mm-174 === Subject: Re: Solution to an inequation 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 canêt 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. === Subject: Re: Who contributed most to mathematics? (robert egri) said: >Please forgive me my ignorance, I missed a few classes in college >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. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Tricky integral > oo ln(x) > S --------------------- dx > 0 (x^n) -1 n=2,3,4,..... a2 := int(log(x)/(x^2-1),x=0..in'nity); 2 a2 := 1/4 Pi > a3 := int(log(x)/(x^3-1),x=0..in'nity); a3 := -1/6 I sqrt(3) ln(2) ln(1 + I sqrt(3)) + 1/6 I sqrt(3) ln(2) ln(1 - I sqrt(3)) 2 + 1/12 I sqrt(3) ln(1 + I sqrt(3)) 2 - 1/6 ln(2) ln(1 + I sqrt(3)) + 1/12 ln(1 + I sqrt(3)) 2 - 1/12 I sqrt(3) ln(1 - I sqrt(3)) 2 - 1/6 ln(2) ln(1 - I sqrt(3)) + 1/12 ln(1 - I sqrt(3)) 2 2 + 1/6 Pi + 1/6 ln(2) > simplify(%); 2 4/27 Pi > a4 := int(log(x)/(x^4-1),x=0..in'nity); 2 a4 := 1/8 Pi > a6 := int(log(x)/(x^6-1),x=0..in'nity); 2 a6 := - 1/12 ln(2) ln(1 + I sqrt(3)) + 1/24 ln(1 + I sqrt(3)) 2 - 1/12 ln(2) ln(1 - I sqrt(3)) + 1/24 ln(1 - I sqrt(3)) 2 - 1/24 I sqrt(3) ln(1 - I sqrt(3)) + 1/12 I sqrt(3) ln(2) ln(1 - I sqrt(3)) + 1/12 ln(2) ln(-1 - I sqrt(3)) 2 - 1/24 I sqrt(3) ln(-1 - I sqrt(3)) 2 2 - 1/24 ln(-1 - I sqrt(3)) + 1/12 Pi 2 + 1/12 ln(2) ln(-1 + I sqrt(3)) - 1/24 ln(-1 + I sqrt(3)) 2 + 1/24 I sqrt(3) ln(-1 + I sqrt(3)) - 1/12 I sqrt(3) ln(2) ln(-1 + I sqrt(3)) 2 + 1/24 I sqrt(3) ln(1 + I sqrt(3)) + 1/12 I sqrt(3) ln(2) ln(-1 - I sqrt(3)) - 1/12 I sqrt(3) ln(2) ln(1 + I sqrt(3)) > simplify(a6); 2 1/9 Pi === Subject: Re: NOVA strings and branes > So I submit that a large crowd slowly circling a big square box > makes a great target point... Oh be still, my wildly beating heart! I have wonderful dreams of bombing the Kaba stone and laying a ten meg egg on the Plains of Arrafat. That is not the smell of burnt Wog I am smelling. It is the scent of victory! Bob Kolker === Subject: Re: NOVA strings and branes [snip lots] > We will have troops in the Mideast forever, dying daily. The 'rst > rational act would be to launch a nuclear salvo over Mecca and Medina > during the hajj - payback for the World Trade Center according to > hypocenter for a one megatonne warhead). A dialog may then commence > after a common language is established. If the Muslims are still > sanctimonious... lose Cairo as a demo (13 million dead). A big > city/week thereafter. > its what must be done.. And Unky hits it on the button... > with this enemy it will require a lesson that shakes the > Muslim world to the core... While a small percentage Sci.physics is not the place to debate politics, but for the record let it be noted that not all of us in the U.S. are quite so anxious to spill blood and level cities. No, I donêt have any easy answer, but I am fairly certain that if we were to make beligerent fantasies such as the above into policy we would quickly come to regret our excesses. Best wishes, Mike === Subject: Re: NOVA strings and branes donêt have any easy answer, but I am fairly certain that > if we were to make beligerent fantasies such as the above > into policy we would quickly come to regret our excesses. And if we donêt we will regret them more. It is only a matter of time before the martyrs of al Qêuaida blow up tunnels and bridges in New York City (that is where the Jews are) or shoot down commercial airlines right here in the U.S. Then what do we do? Bite our lip, or slay Wog on a grand, grand scale. Bob Kolker === Subject: Re: NOVA strings and branes if we were to make beligerent fantasies such as the above > into policy we would quickly come to regret our excesses. > And if we donêt we will regret them more. It is only a matter of time > before the martyrs of al Qêuaida blow up tunnels and bridges in New York > City (that is where the Jews are) or shoot down commercial airlines > right here in the U.S. Then what do we do? Bite our lip, or slay Wog on > a grand, grand scale. Bob Kolker I donêt have good answers, but before we get carried away, note that since Sept.11 2001, the terrors we have envisioned have simply not materialized. Additionally, it is worth noting that it appears that not even all the hi-jackers on 9/11 were aware of the suicidal nature of the plan. It appears that those who would see blood spilled on U.S. soil are having dif'culty recruiting people who are willing to commit both murder and suicide and who have enough wits about them to pose a threat of that magnitude. We should not help with recruiting. I suspect that in the immediate future the best thing the U.S. could do is try to set up some form of reasonably functional governments in Iraq and Afghanistan. There are no easy answers. This is not a repeat of the Second World War. First of all, what nation are we to 'ght with? Second, I donêt think the world can stand that sort of devastation again. As this is off topic, I will attempt to refrain from continuing. I merely wished to point out that the views you have expressed are not those of all U.S. citizens. Best wishes, Mike === Subject: Re: NOVA strings and branes | > donêt have any easy answer, but I am fairly certain that | > if we were to make beligerent fantasies such as the above | > into policy we would quickly come to regret our excesses. | > And if we donêt we will regret them more. It is only a matter of time | > before the martyrs of al Qêuaida blow up tunnels and bridges in New York | > City (that is where the Jews are) or shoot down commercial airlines | > right here in the U.S. Then what do we do? Bite our lip, or slay Wog on | > a grand, grand scale. | > | > Bob Kolker | | I donêt have good answers, but before we get carried away, note | that since Sept.11 2001, the terrors we have envisioned have | simply not materialized. Additionally, it is worth noting | that it appears that not even all the hi-jackers on 9/11 | were aware of the suicidal nature of the plan. It appears | that those who would see blood spilled on U.S. soil are having | dif'culty recruiting people who are willing to commit both | murder and suicide and who have enough wits about them to pose a | threat of that magnitude. We should not help with recruiting. | | I suspect that in the immediate future the best thing the U.S. | could do is try to set up some form of reasonably functional | governments in Iraq and Afghanistan. | | There are no easy answers. This is not a repeat of the Second | World War. First of all, what nation are we to 'ght with? | Second, I donêt think the world can stand that sort of devastation | again. | | As this is off topic, I will attempt to refrain from continuing. | I merely wished to point out that the views you have expressed | are not those of all U.S. citizens. Sun. Blah, blah, blah. Man, what does it take to wake people up? We are at war. We did not start it. We will be at war until every last bloody terrorist is dead or captured. It is called The War on Terrorism. It donêt have to be with any particular country. You are with us or against us. It is that simple. It is only a matter of time before some of us see that great white ¤ash of a nuke going off over here set off by some asshole terrorist. I would much rather it be them than us. Nukes are physics. This is on topic. Terrorism *has* to be wiped out or some of us will be munching on some pretty nasty radioactive stuff. FrediFizzx === Subject: Re: NOVA strings and branes > donêt have any easy answer, but I am fairly certain that > if we were to make beligerent fantasies such as the above > into policy we would quickly come to regret our excesses. > And if we donêt we will regret them more. It is only a matter of time > before the martyrs of al Qêuaida blow up tunnels and bridges in New York > City (that is where the Jews are) or shoot down commercial airlines > right here in the U.S. Then what do we do? Bite our lip, or slay Wog on > a grand, grand scale. Bob Kolker I donêt have good answers, but before we get carried away, note > that since Sept.11 2001, the terrors we have envisioned have > simply not materialized. Additionally, it is worth noting > that it appears that not even all the hi-jackers on 9/11 > were aware of the suicidal nature of the plan. It appears > that those who would see blood spilled on U.S. soil are having > dif'culty recruiting people who are willing to commit both > murder and suicide and who have enough wits about them to pose a > threat of that magnitude. We should not help with recruiting. There has been no dif'culty producing suicidal idiots in the schools and masques... The fact that there have been no large scale action on US soil speaks more of our method to prevent than their method to attack.... I suspect that in the immediate future the best thing the U.S. > could do is try to set up some form of reasonably functional > governments in Iraq and Afghanistan. > Ya suspect ?..... There are no easy answers. This is not a repeat of the Second > World War. First of all, what nation are we to 'ght with? > Second, I donêt think the world can stand that sort of devastation > again. > No its not a repeat... Its on a far larger scale yet to be perceived by many. The Nation we are 'ghting is a sub-set borderless nation of Islam.. It comprises approx 12% to 20% of varied races of people who believe in Wahabisim version of Islam and have vowed to kill you, your children and anyone that does not follow the line of Old Kingdom rules they base their very existence on. And at 15% thatês more foot troops than all the combined armed forces lines against us in WW2 on both the German and Japanese sides. Now add to that the Japanese philosophy of Kamikaze and a seat next to godês right hand and it , if not checked with absolute force, is far more dangerous than what the western lifestyle faced in WW 1 and 2...... As this is off topic, I will attempt to refrain from continuing. > I merely wished to point out that the views you have expressed > are not those of all U.S. citizens. > But I will suggest that its the Majority and will be a much greater majority the day after a attack is allowed through by penthouse living users of the sacri'ces of others. Best wishes, > Mike > Paul R. Mays -------------------------------------------------------------- -------------- - Some where within the Quantum State Http://Paul.Mays.Com/story.html http://paul.mays.com/mayday.html http://paul.mays.com/rainy.html Science is a system of statements based on direct experience, and controlled by experimental veri'cation. Veri'cation in science is not, however, of single statements but of the entire system or a sub-system of such statements. - Rudolf Carnap (18911970) === Subject: Re: NOVA strings and branes That was a rather small island. In Japan itself, the Japanese > military had ten times as many troops (granted, poorly equipped, but > so were those in Okinawa) and many millions of civilians who were > ready to 'ght to the end. Based on the Okinawan experience you can > begin to estimate the cost, in casualties and destruction that would > result from landing in Japan. The estimes for Olympic and Coronet were about 1 million allied casualties, 5 million jap casualites. The military always gets its numbers wrong so the casualties might have gone higher still. The Revisionists love to tell us how these estimates were exagerrations. They should have been on Okinawa, even has you have pointed out. Revisionist Dunces. They are Ann Coulter-class Liberal Traitors who hate America. Bob Kolker === Subject: Re: NOVA strings and branes > That was a rather small island. In Japan itself, the Japanese >> military had ten times as many troops (granted, poorly equipped, but >> so were those in Okinawa) and many millions of civilians who were >> ready to 'ght to the end. Based on the Okinawan experience you can >> begin to estimate the cost, in casualties and destruction that would >> result from landing in Japan. The estimes for Olympic and Coronet were about 1 million allied >casualties, 5 million jap casualites. The military always gets its >numbers wrong so the casualties might have gone higher still. Extrapolating from previous encounters, most probably so. Especially when one adds to direct casualties the results of the (pretty much guaranteed, in case of invasion) mass starvation of Japanese civilians. > The Revisionists love to tell us how these estimates were exagerrations. >They should have been on Okinawa, even has you have pointed out. > Well, they just conveniently ignore it. >Revisionist Dunces. They are Ann Coulter-class Liberal Traitors who hate >America. > Yep. Intellectual offal writing for garbage intelligentsia. Mati Meron | When you argue with a fool, meron@cars.uchicago.edu | chances are he is doing just the same === Subject: complex integral.....?? hello.......... int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- i think ...........answer is 0 ?? right?? um.......please......con'rm..... === Subject: Re: complex integral.....?? > hello.......... int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- i think ...........answer is 0 ?? right?? Well, why do you think itês 0? What theorem are you using, and why do you think itês applicable? === Subject: Re: complex integral.....?? > int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 thus, f(z) is analysis of |z|<1/2 thus.......i think that answer is 0 it is right?? === Subject: Re: complex integral.....?? > int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 thus, f(z) is analysis of |z|<1/2 thus.......i think that answer is 0 it is right?? > All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., |a_{n-1}|, |a_n|) Then, you are right ... -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: complex integral.....?? Ignacio Larrosa Ca.96estro All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., > |a_{n-1}|, |a_n|) I think itês easier to notice that if |z| <= 1/2, then |z^3 + z^5| <= ..., hence z^3 + z^5 could not = -1. === Subject: Re: complex integral.....?? Ignacio Larrosa Ca.96estro escribi.97 int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- > um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 thus, f(z) is analysis of |z|<1/2 thus.......i think that answer is 0 it is right?? > All the roots of a_n*z^n + a_{n-1}z^(n-1) + ... + a_1*z + a0 verify 1/(1 + B/|a_0|) < |z| < 1 +A/|a_n| If a_n * a_0 =/= 0, of course ... > Where A = max(|a_0|, |a_1|, ..., |a_{n-1}|) and B = max(|a_1|, ..., > |a_{n-1}|, |a_n|) Then, you are right ... > -- Ignacio Larrosa Ca.96estro > A Coru.96a (Espa.96a) > ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: complex integral.....?? >int [cosz / {1+(z^3)+(z^5)}] dz = ? contour |z| = 1/2 -------------------------- um...... i think that root of 1+(z^3)+(z^5)}] exist between -1 ~ -0.5 Well, of course there is at least one *real* root between -1 and -1/2, but the question is: what about *complex* roots z such that |z| < 1/2? Thatês what matters here. > thus, f(z) is analysis of |z|<1/2 My guess is that you meant analytic here, not analysis. > thus.......i think that answer is 0 it is right?? Yes, it is right, but it doesnêt seem that you understand why. Jose Carlos Santos === Subject: The Dilbert Matrix The Dilbert matrix is a generalization of the Hilbert matrix. De'ne the Dilbert matrix D(t, s) to be the matrix with (i, j) entry given by D(t,s)_{i,j} = 1/( (i+j+t+s+1)*binomial(i+j+t+s, t) ). (Eq. 1) An alternative expression for the entries of the Dilbert matrix is D(t,s)_{i,j} = 1/( (t+1)*binomial(i+j+t+s+1, t+1) ). (Eq. 2) We prove that the entries of the inverse of the Dilbert matrix are integers. Notation: We index all matrices starting at 0. We assume that all matrices are n x n matrices for some positive integer n. Our proof makes use of the Leibniz matrix Z(t, s), which we de'ne to be the matrix with (i, j) entry given by Z(t, s)_{i,j} = 1/( (i+j+t+s+1)*binomial(i+j+t+s, i+t) ). (Eq. 3) The Dilbert and Leibniz matrices are related by the equation L*D(t, s) = Z(t, s), (Eq. 4) where L is the lower triangular matrix given by L_{i,j} = (-1)^j*binomial(i, j). (Eq. 5) The Leibniz matrix Z(t,s) may be factored as Z(t, s) = A(t)*F(t+s)*A(s), (Eq. 6) where A(t) is the diagonal matrix with A(t)_{i,i} = (i+t)!, (Eq. 7) and F(x) is the Hankel matrix with F(x)_{i,j} = 1/(i+j+x+1)!. (Eq. 8) Thus we can factor Z(t,s) as Z(t, s) = A(t)*A(0)^(-1)*Z(0, t+s)*A(t+s)^(-1)*A(s). (Eq. 9) Combining (Eq. 4) and (Eq.9) gives the equation Z(t, s) = A(t)*A(0)^(-1)*L*D(0, t+s)*A(t+s)^(-1)*A(s). (Eq. 10) Now D(0, t+s) is a generalized Hilbert matrix, and an explicit formula for the entries of the inverse of D(0, x) is D(0, x)^(-1)_{i,j} = (-1)^(i+j)*(i+j+x+1) *binomial(n+i+x, n - j - 1) *binomial(n+j+x, n - i - 1) *binomial(i+j+x, i) *binomial(i+j+x, j). (Eq. 11) Combining Eqs. 5, 7, 10, and 11, and rearranging some terms, gives the equation Z(t, s)^(-1)_{i,j} = sum_(k=j)^(n-1) (-1)^(i+k+j) *multinomial(n+i+t+s, n - k - 1, k - j, i+s, j+t, 1) *binomial(n+k+t+s, n - i - 1) *binomial(i+k+t+s, i). (Eq. 12) Thus Z(t, s)^(-1) has integer entries. Since D(t, s)^(-1) = Z(t, s)^(-1)*L, the matrix D(t, s)^(-1) also has integer entries. Remarks: Another formula[2] for the entries of inverse of D(t, 0) is D(t, 0)^(-1)_{i,j} = sum_k=0^j (-1)^(i+k) *binomial(n+i+t, i+1) *binomial(n, i+1) *binomial(n+k+t-1, k) *binomial(n, k) *(i+1)^2 *(i+j+2)*(i+j+3)*...*(i+j+t) /(i+k+1)/(i+k+2)/.../(i+k+t). (Eq. 13) But (Eq. 13) does not make it clear that the entries are integers, as its summands are not always integers. It has the advantage over the approach of this paper that the formula appears to generalize to a formula for the entries of the inverse of a Fibonomial analog of D(t,0).[2, Conjecture 6.1]. Notice the identity the identity L*D(t, s)*L^t = D(s, t). Some sequences of determinants of D(t,0) (and D(0,t)) are listed in the Online Encylopedia of Integer Sequences. [3, A067689, A069640-A069644]. References [1] Layman, John W. The Hankel transform and some of its properties. J. Integer Seq. 4 (2001), no. 1, Article 01.1.5, 11 pp. (electronic). [2] Richardson, Thomas M. The Filbert matrix. Fibonacci Quart. 39 (2001), no. 3, 268--275. Integer Sequences, http://www.research.att.com/~njas/sequences/. === Subject: Re: Very interesting problem on Real Analysis > (mladensavov) > >there is a problem I 'nd dif'cult. Let f:[0,1]->R be in'nitely many >times differentiable such that for each x from [0,1] there exists n(x) >such that the >n(x) derivative of f in the point x is 0. Show that f is polynom. >>Are you certain this is true? Well letês see ... he is trying to prove it ... what would that imply? >> Not sure what you think youêre contributing here, but to answer >> your question no, the fact that someone is trying to prove something >> does not imply that heês certain itês true. >> I mean, duh... >> ************************ >> David C. Ullrich Your mind doesnêt seem to be working, but I will help you out: If he were certain it was true, he would not be trying to prove it. Uh, no. Happens all the time that people try to prove things even though theyêre certain theyêre true. For example, theyêre certain itês true because they saw the proof years ago but donêt recall how it went. Or theyêre essentially certain itês true because itês an exercise in a reliable textbook. Or whatever. (In particular, it happens a lot around here that an exercise asks the reader to prove or disprove something, and a reader posts the question, asking us how to prove it instead of asking us how to prove or disprove it.) Again: duh... ************************ David C. Ullrich === Subject: Re: Key core error argument, stepped out Readers notice how Rick Decker is playing you for fools as I have a >*stepped* out math proof. >>In fact, you have dozens of posts with your proof stepped >>out. And each one is followed by a number of posts showing >>the error you made around step 6 (or possibly 7, I canêt >>recall). >Now then, if youêve gone through the trying and dif'cult process of >'nding a spectacular proof, and then face people who decide to ignore >your hard work and effort, to trot out their own pet examples, to try >and distract people from your proof, how would you take it? >>Hardly ignored. People go step by step through your stepped >>out proof to show in great stepped-out detail why step 6 >>(or 7) is wrong. Thereês a fair amount of work involved in >>composing those arguments. > Such posters must necessarily be challenging the fact that to get from > 7 to 1, when youêre dividing by something, you have to divide by 7. This indicates a complete failure to understand the arguments. Perhaps you would do better to start responding to some by saying, I donêt understand what youêre trying to say here, rather than assuming you do and dismissing it. >>Nor can a post which includes your proof and analyzes your >>steps in detail be considered to be distracting. Any more >>than if I took a calculus text and spent a page analyzing >>what took one line in the text could be considered to be >>ignoring or distracting from the text. >> - Randy > Well, since you *have* to divide 7 by 7 to get 1, some posters > apparently decide to just run away from the proof altogether and make > up some example where they can hide what theyêre trying to push over > on people. It doesnêt matter much though as theyêre still cranks, trying to > challenge the fact that to get from 7 to 1 by division, you have to > divide by 7. No one has challenged that. You have merely not understood the arguments made. