mm-407 === Subject: Re: 2nd Order ODEs > hello. anyone think they can solve > y'' + y.y' = 0 > i know what the answer looks like, but have no idea how to get there. > any help would be great. Notice that this equation can be rewritten as y'' + (y^2/2)' = 0. Integrate this equation to obtain a first order ODE. === Subject: Re: 2nd Order ODEs > hello. anyone think they can solve > y'' + y.y' = 0 > i know what the answer looks like, but have no idea how to get there. > any help would be great. Since (y^2)' = 2 y.y' , your equation becomes y''+(1/2 y^2)'=0 and you can integrate this to get y' + 1/2 y^2 = C1 and now this is Riccati's equation. === athematicians are here under fire from me because they're betraying > their own discipline. Repeatedly I say I have a proof, and I can step > through it point-by-point, but they ignore that fact, and keep trying > to produce side arguments. Side arguments may be vital -- although there are side arguments and side arguments. For example, one can claim that x^2 + 1 = 0 has no roots. Since all squares are positive in the real line a digression leads to some very important numbers, namely imaginary ones. (Why that particular term was chosen, I don't recall.) Real and imaginary yield complex numbers and now we can demonstrate that the claim is false. Or one can digress down various other pathways; for example, 1^2 + 1 = 0 (modulo 2) so we now have a unique solution, namely 1, modulo 2. Of course this isn't all that interesting a digression unless generalized to modulo 3, 4, 5, etc., but it eventually does lead to some very important developments, among them modern cryptography. Or one can digress down the route that the original poster is/was a bellyaching dumbass bereft of all human decency, intelligence, and such, etc. etc. doing impossibly complex things from a topological standpoint with his mother/a doughnut/the creature from the Black Lagoon/ and insult him to within a hairs-breadth of his life, completely ignoring the actual mathematical points of the argument by drowning them in thick, gooey, murky vitriol (a derivative of the mercaptan family, IIRC). I'm assuming you are complaining about the last, not the former two. > That's not mathematics, that's b.s. and it's a shame to the discipline > for them to behave that way. > I have a proof. I can step through that proof logical step, by > logical step, from beginning to end. > I'm putting it to you plainly: if mathematicians can lie here then you > must consider that they can lie elsewhere about mathematics. > My CHALLENGE to mathematicians is to play by rules and try and meet me > in a point-by-point step through my proof. > It's basic algebra after all. > Stop the running in circles. In order to analyze your proof one must see it, first. Did you have a Website that showed this wonderful proof? Or maybe a PDF document? What exactly are you trying to prove here? > -- === Subject: a couple abstract algebra questions The text I have is Fraleigh's A First Course in Abstract Algebra, 5th edition. My problem might be in understanding the notation, so I've tried to preserve it instead of restating it in my own words. Section 2.1 problem 37 (pg 98): Let A be a set, B a subset of A, and let b be one particular element of B. Determine whether the given set is sure to be a subgroup of SA under the induced operation. Here sigma[B] = { sigma(x) | x is an element in B }. (I'll note, SA is S subscript A, and it is the group of all permutations of A, the symmetric group). 37) { sigma an element of SA | sigma[B] is a subset of B } Now, to restate it, I take this to mean that it's the set of all elements of SA (individually denoted by sigma), where the set of all sigmas of elements of B is a subset of B. That is, the set of all permutations in SA which, applied to any element of B, yield an element in B. I am to find out if this is a group or not. Associativity: this is a given, since SA is a group, and all sigmas are elements of SA. Identity: the identity permutation fulfills the criteria, since applying the identity permutation to an element in B always yields an element in B. Inverse: There are two cases here to consider. (a) permutation sigma where all elements in B map to elements in B and no elements not in B map to elements in B. The inverse of this permutation would be a sigma as well, since it too would map all elements in B to other elements in B. (b) permutation sigma where all elements in B map to elements in B and some elements not in B map to elements in B. This is not possible since the permutation would not be a bijection and thus would not be a permutation, and would not be in the group, so the fact that its 'inverse' would not be in the group does not invalidate the group-ness. So I believe I have proved that indeed it is a subgroup. However the answer in the back of the book says No, it is not a subgroup (no inverse). Can you explain this? Section 2.1 problem 42. Strengthening exercise 41 (where I proved that Sn is a nonabelian group for n>=3), show that if n>=3, then the only element sigma of Sn satisfying sigma*x=x*sigma for all x in Sn is sigma=i, the identity permutation. ly I'm lost as to how to proceed. I was trying, given a permutation in Sn not the identity, to show I could construct a permutation in Sn that would violate commutativity. But now I'm wondering if induction might be a better way to go. Any tips? BTW, this may sound like homework, but I am working through this book on my own (not that this means I am looking to have answers fed to me, of course). === Subject: Re: a couple abstract algebra questions > Let A be a set, B a subset of A, and let b be one particular element > of B. Determine whether the given set is sure to be a subgroup of SA > under the induced operation. Here sigma[B] = { sigma(x) | x is an > element in B }. (I'll note, SA is S subscript A, and it is the group > of all permutations of A, the symmetric group). Huh, even tho B subset A, it's impossible B subset S_A much less subgroup. So you want to show B subset A ==> S_B subgroup S_A ? > 37) { sigma an element of SA | sigma[B] is a subset of B } > Now, to restate it, I take this to mean that it's the set of all > elements of SA (individually denoted by sigma), where the set of all > sigmas of elements of B is a subset of B. That is, the set of all > permutations in SA which, applied to any element of B, yield an > element in B. I am to find out if this is a group or not. Well actually only an isomorphic image of S_B is a subgroup S_A. The isomorphism being sort of like what you just described above. S_B is then the permutations of A that leave the elments of A-B unchanged. > Associativity: this is a given, since SA is a group, and all sigmas > are elements of SA. > Identity: the identity permutation fulfills the criteria, since > applying the identity permutation to an element in B always yields an > element in B. > Inverse: There are two cases here to consider. > (a) permutation sigma where all elements in B map to elements in B and > no elements not in B map to elements in B. The inverse of this > permutation would be a sigma as well, since it too would map all > elements in B to other elements in B. Ok. > (b) permutation sigma where all elements in B map to elements in B and > some elements not in B map to elements in B. This is not possible > since the permutation would not be a bijection and thus would not be a > permutation, and would not be in the group, so the fact that its > 'inverse' would not be in the group does not invalidate the > group-ness. I suppose or simply by the isomorphism, S_B is the elements of S_A that leave A-B unchanged. > So I believe I have proved that indeed it is a subgroup. However the > answer in the back of the book says No, it is not a subgroup (no > inverse). Can you explain this? Likely not as you haven't clearly and accurately stated the problem. However strictly speaking S_B isn't a subgroup of S_A as a permutation of B isn't a permutation of A. You have to convert a permutation of B into a permutation of A as indicated above. > Section 2.1 problem 42. > Strengthening exercise 41 (where I proved that Sn is a nonabelian > group for n>=3), show that if n>=3, then the only element sigma of Sn > satisfying sigma*x=x*sigma for all x in Sn is sigma=i, the identity > permutation. > ly I'm lost as to how to proceed. I was trying, given a > permutation in Sn not the identity, to show I could construct a > permutation in Sn that would violate commutativity. But