeks started it. >> Just about every thing the Greeks did was rediscovered by others >> as it was needed in their time. Much of Archemedies work was for War >> efforts as they were needed. Sure they have being first name >> recognition but played no more role than India and other countries >> to thought today. >> I have heard that Euclid's Elements is the book with more editions >> published than any other except the Bible. But wasn't his school in >> Alexandria? So perhaps he should be assigned to Egypt (even if he was >> Greek by culture and ancestry). >And in fact the MacTutor site does assign Euclid to Egypt, although >there's no reliable information about his birthplace or ancestry. The Greeks in Alexandria seem to have been mainly of Greek ancestry, or to be from those of other groups, mainly not Egyptians, who joined the Greeks. The Ptolemies made no attempt to submerge the other cultures, and the Greeks were sufficiently xenophobic (barbarian was the Greek for those who did not speak Greek, basically) that they made no attempt to do so. When the Seleucids did, they ran into open revolts. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University ==== >and the Greeks were >sufficiently xenophobic (barbarian was the Greek for those >who did not speak Greek, basically) Use of the term barbarian might suggest xenophobia in a modern English speaker, but Herodotus for example seems to use it in a quite neutral sense to mean non-Greeks. -- Richard -- FreeBSD rules! ==== > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University -- Ioannis Galidakis http://users.forthnet.gr/ath/jgal/ ==== >I was just trying to see who in fact borned the most mathematicians. >My SWAG list was not intended to be complete but just a start. Ok, I >blew it after France at least according to MacTutor link below. >However I was close with Italy, Germany and former Russia but totally >underestimated England and the U.S. and way over estimated India and >Greece. This is a quantative analysis. The next step is qualitative >analysis which is why I asked you guys the question. I doubt that anybody would claim that being in the MacTutor list is all that significant as an indication of mathematical importance: it's mainly a question of which people somebody has bothered to write a biography of. >>There sure have been a lot of important French mathematicians; >>Fourier, Galois, d'Alembert... >Fermat - but his last theorem was published after his death. >So during his life time it did not play a major role in mathematical >developement. Fermat's importance in mathematics does not rest on his Last Theorem. He made many contributions in various areas. Although he didn't publish much, he corresponded with many important mathematicians, so he was quite influential in his lifetime. > The same can be said of Da Vinci and other Greats whose >certain works were suppressed during their lifetime. Although da Vinci did study a lot of mathematics, I'm not aware of any really significant contributions he made to mathematics. >>but the Greeks started it. >Just about every thing the Greeks did was rediscovered by others >as it was needed in their time. Much of Archemedies work was for War >efforts as they were needed. Sure they have being first name >recognition but played no more role than India and other countries >to thought today. This is just silliness. The main contribution of the Greeks was to discover the idea of mathematical proof. Nearly all Western mathematicians until very recent times got their first taste of mathematics from studying Euclid. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > at 10:26 AM, rge11x@netscape.net (robert egri) said: >Please forgive me my ignorance, I missed a few classes in college >thanks to some heavily hopped Czechish Pilsner, but could you >elaborate on this list especially regarding Czechoslovakia and >Taiwan? Others have addressed some of the other countries. My recollection is > that Hamilton was from Ireland and that Chern was from Taiwan. Shiingshen Chern. Born 26 Oct 1911 in Chia-hsing (or Jiaxing), Chekiang province(now Zhejiang), China. You, too, need some well-hopped Pilsner to refresh your memory ... ==== > There are lots of metrics. In fact, if d is any bounded > metric, and there are lots of bounded metrics equivalent > to any one, take D(f,g) = int d(f(x),g(x)) dP(x). This > is equivalent to the topology you specified. Noel. ==== -Steven > This question may be too simple, but it is important to me. > In 