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: Key core error argument, stepped out > What if he knew he had a paper that could win him the Fieldês Medal, > do you think heêd be so casual about others trying to 'nd ways around > proper process to discredit him? Which Fieldês Medal would that be? The Wheat Fieldês or the Corn Fieldês or... oh, you mean the Fields Medal, as in J.C. Fields. Never mind... -- Wayne Brown (HPCC #1104) | When your tailês in a crack, you improvise fwbrown@bellsouth.net | if youêre good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) Henri Wilson. See why relativity is wrong: > http://www.users.bigpond.com/HeWn/index.htm It would be nice if you used more text and fewer programs. Iêm not your explanation appears to be contained in them, your website says nothing that I am willing to look at. -- Will Twentyman email: wtwentyman at copper dot net === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) > Henri Wilson. See why relativity is wrong: > http://www.users.bigpond.com/HeWn/index.htm It would be nice if you used more text and fewer programs. Iêm not all your > explanation appears to be contained in them, your website says nothing > that I am willing to look at. But Henriês programs are logically bull-proof: http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ LogicBull.html http://users.pandora.be/vdmoortel/dirk/Physics/Fumbles/ Wilgebra.html There is more :-) Dirk Vdm === Subject: Re: I NEED HELP BADLY (sorry, maths not psych) .95meaningless babbleê is a perfect description of relativity. > Any suf'ciently advanced technology is indistinguishable from magic to those unfamiliar with that technology. (Arthur C. Clarke) This drawing is also relevant to Henryês hearing of meaningless babble when reading relativity: http://www.wonderdog.com/farside.htm === Subject: division by general integer using register shifts Hi all, As most here are already aware (but just in case some arenêt), if you shift the digits of a number to the left or right one time then itês 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 youêre 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 canêt 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 === Subject: Re: division by general integer using register shifts > As most here are already aware (but just in case some > arenêt), if you shift the digits of a number to the left or right one > time then itês 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 youêre 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 canêt 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? Youêre in trouble. Distributivity works for multiplication but it doesnêt for division. n/(a+b) /= n/a + n/b Use n/(a+b) = 1/(a/n + b/n) = 1 / [(1/n)(a + b)] Twice use a fast reciprocal algorithm and multiplication shifting. === Subject: Re: division by general integer using register shifts >Youêre in trouble. Distributivity works for multiplication but >it doesnêt for division. > n/(a+b) /= n/a + n/b Youêre right. In fact, the characteristic properties are Axiom 1: a/(b/c) = c/(b/a), Axiom 2: a/(a/b) = b, Axiom 3: (a+b)/c = a/c + b/c. >Use > n/(a+b) = 1/(a/n + b/n) = 1/ [(1/n)(a + b)] >Twice use a fast reciprocal algorithm and multiplication shifting. In fact, n/(a+b) = z/(z/(n/(a+b))) by Axiom 2 = z/((a+b)/(n/z)) by Axiom 1 = z/(a/(n/z) + b/(n/z)) by Axiom 3 and different zês other than 1 may be more useful. Other properties that follow are: (a/c)/(b/c) = c/(b/(a/c)) = c/(c/(a/b)) = a/b (a/b)/(a/c) = c/(a/(a/b)) = c/b a/(b/b) = b/(b/a) = a (a/a)/(b/c) = c/(b/(a/a)) = c/b a/(b/(c+d)) = (c+d)/(b/a) = c/(b/a) + d/(b/a) = a/(b/c) + a/(b/d) a/a = b/(b/(a/a)) = b/(a/(a/b)) = b/b respectively by Axioms 1,1,2; 1,2; 1,2; 1,(1,2); 1,3,1-twice & 2,1,2. === Subject: Re: Something wrong with this integral (see text) > Hi All >> I calculated two expressions: >> FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] >> and >> FullSimplify[Integrate[(-1)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] >> They gave the same result, but in the 'rst expression when we take >> L->In'nity, itês impossible to take integral, but MAthematica CAN do that. You can determine these analytically by using an appropriate contour on >the complex plane and the residue calculus: for example, for the 'rst, >consider the integral of f(z) = -z^2*exp(iz)/(z^2 + 1). On the real axis, >this reduces to the integral you want, and the integrand is even on the >real axis. Thus I = integral( -L^2/(L^2 + 1) *Cos(L), {L, 0, +oo}) >= 1/2 integral( -L^2/(L^2 + 1) *Cos(L), {L, -oo, +oo}) Consider the contour given by the real axis from z = -R to z = R, >and the semicircle {z = R*exp(iQ) : 0 < Q < pi}, enclosing the simple >pole at z=i. By Cauchyês formula, this is equal to 2*pi*i*(residue of f(z) at z=i) = 2*pi*i* lim{z->i} [(z - i)*f(z)] > = 2*pi*i/(2*e*i) > = pi/e. As R -> oo, we can bound the integral on the semicircle by |integral| <= pi*R*exp(-R) -> 0. Exactly how do you show the integral over the semicircle is bounded by this? >Thus, I = pi/(2*e). Same game for the second, but the residue is now -1/(2*i*e) >so the result is -pi/(2*e). > What the explanation of this? You resorted to the computer before exhausting analytic methods >and got the wrong answer. Um, you got the wrong answer by analytic methods, btw. The 'rst integral clearly does not converge. ************************ David C. Ullrich === Subject: Re: Something wrong with this integral (see text) I calculated two expressions: >>FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] >>and >>FullSimplify[Integrate[(-1)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] >>They gave the same result, but in the 'rst expression when we take >>L->In'nity, itês impossible to take integral, but MAthematica CAN do >>that. Iêm using the most current version of Mathematica. They do not give the >same result. The latter gives -Pi/(2*E), which I believe is correct. The >former gives +Pi/(2*E) together with the warning Unable to check >convergence. Ah - that actually makes perfect sense. Itês the sort of thing for which some contour integration or something would give an answer if we could show that part thatês supposed to vanish vanishes, but it doesnêt here... >> I canêt 'gure out what it means to take L->In'nity in >> Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, In'nity}]. >> Iêm guessing you just mean that >> Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, In'nity}] >> does not exist. >> No, it doesnêt. The explanation is that Mathematica doesnêt >> get everything right. I donêt have Mathematica running >> right now - ask it what the integral of cos(x) is, from 0 to >> in'nity and see what happens. Mathematica says that that integral does not converge. David Cantrell >> (If it thinks that integral >> is 0, as I suspect it may, that would explain why it >> thinks the two integrals above are the same...) ************************ David C. Ullrich === Subject: Re: Something wrong with this integral (see text) Hi All I calculated two expressions: FullSimplify[Integrate[(-L^2)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] and FullSimplify[Integrate[(-1)/(L^2 + 1)*Cos[L], {L,0, In'nity}]] They gave the same result, but in the 'rst expression when we take > L->In'nity, itês impossible to take integral, but MAthematica CAN do that. You can determine these analytically by using an appropriate contour on > the complex plane and the residue calculus: for example, for the 'rst, > consider the integral of f(z) = -z^2*exp(iz)/(z^2 + 1). On the real axis, > this reduces to the integral you want, Correction: .95...the real part of this reduces to the integrand you want...' -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will 'nd these people and we will bring them to justice. === Subject: Re: Is there a factorial (n)!/ sum (n) relationship? > I am mathematically challenged so please excuse my numerical methods > of explaning this problem. Is this solvable? Given four know values, 'nd two unknowns x and y where each x and y > are distinct integers. Just two examples are given below but the number of equations ----> oo > where each x and y are distinct integers for each new (n) Where (n) is odd then --- (x/990)/(y/946) = 42+(2/45) The two known values of k(n+1) = 990 = n+1 = 44 and k(n) = 946 = n = > 43 where k = 1+2+3+... + n = n(n+1)/2. The other two known values of > the quotient 42+2/45 is derived from n-1 = 42 and the fraction 2/45 is > derived from n+2 = the devisor (45) and where the dividend (2)/45 is > constant for all odd (n) where in this case odd (n) = 43 Where (n) is even then --- (x/1035)/(y/990) = 43+(1/23) Again, the two known values of k(n+1) = 1035 = n+1 = 45 and k(n) = 990 > = n = 44 > The other two known values of the quotient 43+1/23 is derived from 43 > = n-1 and the fraction 1/23 where (n+2)/2 = the devisor (23) and the > dividend (1)/23 is constant for all even (n) where in this case even > (n) = 44. Can x and y be found other than by brute force methods? Hint, x and y are always factorials of (n+1)! and (n)! respectively, > Therefore x! / y! must = n. > If x and y can be found other than by brute force, then two factorials > x and y can be found for each and every equation by a closed form > method using the closed form for summations as part of the solving > process. Below are a few more equations produced without knowing x or y. (x/1176)/(y/1128) = 46 + (2/49) Where k(n) = 1128 and (n) = 47. > k(n+1)= 1176 and (n+1)= 48. > Equation built on rules for n = odd. (x/630)/(y/595) = 33 + (1/18) Where k(n) = 595 and (n) = 34. > k(n+1)= 630 and (n) = 35. A correction for the line above. k(n+1)= 630 and (n+1)= 35 > Equation built on rules for n = even. > Dan === Subject: Re: Is there a factorial (n)!/ sum (n) relationship? Trying to clear up any confusion about my original post! Mainly I have created equations out of summations where x and y 't as factorials of n and n+1 As an example, where n+1 = 8. k(8) = 36 and n = 7. k(7) =28 (x/36)/(y/28) = 6+(2/9) Where k(n) is the summation of (n) then the closed form k(n) = n(n+1)/2. k(n+1) = 36 k(n) = 28 Where n = 7 thus (n) is odd the fractional portion of the quotient above will always have 2 as a dividend and the devisor of the fraction is n+2 = 9. The (6+) is just (n-1). So if you insert the factorials in the equation instead of x and y --- (8!/36)/(7!/28) = 6 +(2/9) Duplicating the above equation. Works! I 'nd this interesting because the factorials have nothing to do with building the equations. Thus a direct relationship with factorials and summations. Where summations have a closed form and the factorials do not. See the rules for even (n) below. Iêm just not sure if x and y can be found by a closed form method using these special equations with there odd (n) and even (n) rules. Sorry for any confusion! Dan > I am mathematically challenged so please excuse my numerical methods > of explaning this problem. Is this solvable? Given four know values, 'nd two unknowns x and y where each x and y > are distinct integers. Just two examples are given below but the number of equations ----> oo > where each x and y are distinct integers for each new (n) Where (n) is odd then --- (x/990)/(y/946) = 42+(2/45) The two known values of k(n+1) = 990 = n+1 = 44 and k(n) = 946 = n = > 43 where k = 1+2+3+... + n = n(n+1)/2. The other two known values of > the quotient 42+2/45 is derived from n-1 = 42 and the fraction 2/45 is > derived from n+2 = the devisor (45) and where the dividend (2)/45 is > constant for all odd (n) where in this case odd (n) = 43 Where (n) is even then --- (x/1035)/(y/990) = 43+(1/23) Again, the two known values of k(n+1) = 1035 = n+1 = 45 and k(n) = 990 > = n = 44 > The other two known values of the quotient 43+1/23 is derived from 43 > = n-1 and the fraction 1/23 where (n+2)/2 = the devisor (23) and the > dividend (1)/23 is constant for all even (n) where in this case even > (n) = 44. Can x and y be found other than by brute force methods? Hint, x and y are always factorials of (n+1)! and (n)! respectively, > Therefore x! / y! must = n. > If x and y can be found other than by brute force, then two factorials > x and y can be found for each and every equation by a closed form > method using the closed form for summations as part of the solving > process. Below are a few more equations produced without knowing x or y. (x/1176)/(y/1128) = 46 + (2/49) Where k(n) = 1128 and (n) = 47. > k(n+1)= 1176 and (n+1)= 48. > Equation built on rules for n = odd. (x/630)/(y/595) = 33 + (1/18) Where k(n) = 595 and (n) = 34. > k(n+1)= 630 and (n) = 35. A correction for the line above. > k(n+1)= 630 and (n+1)= 35 Equation built on rules for n = even. > Dan === Subject: Re: How to de'ne a function to be smooth? > ... >> I have seen smooth used for once continuously differentiable, twice >> continuously differentiable, C^infty and presumably anything in >> between, so I agree with Gottschalk on this. I had not seen it used >> for analytic before. In any case, it is not a term that should be >> used without a precise de'nition. >> Indeed, I have seen it used for something like the following function: >> f(x) = sqrt(x) when x >= 0 >> = -sqrt(-x) when x <= 0. >> Has certainly a pretty smooth appearance. There is no standard >> de'nition in analysis. >The graph of this f is certainly a smooth curve. Hmm, can we come up >with an analytic map g: (0,1) -> R^2 whose image is this curve, or is >there necessarily a discontinuity in the second derivative? You can certainly 'nd a smooth g that does the trick. For example, parametrize the right side of the curve by (t^2, t) then pre-compose with a smooth 1-1 function R -> R that has vanishing derivatives at 0. Treat the left side of the curve similarly. The two sides match up to in'nite order since they both vanish to in'nite order at t = 0. Any such g that is at least C^2 must be singular at t = 0, though. On the right half of the curve, x = y^2, so xê(0) = 2y(0)yê(0) = 0. If g is non-singular, yê(0) must therefore be non-zero. But then xêê(0) = 2(yê(0))^2 + 2y(0)yêê(0) = 2(yê(0))^2 > 0. A similar argument applied to the left half of the curve, where x = - y^2, shows that xêê(0) < 0: contradiction. To see that no analytic g does the trick, note that the nêth derivative of x(t) may be expressed in terms of the 'rst n derivatives of y(t) on the right side of the curve, using the fact that x = y^2 there. The same fact applied to the left half of the curve yields the same expressions for the derivatives of x(t), but with opposite sign, since x = - y^2 there. It follows that all derivatives of x(t) vanish at t = 0, and therefore x(t) cannot be analytic (unless it vanishes identically, which it musnêt). In fact, examination of the expressions for the derivatives of x in terms of t (which is based on Pascalês triangle) shows that x(t) and y(t) must vanish to in'nite order at t = 0. John Mitchell === Subject: Re: How to de'ne a function to be smooth? > ... >> I have seen smooth used for once continuously differentiable, twice >> continuously differentiable, C^infty and presumably anything in >> between, so I agree with Gottschalk on this. I had not seen it used >> for analytic before. In any case, it is not a term that should be >> used without a precise de'nition. >> Indeed, I have seen it used for something like the following function: >> f(x) = sqrt(x) when x >= 0 >> = -sqrt(-x) when x <= 0. >> Has certainly a pretty smooth appearance. There is no standard >> de'nition in analysis. >The graph of this f is certainly a smooth curve. Hmm, can we come up >with an analytic map g: (0,1) -> R^2 whose image is this curve, Donêt think so, havenêt thought about it. Probably more important is to point out that the smoothness of a curve is typically not measured by how smooth a map has the curve as its _image_. Thereês a reason for that: for example a square curve, like with four corners, would not typically be regarded as very smooth, but itês the _image_ of an in'nitely differentiable map from R to R^2. Not sure about analytic, but thereês certainly an in'nitely differentiable map that has this curve as its image. >or is >there necessarily a discontinuity in the second derivative? But because we donêt want a curve with sharp corners to count as in'nitely differentiable, _if_ we measure the smoothness of a curve by the smoothness of a map having the curve as its image we usually require that the derivative of the map be non-vanishing. If you require that then the second derivative must be discontinuous. ************************ David C. Ullrich === Subject: Re: How to de'ne a function to be smooth? ... > I have seen smooth used for once continuously differentiable, twice > continuously differentiable, C^infty and presumably anything in > between, ... Indeed, I have seen it used for something like the following function: > f(x) = sqrt(x) when x >= 0 > = -sqrt(-x) when x <= 0. > Has certainly a pretty smooth appearance. There is no standard > de'nition in analysis. >>The graph of this f is certainly a smooth curve. Hmm, can we come up >>with an analytic map g: (0,1) -> R^2 whose image is this curve, Donêt think so, havenêt thought about it. Probably more important >is to point out that the smoothness of a curve is typically not >measured by how smooth a map has the curve as its _image_. > ... >David C. Ullrich True, but a reasonable de'nition of smoothness for an implicitly de'ned curve is the maximal degree of smoothness of a *non-singular* parametrization of the curve [non-singular means the 'rst derivative is nowhere zero). John Mitchell === Subject: Re: {Group Theory} Confusing group theory conundrum algebra > questions. OK, every group has an identity element, right? True by de'nition It seems to me that, given a group, that group therefor determines a > unique identity element. (The uniqueness of a given groupês identity > element is easily proved) Therefor, it seems to me that there exists a function It may seem so; but, as you point out later, there is no such function. > (indeed, a > function which algebraists subtly draw upon without even realizing it) > that associates with each group itês identity element. I donêt agree with this statement. The algebraists donêt need to implicitly draw upon such a function. Other replies try to show that something similar (functor) can be used, but I donêt think such a functor is needed to do group theory. > For instance, an algebraist says: We have a group G. Then it must > have the identity element G_e. But by saying this, she has > implicitly used such a function. I donêt see why she has implicitly used such a function. First, if the group G is just an abstract group given in some theorem about groups, then the identity element is guaranteed by the de'nition of a group. Second, if the group G is some concrete group like the integers under addition, then the identity element 0 must be found and proved to have the required group identity property. This just depends upon propositional statements: it does not depend upon some prede'ned function mapping groups to their identity elements. In fact, you couldnêt even de'ne such a function unless you already had the identity element determined for the given group. In summary, I donêt with your initial premise. -- Bill Hale === Subject: Re: {Group Theory} Confusing group theory conundrum > For instance, an algebraist says: We have a group G. Then it must > have the identity element G_e. But by saying this, she has > implicitly used such a function. > I donêt see why she has implicitly used such a function. > Well, letês look at an analog. We have a number N. Then it must have the square N_s. Admittedly, this is rather peculiar terminology and notation, but I did that purposefully to show how we can thus throw a veil over the fact that we are, in this case, implicitly using the function f(x)=x^2. When we conjure up the identity of an abstract group, the situation is very similar, with the exception that there is no easy formula for the groupês identity like there is for a numberês square. I admit I am probably much less experienced than you. Please know I have much respect for your advanced mathematical knowledge and, if you still believe I am mistaken, I look forward to being corrected and set on the path toward increased understanding of these clandestine matters. Sniz Pilbor === Subject: Re: {Group Theory} Confusing group theory conundrum >> For instance, an algebraist says: We have a group G. Then it must >> have the identity element G_e. But by saying this, she has >> implicitly used such a function. >> I donêt see why she has implicitly used such a function. > Well, letês look at an analog. We have a number N. Then it must > have the square N_s. Admittedly, this is rather peculiar terminology > and notation, but I did that purposefully to show how we can thus > throw a veil over the fact that we are, in this case, implicitly using > the function f(x)=x^2. When we conjure up the identity of an abstract > group, the situation is very similar, with the exception that there is > no easy formula for the groupês identity like there is for a > numberês square. Having an identity is part of the de'nition of a group. Group identity is more basic than the concept of a function de'ned on groups. When we say We have here a person. This person must have had a mother., we are not making use of some abstract function de'ned on the set of all persons. People had mothers long before anyone thought up the concept of a function. Besides, as we have seen, there is no such thing as a function whose domain is all groups, because the domain of a function is necessarily a set, and there is no set large enough to contain all groups. We can use category theory and talk about functors, but the existence of a group identity does not in any way depend on category theory. Groups had identities as soon as the concept of a group was formalized, long before anyone thought up category theory. -- Dave Seaman Judge Yohnês mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: {Group Theory} Confusing group theory conundrum Now before you reply that the identity element isnêt unique because a > set can be a group under many diverse different operations, I already > am taking this into consideration. This is independent of your question, but itês still worth pointing > out that a group can have more than one identity element. For example, in the two-element group, both elements are identities. > In that case, there is an automorphism between them, so that they are > not interestingly different. No that last bit is not right. 1+1 = 0, not 1. 0 is the only identity in Z/2Z. In any event a group has exactly one identity, since if e and eê are both identities, e = eeê = eê. === Subject: Re: {Group Theory} Confusing group theory conundrum Or you can use Category Theory. What you actually have is something > which is technically called a Functor from the category of groups to > the category of sets. Categories do not have to be set-based, and the > collection of objects do not have to be sets. > How exactly do functors work? Is it possible to explain w/out having to have a lot of postgraduate level machinery available? How do these categories work? Iêve only ever seen functions de'ned in terms of sets (unless you count basic college algebra where their de'nition is in terms of handwaving), so when you say that that is not what functions are at all, my curiosity is pleasantly piqued and I desire very much to learn more from your esteemed self. === Subject: Re: {Group Theory} Confusing group theory conundrum Adjunct Assistant Professor at the University of Montana. >> Or you can use Category Theory. What you actually have is something >> which is technically called a Functor from the category of groups to >> the category of sets. Categories do not have to be set-based, and the >> collection of objects do not have to be sets. > >How exactly do functors work? A category consists of two things: Objects, and arrows; to each arrow, there are two objects assigned: the domain and the codomain. Think of them as maps between the objects. You can compose arrows a and b when the domain of a is equal to the codomain of b; in that case, you get a new arrow, called ab (think of them as functions being composed from right to left); composition is associative, and to every object O there is a special arrow, whose domain and codomain are both O, called identity_O. It has the property that whenever compositions are possible, a(identity_O)=a and (identity_O)b = b. A Functor between two categories C and D is a way to associate to every object of C an object of D, and to every arrow of C an arrow of D, with the property that the identity of O goes to the identity of whatever O goes to; and that the composition of two arrows goes to the composition of their counterparts. When the collection of objects is a set, and the collections of arrows between two speci'ed objects is a set, the category is said to be small. But in general they donêt have to be small. You have a category, whose objects are all groups, and whose arrows are group morphisms. You have another category, called pointed sets, whose objects are pairs (S,a), where S is a set and a is an element of S. And where an arrow from (S,a) to (T,b) is a set map from S to T that sends a to b. The forgetful functor from the category of groups to the category of sets is the corerspondence that forgets that you have a group and takes only its underlying set. Map each group G to the pointed set (G,e_G); map each group homomorphism to the set map between the underlying sets of the groups. >Iêve only ever seen functions de'ned in terms of >sets (unless you count basic college algebra where their de'nition is >in terms of handwaving), so when you say that that is not what >functions are at all, my curiosity is pleasantly piqued and I desire >very much to learn more from your esteemed self. Like I said, you can de'ne the function you want in terms of sets as follows: For each set S, whose elements are groups, de'ne a function f_S : S -> P(U S) (the power set of the union of S; that is, all sets whose elements are elements of some element of S), mapping G in S to {e_G}. For each set S, we have a function f_S. Now, suppose that there are two sets S and T, both of which have elements which are groups, and such that G is in both S and T. Then {e_G} is in both P(U S) and in P(U T). And the maps f_S and f_T have the same value in G. Thus, given any group G, pick ANY set S as above which contains G, and use the function f_S to 'nd the identity element of G. What you have is a collection of function f_S, which is a proper class (just as the collection of all groups is a proper class), and which does exactly what you want. = Itês not denial. Iêm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) = Arturo Magidin magidin@math.berkeley.edu === Subject: Re: {Group Theory} Confusing group theory conundrum > Therefor, it seems to me that there exists a function (indeed, a > function which algebraists subtly draw upon without even realizing it) > that associates with each group itês identity element. In terms of category theory, itês a functor from the category of groups to the category of pointed sets, sometimes called a forgetful functor. A pointed set is a set with a distinguished element. The functor is remembers only which particular element was the identity. -- Dave Seaman Judge Yohnês mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: {Group Theory} Confusing group theory conundrum Adjunct Assistant Professor at the University of Montana. > Now before you reply that the identity element isnêt unique because a >> set can be a group under many diverse different operations, I already >> am taking this into consideration. This is independent of your question, but itês still worth pointing >out that a group can have more than one identity element. Not if you are talking about the identity with respect to the multiplication; that is, an element e with the property that for all g in G, eg=ge=g. >For example, in the two-element group, both elements are identities. No, they are not. >In that case, there is an automorphism between them, so that they are >not interestingly different. No, there is no automorphism of the 2 element that group that switches the two elements. To see that, note that one of the elements satis'es x*x=x, while the other one does not. = Itês not denial. Iêm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) = Arturo Magidin magidin@math.berkeley.edu === Subject: Re: {Group Theory} Confusing group theory conundrum constitute permission for an emailed response. 