now I'm > wondering if induction might be a better way to go. Any tips? I'll guess you should consider all the x's that are single transpositions. > BTW, this may sound like homework, but I am working through this book > on my own (not that this means I am looking to have answers fed to me, > of course). Why ask me? ;-) I distain permutation groups. === Subject: Re: a couple abstract algebra questions >The text I have is Fraleigh's A First Course in Abstract Algebra, >5th edition. My problem might be in understanding the notation, so >I've tried to preserve it instead of restating it in my own words. >Section 2.1 problem 37 (pg 98): >Let A be a set, B a subset of A, and let b be one particular element >of B. Determine whether the given set is sure to be a subgroup of SA >under the induced operation. Here sigma[B] = { sigma(x) | x is an >element in B }. (I'll note, SA is S subscript A, and it is the group >of all permutations of A, the symmetric group). >37) { sigma an element of SA | sigma[B] is a subset of B } >Now, to restate it, I take this to mean that it's the set of all >elements of SA (individually denoted by sigma), where the set of all >sigmas of elements of B is a subset of B. That is, the set of all >permutations in SA which, applied to any element of B, yield an >element in B. I am to find out if this is a group or not. >Associativity: this is a given, since SA is a group, and all sigmas >are elements of SA. >Identity: the identity permutation fulfills the criteria, since >applying the identity permutation to an element in B always yields an >element in B. >Inverse: There are two cases here to consider. >(a) permutation sigma where all elements in B map to elements in B and >no elements not in B map to elements in B. The inverse of this >permutation would be a sigma as well, since it too would map all >elements in B to other elements in B. >(b) permutation sigma where all elements in B map to elements in B and >some elements not in B map to elements in B. This is not possible >since the permutation would not be a bijection and thus would not be a >permutation, and would not be in the group, so the fact that its >'inverse' would not be in the group does not invalidate the >group-ness. No, this is your mistake. If both A and A-B are infinite, then there exist permutations sigma of A for which sigma(B) is a proper subset of B. The inverse of such a permutation would not satisfy sigma^-1(B) <= B, so the set you define is not a subgroup. For example, A = integers, B = positive integers, sigma(x) = x+1. >So I believe I have proved that indeed it is a subgroup. However the >answer in the back of the book says No, it is not a subgroup (no >inverse). Can you explain this? >Section 2.1 problem 42. >Strengthening exercise 41 (where I proved that Sn is a nonabelian >group for n>=3), show that if n>=3, then the only element sigma of Sn >satisfying sigma*x=x*sigma for all x in Sn is sigma=i, the identity >permutation. Choose a != b in A with sigma(a)=b. Now choose c in A unequal to a,b, and choose an x in Sn with x(a)=a, x(b)=c. Then sigma*x != x*sigma. Derek Holt. >ly I'm lost as to how to proceed. I was trying, given a >permutation in Sn not the identity, to show I could construct a >permutation in Sn that would violate commutativity. But now I'm >wondering if induction might be a better way to go. Any tips? >BTW, this may sound like homework, but I am working through this book >on my own (not that this means I am looking to have answers fed to me, >of course). === Subject: Re: a couple abstract algebra questions > Section 2.1 problem 37 (pg 98): > Let A be a set, B a subset of A, and let b be one particular element > of B. Determine whether the given set is sure to be a subgroup of SA > under the induced operation. Here sigma[B] = { sigma(x) | x is an > element in B }. (I'll note, SA is S subscript A, and it is the group > of all permutations of A, the symmetric group). > 37) { sigma an element of SA | sigma[B] is a subset of B } > Now, to restate it, I take this to mean that it's the set of all > elements of SA (individually denoted by sigma), where the set of all > sigmas of elements of B is a subset of B. That is, the set of all > permutations in SA which, applied to any element of B, yield an > element in B. I am to find out if this is a group or not. > Associativity: this is a given, since SA is a group, and all sigmas > are elements of SA. > Identity: the identity permutation fulfills the criteria, since > applying the identity permutation to an element in B always yields an > element in B. > Inverse: There are two cases here to consider. > (a) permutation sigma where all elements in B map to elements in B and > no elements not in B map to elements in B. The inverse of this > permutation would be a sigma as well, since it too would map all > elements in B to other elements in B. > (b) permutation sigma where all elements in B map to elements in B and > some elements not in B map to elements in B. This is not possible > since the permutation would not be a bijection and thus would not be a > permutation, and would not be in the group, so the fact that its > 'inverse' would not be in the group does not invalidate the > group-ness. > So I believe I have proved that indeed it is a subgroup. However the > answer in the back of the book says No, it is not a subgroup (no > inverse). Can you explain this? The explanation has already been posted -- but I thought I'd mention that your argument above can be used to show that you get a subgroup of S_A _if_ A (or even just B) is _finite_ ... For instance, if B is finite and sigma[B} is a subset of B, then sigma[B] = B (since injections and surjections are the same thing for _finite_ sets) and you can then conclude that sigma^{-1}[B} = B ... === Subject: Addition & multiplication of Complex numbers. Every complex number, a+bi=(a,b), corresponds to a single point in cartesian plane. Right? What do we mean then, when we add or multiply two complex numbers? When Adding: 1) Are we trying to figure out the sum of the distances of the two points from the origin (in the cartesian plane)? ( I assume not as that would have been the sum of the modulus' of the two complex numbers.) 2) Or are we doing what is taught to us in physics, addition of vectors by head to tail rule? When multiplying: 1) Multiplication of vectors? Well you are allowed to laugh if you feel that my assumptions are dumb. msk === Subject: Re: Addition & multiplication of Complex numbers. msk@linuxmail.org says... >Every complex number, a+bi=(a,b), corresponds to a single >point in cartesian plane. Right? >What do we mean then, when we add or multiply two >complex numbers? > When Adding: > 1) Are we trying to figure out the sum of the distances > of the two points from the origin (in the cartesian plane)? > ( I assume not as that would have been the sum of the modulus' > of the two complex numbers.) > 2) Or are we doing what is taught to us in physics, > addition of vectors by head to tail rule? It's 2). Complex addition is just like vector addition. > When multiplying: > 1) Multiplication of vectors? Multiplication of complex numbers is easier to understand using polar coordinates. Every vector (x,y) can be characterized by a length r (r = square-root(x^2 + y^2)) and an angle theta (the angle that the vector makes with the x-axis). So we can consider a complex number to be a pair (r,theta). Complex multiplication of two vectors (r1, theta1) and (r2, theta2) is that vector (r3, theta3) such that r3 = r1 * r2 theta3 = theta1 + theta2 So you multiply the lengths, but you *add* the angles. -- === Subject: Re: Addition & multiplication of Complex numbers. >Every complex number, a+bi=(a,b), corresponds to a single >point in cartesian plane. Right? >What do we mean then, when we add or multiply two >complex numbers? > When Adding: > 2) Or are we doing what is taught to us in physics, > addition of vectors by head to tail rule? Yup. > When multiplying: > 1) Multiplication of vectors? It's easier to understand multiplication if you look at a complex number as an angle (measured counterclockwise from east) and a distance from the origin. Multiplying two complex numbers means adding their angles and multiplying their distances. -- Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: Addition & multiplication of Complex numbers. > Every complex number, a+bi=(a,b), corresponds to a single > point in cartesian plane. Right? > What do we mean then, when we add or