7D space, I have two hyper planes: > (*1): > (x1 x2 x3 x4 x5 x6 x7)*(a1 a2 a3 a4 a5 a6 a7)T = 0 > (*2): > (x1 x2 x3 x4 x5 x6 x7)*(b1 b2 b3 b4 b5 b6 b7)T = 0 > where T is the transpose sign, * is the matrix multiplication sign. > Now I first get the subspace(sub-hyperplane) S1 which is the > intersection of (*1) and (*2). Then I combine the subspace S1 and a > vector > v1 = (a1 a2 a3 a4 a5 a6 a7)T to form a hyper plane (*3), so that this > new hyperplane is the span of S1 and v1, and S1, v1 are on the > hyperplane (*3). The questions what is the formula of this new > hyperplane? > Let u be a vector such that the induced hyperplane x*(u^T)=0 (like in (*1) > and (*2)) contains S1. Then u is a linear combination of a and b. > Proof: Every linear combination of a and b clearly has this property, this > gives a 2D vector space. On the other hand, the orthogonal complement of S! > (that is the set of all such u) only is 2D because S1 is 5D and 2+5=7. qed. So we make the ansatz u=s*a+t*b with unknown coefficients s and t. Because > v1 also lies in the hyperplane (*3) we have v1*(u^T)=0 which gives a linear > condition on s and t. The result is 1D solution space for s and t. It is 1D > because every multiple of any solution u also is a solution. Just pick any > value where not both s and t are zero and calculate u. ==== >Is it possible to convert a problem where we are trying to maximize a >quantity (profit) into a problem where we are trying to minimize a >(different) quantity (loss)? > Maximizing f is the same as minimizing -f. Or perhaps what you have in > mind is the equivalence in Linear Programming between solving the primal > problem (a max problem) and the dual problem (a min problem). Yes. The equivalence in Linear programming was what I was thinking > about. > If yes, can we do the transform in linear >time? In particular, is it possible to transform a Network flow >problem where we try to maximize the flow between the source and sink >into a shortest path problem where we are trying to find the shortest >between two points (say source and sink)? > Maximum flow is equivalent to minimal cut. If it's on a planar graph, > and the source and sink are on the same face, then this is equivalent > to a shortest-path problem in the dual graph. > Here, I want to know if the Maximum flow in a network can be > formulated as a Linear programming problem. Also, can you please > elaborate as to why the Maximal flow is equivalent to the > shortest-path problem _only_ for planar graphs? Certainly maximum-flow (with source s and sink t) is a linear programming problem: maximize f subject to sum_j (x_{sj} - x_{js}) = f sum_j (x_{tj) - x_{jt)) = -f sum_j (x_{ij) - x_{ji}) = 0 for all other i 0 <= x_{ij} <= c_{ij} for all i,j where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in that arc (make x_{ij} = 0 if there is no arc i -> j). Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >Is it possible to convert a problem where we are trying to maximize a >quantity (profit) into a problem where we are trying to minimize a >(different) quantity (loss)? > Maximizing f is the same as minimizing -f. Or perhaps what you have in > mind is the equivalence in Linear Programming between solving the primal > problem (a max problem) and the dual problem (a min problem). > Yes. The equivalence in Linear programming was what I was thinking > about. > If yes, can we do the transform in linear >time? In particular, is it possible to transform a Network flow >problem where we try to maximize the flow between the source and sink >into a shortest path problem where we are trying to find the shortest >between two points (say source and sink)? > Maximum flow is equivalent to minimal cut. If it's on a planar graph, > and the source and sink are on the same face, then this is equivalent > to a shortest-path problem in the dual graph. > Here, I want to know if the Maximum flow in a network can be > formulated as a Linear programming problem. Also, can you please > elaborate as to why the Maximal flow is equivalent to the > shortest-path problem _only_ for planar graphs? Certainly maximum-flow (with source s and sink t) is a linear programming problem: maximize f > subject to > sum_j (x_{sj} - x_{js}) = f > sum_j (x_{tj) - x_{jt)) = -f > sum_j (x_{ij) - x_{ji}) = 0 for all other i > 0 <= x_{ij} <= c_{ij} for all i,j > > where c_{ij} is the capacity of the arc i -> j, and x_{ij} the flow in > that arc (make x_{ij} = 0 if there is no arc i -> j). > Pradip ==== What