26 4E BF 1A 92 TUESDAY WELD!! >This is independent of your question, but itês still worth pointing >out that a group can have more than one identity element. Not if you are talking about the identity with respect to the > multiplication; that is, an element e with the property that for all g > in G, eg=ge=g. Hrm, why is this? >For example, in the two-element group, both elements are identities. No, they are not. Whoops, embarassing mistake! Yes, you are certainly right. But then I wonder why there cannot be multpile elements ful'lling the identity axioms (both an identity and the existence of inverses). === Subject: Re: {Group Theory} Confusing group theory conundrum >>This is independent of your question, but itês still worth pointing >>out that a group can have more than one identity element. >> Not if you are talking about the identity with respect to the >> multiplication; that is, an element e with the property that for all g >> in G, eg=ge=g. > Hrm, why is this? >>For example, in the two-element group, both elements are identities. >> No, they are not. > Whoops, embarassing mistake! Yes, you are certainly right. > But then I wonder why there cannot be multpile elements ful'lling the > identity axioms (both an identity and the existence of inverses). Suppose e and eê are both identities for the same group (G,*). Then e = e*eê = eê, by the de'nition of an identity. -- Dave Seaman Judge Yohnês mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: {Group Theory} Confusing group theory conundrum Adjunct Assistant Professor at the University of Montana. >This is independent of your question, but itês still worth pointing >>out that a group can have more than one identity element. >> Not if you are talking about the identity with respect to the >> multiplication; that is, an element e with the property that for all g >> in G, eg=ge=g. Hrm, why is this? Suppose e and eê are two elements of G which satisfy the axiom. Since e satis'es the axiom, ex = x for all x; so e*eê = eê. Since eê satis'es the axiom, xeê = x for all x; so e*eê = e. Therefore, e=eê. >>For example, in the two-element group, both elements are identities. >> No, they are not. Whoops, embarassing mistake! Yes, you are certainly right. But then I wonder why there cannot be multpile elements ful'lling the >identity axioms (both an identity and the existence of inverses). Say that g is an element, and there exist elements x and y such that xg = gx = e and yg=gy = e. Then x = ex = (yg)x = y(gx) = ye = y. So x=y. = Itês not denial. Iêm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) = Arturo Magidin magidin@math.berkeley.edu === Subject: impossible integrals Iêve known for a long time that x^x dx and sin(x^2) dx are impossible to integrate in elementary terms, but I have no idea how to prove this. Are there any friendly papers or books which give the proof? And, just out of curiousity, are there any other nice-looking functions which are surprisingly impossible to integrate (in elementary terms)? How about if we consider double integrals -- f(x,y)dx dy ? -Tyler === Subject: Re: impossible integrals > Iêve known for a long time that x^x dx and sin(x^2) dx are > impossible to integrate in elementary terms, but I have > no idea how to prove this. Are there any friendly papers > or books which give the proof? Thereês a short text (in french) with proofs concerning this subject at http://perso.wanadoo.fr/denis.feldmann/PDF/liou.pdf Jose Carlos Santos === Subject: Re: impossible integrals <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h }YL`!h_XXr5Q>_nGsY2_ > Iêve known for a long time that x^x dx and sin(x^2) dx are > impossible to integrate in elementary terms, but I have > no idea how to prove this. Are there any friendly papers > or books which give the proof? And, just out of curiousity, are there any other > nice-looking functions which are surprisingly > impossible to integrate (in elementary terms)? > How about if we consider double integrals -- > f(x,y)dx dy ? > -Tyler *** Integration in Finite Terms *** This theory goes back to Liouville, 1835. Classic text on the subject: J. F. Ritt, _Integration in Finite Terms_ (Columbia Univ Pr, 1948) Introductory papers, aimed at undergraduates: A.D. Fitt & G.T.Q. Hoare, The closed-form integration of arbitrary functions. Mathematical Gazette (1993) 227--236. E. Marchisotto & G. Zakeri, An invitation to integration in 'nite terms. College Math. J. 25 (1994) 295--308. A modern text (omitting the algebraic case) M. Bronstein, _Symbolic Integration I: Transcendental Functions_ (Springer-Verlag 1997) Web site, no proofs... -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: JSH: One out of in'nity What Iêve done is 'nd a way to factor a polynomial in such a way that a constant factor, like 49, can be shown to factor out in a certain way based on a ring property. Thatês a big deal, and itês how I manage to separate one factorization out of an in'nity of others so that I 'nd the result (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 the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That factorization stands out from in'nity because in rings like the ring of algebraic integers 7 is not a factor of 22, which gives one possibility, so Iêve found a factorization based on a property of only certain rings. So I use one polynomial to analyze the roots of another polynomial, the argument is straightforward, but people keep 'ghting it. Why? Well lots of times when you do something that no one else has managed to do, the people who didnêt manage it feel jealous or even scared, so they tend to react negatively so that their world doesnêt change. Itês a human reaction to be afraid of change. However, for those of you who are not jealous who are brave enough to accept a new discovery with big implications, the proof is rather straightforward. And here it is. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers, notice that P(x) has a constant term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic de'ning the aês at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Iêve picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is veri'ed by again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant term is 7, which is NOT a factor of 22. Therefore, *none* of the constant terms of P(x)/49 as they multiply to give 22 can have 7 as a factor. (By saying that 7 is NOT a factor of 22, Iêm making a choice as to where the proof is going. Since Iêve been talking about algebraic integers, where 7 is NOT a factor of 22, itês natural to go with a choice where 7 is NOT a factor of 22.) Given that the constant terms are independent of xês value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. 7. But to divide 7 from those constant terms requires dividing through two of the factors, so (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 from reverse use of the distributive property, which gives constant terms that donêt have 7 as a factor, as required. Notice that itês a rather short and direct argument, where if you accept that 22 does not have 7 as a factor, itês obvious enough what the constant terms of the factors must be as you go from 7, 7 and 22, necessarily to 1, 1, and 22, when you divide P(x) by 49. James Harris http://mathforpro't.blogspot.com/ === Subject: Re: JSH: One out of in'nity fuffy > What Iêve done is 'nd a way to factor a polynomial in such a way that > a constant factor, like 49, can be shown to factor out in a certain > way based on a ring property. Thatês a big deal, and itês how I manage to separate one factorization > out of an in'nity of others so that I 'nd the result (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 the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That factorization stands out from in'nity because in rings like the > ring of algebraic integers 7 is not a factor of 22, which gives one > possibility, so Iêve found a factorization based on a property of only > certain rings. So I use one polynomial to analyze the roots of another polynomial, > the argument is straightforward, but people keep 'ghting it. Why? Well lots of times when you do something that no one else has managed > to do, the people who didnêt manage it feel jealous or even scared, so > they tend to react negatively so that their world doesnêt change. Itês a human reaction to be afraid of change. However, for those of you who are not jealous who are brave enough to > accept a new discovery with big implications, the proof is rather > straightforward. And here it is. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic de'ning the aês at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Iêve picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is veri'ed by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant term is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. (By saying that 7 is NOT a factor of 22, Iêm making a choice as to > where the proof is going. Since Iêve been talking about algebraic > integers, where 7 is NOT a factor of 22, itês natural to go with a > choice where 7 is NOT a factor of 22.) Given that the constant terms are independent of xês value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so (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 from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. Notice that itês a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, itês obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. > James Harris > http://mathforpro't.blogspot.com/ === Subject: Re: JSH: One out of in'nity > fuffy What Iêve done is 'nd a way to factor a polynomial in such a way that > a constant factor, like 49, can be shown to factor out in a certain > way based on a ring property. Thatês a big deal, and itês how I manage to separate one factorization > out of an in'nity of others so that I 'nd the result (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 the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). That factorization stands out from in'nity because in rings like the > ring of algebraic integers 7 is not a factor of 22, which gives one > possibility, so Iêve found a factorization based on a property of only > certain rings. So I use one polynomial to analyze the roots of another polynomial, > the argument is straightforward, but people keep 'ghting it. Why? Well lots of times when you do something that no one else has managed > to do, the people who didnêt manage it feel jealous or even scared, so > they tend to react negatively so that their world doesnêt change. Itês a human reaction to be afraid of change. However, for those of you who are not jealous who are brave enough to > accept a new discovery with big implications, the proof is rather > straightforward. And here it is. 1. Let P(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is > in the ring of algebraic integers, notice that P(x) has a constant > term that is 1078. 2. It can be shown that P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 where the *same* polynomial has been put in a form which allows a > factorization into non-polynomial factors so that I have P(x) = (5 a_1(x) + 7)(5 a_2(x)+ 7)(5 a_3(x) + 7) where the aês are roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). 3. Now let x=0, so P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic de'ning the aês at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and Iêve picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. 4. Further let a_3(x) = b_3(x) + 3, to keep indices matched. Then I > have P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 b_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22). 5. Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 b_3(x) + 22) has a factor of 49. 6. However, the constant term of P(x)/49 is 22, which is veri'ed by > again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant term is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. (By saying that 7 is NOT a factor of 22, Iêm making a choice as to > where the proof is going. Since Iêve been talking about algebraic > integers, where 7 is NOT a factor of 22, itês natural to go with a > choice where 7 is NOT a factor of 22.) Given that the constant terms are independent of xês value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so (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 from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. Notice that itês a rather short and direct argument, where if you > accept that 22 does not have 7 as a factor, itês obvious enough what > the constant terms of the factors must be as you go from 7, 7 and 22, > necessarily to 1, 1, and 22, when you divide P(x) by 49. Any argument worth its salt applies to more than just one situation. Consider Q(x) = 7^2*(125*x^3 - 15*x + 22). This has the essential properties of the JSH polynomial: namely, the constant term Q(0) = 7*7*22 = 1078, and it can be factored in the form (5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22) where b_1, b_2, and c_3 are functions of x and are algebraic integers. Note that when x = 0, b_1 = 0, b_2 = 0, and c_3 = 0. Thus when x = 0, b_1 and b_2 are multiples of 7 (i.e., 7*0 in each case). Now according to the JSH argument, b_1 and b_2 should be divisible by 7 when x <> 0 also. So letês see what happens when x = 1. Note that Q(1) = 7^2*(125 - 15 + 22) = 7^2*132. This can be factored as Q(1) = r * s * t, where r = 5*7^{2/3}*w + 7 s = 5*7^{2/3}*w^2 + 7, and t = 5*(7^{2/3} - 3) + 22, where w is a unit in the algebraic integers, w = (-1 + sqrt(-3))/2. Therefore Q(1) is of the required form, Q(1) = (5*b_1 + 7)*(5*b_2 + 7)*(5*c_3 + 22), where b_1 = 7^{2/3}*w, b_2 = 7*{2/3}*w^2, and c_3 = 7^{2/3} - 3, all of which are algebraic numbers. Note that b_1 and b_2 are *not* divisible by 7. Neither are they *coprime* to 7; each has the factor 7^{2/3} in common with 7. Dividing 7^{2/3} out of each factor yields: Q(1)/49 = (5*w + 7^{1/3})*(5*w^2 + 7^{1/3})*(5*(1 - 3/7^{2/3}) + 22/7^{2/3}) = (5*w + 7^{1/3})*(5*w^2 + 7^{1/3})*(5 + 7^{1/3}). It is a matter of arithmetic to check that the right-hand side equals 132 = Q(1)/49, as it should. It is worth noting also that the product of the three constant terms in the expression just above is 7^{1/3} * 7^{1/3} * (22/7^{2/3}) = 22. Note again, this is the factorization when x = 1. The coef'cients b_1, b_2, and c_3 are all functions of x; the values given above are actually b_1(1), b_2(1), and c_3(1). When x = 0, b_1(0) = 0, b_2(0) = 0, and c_3(0) = 0. For values of x other than 1 or 0, these functions are dif'cult to compute. Note 'nally that the factorization of Q(x) is *not* of the form claimed by JSH, i.e., it is *not* of the form Q(1) = (5*a_1 + 7)*(5*a_2 + 7)*(5*b_3 + 22), where a_1 and a_2 are divisible by 7. Yet all the arithmetic checks out; the constant terms as required are 7, 7, and 22. Where does the JSH logic break down? Nora B. > James Harris > http://mathforpro't.blogspot.com/ === Subject: Re: JSH: One out of in'nity > What Iêve done is 'nd a way to factor a polynomial in such a way that > a constant factor, like 49, can be shown to factor out in a certain > way based on a ring property. through two of the factors, so > (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 > from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. w1(x) = gcd(5 a1(x) + 7, 49) > w2(x) = gcd(5 a2(x) + 7, 49) > w3(x) = gcd(5 b3(x) + 22, 49) > k(x) = w1(x).w2(x).w3(x)/49. > These three are easily shown to be algebraic integers for all x. > We factor as: > [k(x).(5 a1(x)+7)/w1(x)] * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) = > 300125 x^3 - 18375 x^2 - 360(x) + 22. Notice that Dik Winterês argument would apply for functions of y, or m, or g, or h as well. That is, where heês *picked* x for his wês and his variable k, you could just as easily pick, oh, u or q, or y, so youêd have w_1(y), as his problem is that 49 doesnêt care. That is, you could just as easily have k(b) = w_1(b) w_2(b) w_3(b)/49 because 49 is just a number, so it doesnêt attach itself to a particular variable, no matter how desperately Dik Winter might wish it did. You see, Dik Winter is trying to dodge around 7 being a constant, but his problem is that if heês right the same thing would apply for 49 wherever it is. Like you think you just have 7(7) = 49, but according to Dik Winterês argument, you really have these functions of x. Heês trying to limit the number 49 to functions that he picks which is a fascinatingly human thing to do. Itês actually rather funny because it is so silly. As you see, it just so happens that to go from 7 to 1 you need to divide by 7. James Harris === Subject: Re: JSH: One out of in'nity ... > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (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 > from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. w1(x) = gcd(5 a1(x) + 7, 49) > w2(x) = gcd(5 a2(x) + 7, 49) > w3(x) = gcd(5 b3(x) + 22, 49) > k(x) = w1(x).w2(x).w3(x)/49. > These three are easily shown to be algebraic integers for all x. > We factor as: > [k(x).(5 a1(x)+7)/w1(x)] * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) = > 300125 x^3 - 18375 x^2 - 360(x) + 22. Notice that Dik Winterês argument would apply for functions of y, or > m, or g, or h as well. Yes, of course. > That is, where heês *picked* x for his wês and his variable k, you > could just as easily pick, oh, u or q, or y, so youêd have w_1(y), as > his problem is that 49 doesnêt care. Indeed. I picked .95xê because you picked .95xê. See those de'nitions of the wês with a1(x), a2(x) and b3(x) in it? > That is, you could just as easily have k(b) = w_1(b) w_2(b) w_3(b)/49 because 49 is just a number, so it doesnêt attach itself to a > particular variable, no matter how desperately Dik Winter might wish > it did. O, no. I picked .95xê because you appeared to be partial to it, using a1(x), a2(x) and b3(x). If you had used a1(q), a2(q) and b3(q), I would have picked .95qê. > You see, Dik Winter is trying to dodge around 7 being a constant, but > his problem is that if heês right the same thing would apply for 49 > wherever it is. Like you think you just have 7(7) = 49, but according to Dik Winterês > argument, you really have these functions of x. But of course (7^{2/3})(7^{4/3) = 49 is also a valid decomposition of 49 in the algebraic integers? > Heês trying to limit the number 49 to functions that he picks which is > a fascinatingly human thing to do. Itês actually rather funny because > it is so silly. As you see, it just so happens that to go from 7 to 1 you need to > divide by 7. I do not contest that, and I go from 7 to 1, dividing by 7. Pray answer the following quesions: 1. I de'ne: w1(x) = gcd(5 a1(x) + 7, 49) w2(x) = gcd(5 a2(x) + 7, 49) w3(x) = gcd(5 b3(x) + 22, 49) k(x) = w1(x).w2(x).w3(x)/49. Are these algebraic integer for all integers x or not? (Note that a1(x), a2(x) and b3(x) are algebraic integers for all integers x.) (k(x) being an algebraic integer may give you problems, I will prove that at the end.) 2. Is w1(x).w2(x).w3(x)/k(x) = 49 for all integers x or not? 3. Is k(x).(5 a1(x) + 7)/w1(x) an algebraic integer for all integers x or not? 4. Is (5 a2(x) + 7)/w2(x) an algebraic integer for all integers x or not? 5. Is (5 b3(x) + 22)/w3(x) an algebraic integer for all integers x or not? 6. Does the product k(x).(5 a1(x) + 7)/w1(x) * (5 a2(x) + 7)/w2(x) * (5 b3(x) + 22)/w3(x) equal 300125 x^3 - 18375 x^2 - 360(x) + 22. for all integers x or not? 7. So is it a valid factorisation in the algebraic integers, or not? If you state not, please state why. ---- Showing that k(x) is an algebraic integer for all integers x is simple. It uses the fact that gcb(a*b, c) | gcd(a, c) * gcd(b, c), where the vertical bar means that the left part is a divisor of the right part in the ring involved. (This supposes, of course, that there can be given a sensible de'nition of gcd, which is not necessarily true in a ring, but is true in the integers and in the algebraic integers.) Some preliminaries. De'ne: f1(x) = 5 a1(x) + 7, f2(x) = 5 a2(x) + 7 and f3(x) = 5 a3(x) + 22 to get rid of lengthy expressions. Also we have P(x) = f1(x).f2(x).f3(x), where P(x) is divisible by 49. Ok, preliminaries aside: 49 = gcd(P(x), 49) = gcd(f1(x).f2(x).f3(x), 49) | gcd(f1(x), 49).gcd(f2(x), 49).gcd(f3(x), 49) = w1(x).w2(x).w3(x). So k(x) = w1(x).w2(x).w3(x)/49 is an algebraic integer for all integers x. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: One out of in'nity > ... > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (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 > from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. > w1(x) = gcd(5 a1(x) + 7, 49) > w2(x) = gcd(5 a2(x) + 7, 49) > w3(x) = gcd(5 b3(x) + 22, 49) > k(x) = w1(x).w2(x).w3(x)/49. > These three are easily shown to be algebraic integers for all x. > We factor as: > [k(x).(5 a1(x)+7)/w1(x)] * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) = > 300125 x^3 - 18375 x^2 - 360(x) + 22. > Notice that Dik Winterês argument would apply for functions of y, or > m, or g, or h as well. Yes, of course. > That is, where heês *picked* x for his wês and his variable k, you > could just as easily pick, oh, u or q, or y, so youêd have w_1(y), as > his problem is that 49 doesnêt care. Indeed. I picked .95xê because you picked .95xê. See those de'nitions > of the wês with a1(x), a2(x) and b3(x) in it? Yup. > That is, you could just as easily have > k(b) = w_1(b) w_2(b) w_3(b)/49 > because 49 is just a number, so it doesnêt attach itself to a > particular variable, no matter how desperately Dik Winter might wish > it did. O, no. I picked .95xê because you appeared to be partial to it, using > a1(x), a2(x) and b3(x). If you had used a1(q), a2(q) and b3(q), I > would have picked .95qê. Interesting reply. > You see, Dik Winter is trying to dodge around 7 being a constant, but > his problem is that if heês right the same thing would apply for 49 > wherever it is. > Like you think you just have 7(7) = 49, but according to Dik Winterês > argument, you really have these functions of x. But of course (7^{2/3})(7^{4/3) = 49 is also a valid decomposition of > 49 in the algebraic integers? Yes it is. > Heês trying to limit the number 49 to functions that he picks which is > a fascinatingly human thing to do. Itês actually rather funny because > it is so silly. > As you see, it just so happens that to go from 7 to 1 you need to > divide by 7. I do not contest that, and I go from 7 to 1, dividing by 7. The constant term of the factor 5a_1(x) + 7 is 7, and dividing P(x) by 49 means that factor 5a_1(x) + 7 is divided by some value where the result is a factor which has a contant term of 1, do you agree or disagree? > Pray answer the following quesions: Ok. > 1. I de'ne: > w1(x) = gcd(5 a1(x) + 7, 49) Ok, w_1(x) = 7, which works. > w2(x) = gcd(5 a2(x) + 7, 49) Ok, w_2(x) = 7, which works. > w3(x) = gcd(5 b3(x) + 22, 49) Ok, w_3(x) = 1, which works. > k(x) = w1(x).w2(x).w3(x)/49. k(x) = 1 > Are these algebraic integer for all integers x or not? (Note that Yup. > a1(x), a2(x) and b3(x) are algebraic integers for all integers x.) Yup. > (k(x) being an algebraic integer may give you problems, I will prove > that at the end.) No it doesnêt give me problems. > 2. Is > w1(x).w2(x).w3(x)/k(x) = 49 > for all integers x or not? Yes. > 3. Is > k(x).(5 a1(x) + 7)/w1(x) > an algebraic integer for all integers x or not? No. > 4. Is > (5 a2(x) + 7)/w2(x) > an algebraic integer for all integers x or not? No. > 5. Is > (5 b3(x) + 22)/w3(x) > an algebraic integer for all integers x or not? Yes. > 6. Does the product > k(x).(5 a1(x) + 7)/w1(x) * (5 a2(x) + 7)/w2(x) * (5 b3(x) + 22)/w3(x) > equal > 300125 x^3 - 18375 x^2 - 360(x) + 22. > for all integers x or not? Yes. Readers should notice how annoying posters like Dik Winter are as they make really long posts. Iêve at times simply deleted off the extra, as I see it as an annoying tactic. > 7. So is it a valid factorisation in the algebraic integers, or not? No. > If you state not, please state why. The ring of algebraic integers is too small, so there are algebraic integer values for x where a_1(x)/7 and a_2(x)/7 are not algebraic integers. > ---- > Showing that k(x) is an algebraic integer for all integers x is simple. Yes it is. > It uses the fact that gcb(a*b, c) | gcd(a, c) * gcd(b, c), where the > vertical bar means that the left part is a divisor of the right part > in the ring involved. (This supposes, of course, that there can be > given a sensible de'nition of gcd, which is not necessarily true in > a ring, but is true in the integers and in the algebraic integers.) > Some preliminaries. De'ne: > f1(x) = 5 a1(x) + 7, > f2(x) = 5 a2(x) + 7 and > f3(x) = 5 a3(x) + 22 > to get rid of lengthy expressions. Also we have > P(x) = f1(x).f2(x).f3(x), > where P(x) is divisible by 49. Ok, preliminaries aside: > 49 = gcd(P(x), 49) = gcd(f1(x).f2(x).f3(x), 49) | > gcd(f1(x), 49).gcd(f2(x), 49).gcd(f3(x), 49) = w1(x).w2(x).w3(x). > So > k(x) = w1(x).w2(x).w3(x)/49 > is an algebraic integer for all integers x. Notice the effort by the crank poster Dik Winter to avoid the *simple* reality. 