multiply two > complex numbers? A very profound question! > When Adding: > 1) Are we trying to figure out the sum of the distances > of the two points from the origin (in the cartesian plane)? > ( I assume not as that would have been the sum of the modulus' > of the two complex numbers.) > 2) Or are we doing what is taught to us in physics, > addition of vectors by head to tail rule? > When multiplying: > 1) Multiplication of vectors? > Well you are allowed to laugh if you feel that my assumptions are dumb. > msk Numbers are rather abstract entities. A very concrete way to think of addition or multiplication by a given complex number (point in the plane) is as a *transformation* of the plane: i) Addition corresponds to a translation that maps the origin to the given point. Thus two numbers add as vectors. ii) (a) Multiplication by a positive real number r corresponds to dilation of the plane by the factor r (same as for the real line). This dilation maps the point (1,0) to (r,0). (b) Rotation by angle t maps (1,0) to (cos t,sin t) and corresponds to multiplication by the complex number (cos t)+i(sin t). (c) Successive rotations and dilations correspond to multiplication of the corresponding complex numbers. So in general multiplication by z = (r cos t)+ i(r sin t) corresponds to dilation by factor r composed with rotation by angle t. Complex numbers emerge naturally from this point of view. Note, multiplying by -1 corresponds to flipping the real line over. Asking to solve x^2=-1 corresponds to asking: What do you do twice to flip the real line over? This way of thinking of complex multiplication (historically due to Hamilton, perhaps) also has the advantage that it is not hard to glimpse what a lot of higher mathematics is about: group theory, linear algebra, functional analysis and complex function theory are largely about transformation rules of vectors and functions. Compare that to present-day `faith-based' mathematics, where students are simply forced to accept i as some abstract thing that satisfies i^2=-1, and generally quit if they don't believe. Bob Pego === Subject: Re: ATHEISTS lot of talk no action haven't seen any replies with the answer next to the phone number yet! its not that difficult, this whole town of 100,000 people are all in on it. prefix with 61 7 reply here next to the number if anyone confirms or denies it > 47482160 > 47788360 > 47792822 > 47290642 > 47254486 > 47855847 > 47491445 > 47230018 > 47785779 they called it TRUEman, about one man all of media and government ganged up on, i posted in 2000 i'm getting spied on by govt. and I can't reach a girl who looks like laurie holden, Laurie Holden then costars Jim Carrey in a romance...... aus.tv know it! I'm from Townsville and YOU ARE the Truman! http://tinyurl.com/iky5 I was in Townsville over the weekend, and I heard him. Very spooky! http://tinyurl.com/iky8 >phone someone in Townsville, half of you must know someone there, >every day I go out people say THERES THE TRUMAN I'm in Townsville. We're sick of you. http://tinyurl.com/iky9 http://tinyurl.com/iky4 You rule Truman! worth a phone call to give me my life back? Herc === Subject: Re: ATHEISTS |-|erc Dude, the only phone call that would get your life back is one to a > competent psychiatrist who can prescribe you the appropriate > medication. I understand that it's difficult for you to accept this, > but please try to grasp that your problem is not that you're at the > center of some kind of real life Truman Show, but that you're > afflicted with mental illness, most likely schizophrenia, judging from > your posts. It's treatable. Get help. I know 'thinking you're on camera' is symptomatic but its incredibly incredibly rare, there is no other person on usenet saying the truman show is based on them. One : you've got to admit calling it TRUEman is a bit odd. Two : they copied the actress for Jim Carrey's follow up. You said you read those urls, how can you ignore 4 posts from people who agree. *Noone* is saying i'm from townsville , there's no truman here, several are saying i'm from townsville, yes you're the truman. The only people saying a truman setup is not possible or not possibly true for me are nowhere near the 100,000 people all in on it. http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 You know the drill, direct comments on url evidence only please. Herc === Subject: Re: ATHEISTS >|-|erc worth a phone call to give me my life back? >> Dude, the only phone call that would get your life back is one to a >> competent psychiatrist who can prescribe you the appropriate >> medication. I understand that it's difficult for you to accept this, >> but please try to grasp that your problem is not that you're at the >> center of some kind of real life Truman Show, but that you're >> afflicted with mental illness, most likely schizophrenia, judging from >> your posts. It's treatable. Get help. >I know 'thinking you're on camera' is symptomatic but its incredibly >incredibly rare, there is no other person on usenet saying the truman >show is based on them. Which is irrelevant. > One : you've got to admit calling it TRUEman >is a bit odd. Why would that be odd? Do you think that the authors of the film didn't think of that? > Two : they copied the actress for Jim Carrey's follow up. >You said you read those urls, how can you ignore 4 posts from people >who agree. *Noone* is saying i'm from townsville , there's no truman here, >several are saying i'm from townsville, yes you're the truman. No, they aren't. I really hate to break the news to you, but I read the links you gave, and as much as you might want to think otherwise, they're making fun of you, not supporting you. Sorry. >The only people saying a truman setup is not possible or not possibly true >for me are nowhere near the 100,000 people all in on it. And once again, thinking 100,000 people are somehow involved in this massive conspiracy is totally symptomatic of schizophrenia. Get help. === Subject: Re: ATHEISTS > No, they aren't. I really hate to break the news to you, but I read > the links you gave, and as much as you might want to think otherwise, > they're making fun of you, not supporting you. Sorry. I'm from Townsville and YOU ARE the Truman! http://tinyurl.com/iky5 I was in Townsville over the weekend, and I heard him. Very spooky! http://tinyurl.com/iky8 >phone someone in Townsville, half of you must know someone there, >every day I go out people say THERES THE TRUMAN I'm in Townsville. We're sick of you. http://tinyurl.com/iky9 You rule Truman! http://tinyurl.com/iky4 ************* anyone else have an opinion on these links, note these are the only replies from people claiming to be in Townsville and they all support that I *am* the Truman. === Subject: Re: ATHEISTS >> No, they aren't. I really hate to break the news to you, but I read >> the links you gave, and as much as you might want to think otherwise, >> they're making fun of you, not supporting you. Sorry. That's because you can't distinguish your delusions from reality. Get help. === Subject: Re: ATHEISTS >worth a phone call to give me my life back? Dude, the only phone call that would get your life back is one to a competent psychiatrist who can prescribe you the appropriate medication. I understand that it's difficult for you to accept this, but please try to grasp that your problem is not that you're at the center of some kind of real life Truman Show, but that you're afflicted with mental illness, most likely schizophrenia, judging from your posts. It's treatable. Get help. === Subject: Re: ATHEISTS > lot of talk no action > haven't seen any replies with the answer next to the phone number yet! > its not that difficult, this whole town of 100,000 people are all in on it. > No one responded to your first post because you are a COMPLETE ING > LOON, and it made no sense. the others have clarified it, its going up your optic nerve but the signals not registering. 1000s of 1000s of witnesses to I'm the truman, all have known for ___over_1_year___, i'm getting impatient it hasn't gone public yet so what can I do? make sense yet? you got anything to lose by replying 'yes i checked your witnesses they didn't know any truman', you got anything lose 'yes these 5 all said its real, you R the truman'. Herc === Subject: Re: ATHEISTS are gutless >>I pinpoint my affection for someone I state looks like Hollywood Trumans >>next romance costar! >> Any chance you could use your super-powers to write a sentence that actually >> parses in English? >I noticed that one, how about : >Prefix with 61 7, reply here next to the number if anyone confirms or denies it. > 47482160 > 47788360 > 47792822 >Come on , you have NO evidence I'm not the Truman, back up your claim >by proving my witnesses are bogus. 