conditions should be satisfied so that a singular solution envelope for F(x,y,c) = 0 exists for parametric variation of c ? It is known that singular solution should satisfy the above F and del F/ del c = 0 ; perhaps it is a condition using second order partial derivative. In the crank - connecting rod example, we have a crank of radius R centered on origin, angle c with x-axis. The end [R cos(c), R sin(c)] connects to a point on x-axis (reciprocating piston) with connecting rod length L ; It is found that an envelope exists in c limits + - ArcTan(L/R), and there is no envelope outside these limits. ==== I learned all of my linear algebra over the reals or complexes, so this seemingly simply problem has baffled me. X is a vector space over the field F_2 = {0,1} of finite dimension. K is a subspace and K^perp is its orthogonal complement, K^perp := {x in X with =0 for all k in K}. The equation I want to prove is dim K + dim (K^perp) = dim X. I've tried on my own, and noticed many discouraging things. Normally (in R^n, say), we have that K and K^perp only intersect at zero. But this is no longer necessarily true. I don't think that you can extend any orthonormal basis of a subspace to an orthonormal basis of the entire space. For one thing, Gram-Schmidt certainly doesn't work any more. Many of these troubles probably stem from the fact that there are many nonzero vectors v for which =0. -Tyler ==== I think I understand now. The equality dim(Null(A)) + dim(Range(A)) = dim(X) is still true, where A is a linear transformation in the finite- dimensional space X. I say still true because I was not certain that it would hold for vector spaces over arbitrary fields. (I reviewed the proof to confirm it still works.) We want to check that dim(F) + dim(F^perp) = dim(X). Choose a basis of F, and let A have these elements as rows. Then Null(A) = F^perp; and the row rank of A is clearly dim(F). Since row rank = col rank = dim(Range(A)), we are done. ==== in message <43b20b30.0311091614.90ffa88@posting.google.com>: > I think I understand now. The equality dim(Null(A)) + dim(Range(A)) = dim(X) is still true, > where A is a linear transformation in the finite- > dimensional space X. I say still true because > I was not certain that it would hold for vector > spaces over arbitrary fields. (I reviewed the > proof to confirm it still works.) We want to check that dim(F) + dim(F^perp) = dim(X). > Choose a basis of F, and let A have these elements > as rows. Then Null(A) = F^perp; and the row rank > of A is clearly dim(F). Since row rank = col rank > = dim(Range(A)), we are done. You can't assume Null(A) = F^perp without knowing more about the inner product (sesquilinear form) you're using. As others have pointed out, you really need to take into account whether the form is degenerate. For example, if X has a non-zero radical R then dim(R) + dim(R^perp) = dim(R) + dim(X) > dim(X). -- Jim Heckman ==== >I learned all of my linear algebra over the reals or complexes, >so this seemingly simply problem has baffled me. X is a vector space over the field F_2 = {0,1} of >finite dimension. K is a subspace and K^perp is its >orthogonal complement, What is the inner product for the vector space, though? There are canonical inner products in R^n and in C^n, but I don't know if there is one for vector spaces over F_2... >K^perp := {x in X with =0 for all k in K}. >The equation I want to prove is dim K + dim (K^perp) = dim X. >I've tried on my own, and noticed many discouraging things. >Normally (in R^n, say), we have that K and K^perp only >intersect at zero. But this is no longer necessarily true. >I don't think that you can extend any orthonormal basis of >a subspace to an orthonormal basis of the entire space. For >one thing, Gram-Schmidt certainly doesn't work any more. >Many of these troubles probably stem from the fact that there >are many nonzero vectors v for which =0. What is your definition of inner product? What are the axioms you require of an inner product? In some books, an inner product is required to be positive definite (i.e., =0, and =0 if and only if v = 0)... ====================================================================== 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 ==== >X is a vector space over the field F_2 = {0,1} of >finite dimension. K is a subspace and K^perp is its >orthogonal complement, >K^perp := {x in X with =0 for all k in K}. >The equation I want to prove is dim K + dim (K^perp) = dim X. As you noted, inner products