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. 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 in the ring, the former factorization is the *only* allowed way for 49 to divide through. James Harris === Subject: Re: JSH: One out of in'nity > ... > 7. But to divide 7 from those constant terms requires dividing > through two of the factors, so > (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 > from reverse use of the distributive property, which gives constant > terms that donêt have 7 as a factor, as required. w1(x) = gcd(5 a1(x) + 7, 49) > w2(x) = gcd(5 a2(x) + 7, 49) > w3(x) = gcd(5 b3(x) + 22, 49) > k(x) = w1(x).w2(x).w3(x)/49. > These three are easily shown to be algebraic integers for all x. > We factor as: > [k(x).(5 a1(x)+7)/w1(x)] * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) = > 300125 x^3 - 18375 x^2 - 360(x) + 22. Notice that Dik Winterês argument would apply for functions of y, or > m, or g, or h as well. Yes, of course. That is, where heês *picked* x for his wês and his variable k, you > could just as easily pick, oh, u or q, or y, so youêd have w_1(y), as > his problem is that 49 doesnêt care. Indeed. I picked .95xê because you picked .95xê. See those de'nitions > of the wês with a1(x), a2(x) and b3(x) in it? Yup. That is, you could just as easily have k(b) = w_1(b) w_2(b) w_3(b)/49 because 49 is just a number, so it doesnêt attach itself to a > particular variable, no matter how desperately Dik Winter might wish > it did. O, no. I picked .95xê because you appeared to be partial to it, using > a1(x), a2(x) and b3(x). If you had used a1(q), a2(q) and b3(q), I > would have picked .95qê. Interesting reply. You see, Dik Winter is trying to dodge around 7 being a constant, but > his problem is that if heês right the same thing would apply for 49 > wherever it is. Like you think you just have 7(7) = 49, but according to Dik Winterês > argument, you really have these functions of x. But of course (7^{2/3})(7^{4/3) = 49 is also a valid decomposition of > 49 in the algebraic integers? Yes it is. Heês trying to limit the number 49 to functions that he picks which is > a fascinatingly human thing to do. Itês actually rather funny because > it is so silly. As you see, it just so happens that to go from 7 to 1 you need to > divide by 7. I do not contest that, and I go from 7 to 1, dividing by 7. The constant term of the factor 5a_1(x) + 7 is 7, and dividing P(x) by > 49 means that factor 5a_1(x) + 7 is divided by some value where the > result is a factor which has a contant term of 1, do you agree or > disagree? Pray answer the following quesions: Ok. 1. I de'ne: > w1(x) = gcd(5 a1(x) + 7, 49) Ok, w_1(x) = 7, which works. w2(x) = gcd(5 a2(x) + 7, 49) Ok, w_2(x) = 7, which works. w3(x) = gcd(5 b3(x) + 22, 49) Ok, w_3(x) = 1, which works. k(x) = w1(x).w2(x).w3(x)/49. k(x) = 1 Are these algebraic integer for all integers x or not? (Note that Yup. a1(x), a2(x) and b3(x) are algebraic integers for all integers x.) Yup. (k(x) being an algebraic integer may give you problems, I will prove > that at the end.) No it doesnêt give me problems. 2. Is > w1(x).w2(x).w3(x)/k(x) = 49 > for all integers x or not? Yes. 3. Is > k(x).(5 a1(x) + 7)/w1(x) > an algebraic integer for all integers x or not? No. 4. Is > (5 a2(x) + 7)/w2(x) > an algebraic integer for all integers x or not? No. 5. Is > (5 b3(x) + 22)/w3(x) > an algebraic integer for all integers x or not? Yes. 6. Does the product > k(x).(5 a1(x) + 7)/w1(x) * (5 a2(x) + 7)/w2(x) * (5 b3(x) + 22)/w3(x) > equal > 300125 x^3 - 18375 x^2 - 360(x) + 22. > for all integers x or not? Yes. Readers should notice how annoying posters like Dik Winter are as they > make really long posts. Iêve at times simply deleted off the extra, > as I see it as an annoying tactic. YOU SHOULD BE GREATFUL HEêS ASKING THESE QUESTIONS. That should mean to you that heês genuinely interested. And you wonder why posters attack you on here? You are a crank, James. 7. So is it a valid factorisation in the algebraic integers, or not? No. If you state not, please state why. The ring of algebraic integers is too small, so there are algebraic > integer values for x where a_1(x)/7 and a_2(x)/7 are not algebraic > integers. ---- > Showing that k(x) is an algebraic integer for all integers x is simple. Yes it is. It uses the fact that gcb(a*b, c) | gcd(a, c) * gcd(b, c), where the > vertical bar means that the left part is a divisor of the right part > in the ring involved. (This supposes, of course, that there can be > given a sensible de'nition of gcd, which is not necessarily true in > a ring, but is true in the integers and in the algebraic integers.) > Some preliminaries. De'ne: > f1(x) = 5 a1(x) + 7, > f2(x) = 5 a2(x) + 7 and > f3(x) = 5 a3(x) + 22 > to get rid of lengthy expressions. Also we have > P(x) = f1(x).f2(x).f3(x), > where P(x) is divisible by 49. Ok, preliminaries aside: > 49 = gcd(P(x), 49) = gcd(f1(x).f2(x).f3(x), 49) | > gcd(f1(x), 49).gcd(f2(x), 49).gcd(f3(x), 49) = w1(x).w2(x).w3(x). > So > k(x) = w1(x).w2(x).w3(x)/49 > is an algebraic integer for all integers x. Notice the effort by the crank poster Dik Winter to avoid the *simple* > reality. 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. 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 in the ring, the former > factorization is the *only* allowed way for 49 to divide through. > James Harris David Moran === Subject: Re: JSH: One out of in'nity Nntp-Posting-Host: apps.cwi.nl ... > Pray answer the following quesions: Ok. 1. I de'ne: > 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. 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. Do you understand how the gcd function is de'ned? It appears not. A simpli'ed (for you) de'nition here. Given two algebraic integers p and q. A common divisor of p and q is an algebraic integer r, such that there exist algebraic integers s and t with p = s.r and q = t.r. r is a greatest common divisor if s and t are co-prime. w2(x) = gcd(5 a2(x) + 7, 49) Ok, w_2(x) = 7, which works. Nope, w2(0) = 7 is the only thing we know at this point. w3(x) = gcd(5 b3(x) + 22, 49) Ok, w_3(x) = 1, which works. No, w3(0) = 1 is the only thing we know at this point. k(x) = w1(x).w2(x).w3(x)/49. k(x) = 1 No, k(0) = 1 is the only thing we know at this point. > Are these algebraic integer for all integers x or not? (Note that Yup. a1(x), a2(x) and b3(x) are algebraic integers for all integers x.) Yup. (k(x) being an algebraic integer may give you problems, I will prove > that at the end.) No it doesnêt give me problems. 2. Is > w1(x).w2(x).w3(x)/k(x) = 49 > for all integers x or not? Yes. 3. Is > k(x).(5 a1(x) + 7)/w1(x) > an algebraic integer for all integers x or not? No. Why not? w1(x) is de'ned as gcd(5 a1(x) + 7, 49), so it is *by de'nition of the greatest common divisor* a divisor of (5 a1(x) + 7), and so the quotient is an algebraic integer. k(x) is an algebraic integer, and so the product is an algebraic integer. Your no is similar to saying that in the integers: f(x)/gcd(f(x),49) is not necessarily an integer for all x. > 4. Is > (5 a2(x) + 7)/w2(x) > an algebraic integer for all integers x or not? No. Same comment applies. 5. Is > (5 b3(x) + 22)/w3(x) > an algebraic integer for all integers x or not? Yes. 6. Does the product > k(x).(5 a1(x) + 7)/w1(x) * (5 a2(x) + 7)/w2(x) * (5 b3(x) + 22)/w3(x) > equal > 300125 x^3 - 18375 x^2 - 360(x) + 22. > for all integers x or not? Yes. Readers should notice how annoying posters like Dik Winter are as they > make really long posts. Iêve at times simply deleted off the extra, > as I see it as an annoying tactic. You are also prone to make long posts... 7. So is it a valid factorisation in the algebraic integers, or not? No. If you state not, please state why. The ring of algebraic integers is too small, so there are algebraic > integer values for x where a_1(x)/7 and a_2(x)/7 are not algebraic > integers. You are saying something different. You say that the greatest common divisor of two numbers is not necessarily a divisor of one of the numbers. ---- > Showing that k(x) is an algebraic integer for all integers x is simple. Yes it is. It uses the fact that gcb(a*b, c) | gcd(a, c) * gcd(b, c), where the > vertical bar means that the left part is a divisor of the right part > in the ring involved. (This supposes, of course, that there can be > given a sensible de'nition of gcd, which is not necessarily true in > a ring, but is true in the integers and in the algebraic integers.) > Some preliminaries. De'ne: > f1(x) = 5 a1(x) + 7, > f2(x) = 5 a2(x) + 7 and > f3(x) = 5 a3(x) + 22 > to get rid of lengthy expressions. Also we have > P(x) = f1(x).f2(x).f3(x), > where P(x) is divisible by 49. Ok, preliminaries aside: > 49 = gcd(P(x), 49) = gcd(f1(x).f2(x).f3(x), 49) | > gcd(f1(x), 49).gcd(f2(x), 49).gcd(f3(x), 49) = w1(x).w2(x).w3(x). > So > k(x) = w1(x).w2(x).w3(x)/49 > is an algebraic integer for all integers x. Notice the effort by the crank poster Dik Winter to avoid the *simple* > reality. 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. 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 in the ring, the former > factorization is the *only* allowed way for 49 to divide through. Getting repetetive James? Se my factorisation above, which also works. In the factorisation [k(x)(5 a1(x)+7)]/w1(x) * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) the constant terms are: k(0)(5 a1(0) + 7)/w1(0) = 1 (5 * 0 + 7)/7 = 1, (5 a2(0) + 7)/w2(0) = (5 * 0 + 7)/7 = 1, and (5 b3(0) + 22)/w3(0) = (5 * 0 + 22)/1 = 22. See how they match? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: JSH: One out of in'nity let me guess that the call & response method, taht seems to be practiced by oldtime users of Usenet-cum- the-googolplex, will never work with Jimmy Dean Harrass; why, I canêt say ... if I knew, I probably *should* not say, any more than he can .95give an actual Object from the Ring of Should.ê > Do you understand how the gcd function is de'ned? It appears not. > In the factorisation > [k(x)(5 a1(x)+7)]/w1(x) * (5 a2(x)+7)/w2(x) * (5 b3(x)+22)/w3(x) > the constant terms are: > k(0)(5 a1(0) + 7)/w1(0) = 1 (5 * 0 + 7)/7 = 1, > (5 a2(0) + 7)/w2(0) = (5 * 0 + 7)/7 = 1, and > (5 b3(0) + 22)/w3(0) = (5 * 0 + 22)/1 = 22. --ils duces dêEnron! http://larouchepub.com/radio/index.html === Subject: Re: JSH: One out of in'nity > ... > Pray answer the following quesions: > Ok. > 1. I de'ne: > 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. Thatês 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. Thatês because Iêve 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 doesnêt. Understand Dik Winter? You do understand factors, right? So you know where I got (7/w_1(x)) (7/w_2(x)) (22/w_3(x)) = 22 from, right? Now adding in more variables, like sticking in wês may look like a good way to hide the truth Dik Winter, but itês childish. Besides, Iêm bored now that Iêve shown that math society is so stupid as to try and 'ght such a basic result. Now the only question is how to start yanking funds so that they can be redirected to real research. James Harris === Subject: Re: JSH: One out of in'nity > As you see, it just so happens that to go from 7 to 1 you need to > divide by 7. ...OR if you just subtract 6. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Array Recursion Puzzle We have a triangular array of integers {a(m,n)}. a(1,1) = 0; And, for 1 <= n <= m, m >= 2, recursively: a(m,n) = m-1 --- --- 1 + > ( > mu(j) ) a(m-1,k) / / --- --- k=1 j|k j>= kn/m Ascii-mode: a(m,n) = 1 + sum{k=1 to m-1} (sum{j|k, j >= kn/m} mu(j)) a(m-1,k) The inner-sum is over the divisors, j, of k, where each j is >= kn/m. And mu() is the Mobius (Moebius) function. Now, the arrayês elements can be easily described with a closed-form (ie. non-recursive) de'nition. What is this closed-form for {a(m,n)}? I will 'rst give a clue in a few days, then the answer a few days after that, if no one posts an answer before I do this. Leroy Quet === Subject: Re: Smooth non-analytic manifold > In <1068219052.224874@ca'rewall.fc.up.pt> schrieb Jos.8e Carlos Santos: Could someone please describe me or tell me where can I 'nd an > example of a smooth but non-analytic manifold? Nowhere! > The topic has been discussed in this newsgroup before. - > I understand that by one of Whitneyês theorems > every (e.g. closed) smooth manifold admits a unique analytic structure. Here is a short thread on the subject: 40lyra.csx.cam.ac.uk question, but found nothing. Jose Carlos Santos === Subject: Re: A 3rd Grade Word Problem---HELP The answer is, of course, that the question is incorrectly posed. Day #of new people ----------------------- Sunday 1 Monday 2 Tuesday 4 Wednesday 8 Thursday 16 Friday 32 Sunday 64 ------------------------ Sum to 127 Thus, the problem should have stated Sunday, not Thursday. I also worked the problem based on the assumption that each of the people you told the story to told it to more people in a similar progression, but the sum grew past 127 on the Sund-Thursday range. Mathematically, the only remotely intersting thing is the common observation that: $$ sum_{i=0}^{n-1} 2^i = 2^n -1 $$ -Daniel Simeone > My 3rd grade son brought this word problem home the other day and he was > given the answer (127). His job, for extra credit, was to 'gure out how to > get 127. Iêm no genius but not a dope either. I couldnêt 'gure out how to > get 127. Nobody in the whole neighborhood could 'gure out how to get 127. > Is this something of a trick question or is there something in the wording > that I am missing? Any Help???? > The Problem: You know a very good story. On Sunday you tell the story to a friend. On > Monday you tell it to two new people. (So far, a total of three people have > heard the story). Each day after Monday, you double the number of new > people you tell the story to. What will be the total number of people that > will have heard your story after you tell it on Thursday? === Subject: Re: A 3rd Grade Word Problem---HELP === >Subject: Re: A 3rd Grade Word Problem---HELP >Message-id: The answer is, of course, that the question is incorrectly posed. The answer is, of course, that the answer is incorrectly posed. Day #of new people >----------------------- >Sunday 1 >Monday 2 >Tuesday 4 >Wednesday 8 >Thursday 16 >Friday 32 Wormhole! >Sunday 64 >------------------------ >Sum to 127 Thus, the problem should have stated Sunday, not Thursday. Thus, the answer should have stated Saturday, not Sunday. I also worked the problem based on the assumption that each of the >people you told the story to told it to more people in a similar >progression, but the sum grew past 127 on the Sund-Thursday range. Mathematically, the only remotely intersting thing is the common >observation that: $$ sum_{i=0}^{n-1} 2^i = 2^n -1 $$ >-Daniel Simeone >> My 3rd grade son brought this word problem home the other day and he was >> given the answer (127). His job, for extra credit, was to 'gure out how >to >> get 127. Iêm no genius but not a dope either. I couldnêt 'gure out how to >> get 127. Nobody in the whole neighborhood could 'gure out how to get 127. >> Is this something of a trick question or is there something in the wording >> that I am missing? Any Help???? >> The Problem: >> You know a very good story. On Sunday you tell the story to a friend. On >> Monday you tell it to two new people. (So far, a total of three people >have >> heard the story). Each day after Monday, you double the number of new >> people you tell the story to. What will be the total number of people that >> will have heard your story after you tell it on Thursday? -- Mensanator Ace of Clubs === Subject: Fulton/Harris Representation Theory good? I am trying to begin a masterês thesis on representations of lie algebras, and my background is two semesters of algebra (introduction to group, ring, module, 'eld, and a little galois theory), as well as analysis, linear algebra, taking topology now, intro combinatorics, some measure theoretic advanced probability... Is this a good enough background to read a good chunk Fulton/Harrisês Representation theory yellow book and to write a Masterês thesis on representations of lie algebras? I noticed that tensor products are introduced on the second page, and I would have to review alternating powers, symmetric powers, but I think that my other algebra skills are fairly good. I learned Algebra from Dummit and Foote... Any opinions are appreciated, Michael === Subject: Re: Fulton/Harris Representation Theory good? I am trying to begin a masterês thesis on representations of lie algebras, > and my background is two semesters of algebra (introduction to group, ring, > module, 'eld, and a little galois theory), as well as analysis, linear > algebra, taking topology now, intro combinatorics, some measure theoretic > advanced probability... Is this a good enough background to read a good chunk Fulton/Harrisês > Representation theory yellow book and to write a Masterês thesis on > representations of lie algebras? I noticed that tensor products are > introduced on the second page, and I would have to review alternating > powers, symmetric powers, but I think that my other algebra skills are > fairly good. I learned Algebra from Dummit and Foote... It seems to me that the only background knowledge that you need is some basic knowledge concerning differentiable manifolds. Take a look, for instance, at the 'rst volume of Spivakês Comprehensive Introduction to Differential Geometry. Apart from that, you should be OK. Jose Carlos Santos === Subject: Re: Fulton/Harris Representation Theory good? I am trying to begin a masterês thesis on representations of lie algebras, > and my background is two semesters of algebra (introduction to group, ring, > module, 'eld, and a little galois theory), as well as analysis, linear > algebra, taking topology now, intro combinatorics, some measure theoretic > advanced probability... Is this a good enough background to read a good chunk Fulton/Harrisês > Representation theory yellow book and to write a Masterês thesis on > representations of lie algebras? I noticed that tensor products are > introduced on the second page, and I would have to review alternating > powers, symmetric powers, but I think that my other algebra skills are > fairly good. I learned Algebra from Dummit and Foote... > Hungerford is pretty good on tensor products. === Subject: Re: Science and Vectors Iêm not sure I understand your question. It seems like youêre talking about a dihedral angle. On the other hand, you give cyclohexane as an example, which typically we donêt really think of as having a dihedral angle. If you can be a bit more clear, I think that I can help you. MB One of my students has given me a chemistry math problem which he > found and I canêt see how to solve it. > It reads like this: Consider a molecule such as cyclohexane in the trans con'guration in > which as the bonds are the same length > | c | | b > d | | > <---------------< | > | | > | >---------------- e | | a > | | f > < | > (Sorry for the bad diagram but the arrows chase each other round > a->b->c->d->e->f) and the angle theta between two consecutive bonds is constant > throughout the molecule. (Notice also that a and d are parallel). The > angle of skew alpha between two bonds are separate by a single > intermediate one is de'ned as the angle which these two bonds appear > to form when viewed along the intermedite one. Show that this is given > by cos alpha = - cos theta / 2 (cos (theta/2))^2 Does any one know how to solve this? I would really appreciate it! > Sarah === Subject: Lipschitz, derived numbers & all that by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA7EQ0t11619; If the derived numbers of a function f on an interval [a,b] are all bounded by a common constant K show that f is Lipschitz, hence of bounded variation. Is this okay? : let h(x) be the derived number. So |h(x)| =< K for all x in [a,b] given. => there exists a seq. g_n -> 0 s.t. |lim n-> in'nity (f(x+g_n) - f(x) )/g_n| =< K for all x in [a,b] Find N s.t |( f(x+g_n) - f(x) )/g_N| =< K => |f(x+g_n) - f(x)| =< K*|g_N| => |f(x+g_n) - f(x)| =< K*|x + g_N -x| => |f(y) - f(x)| =< K*|y - x| if y = x + g_N This is for any x in [a,b] but y depends on x. But some |(f(x+g_n) - f(x))/g_n| =< K for all n >= N then the above is true for any x an any y within g_N of x (since h_n -> 0) => f is Lipschitz on [x-g_N, x+g_N] for all x in [a,b] => f is Lipschitz on [a,b] === Subject: Re: Lipschitz, derived numbers & all that > If the derived numbers of a function f on an interval [a,b] are all > bounded by a common constant K show that f is Lipschitz, hence of > bounded variation. > Whatês the derived numbers of f? > Is this okay? : let h(x) be the derived number. So |h(x)| =< K for all x in [a,b] given. > => there exists a seq. g_n -> 0 s.t. > |lim n-> in'nity (f(x+g_n) - f(x) )/g_n| =< K for all x in [a,b] Find N s.t |( f(x+g_n) - f(x) )/g_N| =< K > => |f(x+g_n) - f(x)| =< K*|g_N| > => |f(x+g_n) - f(x)| =< K*|x + g_N -x| > => |f(y) - f(x)| =< K*|y - x| if y = x + g_N This is for any x in [a,b] but y depends on x. > But some |(f(x+g_n) - f(x))/g_n| =< K for all n >= N > then the above is true for any x an any y within g_N of x > (since h_n -> 0) => f is Lipschitz on [x-g_N, x+g_N] for all x in [a,b] > => f is Lipschitz on [a,b] === Subject: Re: NOVA strings and branes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA7IvhH31128; === >Subject: NOVA Look at Elegant Universe NOVA tonight. we might know the ultimate truth of reality, as these researchers quest for, seems bogus to me. Any system we conjure up must, it seems to me, be subject to Godelês or Cantorês incompleteness constraint. Even though we seem able intellectually to extend, rise above or jump outside the formal theoretical systems we create; once we do so, in each paricular case, it seems we can, in-turn, then formalize a new system that includes extending theories to 'll known gaps, but which result nevertheless, must still remain incomplete. Thus, it seems to me, we can never quite get to the ultimate theory The Holy Grail that explains everything. I would like to be persuaded otherwise. Please lend me a hand, if you can. Damscot === Subject: On Solving FLT by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA7IvpP31154; Using the conventional symbols, Suppose one proved that z^p (not=) x^p + y^p, where z = pK, (K,p) = 1. 