50 cents. I couldn't give a rat's ass if you are Truman or not (my suspicions, I must admit, lie in the you are a complete loony area, but that's just a hunch). I'm just asking you to make sense once in a while. If you think I'm calling random people in Australia to ask if you are a nutball or not then you are as crazy as I think you are. -- === Subject: Re: ATHEISTS are gutless posted in alt.atheism: >apart from alt.business I have 100s of ontopic posts in each group , some >over years Bull. -- === Subject: Re: ATHEISTS are gutless > posted in alt.atheism: >apart from alt.business I have 100s of ontopic posts in each group , some >over years > Bull. http://www.google.com.au/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&q=group%3Asci.ma th+author%3A-%7Cerc&btn G=Google+Search group=sci.math author=|-|erc results=546 group=rec.org.mensa author=|-|erc results=270 group=alt.politics.democrats author=|-|erc results=177 I didn't mean alt.atheism since the subject is clearly on topic. Anyone going to make a random phone call to see that God is alive? If you call people crazy for suggesting a 50c call then I don't have to worry about your other quackery then. Herc Re: ATHEISTS are gutless === Subject: Re: ATHEISTS are gutless > Anyone going to make a random phone call to see that God is alive? He must be alive because there was nobody home when I called. === Subject: Re: ATHEISTS are gutless > Anyone going to make a random phone call to see that God is alive? > He must be alive because there was nobody home > when I called. If God was one of us.... just tryna make his way home....nobody calling on the phone.... http://tinyurl.com/iky5 http://tinyurl.com/iky8 http://tinyurl.com/iky9 http://tinyurl.com/iky4 Atleast post up here these check out, few people from Townsville admit it. Herc === Subject: Re: ATHEISTS are gutless >>I pinpoint my affection for someone I state looks like Hollywood Trumans >>next romance costar! >> Any chance you could use your super-powers to write a sentence that actually >> parses in English? >I noticed that one, how about : >Prefix with 61 7, reply here next to the number if anyone confirms or denies it. > 47482160 > 47788360 > 47792822 >Come on , you have NO evidence I'm not the Truman, back up your claim >by proving my witnesses are bogus. 50 cents. >Someone in Australia do it for 20 cents. 100,000 people torment me for over 1 year >every day I go out, it will take another 5 years to go public unless someone gives >me 1 minute of benefit of the doubt. Laurie Holden is Hollywoods model for Eve, >the perfect 10, she was 21 when I last saw her 5 years ago, 5 years of the Truman >company tormenting me, the last 2 its been in public. They will honestly squeeze >every bit of value out of me they can, one of intel just said then yeah no joking. >If this was a video converence link you'd hear the Truman company constantly in >the background showing off their laser to sound satelite spy systems. >Its actually on topic for all the groups, a variant of my myriad of proposals to each for >benefit of the doubt that a man *can* be unique. Ahem, with all due respect sir, just what the are you talking about? Robyn Resident Witchypoo & EAC Spellcaster #1557 === Subject: Re: ATHEISTS are gutless X-EAC: does not exist. >I pinpoint my affection for someone I state looks like Hollywood >Trumans next romance costar! > Any chance you could use your super-powers to write a sentence that > actually parses in English? >>I noticed that one, how about : >>Prefix with 61 7, reply here next to the number if anyone confirms or >>denies it. >> 47482160 >> 47788360 >> 47792822 >>Come on , you have NO evidence I'm not the Truman, back up your claim >>by proving my witnesses are bogus. 50 cents. >>Someone in Australia do it for 20 cents. 100,000 people torment me for >>over 1 year every day I go out, it will take another 5 years to go public >>unless someone gives me 1 minute of benefit of the doubt. Laurie Holden >>is Hollywoods model for Eve, the perfect 10, she was 21 when I last saw >>her 5 years ago, 5 years of the Truman company tormenting me, the last 2 >>its been in public. They will honestly squeeze every bit of value out of >>me they can, one of intel just said then yeah no joking. >>If this was a video converence link you'd hear the Truman company >>constantly in the background showing off their laser to sound satelite >>spy systems. >>Its actually on topic for all the groups, a variant of my myriad of >>proposals to each for benefit of the doubt that a man *can* be unique. > Ahem, with all due respect sir, just what the are you talking about? He believes he lives the Truman Show. No, really. This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........ -- Mark K. Bilbo #1423 EAC Department of Linguistic Subversion ________________________________________________________________ You are not a fiscal conservative when the deficit runs to $400 ion on your watch. - Stan Collender, Fleishman-Hillard Inc. === Subject: Re: ATHEISTS are gutless >>I pinpoint my affection for someone I state looks like Hollywood >>Trumans next romance costar! >> Any chance you could use your super-powers to write a sentence that >> actually parses in English? >I noticed that one, how about : >Prefix with 61 7, reply here next to the number if anyone confirms or >denies it. > 47482160 > 47788360 > 47792822 >Come on , you have NO evidence I'm not the Truman, back up your claim >by proving my witnesses are bogus. 50 cents. >Someone in Australia do it for 20 cents. 100,000 people torment me for >over 1 year every day I go out, it will take another 5 years to go public >unless someone gives me 1 minute of benefit of the doubt. Laurie Holden >is Hollywoods model for Eve, the perfect 10, she was 21 when I last saw >her 5 years ago, 5 years of the Truman company tormenting me, the last 2 >its been in public. They will honestly squeeze every bit of value out of >me they can, one of intel just said then yeah no joking. >If this was a video converence link you'd hear the Truman company >constantly in the background showing off their laser to sound satelite >spy systems. >Its actually on topic for all the groups, a variant of my myriad of >proposals to each for benefit of the doubt that a man *can* be unique. >> >> Ahem, with all due respect sir, just what the are you talking about? >He believes he lives the Truman Show. Saaaaaaaaay WHAT?!? >No, really. Really, really? >This one's waaaaaaaaaaaaaaaaaaaaaaaaay out there........ Indeed - That movie was terrible ;) Robyn Resident Witchypoo & EAC Spellcaster #1557 === Subject: Re: ATHEISTS are gutless I don't believe that saying that ATHEISTS are gutless, is accurate. They are like white washed tombs, clean on the outside and with dead man's bones on the inside. :-D. === Subject: Re: ATHEISTS are gutless >They are like white washed tombs, clean on >the outside and with dead man's bones on the inside. :-D. These and other fine sayings can be found in my latest book 1001 dumbass glib preacherisms for your church's marquee. Look for it in your local Christian bookstore. --- Hattan Grand High UberPope - First Church of Shatnerology @thecodezone.com http://www.shatnerology.com === Subject: Re: ATHEISTS are gutless >They are like white washed tombs, clean on >the outside and with dead man's bones on the inside. :-D. > These and other fine sayings can be found in my latest book 1001 > dumbass glib preacherisms for your church's marquee. Look for it in > your local Christian bookstore. > --- > Hattan Grand High UberPope - First Church of Shatnerology > @thecodezone.com http://www.shatnerology.com you have no right to speak, your TJ Hooker links are down, hang your head in shame.... Herc === Subject: Re: Axiomization of Number Theory diagonalize in the sense of Cantor? > requirements. For all of these, there exists a primitive recursive > function f G, s.t. when it is fed with a p.r. axiomatisation of a theory > satisfying suitable conditions (namely being capable of representing the > above mentioned functions and having the property that we can > diagonalise the resulting provability formula) it results in a formula > which is as true as the original theory was, but