don't always behave like the standard one in R^n. But this result still holds if, for example, is still defined as the dot product. In that case, you can describe K^perp as the solution to a system of linear equations. Do you know how to compute the dimension of a nullspace (kernel)? Note that the result is _false_ if for example =0 for all u and v. You'll need to look to see what properties you're assuming about the inner product before you try to prove something about it! (Nondegeneracy, for starters.) dave ==== > I learned all of my linear algebra over the reals or complexes, > so this seemingly simply problem has baffled me. X is a vector space over the field F_2 = {0,1} of > finite dimension. K is a subspace and K^perp is its > orthogonal complement, > K^perp := {x in X with =0 for all k in K}. > The equation I want to prove is dim K + dim (K^perp) = dim X. I've tried on my own, and noticed many discouraging things. > Normally (in R^n, say), we have that K and K^perp only > intersect at zero. But this is no longer necessarily true. > I don't think that you can extend any orthonormal basis of > a subspace to an orthonormal basis of the entire space. For > one thing, Gram-Schmidt certainly doesn't work any more. > Many of these troubles probably stem from the fact that there > are many nonzero vectors v for which =0. -Tyler It is best to think of K^perp not as a subspace of X but as a subspace of the dual space X*. (Now via your pairing , it is true that X* is identified with X, but that just confuses things.) -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== > I learned all of my linear algebra over the reals or complexes, > so this seemingly simply problem has baffled me. X is a vector space over the field F_2 = {0,1} of > finite dimension. K is a subspace and K^perp is its > orthogonal complement, > K^perp := {x in X with =0 for all k in K}. > The equation I want to prove is dim K + dim (K^perp) = dim X. What do you mean by ? You have not defined it. Jose Carlos Santos ==== Sorry. Fix a basis of X (suppose dim X = m). Then let the coordinates/coefficients of x and k be x1...xm and k1...km respectively. Now we can define :=x1k1 + x2k2 + ... + xmkm, in a manner analogous to the standard inner product. It is linear and symmetric, but does not really give a norm. ==== >Sorry. Fix a basis of X (suppose dim X = m). Then let the >coordinates/coefficients of x and k be >x1...xm and k1...km respectively. Now we can >define :=x1k1 + x2k2 + ... + xmkm, >in a manner analogous to the standard inner >product. It is linear and symmetric, but does >not really give a norm. So? Not all inner products are positive definite. ====================================================================== 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 ==== >Sorry. Fix a basis of X (suppose dim X = m). Then let the >coordinates/coefficients of x and k be >x1...xm and k1...km respectively. Now we can >define :=x1k1 + x2k2 + ... + xmkm, >in a manner analogous to the standard inner >product. It is linear and symmetric, but does >not really give a norm. So? Not all inner products are positive definite. This is, of course, depending on your definition of inner product. The one usually given in linear algebra and hilbert spaces insists upon it. Take the usual definition of inner product for a vector space over the real or complex field. For vectors x,y,z and scalars a,b. We require 1) linear = a+b 2) (hermitian) symmetric = conj 3) positive definite = 0 and iff x = 0. However, you are right, some, physics in Lorentz space and this topic of finite fields, allow a broader definition. Imho we should have given it a different name since not all the properties of the positive definite inner product (PD IP) carry over to a non-PD IP. For example, for finite fields, Gramm-Schmidt fails to generate a set of orthogonal vectors from a set of indendent ones. -- Johan KULLSTAM sysengr ==== > Well, I discovered something rather very odd (at least to me) today. > Let n be a positive integer, and let s(n) denote the sum of all the > divisors of n. So, for example, s(6) = 1+2+3+6 = 12. It turns out there is a class of numbers with the property that a) n is a perfect square > b) s(n) is a perfect square This one is in Sloane's online database of offbeat sequences, number A008847. Go to http://www.research.att.com/~njas/sequences/ and enter 1, 9, 20, 180, 1306 and click Search. The search will cough up more terms of the table, and some references in the lit. Larry