1. Has this already been done using simple algebra? If not, 2. Has such proof any value toward solving FLT completely? Damscot === Subject: Re: On Solving FLT by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hA7K7lB04772; >Using the conventional symbols, Suppose one proved that >z^p (not=) x^p + y^p, where z = pK, (K,p) = 1. >1. Has this already been done using simple algebra? If not, >2. Has such proof any value toward solving FLT completely? >Damscot Thatês the so-called 'rst case of Fermatês Last Theorem. See the write-up at http://mathworld.wolfram.com/FermatsLastTheorem.html to get you started. === Subject: Proof of Loan Amortization Formula Great Oracle of Mathmaticians, I know you guys get tired of guys like me asking for proofs but, this will be the last time I ask. I think seeing the proof actually helps me visualize and learn how to apply it in real life. When I look at the loan amortization formula, I just dont see that link right away. Could you take some time to explain or recommend a book that has the P.S Here is the demon below: L t(1+t)^n p = _________________ [ (1+t)^n - 1 ] === Subject: Re: Proof of Loan Amortization Formula In sci.math, Jay I know you guys get tired of guys like me asking for proofs but, this > will be the last time I ask. I think seeing the proof actually helps > me visualize and learn how to apply it in real life. When I look at > the loan amortization formula, I just dont see that link right away. > Could you take some time to explain or recommend a book that has the P.S Here is the demon below: L t(1+t)^n > p = _________________ [ (1+t)^n - 1 ] The simplest way of proving it would be to posit the problem in this fashion, perhaps. Assume one has applied for loan of a principal L, paid to him at the start of the loan. He is to pay back this loan in n months, at a constant (t * 100) % interest rate (per month). [*] Basically, one pays interest on any outstanding principal at the start of the month. (We assume that his 'rst payment is a month after the loan, as well. Note that this is a *compound interest* loan, as opposed to a *simple interest* loan, which is rarely used nowadays. Another variant is a continuously compounded interest loan, which computes the interest on the outstanding loan amount using a formula such as L * exp(k * ln(1+t)). Still other variants are possible, such as ARMs, which posit a variable t; usually the variations are such that one can use a table lookup, as they are adjusted every 6 months or so.) At month 0 his balance sheet (relative to the loan company) looks like: B(0) = L At month 1, we assume the payment is promptly credited and one has interest on the loan: B(1) = L * (1+t) - P At month 2: B(2) = (L * (1+t) - P) * (1+t) - P At month 3: B(3) = ((L * (1+t) - P) * (1+t) - P) * (1+t) - P At this point you might want to gather terms, as I for one smell a possible induction hypothesis here: B(3) = L * (1+t)^3 - P * ( (1+t)^2 + (1+t) + 1) = L * (1+t)^3 - P * ((1+t)^3 - 1) / ((1+t) - 1) Letês see if this works for B(4): B(4) = L * (1+t)^4 - P * (1+t) * ((1+t)^3 - 1) / ((1+t) - 1) - P = L * (1+t)^4 - P * ((1+t)^4 - (1+t) + (1+t) - 1) / ((1+t) - 1) = L * (1+t)^4 - P * ((1+t)^4 - 1) / ((1+t) - 1) Oooh! Now letês set up a more formal hypothesis. We assume B(k) = L * (1+t)^k - P * ((1+t)^k - 1) / ((1+t) - 1) for an integer k, and prove that B(k+1) = L * (1+t)^(k+1) - P * ((1+t)^(k+1) - 1) / ((1+t) - 1) using similar algebraic manipulation (which I leave to the interested reader, as itês very similar to my B(3)=>B(4) transition above). We also trivially verify that B(0) = L * (1+t)^0 - P * ((1+t)^0 - 1) / ((1+t) - 1) = L so weak induction follows; we now have a closed form for the balance after the kêth month. The terms of the loan are that after n months the loan is paid off, so B(n) = 0, and therefore B(n) = 0 = L * (1+t)^n - P * ((1+t)^n - 1) / ((1+t) - 1) or L * (1+t)^n = P * ((1+t)^n - 1) / ((1+t) - 1) or P = L * t * (1+t)^n /((1+t)^n - 1). QED In practice, the equality is not quite exact because of rounding of P to the nearest penny, but the amount of variation is at most 0.01 * ((1+t)^n - 1) / ((1+t) - 1) regardless of the value of L. If we substitute n = 360 (a 30 year house mortgage) and t = 0.005 (about a 6% per annum rate), we get $10.045 ..., which is miniscule compared to most house loans nowadays; the last payment is increased by at most this amount, or perhaps the mortgage company remits a check. [*] just to confuse things: the payments in many loans are *per month*, but the rate is quoted as an interest rate *per year*, which may complicate the analysis in real life, even for this relatively simple loan. -- #191, ewill3@earthlink.net Itês still legal to go .sigless. === Subject: Re: Continuum > Besides, to comment on you > awareness, what languages do you ¤uently read in except english ? You presume that I am ¤uent in English? === Subject: Re: Continuum >>[...] >Besides, to comment on you >awareness, what languages do you ¤uently read in except english ? Duh! Latin! === Subject: Re: Applications of mathematics >Suppose you were to tell senior highschool students about applications >of mathematics that would be interesting and understandable to them. >What applications would you talk about? Knot theory and sex, because sex is the only thing interesting to them. === Subject: Re: Applications of mathematics >>Suppose you were to tell senior highschool students about applications >>of mathematics that would be interesting and understandable to them. >>What applications would you talk about? Knot theory and sex, because sex is the only thing interesting to them. That reminds of a quote from Louis-Ferdinand C.8elineês Voyage au bout de la nuit found on the Mathematical Quotations Server (http://math.furman.edu/~mwoodard/mquot.html): -------------------------------------------------------------- -------------- ------------------------ Entre le p.8enis et les math.8ematiques... il nêexiste rien. Rien! Cêest le vide. [Between the penis and mathematics there is nothing. Nothing! The void!] -------------------------------------------------------------- -------------- ------------------------ John Mitchell === Subject: Re: Applications of mathematics There are some beautiful applications if mathematics to biology that are often overlooked. Everything from the shape of the nautilus, to the Fibonacci pattern of sun¤ower seeds, to the mathematics of genetics, to the structure of proteins, to the use of math in MRI, structural biology, to the study of chaos (which came in part from Mayês study of population dynamics), to the analysis of how diseases are spread, to the shape of viruses, etc,etc,etc. MB > Suppose you were to tell senior highschool students about applications > of mathematics that would be interesting and understandable to them. > What applications would you talk about? TIA, > Felix. === Subject: Re: More symmetry between derivative and antiderivative? > If mathematics were satis'ed with such slovenly thinking, then it > would never have been able to develop most of those mathematical > tools that are so useful outside of (formal) mathematics. I see it the other way round. Final success in cooperation between > matthematics and its application requires both the necessary degree of > mathematical rigorosity and careful and comprehensive analysis of the > practical circumstances. If pure mathematics is found to be useful outside of pure mathematics, that is only of incidental interest. If a mathematical model does not adequately describe what is is being applied to, 'nd a different model. > Let me give an example for blind mathematics: > Mathematicians consider a function a causal one if it equals zero for > t<0. They are applying this to freqency too. So called causal > frequency vanishes for negative frequency. It is only misleading to those who mistakenly apply mathematics > This is highly misleading because negative frequency makes only > sense in complex Fourier analysis. However, causal frequency > does not at all correspond to a causal time signal but to an > unreal so called analytical signal. If this offends you, 'nd a better model! > What was sloppy in my thinking? I argue that mathematics cannot decide > whether or not its applications are useful to the costumer. It does not presume to. === Subject: Re: More symmetry between derivative and antiderivative? > I see it the other way round. Final success in cooperation between > matthematics and its application requires both the necessary degree of > mathematical rigorosity and careful and comprehensive analysis of the > practical circumstances. Some languages have several sets of words for numbers, with different sets used for counting different things. Analogous too close to the 're pragmatic redundancy is seen a typical C program (even more in C++), where the very same thing is coded 5 different ways, each named in accordance with its application (as good naming convention dictates), thus totally obscuring the fact that it was the same thing. === Subject: Question on 7d geometry Hello list, 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 'rst 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? advance. -Steven === Subject: Re: Question on 7d geometry > Hello list, > 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 'rst 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 coef'cients 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. -- reverse my forename for mail! === Subject: re: The Emperorês New G-String Jack you ask for ... a good discussion of the Lie algebra of ... and the physical meaning of the special conformal transformations .... That was the main part of the life work of Irving Ezra Segal. Is there a book? He found that if you included the special conformal transformations, the effective ground-state metric of spacetime changed from ¤at Minkowski spacetime without special conformal generators to a curved spacetime that sort of acted like an expanding universe. I am not surprised because in my theory it appears that the dark energy comes from locally gauging that 4-parameter subgroup of the conformal group. I am not sure of that, just a hunch at this point. However, due to the tenor of his times, he did not interpret the curved spacetime as Dark Energy making the universe expand, The Universe expands without the repulsive dark energy of course, but the dark energy accelerates the expansion rate not letting it decelerate once it builds up to an amount greater than the dark matter. Apparently this happened in our Universe about 7 billion years ago. but rather as a large-scale metric (explaining redshift, etc) that somehow was the effective metric on cosmological scales while he viewed the Minkowski metric as being the effective ground-state metric only on local scales, like the earth or solar system or so. Had he lived to see the sort of interpretations that todayês cosmologists use, I suspect that he might have happily said that his conformal curved metric special WAS evidence of the Dark Energy, Note for the historical record, I came to that tentative conclusion entirely on my own without knowing anything about Segalês ideas. Indeed, it was the very recent message from Thomas Angelidis that put me on that path. whose carrier bosons are the special conformal generators (which is pretty much the way I think that you and I see it, and which leads us to be pretty sure that it is mostly an engineering problem of how to bottle it or control it.) I never really understood what your conformal graviphotons were all about until the past few days. Now I grok it. :-) I have some web pages about his work at http://www.innerx.net/personal/tsmith/SegalConf.html and related pages, and I include references to some of his papers which describe in some detail the Lie algebra generators of the special conformal transformations (I like to call them Conformal Graviphotons to suggest connections among the three concepts of conformal group, gravity, and electromagnetism). My apologies for being a pretty crummy expository writer, but a lot of interesting stuff is on my web pages if anybody goes to the trouble to work through it, follow references, and generally work to dig it out. That brings me to Chiao gravity antennas, and Chiaoês experimental efforts to see conversion/transduction from gravity to electromagnetism and vice versa. Yes, well that would allow us to directly communicate between brane worlds if the current ideas of the Witten Cavalcade are correct - and they probably are roughly speaking. Itês amazing that so much speculation is hyped in prime time on PBS with money from big corporations. Actually, I think that is a good thing all in all. Some of my web pages, with links to some of Chiaoês work, are chiao and http://www.innerx.net/personal/tsmith/coscongraviton.html# ChiaoEMGR and http://www.innerx.net/personal/tsmith/QuanConResonance.html# elegravdyn ... Anyhow, in his paper at http://xxx.lanl.gov/abs/gr-qc/0303089 which is quoted in part on the third listed URL about Chiaoês work, Chiao and Fitelson say: ... ... How then do we account for the lack of any observable quantum transducer conversion in our experiment? There are several possible reasons, the most important ones probably having to do with the material properties of the YBCO medium. ... the choice of YBCO as the material medium for the Hertz-type experiment may not have been a good one. ... That brings me to another point in the message you received from Constantin Ivanenko. In it, he said ... importance of barium titanate - key component of psi-genome weapons - was intuitively foreseen by SF author H. Kuttner; contained in the chip of barium titanate crystal. .... Barium titanate, like some high-temperature superconductors, has a perovskite-type crystal structure. With some such things, such as YBCO, you get superconductivity. With barium titanate, you get interesting dielectric and pizeoelectric properties. My guess is that successful Dark Energy engineering will involve a similar crystal structure with BOTH superconductivity AND properties such as pizeoelectricity to get useful transduction from gravity to EM and vice versa, and that Mead resonance will be another necessary interesting piece of engineering. Whatês the Mead resonance? As for references about Mead resonance, I have some quotes from his book on my web page at http://www.innerx.net/personal/tsmith/QuanConResonance.html# resonance Of course, there are a lot of papers about perovskite and related structures, but one pretty good elementary introductory page is at http://www.molecularuniverse.com/Building_In_3D/bld5.htm Search engine searches can 'nd many more papers at varying technical levels. In my opinion, if we lived in a rational USA, there would be a Manhattan project on this stuff. However, it is possible that the money-lords of the USA might 'ddle away like Nero, leaving the advance to somebody else. Yeah, like the Chinese who are forging ahead. http://www.arxiv.org/abs/hep-th/0307102 who write from Peking: I. INTRODUCTION Although the energy density of a 'eld in classical physics is strictly positive, the local energy density in quantum 'eld theory can be negative due to quantum coherence effects [1]. The Casimir effect [2] and squeezed states of light [3] are two familiar examples which have been studied experimentally. As a result, all the known pointwise energy conditions in classical general relativity, such as the weak energy condition and null energy condition, are allowed to be violated. However, if the laws of quantum 'eld theory place no restrictions on negative energy, then it might be possible to produce gross macroscopic effects such as violation of the second law of thermodynamics [4,5], traversable wormholes [6,7], warp drive [8], and even time machines [7,9]. If you believe the writings of people like Lev Navrozov at http://www.levnavrozov.com/ it might be China that brings humanity to a new level. After all, the New Scientist (pages 36-43), it was Japan and China that gave jobs to people let go from Navy jobs when the US Navy closed out its (at least partially successful) cold fusion program about 6 years ago. Tony === Subject: Re: (SORRY) discrete logarithmic & Dif'e Hellman Nntp-Posting-Host: apps.cwi.nl Sorry, that iêm 'rtsly posted in german, here the english post: 1. Do anyone of you know, when the dicrete log was invented ? >2. Knowês anyone of you a prove of the diffe - hellman algorithm ? > (or can anyone of you explain me, why it does function ?) > Discrete logarithms are very old, but were called indices (plural of index) until about 1980. Gute Sonntag -- After Californiaês recall election, 'res Schwartz-en-ed the Bush-lands on its geographic right. pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: I canêt stand it anymore > amanda replied: > meplace.com>... >amanda replied: > 745$of7.2236759@wagner.videotron.net>... >What gave you the idea that I was upset about the term Asian? My only point was that IQ tests to determine peopleês intelligence a > re >>bs. >>Ah, of course. It would be far better indeed to administer >>intelligence tests rather than IQ tests to determine peopleês >>intelligence, right? >>Really, you sound like someone who is disappointed by their >>own results on such a test. I knew that someone would come up with that stupid opinion. It just >happens to be you. Sorry about that Donêt you realize that us Asians donêt need the tests invented by >non-Asians to 'nd out whether we are intellgent or not? It was only at the turn of 2oth century that the Western World >started to reliaze that Asians are not inferior to them >>Itês interesting that in the Asian countries, the inferiority of >>non-Asians is still widely accepted and believed. Now, where did you get that idea? Everything Iêve read my and about various Asian cultures. I didnêt say that, did I? I didnêt say you did, did I? >There is no attempt for or desire for reciprocity. >as they have been told by the 19th century European racists >>Is there an Asian country which is not racist? Whya re you talking abput cpuntries. Why not individuals? Cause itês not just individuals, itês a cultural thing. > Why donêt you talk about individual Americans? You are the one who accused all Asian countries as racists. You do not understand how Western culture (as they see from the movies) scare the parents in those countries. These governments feel obligated to deter the bad part of western culture. In doing so, sometimes they go extreme. When the movie Flash Dance came out, any kid who twist their body would be taken away by the police, at least for a moment. My cousin, about 13, was really good at it. his father never knew that he could do that. If the father knew, he would probably scold the son. >>Just look at the nice >>things high Japanese govt of'cials have said about Americans. >who believed that they were superior to all other humans. >>As do all Asian races currently. Currently is the keyword but you are wrong to generalize all Asians. So you say. Nice that you are using what Mlaysia said as the ruler. Never mind that Malaysian government a Muslim government. Show me any Muslims who thinks western culture is suitable to their children. http://news.bbc.co.uk/1/hi/entertainment/showbiz/2582833.stm .95Anti-Asianê Brad Pitt advert banned A car advert featuring Brad Pitt has been banned in Malaysia > because it is an insult to Asians, according to reports > in the country. The government ordered the ads to be pulled because non-Asian > faces would plant a sense of inferiority among Asians, the > countyês deputy information minister Zainuddin Maidin told > national news agency Bernama. Arenêt our own people handsome enough? > Zainuddin Maidin Deputy information minister The adverts, for Toyota Altis cars, ran on television, billboards > and in newspapers across Asia, and it was reported that the > posters became a target for fans who wanted them for their > bedroom walls. Why must we use their faces in our advertisements? Mr Zainuddin > said, according to the agency. Arenêt our own people handsome enough? > an insult to Asians. Pitt, one of the biggest stars in Hollywood, was last seen on > the big screen as a smooth-talking criminal in Oceanês Eleven. He is married to Friends star Jennifer Aniston, who fell foul > of Malaysian censors earlier in the year when they banned an > episode of the comedy because they said it would encourage > promiscuity. Aniston has said she expects the show to end, and is planning > start a family with Pitt. In reaction to speculation that the next series of Friends > would be the last, Aniston said: In my mind, Iêm done. I want to start my family, she told United States magazine > Entertainment Weekly. The Hollywood couple wed in a million-dollar ceremony in Malibu > in July 2000. She won a prestigious Emmy award for best comedy actress for > her role in the hit series earlier in the year - but there is > talk that she may get more awards for her latest 'lm, The Good Girl. In it, she plays an adulterous and depressed Texan shop assistant, > and the 'lm has taken $14m ( 8.75m) at US box of'ces. Aniston is getting the best reviews of her career - the kind of > reviews that can lead to Oscar nominations, Time magazine said. Others say different. >But somehow you donêt seem to see anything wrong with this. Thatês your opinion. I see what you objected to, and I see what you have not yet objected to. Twist as you wish. I just didnêt want to spend time refuting silly accusations. My saying that Asian do not need a test invented by someone else See? Youêre into .95usê and .95themê. Obviously for you .95usê is Asians, and > .95themê is everyone else. Again, another silly accustaions that I am into us and them. Beside, we both know who invented IQ tests. So donêt make it look like I meant everyone else when Is aid them. > And you 'nd .95themê wanting in the same fashion > as Zainuddin Maidin above. Pretty bold accusation. You think you know but you do not know how Asians see western culture very well. to determine their intelligence level, You leave you own culture you live by the rules of the culture you move > to. it didnêt imply that I think those who invented that test were inferior > . Itês a bit late for that. Why? because you say so? If you canêt understand that, thatês your problem. Thatês the problem. I see what you actually write and object to, and > itês no different from the stuff others post. You 'nd things wanting > just because they are not Asian in origin. You canêt see it, I understand that. Whatever? I do not know how you concluded that I belive the Asians are superior to anyone else. You seem to be upset that I implied that Asians are superior to you. May eb you had have some experience with soem Asians acting like that. Heck..I have met Chinese who claimed to be superior to anyone else. In fact, one time, when this friend of mine said that America is doing well because of its Chinese population. Immediately after I smiled and replied I donêt think so, I realized that that moment was the beginning of the end of our friendship. And she wasnêt the 'rst Chinese I met who put down on Amercans (talking about white Americans). Everytime, I have defended from the Americansê side. For what? To be accused by people like you that I think white American are inferior to Asians. Very nice! >This entire post would be ironic were it not so hypocritical. That also is your opinion. Which youêve been good enough to substantiate. >Why donêt you also complain about Asian racism? Why should I complain about it here? Because Iêm asking you to, just to show that you care about racism, not > just Asians. If I didnêt care about racism, I would not have started this thread. Why are you clueless about that? I have complained many times about Indian caste system somewhere else. Elsewhere? What are your complaints? Long time ago - it was soon after 9/11 - I happened to go into an egpytian group thinking I would learn more about the ancient egyptian culture. Instead, some Indian Hindu hate mongers were trashing the Muslims (mainly Pakistanis) pickign on their religion and treatment on women. So I borught up suttee, caste system, and treatment of women in their society. Caste systen has been abolished but has it really gone? I told them to clean up their hosue before picking on others. I didnêt use the same user name. >Probably because you support it. Now, now. Probably is not the same as you support it, does it? No. Itês a word that implies speculation. As a minority in the country I was born (my parent were born and > raised thare too) who suffered discrimination and were not considered > as full-citizens, you must be joking accusing me as a racist. What does that have to do with whether you are racist or not? Because it was due to race that we were discriminated. > I donêt > think you really understand the concept. You were the one who didnêt understand what I said. >Itês like saying you are black, > you canêt be racist. > I got news for ya, if you can say that, you are. I got news for you. You underestimated me in assuming that I think black cannot be racists. Beside, as a person whose blood covers basically almost all races of > the world (not through Caucasoid and Negroid per se but through > Indo-Aryan which you probably have no clue about), including semite > (not jewish), you gotta be kidding to accuse me as a racist. Youêve gotta be kidding if you think this denies that you can be > a racist. >Itês clear that you will pro't from any pro-Asian racism. What are you talking about? Think Af'rmative Racism. Think .95diversityê. When you said Asian, are you thinking of the slant-eyes, ¤at nose, > dark, straight haired people OR are you referring to ALL THOSE FROM > ASIAN COUNTRIES which would include Arabs, Persians, people from > Afghanistan some of who are Caucasians, the Mongoloid group, i.e > Chinese and Japanese and more, the Indo-Aryans of India, etc.? Note > that I have excluded the people of Turkey. And what difference does this make? You have no clue what I was talkign about, are you? I was/am going with the second one, WHICH WAS MY WHOLE POINT OF > criticizing the use of the term Asian in IQ test, duh... That is was not invented by an Asian. That was your objection, how > date a non-Asian even imply that they can say anything about an Asian. My objection was that I as an Asian did not need to run and take the IQ test to 'nd out my intelligence level when, I repeat WHEN *that* guy told me that I must not have done well blah..blah..blah and ehnce against the IQ tests. >Itês almost certain that you have. The most stupid thing I have ever heard. What I have had was getting > pulled over by the red-neck police Racism. Red neck is red neck. Simple as that. Here in CA, I approached a police car once to ask for some direction. The 'rst things he said, with a very pleasant manner was what can I do for you?. In Houston, it would be like Whatês your problem?. Another time, as the police blocked the route to my place, I asked the police standing there what I should do and he was really nice. In Houston, it would be like I am busy; get lost. I even told that (that most of them are red neck) to a police of'cer once. It was 11 pm at night, not on the highway but he happened to be really closed by and I didnêt see his car because I wanted to get home which was not too far and did not slow down as he expected. I was the only car around there. I told him to just give me the ticket and spare the lecture. He asked me why I talked like that. I told him that I havenêt met a police of'cer who did not give me a ticket. I told him that most are red necks. He just laughed. May be he didnêt get mad because I was not far from the campus and he knew that I was a student. (I did tell him that I was tired and didnêt