which is not provable > in the theory. Thus for a given axiomatisation alpha, for all G > f G(alpha) is unprovable from alpha. > Now the formula Ey(x = 2*y) defines the set of true statements under > this coding. Obviously G' and its ilk are pathological in the sense that > they leave various very basic operatiosn on formulae hugely > non-recursive. In fact, for the coding G', the operations are not even > arithmetic, as follows from Tarski's result of undefinability of truth. --les ducs d'Enron! http://www.tarpley.net === Subject: Re: Axiomization of Number Theory > Neither word appears in my dictionary here at home (I don't have the > OED). Searching Google for a clue to actual usage, axiomize gives > 84 hits while axiomatize gives 4,580. The fact that axiomatize is > 50 times more popular doesn't quite prove that there is no such > word as axiomize, but it's certainly sufficient to make no, it's > axiomize seem a little silly... So, what you're saying is that something silly was posted, to Usenet, by someone styling himself Charlie-Boo. What are the odds? === Subject: Re: Axiomization of Number Theory >> Supposing for the sake of argument that Keith was correct >> when he said that there's no formal system for second-order >> logic corresponding to the well-known formal systems for >> first-order logic, exactly what does it mean for something >> to go through in second-order PA? > It means, can it be proved in the system PA2, second-order Peano > Arithmetic? Briefly, PA2 is a system of second-order logic with full > comprehension and the Peano Axioms (where the Induction Schema may be > replaced by an axiom). This is a spot where it's easy to get confused over terminology, and I'm not even sure exactly what the standard terminology is. But in any case what Andrew is describing (which I snipped) is actually a theory of two-sorted *first*-order logic. Its models have (in one way of describing them) two universes: One over which the natural-number variables range, and another over which the set-of-naturals variables range. I think these may be called Henkin models or some such. PA2, interpreted *this* way, *does* admit a formal system that mechanically verifies or rejects putative proofs, and it does *not* fully determine the natural numbers up to isomorphism (or even decide all first-order sentences about them). When you're talking about true second-order logic, you can take exactly the same axioms as above, and now they *do* determine the natural numbers up to isomorphism. But a model is a different sort of gadget: It has only one universe. The natural-number variables range over elements of that universe, and the set variables range over subsets of the universe. === Subject: Re: Axiomization of Number Theory > And, sort of getting to my real question: given that models > of set theory can be confused about what subsets of N > exist (in a sense described more precisely above) do > you see anything fuzzy or confusing about exactly what > second-order PA actually means? Good thing I didn't answer this right away. I guess there *is* something confusing, given the subthread with Andrew Boucher. It's not that hard to keep straight, just easy to miscommunicate. === Subject: Re: Axiomization of Number Theory > >> >Supposing for the sake of argument that Keith was correct >when he said that there's no formal system for second-order >logic corresponding to the well-known formal systems for >first-order logic, exactly what does it mean for something >to go through in second-order PA? > >>It means, can it be proved in the system PA2, second-order Peano >>Arithmetic? Briefly, PA2 is a system of second-order logic with full >>comprehension and the Peano Axioms (where the Induction Schema may be >>replaced by an axiom). >> >This is a spot where it's easy to get confused over terminology, and >I'm not even sure exactly what the standard terminology is. But >in any case what Andrew is describing (which I snipped) is actually >a theory of two-sorted *first*-order logic. Its models have (in one >way of describing them) two universes: One over which the >natural-number variables range, and another over which the set-of-naturals >variables range. I think these may be called Henkin models or >some such. >PA2, interpreted *this* way, *does* admit a formal system that mechanically >verifies or rejects putative proofs, and it does *not* fully determine >the natural numbers up to isomorphism (or even decide all first-order >sentences about them). >When you're talking about true second-order logic, you can take exactly >the same axioms as above, and now they *do* determine the natural numbers >up to isomorphism. But a model is a different sort of gadget: It >has only one universe. The natural-number variables range over elements >of that universe, and the set variables range over subsets of the universe. I've seen a distinction made between semantical second-order logic and deductive second-order logic. Here on FOM for instance (the thread is appropriately entitled SOL Confusion): http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html I'll leave others to say what they're talking about, but I'm referring to deductive second-order Peano Arithmetic - the proof system and the theorems which can be generated from PA2. So still, shouldn't Wiles proof go through in PA2? > === Subject: Re: Axiomization of Number Theory >I've got a little question. Is the word axiomize? Because of some reason I thought it was axiomatize. > Searching Google axiomize gives > 84 hits while axiomatize gives 4,580. The fact that axiomatize is > 50 times more popular doesn't quite prove that there is no such > word as axiomize, but it's certainly sufficient to make no, it's > axiomize seem a little silly... > I posted my question on 83 boards. I wonder who is was that agreed with me? Nothing wrong with being silly. At last a meaningful conversation! CSV === Subject: Re: Axiomization of Number Theory > I've seen a distinction made between semantical second-order logic and > deductive second-order logic. > Here on FOM for instance (the thread is appropriately entitled SOL > Confusion): > http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html > I'll leave others to say what they're talking about, but I'm referring > to deductive second-order Peano Arithmetic - the proof system and the > theorems which can be generated from PA2. Well, it seems to me that it's no longer obvious what one means by the proof system in this case. Deductive second-order logic could plausibly refer to any r.e. way of generating sematically correct second-order conclusions, whereas I gather that you want to restrict it to systems that generate conclusions that hold in all two-sorted first-order models of the axioms. So I really think two-sorted first-order logic is a more accurate description than deductive second-order logic. > So still, shouldn't Wiles proof go through in PA2? You'd have to ask someone who knows a whole heck of a lot more about Wiles' proof than I do. === Subject: Re: Axiomization of Number Theory >>I've seen a distinction made between semantical second-order logic and >>deductive second-order logic. >>Here on FOM for instance (the thread is appropriately entitled SOL >>Confusion): >>http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html >>I'll leave others to say what they're talking about, but I'm referring >>to deductive second-order Peano Arithmetic - the proof system and the >>theorems which can be generated from PA2. > Well, it seems to me that it's no longer obvious what one means by > the proof system in this case. Deductive second-order logic could > plausibly refer to any r.e. way of generating sematically correct > second-order conclusions, whereas I gather that you want to restrict > it to systems that generate conclusions that hold in all two-sorted > first-order models of the axioms. So I really think two-sorted first-order > logic is a more accurate description than deductive second-order logic. One usually restricts attention to only those two-sorted models which model various second order truths, such as comprehension, choice, &c. Then logical validity is defined with respect to that class of models, not all two-sorted models. These still admit an r.e. deductive system. Otherwise I agree with you. I think the important thing about second order logic is its semantics, not any particular necessarily non-complete and arbitrary deductive system. There are, however, some interesting things that happen when going from, say, first order PA to a two-sorted version (of the type described) above. G.9adel's speed-up theorem shows that for any recursive function f, there are an infinite number of statements phi provable in first order PA and also in two-sorted PA, but if the shortest proof of phi in two-sorted PA has m steps, then the shortest proof in first order PA takes f(m) steps. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Axiomization of Number Theory >> I've got a little question. >> Is the word axiomize? Because of some reason I thought it was > axiomatize. >> No, it's axiomize. So Charlie-Poo is right and the rest of the world is wrong :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Axiomization of Number Theory >>Do you really believe that Godel showed that >>... to derive theorems from Number Theory >>was impossible? > Well first, it's clear that the from Number Theory > meant of Number Theory - we want to derive theorems > of number theory from the formal system. > Of course Godel didn't show that it's impossible to > derive theorems of number theory from a formal system, > if by that we mean derive _some_ theorems of number > theory. Giving a formal system from which it's possible > to derive some theorems of number theory is so easy > that it seems it can't be what we really wanted (for > example, consider the formal system with Fermat's > Last Theorem as the one and only axiom, and no > inference rules. It's not hard to derive Fermat's Last > Theorem from that formal system, but it doesn't seem > very interesting.) I and the person who claimed Godel > showed it was impossible were assuming that it > was to derive _all_ the theorems of number theory > from a formal system. > If the it in your didn't Peano do it meant give > a formal system from which one could derive > _some_ theorems of number theory then yes, > of course Peano did it (but there are much easier > ways to do it, as above.) Who's we again? and by it I meant the OP's request. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Axiomization of Number Theory |I've been told that Wiles' proof rests on a result that was proven using |the consequence of the axiom of choice that a certain extension of the |p-adics which is both algebraically closed and complete in the p-adic |metric is isomorphic as a field to (a subset of?) the complex numbers. |Now, we know that any proof of Fermat's Last Theorem (or even of the |modularity conjecture) which uses the axiom of choice doesn't really need |it. But I don't know whether there's any non-messy automatic way of |extracting it from the proof. [...] |>>Shouldn't Wiles proof goes through in second-order PA ? |> |>Supposing for the sake of argument that Keith was correct |>when he said that there's no formal system for second-order |>logic corresponding to the well-known formal systems for |>first-order logic, exactly what does it mean for something |>to go through in second-order PA? |> |>>If it doesn't, then I'm afraid here's one vote for saying he hasn't |>>really proved FLT, and JH still has a chance. | |It means, can it be proved in the system PA2, second-order Peano |Arithmetic? Saying that if it didn't, you wouldn't count it as a proof strikes me as unduly paranoid about PA_3 or PA_4 (say). Do you really think that the impredicativity involved in talking about sets of sets of natural numbers (defined by phrases which casually quantify over sets of sets of natural numbers) poses a meaningful possibility of introducing an error? Of course we would enjoy seeing an elementary proof, but not for this reason. I would be very surprised if it couldn't be proven in PA_2. The objects it mainly is concerned with encode pretty straightforwardly either as integers or as sequences of integers. A modular form is essentially an infinite series in this context, for example. A sequence of complex numbers (coefficients) can be encoded as a sequence of integers. I believe the mentioned workaround (which I don't happen to know) shows a way out of using AC, which doesn't require referring to w_1 or whatever. On the other hand, I think such a translation has a certain messiness to it. I think it could be fairly tedious to verify with any care. Here's an example. I've had described to me a place in the deformation theory where the Freyd special adjoint functor theorem is applied. The Freyd special adjoint functor theorem gives a sufficient condition for a functor to have an adjoint. Half of the condition is one of the usual consequences of a functor having an adjoint. The other half is one of these conditions which hinges on something being a set rather than a proper class. For example, when using Freyd to show that there's such a thing as the free group on a set of generators, the condition being used is that given a group G, there exists a _set_ of maps {G->H_i} with the property that any homomorphism G->K factors through one of them, G->H_i->K. One choice for that is a set of groups including each isomorphism class of groups of cardinality no greater than G, with all possible homomorphism from G to them. (We could consider alternative group laws on subsets of the underlying set of G.) Obviously, a *class* of such maps exists (take all of them), but for the theorem to work we need there to be a *set*. This abstraction is what we get for having had Grothendieck work on algebraic geometry. :-) I believe the place where my acquaintance said it was being used is where there are functors from the category of rings to the category of sets: a functor giving the set of either elliptic curves or modular forms with certain properties, having coefficients in the given ring. Luckily, only certain appropriate rings need to be considered, or we'd have to cope somehow also with the issue of what a functor from one big category (i.e., whose objects form a proper class) to another one really is. Anyhow, I don't think there's any actual gotcha involved in the way this lemma is used, but to be rigorously confident that it can all be done with PA_2 one would have to think it all over carefully in the context where it's used. And to return to my original point, there's some value in trying to make an axiom system within which some proofs can be translated *and still look like the same proof* (unlike a complete translation of this proof into PA_2). I admit I had forgotten the O.P. wanted to do automatic proof generation, but even so, trying to get an automatic proof generator to prove results in an unsuitable language might cause it to thrash. Keith Ramsay === Subject: Re: Axiomization of Number Theory > I've seen a distinction made between semantical second-order logic and > deductive second-order logic. > Here on FOM for instance (the thread is appropriately entitled SOL > Confusion): > http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html I'll leave others to say what they're talking about, but I'm referring > to deductive second-order Peano Arithmetic - the proof system and the > theorems which can be generated from PA2. > Well, it seems to me that it's no longer obvious what one means by > the proof system in this case. Deductive second-order logic could > plausibly refer to any r.e. way of generating sematically correct > second-order conclusions, whereas I gather that you want to restrict > it to systems that generate conclusions that hold in all two-sorted > first-order models of the axioms. So I really think two-sorted first-order > logic is a more accurate description than deductive second-order logic. My impression was that I was using standard terminology, e.g Simpson speaks of Second-Order Arithmetic and Shapiro makes a case for Second-Order Logic. Boolos uses this terminology as well. While deductive second-order Peano Arithmetic *could* refer to any r.e. way as you say, in practice it is not widely used that way, to my knowledge anyway. Instead it refers (again in practice) to the specific deductive system I have described, with little- and big-letters (i.e. two sorts), with comprehension. In any case, I can't think of any other interpretation of what people mean should they ask whether PA2 can prove such or such theorem. But I could well be wrong, since I may well not have a standard view of what is standard. === Subject: Re: Axiomization of Number Theory > My impression was that I was using standard terminology, e.g Simpson > speaks of Second-Order Arithmetic and Shapiro makes a case for > Second-Order Logic. Boolos uses this terminology