see him when he asked why I didnêt slow down after seeing him). Or may be I looked cute to him (he wasnêt very old) when i said to him that most of them are red necks. Who knows? As he gave me the ticket, he said that he would not have given me the ticket if I didnêt act the way I did. I replied - silently - yeah, right! in Houston for driving a beat-up > car in my student days because I look a bit Hispanic or South > Americans. Odd, I was always pulled over by the cops when I had my old 1965 > Mustang convertible. Racism obviously, right? In my case, there were other cars speeding but they didnêt get stopped. >Rich Note: I will not be wasting any more time defending groundless > accusations. Youêve already proven that they are not groundless. Whatever. Rich === Subject: Re: I canêt stand it anymore amanda replied: [...] >>Nevertheless, the notion that they are totally worthless is also worth >>examining. The poster .95Uncle Alê (I think it was him) said that they measure >>your ability to take IQ tests but thatês naught but a tautology. IQ tests >>donêt tale place in a vacuum and students should have learned something from >>all those years in school. >>Do you think that all tests are worthless? some are worthless? Somewhere >>in-between? > Because of the way the IQ tests are being used today, I didnêt bother > to check the worthiness of it. I guess it is as worthy as those standardized tests, namely, GRE, > GMAT, LSAT, MCAT, etc.. In another word, just like those standardized > tests, it is designed for a speci'c group of people. Explain to me how algebra is designed for a certain group of people. Then explain why Asians do better on these tests. Were these tests designed for Asians? Is that your claim? Rich >>Rich > === Subject: Re: I canêt stand it anymore > amanda replied: [...] >Nevertheless, the notion that they are totally worthless is also worth >>examining. The poster .95Uncle Alê (I think it was him) said that they measure >>your ability to take IQ tests but thatês naught but a tautology. IQ tests >>donêt tale place in a vacuum and students should have learned something from >>all those years in school. >>Do you think that all tests are worthless? some are worthless? Somewhere >>in-between? > Because of the way the IQ tests are being used today, I didnêt bother > to check the worthiness of it. I guess it is as worthy as those standardized tests, namely, GRE, > GMAT, LSAT, MCAT, etc.. In another word, just like those standardized > tests, it is designed for a speci'c group of people. Explain to me how algebra is designed for a certain group of people. If you want to sretch, be my guest. > Then explain why Asians do better on these tests. Were these tests > designed for Asians? May be non-Asians are not aaeching their kids Algebra the best way. > Is that your claim? I didnêt claim anything about Algebra tests. Rich >Rich > === Subject: Re: help me, please! ha scritto nel messaggio > Is there a fast way to determine X and Y both integer such that: > X^2 - Y^2 = A 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 'nd. To make sure you understand the process > solve x^2 - y^2 = 12 for integers, x,y. > Clearly you skipped this step because of the question below you asked. So show us some work and what answers you get. If you just guess instead of following the process then to learn whatês to be know, solve for integers x^2 - y^2 = 4972 > First list all the ways of factoring 12. > Letês see what you try. > What are your answers? x^2 - y^2 = 66 has no solution. Why? bacause it has tre factors. > No. The number of factors has nothing to do with it. === Subject: Re: differential equation >>It could be a constant vector, but my guess is that the equation is >>actually >rêê = -k r/|r|^3 Well of course that was my guess as well, which is why I >didnêt decide right away that k was a constant vector... >Well, of course it has many closed-form solutions. Itês >>not possible to express the general solution in closed >>form, is it? Not as far as I know (for r as a function of t). Of course >you can eliminate t to get Keplerês conics etc. >(...) Once you have |r| in terms of the angle w, you also have r^2 dw/dt = c dt = (c/r^2)dw So integration gives t in tems of w, rather than vice versa. Still, this can be considered a solution, no? Also, it might be possible that Maple can integrate and then 'nd an inverse function - does anybody know? === Subject: integration Is integration a one way function? More exactly Is integration a one way functional? So if I ask how to integrate complicated expression am I actually asking for help to break a code? ;-) === Subject: Re: How do you integrate the function f(x) = x/(tanx) [0,pi/2]? Ray Steiner > I came up with another way of showing that x/tan x has no elementary antiderivative. > We need only one result from Wienerês 1997 paper: > arcsin(x)/x does not have an elementary antiderivative. Let I = int(x/tan x dx)= int (x cot x dx) > Use integration by parts to get > I = x ln(sin x) - int( ln(sin x) dx) > Let I2= int( ln(sin x) dx) Let u= sin x, x = arcsin(u), dx = 1/sqrt(1-u^2) du > Then > I2= int ( ln(u)/sqrt(1-u^2) du) Finally, use parts again to get > I2= ln(u) arcsin(u) - int(arcsin(u)/u du). > So, by Wienerês result, the original integral is not elementary. More results: > By exactly the same method one can show that > I3 = int (x tan x dx) is not elementary. > Now, letês substitue u= tan x, x = arctan u, dx= 1/ (u^2 + 1) du in I3. > Then it reduces to > I4 = int( u*arctan(u)/(1+u^2) du). > so the second integral of my previous post is non-elementary. > Finally, consider > I5= int ( (arctan(x))^2 dx). > By parts, one can reduce it to integrating I4, so I5 is also non-elementary. LH === Subject: Re: torque T = r x F and basic tensors >I think you might be wrong here. A vector transforms uê = Qu, and thus >so should torque if it were really a (polar) vector.But the cross product >makes it a psuedo-vector, or so all the books say. >So if torque is really the 2nd rank anti-symm >matrix I described above, it must transform as Tê = Q^{-1} T Q, right? > okay okay okay... it is an _accident_ of working in three dimensions that the cross product of two vectors is also a vector... I.e. There is no reason why the cross product (de'ned by: C = A X B de'ned by C_i = eps_{ijk}A_jB_k (eps is the epsilon tensor)) should necessarily transform like a vector. but, it does... although it sure is hard to show that it does...it took me like 45 minutes to remember how to work it out. you need to use the fact that: eps_{ijk}R_{im}R_{jn}R_{kp} = eps_{mnp} this isnêt too hard to show since the LHS obviously is zero when m=n or m=p or p=n. A for the case m=1 n=2 p=3 you just have the de'nition of the determinint which is convieniently equal to one. the other cases follow directly. then use the fact that R is orthogonal to show that: eps_{rjk}R_{jb}R_{kg} = eps_{abg}R_{ra} Finally...'nally!... lets consider torque: T = r x F T_i = eps_{ijk} r_j F_k since we know how r and F transform we know T transforms to: T_i --> Tê_i = eps_{ijk} R_{jm} R_{kp} r_m F_p = eps_{qmp} R_{iq} r_m F_p = R_{iq} eps_{qmp} r_m F_p = R_{iq} T_i whoo hooo. it transforms like a vector... a pseudo-vector. But yeah, itês a tensor too which means nothing less and nothing more than: T_{i,j} = (r_i F_j) - (r_j F_i) T_{i,j} ---(coordinate transformation R)--> Tê_{i,j} where Tê_{i,j} = R_{i,k} R_{j,m} T_{k,m} (sum on k,m) = R_{i,k} T_{k,m} R^transpose_{m,j} adam === Subject: Re: Two questions in propositional logic > Weêre working in propositional logic, with our basic symbols being > implies, falsehood and p_1, ... . The axioms are p -> (q -> p) > [ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] > p -> p > Ah, you didnêt expect us be psychic, you included your logical system. > (Where p = ( p -> F ) ). > ~p shows up better than your 1/4 p with 1/4 squeezed into one character. > 1. Write down an explicit function f(n) such that every tautology of > length n has a proof of at most f(n) lines in length. I wasnêt entirely sure what he meant by length here, > Including parenthesis as wf construction with -> is (p->q) len(.95(p->q)') = 3 + len(.95pê) + len(.95qê) Random thoughts aspiring unto slight of hand beyond mere hand waving: How long does it take to convert a statement into itês disjunctive normal form? How long does it take to convert the disjunctive normal form of a tautology to t, ie f->f ? > 2. Let C be a set of propositions. We say C is a chain if for every p, q > in C either p proves q or q proves p (and not both). If the set of > primitive propositions is allowed to be uncountable, can there be an > uncountable chain? > No. For C = { P_r | 0 < r in R } to be a chain you have an uncountable number of P_r |- P_s that is, an uncountable number of |- P_r -> P_s But as the number of theorems are countable... === Subject: Re: Two questions in propositional logic >Weêre working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>p -> p >> Ah, you didnêt expect us be psychic, you included your logical system. >>(Where p = ( p -> F ) ). >> ~p shows up better than your 1/4 p with 1/4 squeezed into one character. >>1. Write down an explicit function f(n) such that every tautology of >>length n has a proof of at most f(n) lines in length. >>I wasnêt entirely sure what he meant by length here, >> Including parenthesis as wf construction with -> is (p->q) > len(.95(p->q)') = 3 + len(.95pê) + len(.95qê) Random thoughts aspiring unto slight of hand beyond mere hand waving: > How long does it take to convert a statement into itês > disjunctive normal form? > How long does it take to convert the disjunctive normal form > of a tautology to t, ie f->f ? Yeah, we wondered about that. Unfortunately it doesnêt appear to be signi'cantly simpler to do it that way. Or if it is, then we all missed why it was. Another approach we tried was to use the deduction theorem to reduce it to the problem of 'nding an upper bound on the length of a proof of p_i from X with p_i a primitive proposition and |X| <= n (where n may been relabelled from the previous one here). >2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? >> No. For C = { P_r | 0 < r in R } to be a chain > you have an uncountable number of > P_r |- P_s > that is, an uncountable number of > |- P_r -> P_s > But as the number of theorems are countable... > Not so. You missed the hypothesis that the number of primitive propositions was allowed to be uncountable. So, for example, { p_i -> p_i : i in I} is an uncountable set of theorems (although of course not a chain). David === Subject: Re: Two questions in propositional logic in C either p proves q or q proves p (and not both). If the set of > primitive propositions is allowed to be uncountable, can there be an > uncountable chain? This one really annoyed me, as every time I tried to prove the answer > was no I thought I saw a way to construct one, and every time I tried to > construct one it failed in such a way I thought I might see a way to > prove it. No joy though. My friend John does have a proof, but itês > rather long and horrible, so I was hoping for a nicer one. > Never mind about this one - a quick look through Johnês proof and I think I see a way to do it in a shorter, friendlier form. :) David === Subject: Re: Two questions in propositional logic >Iêve just 'nished being supervised on an example sheet for our .95Logic, >Computation and Set Theoryê course, and there were two questions which >neither I, my partner, nor my superviser could actually answer. I was >wondering if someone could give me some hints so that I could try and >sort these out. (Preferably not more than hints, as some people havenêt >been supervised on this yet. Donêt want to spoil it for them. :) Weêre working in propositional logic, with our basic symbols being >implies, falsehood and p_1, ... . The axioms are p -> (q -> p) >[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >p -> p (Where p = ( p -> F ) ). Fairly standard, but people use some slightly different axioms so I just >thought I should say which ones Iêm using. You also need to specify what the inference rules are... >The two questions were as follows: 1. Write down an explicit function f(n) such that every tautology of >length n has a proof of at most f(n) lines in length. For some systems of axioms and rules this is easy, but it can be much harder for others. >I wasnêt entirely sure what he meant by length here, but it doesnêt >really matter. Given two sensible de'nitions of length you can move >from such an f in one to one in the other. The one I thought he meant, >because of something mentioned in the lectures, was that each primitive >statement has length one, and (p -> q) has length equal to 1 + the >maximum length of p and q. I was completely stumped on this one. I saw two possible approaches, but >neither of them looks very helpful. One was using the deduction theorem >it is equivalent to 'nding an upper bound on the size of a proof of an >atomic proposition from a set of size n. The other was that because we >can de'ne such an f(n) by maximising/minimising over proofs of length ><= n (because there are only 'nitely many, up to relabelling of >primitive propositions). Then, because (mumble mumble) f de'ned that >way must be primitive recursive (which Iêm not sure I believe anyway) >eventually A(n), the ackerman function, bounds it above. So if you scale >A by some factor then you get an upper bound for f(n). I donêt think much of that last one though. Itês more of a .95throwing >hands into the air and giving upê approach. :) 2. Let C be a set of propositions. We say C is a chain if for every p, q >in C either p proves q or q proves p (and not both). If the set of >primitive propositions is allowed to be uncountable, can there be an >uncountable chain? I doubt it. Say p < q if p |- q but not q |- p; then your chain is totally ordered by <. First, either there exists an increasing sequence with at least two upper bounds or a decreasing sequence with at least two lower bounds (by the traditional proof that any sequence of reals contains a monotone subsequence: Say you have elements p_a, indexed by the countable ordinals. Say a is dominant if p_a > p_b for all b > a. Either there are uncountably many dominant a or not. If there are uncountably many dominant a then the p_a with a dominant give a co'nal decreasing set, while if there are only countably many dominant a then the dominant a are bounded above and you get a co'nal increasing subset.) So, changing the notation and replacing all the wffs by their negations if needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a tautology (it seems like this is by compactness or something, but I canêt quite prove it this second) and then q_2 is weaker than a tautology, contradiction. ??? >This one really annoyed me, as every time I tried to prove the answer >was no I thought I saw a way to construct one, and every time I tried to >construct one it failed in such a way I thought I might see a way to >prove it. No joy though. My friend John does have a proof, but itês >rather long and horrible, so I was hoping for a nicer one. Another thing from the example sheet was providing a direct (i.e. not >using completeness) proof of the compactness theorem for propositional >logic. I came up with a topological argument that did the trick, but >couldnêt see a more natural approach than that. Any suggestions? David > ************************ David C. Ullrich === Subject: Re: Two questions in propositional logic >>Iêve just 'nished being supervised on an example sheet for our .95Logic, >>Computation and Set Theoryê course, and there were two questions which >>neither I, my partner, nor my superviser could actually answer. I was >>wondering if someone could give me some hints so that I could try and >>sort these out. (Preferably not more than hints, as some people havenêt >>been supervised on this yet. Donêt want to spoil it for them. :) >>Weêre working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>p -> p >>(Where p = ( p -> F ) ). >>Fairly standard, but people use some slightly different axioms so I just >>thought I should say which ones Iêm using. > You also need to specify what the inference rules are... > Sorry, of course. The only rule of inference is modus ponens. >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds This is presumably under the assumption that C is uncountable? > or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a co'nal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a co'nal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I canêt quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. ??? > Unless youêre using some extra property about the p_i, q_i in saying this that Iêm not seeing (which is very possible, on account of it being Early here and I just got up. :), that neednêt be true. For example take the following sequence: p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. I thought at one point that you could have a chain isomorphic to any countable ordinal, although in retrospect Iêm not entirely convinced by my argument (Iêm slightly worried about what happens with some of the bigger limit ordinals), but this doesnêt really matter. Of course not every chain is isomorphic to an ordinal, as the reverse of a chain is also a chain. Anyway, as I said in the other post I think Iêve got this sorted out now (although Iêll want to write my own proof of it before Iêm fully happy with it). If youêre interested, Johnês proof went roughly as follows: De'ne an equivalence relation on C by p ~ q if there is a bijection f from the set of primitive propositions to itself, such that q = p with all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence class in C, and that there are countably many equivalence classes greater than p (because you can take some countable subset of the whole set of primitive propositions and represent every formula in C as a formula using only this countable subset and the primitives that appear in p). He then went on to show there were only countably many elements less than p in a similar manner. I pointed out that this was rather easier, as you could just reverse the chain by negation and use the previous result. I think I can make that proof slightly shorter by considering the set of propositions in the chain provable in <= n lines and considering how .95newê primitives are introduced. Then, up to equivalence, there should be only countably ('nitely?) many of these. Then a countable union of countable sets is countable. Not sure if thatês going to work, but it looks plausible. David === Subject: Re: Two questions in propositional logic > Anyway, as I said in the other post I think Iêve got this sorted out now > (although Iêll want to write my own proof of it before Iêm fully happy > with it). If youêre interested, Johnês proof went roughly as follows: De'ne an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence > class in C, and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. I found this the most natural approach. However (and IBL will probably kill me for saying anything at all about this sheet in public) you should also be aware there is a one word answer to this question. That is, there is one particular word (familiar to everyone) which instantly makes this question a triviality. Iêll leave you to guess what it is. By the way you didnêt ask about the last question (if the set of primitive propositions is allowed to be uncountable is it true given a set S of propositions that you can 'nd an independent set of propositions equivalent to S); does that mean youêve solved it? I still donêt have it, one year after taking the course. No hints please! Michael === Subject: Re: Two questions in propositional logic >>Anyway, as I said in the other post I think Iêve got this sorted out now >>(although Iêll want to write my own proof of it before Iêm fully happy >>with it). >>If youêre interested, Johnês proof went roughly as follows: >>De'ne an equivalence relation on C by p ~ q if there is a bijection f >>from the set of primitive propositions to itself, such that q = p with >>all the p_i in p replaced with f(p_i). >>He then showed that there can be at most one element of each equivalence >>class in C, and that there are countably many equivalence classes >>greater than p (because you can take some countable subset of the whole >>set of primitive propositions and represent every formula in C as a >>formula using only this countable subset and the primitives that appear >>in p). >>He then went on to show there were only countably many elements less >>than p in a similar manner. I pointed out that this was rather easier, >>as you could just reverse the chain by negation and use the previous >>result. > I found this the most natural approach. However (and IBL will probably > kill me for saying anything at all about this sheet in public) you > should also be aware there is a one word answer to this question. That > is, there is one particular word (familiar to everyone) which > instantly makes this question a triviality. Iêll leave you to guess > what it is. Hmm. ... Oh hell. Is it the one-word that you always use to prove obvious statements? It is, isnêt it... ::sketches proof in his head:: Excuse me while I go sulk. (To anyone reading this who doesnêt understand that line, itês a Leaderism. :) > By the way you didnêt ask about the last question (if the set of > primitive propositions is allowed to be uncountable is it true given a > set S of propositions that you can 'nd an independent set of > propositions equivalent to S); does that mean youêve solved it? I > still donêt have it, one year after taking the course. No hints > please! Well... yes and no. I thought I had a solution. There was a tiny problem in it contained right at the end which I think I can 'x (but it may be a way bigger problem than I thought it was. :) That being said, my solution to the last question was only one line. My solution to the previous question did not assume countability. I was lazy and didnêt feel like proving it twice... David === Subject: Re: Two questions in propositional logic > Well... yes and no. I thought I had a solution. There was a tiny problem > in it contained right at the end which I think I can 'x (but it may be > a way bigger problem than I thought it was. :) > Hmm... Yes. It does indeed appear to be a bigger problem than I initially anticipated. Thereês a surprise. ::grumble grumble evil frigging example sheets grumble:: David === Subject: Re: Two questions in propositional logic >[...] >2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? [...] If youêre interested, Johnês proof went roughly as follows: De'ne an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence > class in C, Because if p ~ q and p |- q then a permutation of the variables in the proof shows that q |- p. >and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). ??? Isnêt it clear that there are only countably many equivalence classes altogether, and so weêre done? I mean any wff involving only n variables is ~ - equivalent to one involving only p_1,... p_n, and up to logical equivalence there are only 'nitely many of those. Oh. Thatês not right with the ~ you de'ned. But why not instead say p ~ q if there exists a permutation of the variables which makes q into a wff logically equivalent to p, instead of requiring that it literally become p? > He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. I think I can make that proof slightly shorter by considering the set of > propositions in the chain provable in <= n lines and considering how > .95newê primitives are introduced. Then, up to equivalence, there should > be only countably ('nitely?) many of these. Then a countable union of > countable sets is countable. Not sure if thatês going to work, but it > looks plausible. David === Subject: Re: Two questions in propositional logic >>Iêve just 'nished being supervised on an example sheet for our .95Logic, >>Computation and Set Theoryê course, and there were two questions which >>neither I, my partner, nor my superviser could actually answer. I was >>wondering if someone could give me some hints so that I could try and >>sort these out. (Preferably not more than hints, as some people havenêt >>been supervised on this yet. Donêt want to spoil it for them. :) >>Weêre working in propositional logic, with our basic symbols being >>implies, falsehood and p_1, ... . The axioms are >>p -> (q -> p) >>[ p -> (q -> r) ] -> [ (p -> q) -> (p -> r) ] >>p -> p >>(Where p = ( p -> F ) ). >>Fairly standard, but people use some slightly different axioms so I just >>thought I should say which ones Iêm using. > You also need to specify what the inference rules are... > Sorry, of course. The only rule of inference is modus ponens. >2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds This is presumably under the assumption that C is uncountable? or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a co'nal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a co'nal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I canêt quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. ??? > Unless youêre using some extra property about the p_i, q_i in saying > this that Iêm not seeing (which is very possible, on account of it being > Early here and I just got up. :), that neednêt be true. For example take the following sequence: p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. Pointed out a few minutes ago that this is not a counterexample. Of course what I said is wrong - a counterexample is p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. (Realized I wanted to show that the intersection of a strictly decreasing sequence of open subsets of the Cantor set had empty interior. Realized that that was false. Now, I bet that thereês no strictly decreasing omega_1-sequence of subsets of the Cantor set, but never mind, the proof you already have is simpler.) Sorry if I already said this - I just had a PC go up in smoke, not sure where I was when that happened... > I thought at one point that you could have a chain isomorphic to any > countable ordinal, although in retrospect Iêm not entirely convinced by > my argument (Iêm slightly worried about what happens with some of the > bigger limit ordinals), but this doesnêt really matter. Of course not > every chain is isomorphic to an ordinal, as the reverse of a chain is > also a chain. Anyway, as I said in the other post I think Iêve got this sorted out now > (although Iêll want to write my own proof of it before Iêm fully happy > with it). If youêre interested, Johnês proof went roughly as follows: De'ne an equivalence relation on C by p ~ q if there is a bijection f > from the set of primitive propositions to itself, such that q = p with > all the p_i in p replaced with f(p_i). He then showed that there can be at most one element of each equivalence > class in C, and that there are countably many equivalence classes > greater than p (because you can take some countable subset of the whole > set of primitive propositions and represent every formula in C as a > formula using only this countable subset and the primitives that appear > in p). He then went on to show there were only countably many elements less > than p in a similar manner. I pointed out that this was rather easier, > as you could just reverse the chain by negation and use the previous > result. I think I can make that proof slightly shorter by considering the set of > propositions in the chain provable in <= n lines and considering how > .95newê primitives are introduced. Then, up to equivalence, there should > be only countably ('nitely?) many of these. Then a countable union of > countable sets is countable. Not sure if thatês going to work, but it > looks plausible. David === Subject: Re: Two questions in propositional logic Unless youêre using some extra property about the p_i, q_i in saying >>this that Iêm not seeing (which is very possible, on account of it being >>Early here and I just got up. :), that neednêt be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. > Pointed out a few minutes ago that this is not a counterexample. > Of course what I said is wrong - a counterexample is p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. Oops. Should have read this post before my last one. Thatês exactly the counterexample I had in mind when I made the (wrong) counterexample this morning, I was just half asleep and misremembering. :) > (Realized I wanted to show that the intersection of a strictly > decreasing sequence of open subsets of the Cantor set had > empty interior. Realized that that was false. Now, I bet > that thereês no strictly decreasing omega_1-sequence of > subsets of the Cantor set, but never mind, the proof you > already have is simpler.) > Are you sure that the cantor set approach works? Also, do you mean {0,1}^X for some possibly uncountable X rather than the cantor set? (which is homeomorphic when X is countable). Because I tried that, and I couldnêt come to the conclusion that you could get the association to work both ways - you could get a clopen subset of {0, 1}^X for every proposition, but I wasnêt sure you could go the other way... it looked like if you chose some suf'ciently nasty clopen set it wouldnêt correspond to a proposition. (I didnêt prove that you couldnêt, but it looked like it was going to go badly wrong if I tried to prove that you could, so I gave up on that approach). David === Subject: Re: Two questions in propositional logic >[...] >2. Let C be a set of propositions. We say C is a chain if for every p, q >in C either p proves q or q proves p (and not both). If the set of >primitive propositions is allowed to be uncountable, can there be an >uncountable chain? >> I doubt it. Say p < q if p |- q but not q |- p; then your chain is >> totally ordered by <. First, either there exists an increasing >> sequence with at least two upper bounds This is presumably under the assumption that C is uncountable? Yes. >> or a decreasing sequence with at least two lower bounds >> (by the traditional proof that any sequence of reals contains >> a monotone subsequence: Say you have elements p_a, indexed by the >> countable ordinals. Say a is dominant if p_a > p_b for all b > a. >> Either there are uncountably many dominant a or not. If there >> are uncountably many dominant a then the p_a with a dominant >> give a co'nal decreasing set, while if there are only countably >> many dominant a then the dominant a are bounded above >> and you get a co'nal increasing subset.) So, changing the >> notation and replacing all the wffs by their negations if >> needed, you have >> p_1 < p_2 < ... < q_1 < q_2. >> It seems clear to me that this implies q_1 must be a >> tautology (it seems like this is by compactness or >> something, but I canêt quite prove it this second) >> and then q_2 is weaker than a tautology, contradiction. >> ??? > >Unless youêre using some extra property about the p_i, q_i in saying >this that Iêm not seeing No. >(which is very possible, on account of it being >Early here and I just got up. :), that neednêt be true. For example take the following sequence: p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. ??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I donêt think so... Not that I can see how to prove what I said, and it may well be false, but this is not a counterexample, unless _Iêm_ missing something - I did think about simple examples like this. In fact assuming that the p_j are propositional _variables_ then if p_3 v ... v p_n |- w for all n then it does follow that w is a tautology: If w is not a tautology then it is falsi'ed by some truth assignment. Since w contains only 'nitely many variables, it is falsi'ed by some truth assignment that sets some irrelevant variable to T, and that shows that p_3 v ... v p_n |- w is false for some n. I actually 'gured that out last night - realizing that _that_ chain cannot be followed by anything but a tautology is why I conjecture that no increasing sequence can be followed by anything but a tautology. (Any increasing sequence has the same form as above (identifying wffs that are logically equivalent), except that the p_j are wffs instead of variables. It seems like one should be able to use more or less the same argument, sort of, although I havenêt worked out _exactly_ how it goes...) >I thought at one point that you could have a chain isomorphic to any >countable ordinal, although in retrospect Iêm not entirely convinced by >my argument (Iêm slightly worried about what happens with some of the >bigger limit ordinals), but this doesnêt really matter. Of course not >every chain is isomorphic to an ordinal, as the reverse of a chain is >also a chain. Anyway, as I said in the other post I think Iêve got this sorted out now >(although Iêll want to write my own proof of it before Iêm fully happy >with it). If youêre interested, Johnês proof went roughly as follows: 'rst. (Sorry if thereês something below I should have replied to that Iêm appearing not to notice - youêve forced me to stop reading at this point..) [snipped with eyes shut] ************************ David C. Ullrich === Subject: Re: Two questions in propositional logic > >>[...] >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? >I doubt it. Say p < q if p |- q but not q |- p; then your chain is >totally ordered by <. First, either there exists an increasing >sequence with at least two upper bounds >>This is presumably under the assumption that C is uncountable? > Yes. >or a decreasing sequence with at least two lower bounds >(by the traditional proof that any sequence of reals contains >a monotone subsequence: Say you have elements p_a, indexed by the >countable ordinals. Say a is dominant if p_a > p_b for all b > a. >Either there are uncountably many dominant a or not. If there >are uncountably many dominant a then the p_a with a dominant >give a co'nal decreasing set, while if there are only countably >many dominant a then the dominant a are bounded above >and you get a co'nal increasing subset.) So, changing the >notation and replacing all the wffs by their negations if >needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a >tautology (it seems like this is by compactness or >something, but I canêt quite prove it this second) >and then q_2 is weaker than a tautology, contradiction. ??? >Unless youêre using some extra property about the p_i, q_i in saying >>this that Iêm not seeing > No. >>(which is very possible, on account of it being >>Early here and I just got up. :), that neednêt be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. > ??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I donêt think so... Gah. Youêre right. I was trying to construct that counterexample from memory way too early in the morning and it didnêt work. :) What it is *meant* to say is the following: p_1 ^ p_3, p_1 ^ (p_3 v p_4), ... p_1, p_1 v p_2 If youêre interested, Johnês proof went roughly as follows: 'rst. > (Sorry if thereês something below I should have replied > to that Iêm appearing not to notice - youêve forced me to > stop reading at this point..) [snipped with eyes shut] > Heh. No problem. :) There wasnêt anything below it except a sketch of Johnês proof and my possible proof approach based loosely on what he did. David === Subject: Re: Two questions in propositional logic >[...] >>2. Let C be a set of propositions. We say C is a chain if for every p, q >>in C either p proves q or q proves p (and not both). If the set of >>primitive propositions is allowed to be uncountable, can there be an >>uncountable chain? > I doubt it. Say p < q if p |- q but not q |- p; then your chain is > totally ordered by <. First, either there exists an increasing > sequence with at least two upper bounds >>This is presumably under the assumption that C is uncountable? Yes. or a decreasing sequence with at least two lower bounds > (by the traditional proof that any sequence of reals contains > a monotone subsequence: Say you have elements p_a, indexed by the > countable ordinals. Say a is dominant if p_a > p_b for all b > a. > Either there are uncountably many dominant a or not. If there > are uncountably many dominant a then the p_a with a dominant > give a co'nal decreasing set, while if there are only countably > many dominant a then the dominant a are bounded above > and you get a co'nal increasing subset.) So, changing the > notation and replacing all the wffs by their negations if > needed, you have p_1 < p_2 < ... < q_1 < q_2. It seems clear to me that this implies q_1 must be a > tautology (it seems like this is by compactness or > something, but I canêt quite prove it this second) > and then q_2 is weaker than a tautology, contradiction. ??? >Unless youêre using some extra property about the p_i, q_i in saying >>this that Iêm not seeing No. >(which is very possible, on account of it being >>Early here and I just got up. :), that neednêt be true. >>For example take the following sequence: >>p_3, p_3 v p_4, p_3 v p_4 v p_5 ... , p_1 v p_3 , p1 v p_2 v p_3. ??? Is it true that p_3 v p_4 |- p_1 v p_3 ? I donêt think so... Not that I can see how to prove what I said, and it may well >be false, but this is not a counterexample, unless _Iêm_ >missing something - I did think about simple examples >like this. In fact assuming that the p_j are propositional >_variables_ then if p_3 v ... v p_n |- w for all n then it >does follow that w is a tautology: If w is not a tautology then >it is falsi'ed by some truth assignment. Since w contains >only 'nitely many variables, it is falsi'ed by some >truth assignment that sets some irrelevant variable to T, >and that shows that p_3 v ... v p_n |- w is false for some n. I actually 'gured that out last night - realizing that _that_ >chain cannot be followed by anything but a tautology is >why I conjecture that no increasing sequence can be >followed by anything but a tautology. (Any increasing >sequence has the same form as above (identifying >wffs that are logically equivalent), except that the p_j >are wffs instead of variables. It seems like one should >be able to use more or less the same argument, >sort of, although I havenêt worked out _exactly_ >how it goes...) Never mind - ths above is not a counterexample, but of course what I conjectured is false. Realized that I wanted to show that the intersection of any strictly decreasing sequence of open subsets of the Cantor set had empty interior, realized immediately that that was false, hence the example p_1 & (p_2), p_1 & (p_2 v p_3), ... p_1. (Well, if you had an uncountable chain you could get an increasing chain isomorphic to omega_1 as above, and I bet itês impossible to have a strictly decreasing omega_1-sequence of open subsets of the Cantor set... but never mind, you say you have a proof.) >>I thought at one point that you could have a chain isomorphic to any >>countable ordinal, although in retrospect Iêm not entirely convinced by >>my argument (Iêm slightly worried about what happens with some of the >>bigger limit ordinals), but this doesnêt really matter. Of course not >>every chain is isomorphic to an ordinal, as the reverse of a chain is >>also a chain. >>Anyway, as I said in the other post I think Iêve got this sorted out now >>(although Iêll want to write my own proof of it before Iêm fully happy >>with it). >>If youêre interested, Johnês proof went roughly as follows: > 'rst. >(Sorry if thereês something below I should have replied >to that Iêm appearing not to notice - youêve forced me to >stop reading at this point..) [snipped with eyes shut] >************************ David C. Ullrich ************************ David C. Ullrich === Subject: Re: teaching descriptive statistics / proposal > The bene'ts, you say, are that the chiral index is > (a) simpler to compute by hand, and > (b) equal to zero *only* for symmetry. And, a .95drawbackê of skewness is that in some conceivable > distributions, which no one has ever seen in the wild(?), it is > possible to have a skewness of zero while lacking symmetry. To me, that does not seem like much to gain. > Any parameter-estimation, which is derived by aggregation/summation of some values can be cheaten by appropriate data. With skewness one outlier on the one hand can compensate a cluster of data on the other hand making the additive parameter zero. With for appropriately elaborated cases. So it may be of more interest, *in which way* our parameter can be cheaten, and if the one or the other re¤ects my intuition better, as I understand you, too: > Now, what I would 'nd valuable is an index that tells me > how badly the outliers are going to disrupt my oneway > ANOVA, and another that shows how the scaling will > effect analyses that are more complicated. - I am pretty > sure that those are different, because the simple ANOVA > is robust to analyzing *ranks* ; whereas ranks are *not* > very compatible with the multi-way ANOVA or regression > where the R-squared is large. > Since I didnêt 'nd powers of 3 in that computation I assume, that the coef'cient is less sensible against outliers. But I didnêt understand much and would like to get more information beforehand. Gottfried Helms === Subject: Extremely bizarre number theoretical property 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 Hereês some examples: s(9^2) = 11^2 s(20^2) = 31^2 s(180^2) = 341^2 s(1306^2) = 1729^2 s(1910^2) = 2821^2 s(11754^2) = 19019^2 s(17190^2) = 31031^2 s(32486^2) = 43617^2 s(38423^2) = 43491^2 s(47576^2) = 68961^2 s(48202^2) = 72219^2 s(50920^2) = 82677^2 s(51590^2) = 86583^2 s(83884^2) = 117831^2 (due to Jose Luis Gomez Pardo) Whatês VERY VERY VERY REALLY SUPER interesting, is that there are numbers x and y such that a) x and y are perfect squares b) s(x) and s(y) are perfect squares c)!!! s(xy)=s(x)s(y). Anyone who has taken some algebra should recognize this property immediately. Hereês an example of some numbers with this property (calculation can be veri'ed by using the above table) s(9^2*20^2) = 341^2 = 11^2*31^2 = s(9^2)*s(20^2) It turns out that just using the above table you can easily 'nd more numbers like this. If you write a maple function to calculate the Sqrt(DivisorSum(a^2*b^2)), you can easily 'nd more numbers like this that are not on the table above. This is one of the most amazing things I have ever seen. I am going to investigate this further, but the problem is quite complex, and I doubt Iêll discover anything. Itês a shame it doesnêt always work though, because that would be utterly remarkable. But does anybody have any ideas what might be going on here? === Subject: Re: Extremely bizarre number theoretical property > 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 Hereês some examples: annotated, with factors of sqrt(n): > s(9^2) = 11^2 3 3 > s(20^2) = 31^2 2 2 5 > s(180^2) = 341^2 [non-primitive - 9*20] > s(1306^2) = 1729^2 2 653 > s(1910^2) = 2821^2 2 5 191 > s(11754^2) = 19019^2 [non-primitive - 9*1306] > s(17190^2) = 31031^2 [non-primitive - 9*1910] > s(32486^2) = 43617^2 2 37 439 > s(38423^2) = 43491^2 7 11 499 > s(47576^2) = 68961^2 2 2 2 19 313 > s(48202^2) = 72219^2 2 7 11 313 > s(50920^2) = 82677^2 2 2 2 5 19 67 > s(51590^2) = 86583^2 2 5 7 11 67 > s(83884^2) = 117831^2 2 2 67 313 104855 117831 5 67 313 132682 187131 2 11 37 163 198534 347529 2 3 7 29 163 247863 347529 3 7 11 29 37 292374 479787 [non-primitive - 9*32486] 300876 503347 2 2 3 25073 312374 414309 2 313 499 313929 436107 3 3 3 7 11 151 334330 496713 2 5 67 499 345807 478401 [non-primitive - 9*38423] 376095 503347 3 5 25073 428184 758571 [non-primitive - 9*47576] 433818 794409 [non-primitive - 9*48202] 458280 909447 [non-primitive - 9*50920] 464310 952413 [non-primitive - 9*51590] 469623 658749 3 7 11 19 107 498892 696787 2 2 191 653 contains 1306=2*653 623615 696787 5 191 653 754956 1296141 [non-primitive - 9*83884] 768460 1348221 [non-primitive - 20*38423] 787127 828723 11 163 439 943695 1296141 [non-primitive - 9*104855] 985369 1125579 7 11 67 191 1194138 2058441 [non-primitive - 9*132682] 1276880 2106853 2 2 2 2 5 11 1451 1378608 2650557 2 2 2 2 3 7 11 11 373 1547052 2728713 2 2 3 13 47 211 1606281 1993719 3 29 37 499 1676904 2854579 2 2 2 3 107 653 contains 1306=2*653 1754039 1903629 7 83 3091 1933815 2728713 3 5 13 47 211 2015094 3310671 2 3 29 37 313 2034423 2501877 3 3 3 151 499 2156730 3969147 2 3 5 29 37 67 2181409 2322957 19 29 37 107 2452440 4657471 2 2 2 3 5 107 191 2552202 4154493 2 3 3 3 151 313 2731590 4980801 2 3 3 3 5 67 151 2811366 4557399 [non-primitive - 9*312374] 3008970 5463843 [non-primitive - 9*334330] 3043401 3779139 3 19 107 499 3817974 6275451 2 3 19 107 313 4086330 7523607 2 3 5 19 67 107 4490028 7664657 [non-primitive - 9*498892] 4957260 10773399 [non-primitive - 20*247863] 5147241 6540807 3 11 61 2557 5545617 7536711 3 7 11 24007 5570344 8390001 2 2 2 13 19 2819 5612535 7664657 [non-primitive - 9*623615] 5643638 8786379 2 7 11 13 2819 5805336 10287381 2 2 2 3 19 29 439 5881722 10773399 2 3 7 11 29 439 6278580 13519317 [non-primitive - 20*313929] 6385703 6457269 67 191 499 6403664 9346701 2 2 2 2 29 37 373 6416984 10185273 2 2 2 7 19 37 163 6916140 14830431 [non-primitive - 9*20*38423] 7084143 9115953 [non-primitive - 9*787127 7542184 11178921 2 2 2 13 47 1543 7642712 12516231 2 2 2 7 11 19 653 contains 1306=652*2 8010922 10722621 2 67 191 313 8868321 12381369 [non-primitive - 9*985369] 8934096 15205827 2 2 2 2 3 373 499 9392460 204212197 [non-primitive - 20*469623] 9821396 14335671 2 2 13 67 2819 Some prime, such as 37, 67, 107, 191, 313, 373, 439, 499, 653 are a factor of n quite frequently? I wonder why. Their squaresê sigma contributions are as follows: 107 7 13 127 191 7 31 13^2 373 3 13 73 7^2 439 3 67 31^2 499 3 7 109^2 And some other primes: (2 gives 7 more easily than 653, and 5 gives 31 more easily than 67.) I guess that when looked at in the simplest possible terms you can just create a binary matrix of non-square terms in sigma(p^2) for prime p, and 'nd combinations that sum to zero in exactly the same way as you do in QS (in fact, the similarity with QS is massive, as youêd want to sieve over a quadratic expression to form the vectors in the 'rst place). Iêd conjecture that there is an in'nite quantity of such numbers, and that itês not hard to construct one with pretty massive numbers of factors. Iêll leave that for someone else. Phil -- Unpatched IE vulnerability: Basic Authentication URL spoo'ng Description: Spoo'ng the URL displayed in the Address bar Reference: http://msgs.securepoint.com/cgi-bin/get/bugtraq0306/15.html === Subject: Re: Extremely bizarre number theoretical property > 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 ARe there in'nitely many? Whatês VERY VERY VERY REALLY SUPER interesting, is that there are > numbers x and y such that a) x and y are perfect squares > b) s(x) and s(y) are perfect squares > c)!!! s(xy)=s(x)s(y). Donêt forget that s is a multiplicative function: s(ab) = s(a)s(b) whenever a and b are coprime. Your VERY VERY VERY REALLY SUPER interesting fact follows from the fact that the set {n in N: n, s(n) are square} contains two coprime numbers. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Re: A Question for James Harris > Giggle. James Harris >http://mathforpro't.blogspot.com/ ************************ David C. Ullrich Nothing makes a pig happier than a roll in the wallow, as this giggling porker knows! http://www.shop4egifts.com/target.asp?item=/jpg/25112.jpg& width=314&height=4 00# http://www.fetch'do.co.uk/sound_'les/giggle.wav === Subject: Higher Dimensional Knot Theory Hi! Iêve recently become interested in knots, and I was just wondering if anyone was able to help me out... Iêve read that itês possible to generalise knot theory to higher dimensions by embedding n-manifolds in n+2 dimensional space. Is this purely topological, or is there dependence on the metric structure? What I mean is: do you need more dimensions to embed a pseudo-Riemannian manifold in a knotted way? The only reason I ask is that Campbellês theorem from differential geometry says you generally do need more dimensions for embedding pseudo-Riemannian manifolds. I wasnêt sure if this was the case for obtaining knottedness. Can anyone recommend a good book on higher dimensional knot theory? from Jonny Evans! === Subject: The Mathematics of the Human Soul boundary=----=_NextPart_000_005B_01C3A5E8.C601AC60 -------------------------------------------------------------- ------- General strategies of behavior acquisition with mathematically de'nable 'lters: What can this hypothesis contribute to a better understanding of man? Do you understand German? Click www2.arnes.si/~mkramb5 and choose the length of the answer (4 levels, from 1 to 270 book pages). You don?t understand German? Click www2.arnes.si/~mkramb5 and go to Level 2 for a brief presentation of the theory in English. === Subject: Re: Equation Object converting? Youêll need to write a macro using VBA. Most, if not all of the info you need, should be in one of the Word VBA books I have listed in the list of WordVBA books at my URL below. I suggest getting Steve Romanês Writing Word Macros. -- http://www.standards.com/; See Howard Kaikowês web site. Probably can do so with a quick macro. > To what version of Word are you converting? Please, give me more details how to do with a quick macro. > === Subject: Differentiation Is differentiation a one way function? More exactly Is differentiation a one way functional? So if I ask how to integrate complicated expression am I actually asking for help to break a code? ;-) ---- === Subject: Re: Differentiation Is differentiation a one way functional? Assuming you mean is solving the analytical form for the anti-derivative hard, I think this depends on your function space. For example, in P_n, the polynomial function space of degree n, itês clearly not one-way since you can easily differentiate and integrate any polynomial. === Subject: Re: torque T = r x F and basic tensors 2. How is torque a tensor? Following Feynman vol 2 ch 31, a tensor is a > linear map A relating 2 vector quantities, for example p = Ae (where I am > using p for the polatization vector of a dielectric and e for the electric > 'eld vector) and hence if we change the basis of R^3, A must transform as > Aê = Q^{-1} A Q where V is the new ordered basis, U is the old and > V = UQ^{-1} I can see Feynman gets the 3 by 3 anti-symmetric matrix > T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3 How would this matrix object ever be used for torque? If it has any relevance, I can derive the formula > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but Iêm not quite seeing > how that matches the Aê = Q^{-1} A Q form. As you say, under the change of orthonormal basis, the vectors a and b go to aê = Qa and bê = Qb. In component form, a_i = (sum over m) Q_(im) a_m and b_i = (sum over n) Q_(in) b_n Iêve 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. De'ne matrix A by A_(ij) = a_i b_j - a_j b_i. Under a transformation, Aê_ij = aê_i bê_j - aê_j bê_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 de'nes things, aê_i = (sum over n) Q^(-1)_(in) a_n. Itês a good exercise to show this. George === Subject: Re: torque T = r x F and basic tensors Here are 2 questions which perhaps are related: 1. Why is there a cross product in the de'nition of torque T = r x F ? > In Feynmanês Lectures on Physics, he derives torque in the xy plane > as T = xF_x - yF_y (where _x means subscript) in Vol 1 Ch 18. But in Ch 20 he generalizes it into 3 dimensions, and I canêt quite follow > that, especially how we would know, a priori, to permute the x_1, x_2, x_3 > symbols in the way for a right handed co-ord system. (Of course I see that r x F in 3 dimensions reduces to what Feynman gives in > the xy plane). >2. How is torque a tensor? Following Feynman vol 2 ch 31, a tensor is a > linear map A relating 2 vector quantities, for example p = Ae (where I am > using p for the polatization vector of a dielectric and e for the electric > 'eld vector) and hence if we change the basis of R^3, A must transform as > Aê = Q^{-1} A Q where V is the new ordered basis, U is the old and > V = UQ^{-1} I can see Feynman gets the 3 by 3 anti-symmetric matrix > T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3 How would this matrix object ever be used for torque? If it has any relevance, I can derive the formula > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but Iêm not quite seeing > how that matches the Aê = Q^{-1} A Q form. Jeff > I get the impression you think the torque is in the xy plane > (he derives torque in the xy plane) > as you have not attached a z to your torque equation. Howês this: > T_z = F_x x R_y, or > T3 = F1 x R2. > The torque is always normal to both the other vectors. The tensor equation is T_m = Eps_ikm F_i R_k, where Eps_ikm is a third > rank tensor. Mr. Dual Space > (If you have something to say, write an equation. > If you have nothing to say, write an essay). the twisting about an axis, so we would need a vector for the axis and a magnitude for the amount of twisting. I also know what the permutation symbol Eps_ikm is, so I understand the equation, I just donêt see how one would derive it. Here is what I am thinking now: If the moment arm r and the force F are both in the xy plane, then it is easy to see that the twisting is about the z axis, and we have the formula xF_y - yF_x for the magnitude. Now we want to derive a formula for an arbitrary plane in R^3. Maybe the matrix representation of torque gives the amount of twisting about an arbitrary axis? ( which would de'ne an arb plane ). In that sense, T is an operator...? I havenêt thought this through yet... --JEff === Subject: Re: torque T = r x F and basic tensors Here are 2 questions which perhaps are related: 1. Why is there a cross product in the de'nition of torque T = r x F ? > In Feynmanês Lectures on Physics, he derives torque in the xy plane > as T = xF_x - yF_y (where _x means subscript) in Vol 1 Ch 18. I believe you have the subcripts backwards. That should be xF_y - > yF_x. But in Ch 20 he generalizes it into 3 dimensions, and I canêt quite follow > that, especially how we would know, a priori, to permute the x_1, x_2, x_3 > symbols in the way for a right handed co-ord system. When we talk about elements of rotational motion in general, we > usually talk in terms of cross products. A cross product of two > vectors A and B is de'ned as A x B = (A_y B_x - A_x B_y) i + > (A_z B_y - A_y B_z) j + (A_x B_y - A_y B_x) k , where i, j, and k are > the unit vectors along the x,y, and z axes respectively. Maybe youêve > noticed the pattern there. Anyway, itês relatively easy to remember > once you see the pattern. Yes, as I said to another poster, I know what a cross product is and how to compute one, but I donêt quite see *why* torque is de'ned by a cross product. > (Of course I see that r x F in 3 dimensions reduces to what Feynman gives in > the xy plane). >2. How is torque a tensor? Following Feynman vol 2 ch 31, a tensor is a > linear map A relating 2 vector quantities, for example p = Ae (where I am > using p for the polatization vector of a dielectric and e for the electric > 'eld vector) and hence if we change the basis of R^3, A must transform as > Aê = Q^{-1} A Q where V is the new ordered basis, U is the old and > V = UQ^{-1} Yes, torque is a second rank antisymmetric tensor. A tensor is a > mathematical set of objects that transforms in a one-to-one manner. > In other words, no matter how I rotate my system or my coordinates, > each component of the torque will map into a linear combination of the > three original components given above. > I can see Feynman gets the 3 by 3 anti-symmetric matrix > T_{ij} = x_i F_j - x_j F_i i,j = 1,2,3 How would this matrix object ever be used for torque? The indices i and j denote coordinate axes here. x is represented by > 1, y by 2, and z by 3. That tends to be standard notation in tensor > analysis. It allows for brevity when youêre dealing with several > equations at once. If it has any relevance, I can derive the formula > Q( a x b ) = 1/(detQ) ( Qa x Qb ) > where Q is the matrix of the change of basis, but Iêm not quite seeing > how that matches the Aê = Q^{-1} A Q form. It wonêt match it. That latter equation is only true for vectors > ('rst rank tensors). I think you might be wrong here. A vector transforms uê = Qu, and thus so should torque if it were really a (polar) vector. But the cross product makes it a psuedo-vector, or so all the books say. So if torque is really the 2nd rank anti-symm matrix I described above, it must transform as Tê = Q^{-1} T Q, right? === Subject: Re: torque T = r x F and basic tensors > Here are 2 questions which perhaps are related: 1. Why is there a cross product in the de'nition of torque T = r x F ? > In Feynmanês Lectures on Physics, he derives torque in the xy plane > as T = xF_x - yF_y (where _x means subscript) in Vol 1 Ch 18. I think you got the subscripts mixed up there. Yes, thank you. I meant xF_y - yF_x. But in Ch 20 he generalizes it into 3 dimensions, and I canêt quite > follow > that, especially how we would know, a priori, to permute the x_1, x_2, > x_3 > symbols in the way for a right handed co-ord system. r x F can be easily calculated in Cartesian components as the determinant of > a 3x3 matrix whose rows/columns are the unit vectors, r, and F. Expansion > by minors gives the components of the cross product. Expanding r x F in > component form gives the same result if you assume i x j = k, j x k = i, > etc. and basic properties of the cross product (distributivity). Yes, I do know what a cross product is and how to compute it. I am wondering *why* torque is de'ned in terms of a cross product. --Jeff === Subject: Re: torque T = r x F and basic tensors > Yes, I do know what a cross product is and how to compute it. I am wondering > *why* torque is de'ned in terms of a cross product. --Jeff Because of the geometrical interpretation of cross product as the product of force and moment arm, or equivalently as the product of displacement and force perpendicular to displacement. This de'nition of torque follows intuition and is mathematically simple. Halliday and Resnick discusses this. === Subject: Re: torque T = r x F and basic tensors Yes, I do know what a cross product is and how to compute it. I am > wondering > *why* torque is de'ned in terms of a cross product. --Jeff Because of the geometrical interpretation of cross product as the product of > force and moment arm, or equivalently as the product of displacement and > force perpendicular to displacement. This de'nition of torque follows > intuition and is mathematically simple. Halliday and Resnick discusses > this. 1. The 2 vectors r and F determine a plane in R^3 2. In this plane, let F_p be the component of F perpendicular to r. Then it makes perfect sense that |T| = |r||F_p| = |r||F| sin(a) where a is the angle between r and F. 3. If we adopt the convention that turning according to the right hand rule is positive, then we have the vector equation T = |r||F| sin(a) n where n is the unit normal to plane and this is exactly r x F. I also see how any cross product of vectors transforms as an axial vector. I have no idea why anyone would want to form a 3 by 3 anti-symmetric matrix from the components of an array of 3 numbers, like the components of the T above... What good does it do, and why does Feynman seem to claim that this is the real quantity of interest? vol 2 p. 31-8 === Subject: parallelizability of manifolds In my self-studies of differential topology, I am not sure if I understood the concept of parallelizability correctly. A n-dim. manifold M is said to be parallelizable iff its tangent bundle TM is diffeomorphic to (M x R^n). Is this equivalent to the following: two tangent vectors (at maybe different points on the manifold) are identical in one chart if and only if they are identical in all charts (with the appropriate range). The following argument makes me think that this is false, but I do not quite understand why. I have read that S^1, S^3 and S^7 are parallelizable. Because there are no speci'c charts mentioned, I assume that this is the case for every atlas equivalent to the atlas consisting of the two stereographic projection charts. But this is not true; even for S^1, it is easy to show that, given any non-constant differentiable curve on S^1, tangent vectors in one chart may be parallel while they are not in the other one. So how to imagine parallelizability? And how can it be proven? Are the tori T^n:=(S^1)^n parallelizable? TIA, Tobias -- reverse my forename for mail! === Subject: Re: parallelizability of manifolds >In my self-studies of differential topology, I am not sure if I understood >the concept of parallelizability correctly. >A n-dim. manifold M is said to be parallelizable iff its tangent bundle TM >is diffeomorphic to (M x R^n). Is this equivalent to the following: two >tangent vectors (at maybe different points on the manifold) are identical >in one chart if and only if they are identical in all charts (with the >appropriate range). Itês hard for me to make sense of this last sentence, but in the best rendering I can give it, the answer is, no, thatês not what parallelizable means. >So how to imagine parallelizability? And how can it be proven? Surely the simplest way to understand the concept of parallelizability is this: the n-manifold M is parallelizable if and only if there exist n vector'elds V_i on M with the property that, at each point x of M, the n vectors V_i(x) are a basis of the tangent space of M at x. No imagination required. >Are the >tori T^n:=(S^1)^n parallelizable? Yes, indeed. (And, more generally, the product of parallelizable manifolds is parallelizable.) Lee Rudolph === Subject: Re: parallelizability of manifolds Surely the simplest way to understand the concept of parallelizability > is this: the n-manifold M is parallelizable if and only if there exist > n vector'elds V_i on M with the property that, at each point x of M, > the n vectors V_i(x) are a basis of the tangent space of M at x. > No imagination required. > Ah, thatês interesting! So why are there no such vector 'elds for S^2? Yes, indeed. (And, more generally, the product of parallelizable > manifolds is parallelizable.) > Ok, this is clear because the tangent bundle of a product is the product of the tangent bundles. -- reverse my forename for mail! === Subject: Re: parallelizability of manifolds >> Surely the simplest way to understand the concept of parallelizability >> is this: the n-manifold M is parallelizable if and only if there exist >> n vector'elds V_i on M with the property that, at each point x of M, >> the n vectors V_i(x) are a basis of the tangent space of M at x. >> No imagination required. >Ah, thatês interesting! So why are there no such vector 'elds for S^2? Thereês not even the *beginning* of such a sequence of vector'elds for S^2: no matter what the vector'eld V_1 on S^2, there is always some point x of S^2 at which V_1(x) does not belong to any basis of the tangent space of S^2 at x. Thatês just a (seemingly more complicated, but ultimately worthwhile) way of saying that every (continuous!) vector'eld on S^2 has at least one zero. And that, in turn, follows (after building the appropriate machinery relating differential topology to algebraic topology) from the fact that the Euler characteristic of S^2 is non-zero. In general, you could (and people do) ask, given an n-manifold M, what is the smallest k such that there is some sequence of k vector'elds V_i on M such that, at every point x of M, the vectors V_i(x) span the tangent space of M at x. Alternatively (somewhat dually) you could (and people do) ask, given a generic sequence of k vector'elds V_i on M, what is the nature of the subset of M at every point of which the vectors V_i(x) are linearly dependent. If you pursue such questions, you eventually 'nd yourself studying characteristic classes (in homology as well as in cohomology); the Euler class is just the tip of that particular iceberg. Lee Rudolph === Subject: Re: parallelizability of manifolds | |>> Surely the simplest way to understand the concept of parallelizability |>> is this: the n-manifold M is parallelizable if and only if there exist |>> n vector'elds V_i on M with the property that, at each point x of M, |>> the n vectors V_i(x) are a basis of the tangent space of M at x. |>> No imagination required. |>> |>Ah, thatês interesting! So why are there no such vector 'elds for S^2? | |Thereês not even the *beginning* of such a sequence of vector'elds |for S^2: no matter what the vector'eld V_1 on S^2, there is always |some point x of S^2 at which V_1(x) does not belong to any basis of |the tangent space of S^2 at x. Thatês just a (seemingly more |complicated, but ultimately worthwhile) way of saying that every |(continuous!) vector'eld on S^2 has at least one zero. you should at least mention something about hedgehogs or coconuts here. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: parallelizability of manifolds Thereês not even the *beginning* of such a sequence of vector'elds > for S^2: no matter what the vector'eld V_1 on S^2, there is always > some point x of S^2 at which V_1(x) does not belong to any basis of > the tangent space of S^2 at x. Thatês just a (seemingly more > complicated, but ultimately worthwhile) way of saying that every > (continuous!) vector'eld on S^2 has at least one zero. And that, > in turn, follows (after building the appropriate machinery relating > differential topology to algebraic topology) from the fact that the > Euler characteristic of S^2 is non-zero. > Indeed, algebraic topology seems to be an extremely powerful tool: S^2 is simply connected, so every continuous image of it also is; in contrast, R^2{(0,0)} is not. Realizing that this gives the proof for S^2 was a pretty cool heureka-moment! Maybe it is possible to use higher homotopy groups for the n-spheres with n>2, but one has to be careful for n=7 :-) > In general, you could (and people do) ask, given an n-manifold M, > what is the smallest k such that there is some sequence of k > vector'elds V_i on M such that, at every point x of M, the > vectors V_i(x) span the tangent space of M at x. Alternatively > (somewhat dually) you could (and people do) ask, given a generic > sequence of k vector'elds V_i on M, what is the nature of the > subset of M at every point of which the vectors V_i(x) are > linearly dependent. If you pursue such questions, you eventually > 'nd yourself studying characteristic classes (in homology as well > as in cohomology); the Euler class is just the tip of that particular > iceberg. Sounds very interesting, but Iêm afraid that it is a lot ahead of me now... -- reverse my forename for mail! === Subject: Re: parallelizability of manifolds > Thereês not even the *beginning* of such a sequence of vector'elds >> for S^2: no matter what the vector'eld V_1 on S^2, there is always >> some point x of S^2 at which V_1(x) does not belong to any basis of >> the tangent space of S^2 at x. Thatês just a (seemingly more >> complicated, but ultimately worthwhile) way of saying that every >> (continuous!) vector'eld on S^2 has at least one zero. And that, >> in turn, follows (after building the appropriate machinery relating >> differential topology to algebraic topology) from the fact that the >> Euler characteristic of S^2 is non-zero. >Indeed, algebraic topology seems to be an extremely powerful tool: S^2 is >simply connected, so every continuous image of it also is; in contrast, >R^2{(0,0)} is not. Realizing that this gives the proof for S^2 was a >pretty cool heureka-moment! ... Iêm not sure your eureka! moment was authentic. For example, S^3 is both simply connected and parallelizable. John Mitchell === Subject: Factorization dispute 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. Iêll 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 aês are, see http://mathforpro't.blogspot.com/ where more is explained.) I have a result that shows a problem with a de'nition that mathematicians have used for over a hundred years, 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 Iêve gathered, are at least trying to stand and 'ght, 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. Letês 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. Letês take .95em out. James Harris http://mathforpro't.blogspot.com/ === Subject: Re: Factorization dispute binomial factorizations, a fruitful 'eld of endeavor; whatês that one method, Partial Quotients, used in calculus?... I never really got the hang of it, but it was interesting. the funny thing about the Army (and all of the services, and the NSA etc.) is that they have lots of applied mathematicians ... and generals can ask them to look at stuff ... perhaps, they already have looked at monsieur Harris and his ordnances! > Letês take .95em out. > http://mathforpro't.blogspot.com/ --ils duces dêEnron! http://larouchepub.com/radio/index.html === Subject: Re: Factorization dispute > 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. Iêll 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 aês are, see http://mathforpro't.blogspot.com/ where more is explained.) I have a result that shows a problem with a de'nition that > mathematicians have used for over a hundred years, 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 Iêve gathered, > are at least trying to stand and 'ght, but because what I have is a > math proof, their claims are necessarily irrational. Thatês not really true. The people with brains stopped 'ghting mathematicians and their holistic continuum, the day after they invented it. Since itês still well-known throughtout the universe that theyêre the only people who believe that there are an in'nite number of dimensions in a universe that quite obviously only has three. I need you to stand up for the truth, here and now. Itês still impossible to stand up in a mathematics class, since theyêre only people who actually believe in in'nite value logic. And if you did stand up, all they would do is start chanting some some sort of political spirituals about the evil of things that arenêt positive. > Letês 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. Thatês not true. The only statement that has ever universally about science is that mathematicians do Gês, physicists do Eês, and the people with brains do brainy things. Fission is old, Fusion is new, the NSA is cold, and the U.N. is Blue. Letês take .95em out. > James Harris > http://mathforpro't.blogspot.com/ === Subject: Re: Factorization dispute > I have a result that shows a problem with a de'nition that > mathematicians have used for over a hundred years, No, you donêt. Your claim is false. > and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. They are challenging your (false) claim. No one is running, except in the sense that anyone would run from a maniac ¤inging manure. > Some posters, not professional mathematicians from what Iêve gathered, > are at least trying to stand and 'ght, but because what I have is a > math proof, their claims are necessarily irrational. Your proof is ¤awed. It has been thoroughly refuted many times. That makes the challenges rational, and your defense irrational. > I need you to stand up for the truth, here and now. Been there, done that. You ignore, or simply repost. > Letês chase these mathematician cowards down, and make them face the > music. ..and dance? > After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. What mathematicians seem to believe that they can just ignore problems? Iêve seen no evidence of that.You, on the other hand, have a consistent prior record of ignoring counter-examples and refutations of your arguments. > Letês take .95em out. James Harris > http://mathforpro't.blogspot.com/ Take .95em 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? -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Factorization dispute Letês take .95em out. James Harris Take .95em 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 thereês an ulterior motive. === Subject: Re: Factorization dispute Nothing. http://www.crank.net/harris.html Itês not every braying jackass that gets a whole page at crank.net -- Uncle Al http://www.mazepath.com/uncleal/ (Toxic URL! Unsafe for children and most mammals) Quis custodiet ipsos custodes? The Net! === Subject: Re: Factorization dispute [all newsgroups except sci.math snipped] mathematicians have used for over a hundred years, and rather than > face the result which follows from rather basic algebra mathematicians > are being pussies and running like scared cowards from the result. No you donêt. And nobodyês running. You canêt refute the errors others have pointed out and have resorted to ad hominem attacks and cut and pasting of your entire argument assuming that if you repeat it often enough, then people will fail to respond to one of you posts and you can then assume you must 'nally be right. > Some posters, not professional mathematicians from what Iêve gathered, > are at least trying to stand and 'ght, but because what I have is a > math proof, their claims are necessarily irrational. Nonsense. Maybe if your proof was ¤awless you could claim something like that, but you have a serious error in step 6 of your argument. (think about this - *everybody* who has responded to you has pointed this out). You assume that because you can divide the terms of your polynomial by 49 that you can show that two of the factors must be divisible by 7, but thatês just not true. You havenêt come close to proving that. And in your new and improved short form argument you start 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 You start off with your conclusion and work backward, proving what? Nothing at all. You *assume* that the variable terms of the 'rst two factors are divisible by 7. Nice trick. > I need you to stand up for the truth, here and now. Making appeals to the gallery again... Iêm in the gallery, and Iêll stand with the mathematicions on this one. Youêre argument is ¤awed. > Letês chase these mathematician cowards down, and make them face the > music. Hilarious. Make my music King Crimsonês 21st Century Schizoid Man. > After all physics had its challenges and physicists faced them, while > mathematicians seem to believe that they can just ignore problems. HIlarious. > Letês take .95em out. To do that you will need to correct your errors (if you can...). I will echo the advice another poster gave you yesterday. Get a text on Abstract Algebra and work through the theorems. Ask for help here when you get stuck. In a few months you will be a better amateur mathematician. Youêve already given 6-7 years to this effort. Taking a few months off for education will give you the potential for not wasting the next 7 years. --Stan Gula === Subject: Re: Factorization dispute 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 3Ausers.pandora.be === Subject: Re: Factorization dispute 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 > 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. Thatês because Iêve isolated the factors of 22, which is obvious by inspection. James Harris http://mathforpro't.blogspot.com/