as well. While > deductive second-order Peano Arithmetic *could* refer to any r.e. > way as you say, in practice it is not widely used that way, to my > knowledge anyway. Instead it refers (again in practice) to the > specific deductive system I have described, with little- and > big-letters (i.e. two sorts), with comprehension. Well, but you didn't actually specify a deductive system. You just gave the axioms, not the rules. I assume you mean that some standard Hilbert- or Gentzen-type thing is to be done with the axioms, but you didn't actually say so. If you *are* doing the usual Hilbert or Gentzen thing, then it seems clear to me that this is *first*-order logic, though I absolutely agree that it's second-order *arithmetic*. It's second-order arithmetic because the intended interpretation involves sets of naturals, not just naturals, but the *logic* is first-order (though two-sorted). > In any case, I > can't think of any other interpretation of what people mean should > they ask whether PA2 can prove such or such theorem. I agree that the term PA2 is standardly used for the first-order theory of second-order arithmetic. === Subject: Re: Axiomization of Number Theory >I've seen a distinction made between semantical second-order logic and >deductive second-order logic. >Here on FOM for instance (the thread is appropriately entitled SOL >Confusion): >http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html >I'll leave others to say what they're talking about, but I'm referring >to deductive second-order Peano Arithmetic - the proof system and the >theorems which can be generated from PA2. >>Well, it seems to me that it's no longer obvious what one means by >>the proof system in this case. Deductive second-order logic could >>plausibly refer to any r.e. way of generating sematically correct >>second-order conclusions, whereas I gather that you want to restrict >>it to systems that generate conclusions that hold in all two-sorted >>first-order models of the axioms. So I really think two-sorted first-order >>logic is a more accurate description than deductive second-order logic. > My impression was that I was using standard terminology, e.g Simpson > speaks of Second-Order Arithmetic and Shapiro makes a case for > Second-Order Logic. Boolos uses this terminology as well. While > deductive second-order Peano Arithmetic *could* refer to any r.e. > way as you say, in practice it is not widely used that way, to my > knowledge anyway. Instead it refers (again in practice) to the > specific deductive system I have described, with little- and > big-letters (i.e. two sorts), with comprehension. Shapiro makes a case for second order logic *with* standard semantics, not for the specific deductive system. And Shapiro includes a form of axiom of choice in at least one of the deductive systems considered in his book. > In any case, I > can't think of any other interpretation of what people mean should > they ask whether PA2 can prove such or such theorem. This is true. However, I don't think there's really any agreement how many second order validities should one accept as axioms. Even if there is, it's utterly arbitrary as far as I can see. -- Aatu Koskensilta (aatu.koskensilta@xortec.fi) Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Logico-Philosophicus === Subject: Re: Axiomization of Number Theory >> Neither word appears in my dictionary here at home (I don't have the >> OED). Searching Google for a clue to actual usage, axiomize gives >> 84 hits while axiomatize gives 4,580. The fact that axiomatize is >> 50 times more popular doesn't quite prove that there is no such >> word as axiomize, but it's certainly sufficient to make no, it's >> axiomize seem a little silly... >So, what you're saying is that something silly was posted, to Usenet, >by someone styling himself Charlie-Boo. >What are the odds? There's a first time for everything. === Subject: Re: Axiomization of Number Theory Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >The fact that axiomatize is >50 times more popular doesn't quite prove that there is no such >word as axiomize, but it's certainly sufficient to make no, it's >axiomize seem a little silly... Yes. However, OneLook does recognize axiomize. http://www.onelook.com/?w=axiomize&ls=a -- Tim Chow tchow-at-alum-dot-mit-dot-edu The range of our projectiles---even ... the artillery---however great, will never exceed four of those miles of which as many thousand separate us from the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences === Subject: Re: Axiomization of Number Theory > Yes. However, OneLook does recognize axiomize. Then I suppose both words exist. === Subject: Re: Axiomization of Number Theory >[...] >|>>Shouldn't Wiles proof goes through in second-order PA ? >||>Supposing for the sake of argument that Keith was correct >|>when he said that there's no formal system for second-order >|>logic corresponding to the well-known formal systems for >|>first-order logic, exactly what does it mean for something >|>to go through in second-order PA? >||>>If it doesn't, then I'm afraid here's one vote for saying he hasn't >|>>really proved FLT, and JH still has a chance. >| >|It means, can it be proved in the system PA2, second-order Peano >|Arithmetic? >Saying that if it didn't, you wouldn't count it as a proof strikes me as >unduly paranoid about PA_3 or PA_4 (say). Well, actually, it's more in the reverse direction, but this is getting off-topic and I'm going on vacation tomorrow for three weeks (I do live in France and it will be August), so I'm not able to sustain a conversation on this. But just to state my position: I wouldn't accept Wiles' proof if it doesn't go through in PA_2, but I don't feel obliged to accept it *even if it does*. I would accept it only if it goes through in PA_2 minus the ad infinitum assumption that there is always a next natural number. Mind you, if the proof were to go through in PA_2, then it should also go through in the restricted system, so I don't expect any surprises. >I would be very surprised if it couldn't be proven in PA_2. Ok. (And thanks for your subsequent comments.) I think you have good company in your belief, but it does seem to be an open question. > === Subject: Re: Axiomization of Number Theory > >>My impression was that I was using standard terminology, e.g Simpson >>speaks of Second-Order Arithmetic and Shapiro makes a case for >>Second-Order Logic. Boolos uses this terminology as well. While >>deductive second-order Peano Arithmetic *could* refer to any r.e. >>way as you say, in practice it is not widely used that way, to my >>knowledge anyway. Instead it refers (again in practice) to the >>specific deductive system I have described, with little- and >>big-letters (i.e. two sorts), with comprehension. >> >Well, but you didn't actually specify a deductive system. Details! > You >just gave the axioms, not the rules. Well, this is the way Simpson puts it: By second order arithmetic we mean the formal system in the language L2 consisting of the axioms of second order arithmetic, together with all formulas of L2 which are deducible from those axioms by means of the usual logical axioms and rules of inference. OK so maybe I'm out of it, but there do seem to be usual axioms and rules of inference - and I'm sure you know the ones he and I are thinking about. >I assume you mean that some >standard Hilbert- or Gentzen-type thing is to be done with the >axioms, but you didn't actually say so. OK sorry... >If you *are* doing the usual Hilbert or Gentzen thing, then it >seems clear to me that this is *first*-order logic, though I absolutely >agree that it's second-order *arithmetic*. It's second-order arithmetic >because the intended interpretation involves sets of naturals, not >just naturals, but the *logic* is first-order (though two-sorted). Well ok, I'm guess I'm non-standard on this. Say Simpson doesn't put in the axioms of arithmetic, but just comprehension - I would call that second-order logic. (Second-order arithmetic minus the axioms of arithmetic is second-order logic. Doesn't that sound logical ?) Is it really misleading you? Because if it does (and you are not alone), then obviously I should stop, just for the sake of clear communication. === Subject: Re: Axiomization of Number Theory >> I've seen a distinction made between semantical second-order >> logic and >> deductive second-order logic. >> Here on FOM for instance (the thread is appropriately entitled SOL >> Confusion): >> http://www.cs.nyu.edu/pipermail/fom/2000-September/004354.html >> I'll leave others to say what they're talking about, but I'm referring >> to deductive second-order Peano Arithmetic - the proof system and the >> theorems which can be generated from PA2. > Well, it seems to me that it's no longer obvious what one means by > the proof system in this case. Deductive second-order logic could > plausibly refer to any r.e. way of generating sematically correct > second-order conclusions, whereas I gather that you want to restrict > it to systems that generate conclusions that hold in all two-sorted > first-order models of the axioms. So I really think two-sorted > first-order > logic is a more accurate description than deductive second-order > logic. >> My impression was that I was using standard terminology, e.g Simpson >> speaks of Second-Order Arithmetic and Shapiro makes a case for >> Second-Order Logic. Boolos uses this terminology as well. While >> deductive second-order Peano Arithmetic *could* refer to any r.e. >> way as you say, in practice it is not widely used that way, to my >> knowledge anyway. Instead it refers (again in practice) to the >> specific deductive system I have described, with little- and >> big-letters (i.e. two sorts), with comprehension. > Shapiro makes a case for second order logic *with* standard semantics, > not for the specific deductive system. And Shapiro includes a form of > axiom of choice in at least one of the deductive systems considered in > his book. It was a poor choice of wording on my part if you think I meant that Shapiro makes a case for a specific deductive system. But he does introduce a deductive system, indeed the standard one (with one exception!), and while I can't find him using the words second-order logic to refer to the deductive system, he does say things like second-order version of existential introduction. And you are right (this is the exception) that he adds a version of the axiom of choice. What can I say? I don't like it and I don't want it. > In any case, I >> can't think of any other interpretation of what people mean should >> they ask whether PA2 can prove such or such theorem. > This is true. However, I don't think there's really any agreement how > many second order validities should one accept as axioms. Even if > there is, it's utterly arbitrary as far as I can see. I'm not sure about the arbitrariness, but I don't have an argument at this point. === Subject: Re: Axiomization of Number Theory > Well ok, I'm guess I'm non-standard on this. Say Simpson doesn't put > in the axioms of arithmetic, but just comprehension - I would call that > second-order logic. (Second-order arithmetic minus the axioms of > arithmetic is second-order logic. Doesn't that sound logical ?) Is it > really misleading you? Because if it does (and you are not alone), then > obviously I should stop, just for the sake of clear communication. To me the important aspect of the logic is how inferences are made, not the axioms. Hilbert-style or Gentzen-style derivations or whatever fancy improvements on them there may have been, all give you first-order logic, no matter what axioms you feed in the front end. Throw in omega-rule and you're outside of first-order logic but you haven't gotten to second-order yet. (omega-rule is obviously semantically valid -- too bad journals are reluctant to publish proofs that use it, paper prices being what they are today.) === Subject: Re: Axiomization of Number Theory > >>Well ok, I'm guess I'm non-standard on this. Say Simpson doesn't put >>in the axioms of arithmetic, but just comprehension - I would call that >>second-order logic. (Second-order arithmetic minus the axioms of >>arithmetic is second-order logic. Doesn't that sound logical ?) Is it >>really misleading you? Because if it does (and you are not alone), then >>obviously I should stop, just for the sake of clear communication. >> >To me the important aspect of the logic is how inferences are >made, not the axioms. Hilbert-style or Gentzen-style derivations >or whatever fancy improvements on them there may have been, >all give you first-order logic, no matter what axioms you feed >in the front end. >Throw in omega-rule and you're outside of first-order logic but >you haven't gotten to second-order yet. (omega-rule is obviously >semantically valid -- too bad journals are reluctant to publish >proofs that use it, paper prices being what they are today.) > Well, I certainly don't want the omega-rule ! Anyway, here's Boolos (p. 7, Logic, Logic, and Logics). There is a standard extension of the proof theory for first-order logic to second-order logic. The notion of derivation is changed only by the addition of new axioms, most importantly by the scheme of comprehension... === Subject: Re: Axiomization of Number Theory > Anyway, here's Boolos (p. 7, Logic, Logic, and Logics). There is a > standard extension of the proof theory for first-order logic to > second-order logic. The notion of derivation is changed only by the > addition of new axioms, most importantly by the scheme of comprehension... Hm. Well, it looks like we'll both have to watch out and make sure people don't misunderstand us. You can believe that if I'd been Boolos' editor, I'd have objected to that passage. === Subject: Re: Axiom of choice: is it wrong? X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >The axiom of choice, that says it's always possible to choose one >element of each member of a family of sets, That's a restatement into English, which is replete with ambiguity. The AOC makes a statement about the existence of certain entities, not a statement about human psychology. >Now, let's consider the subset of all the undefinable real numbers. >There is no way of referring specifically to any one of them. How >could one possibly say that you have chosen one of them? But that gas nothing to do with what the AOC actually says. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any 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 not reply to spamtrap@library.lspace.org === Subject: Re: Axiom of choice: is it wrong? <1nidnaNkWNfZ67uiU-KYgg@comcast.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >I'm fuzzy on the undefinable real thing. I can't find a proper >definition for this phrase. Also, if a real number is thought to be >an equivalence class of Cauchy sequences* of rational numbers then I >would think if x is a real number there must be at least one Cauchy >sequence of rationals that converges to it... Of course. >If not, is it reasonable to conclude that x isn't a real number? Yes, because then x is *not* an equivalence class of Cauchy sequences. >* Or Dedekind cuts which I don't really understand yet... They're actually simpler than equivalence classes of Cauchy sequences. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do not reply to spamtrap@library.lspace.org === Subject: Re: Axiom of choice: is it wrong? <819f2d19.0307290941.5afd5ef6@posting.google.com> X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS >I don't deny that a nonempty set has at least one member. But the >axiom of choice says that we can define a function. No, it only says that the function exists. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any 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 not reply to spamtrap@library.lspace.org === Subject: Re: Axiom of choice: is it wrong? There is a big difference between set A can be well-ordered and > formula phi(x, y) well-orders A (phi being a formula in ZFC, or > whatever your favourite set theory is). My point is: what if it is impossible to define any such formula? What > is the meaning then of set A can be well-ordered? > It is certainly possible for a set to be well-orderable without there > being a formula which defines the well-ordering. For an example or > proof you'll need one of the experts. > GC There is something I found as I thought about these undefineable numbers; I am no expert in set theory, but this sounds VERY weird to me: suppose you have a well-ordering of the real numbers (we know that there exists one) and now consider the set of all these undefinable real numbers. We know that this set has a least element, so we have defined the undefinable number... Can you explain this? Does it have something to do with `s paradox? === Subject: Re: Axiom of choice: is it wrong? <819f2d19.0307290936.23a217f@posting.google.com> <3F26B9A2.1A2063D6@btinternet.com> <6bd16859.0307311248.3580f017@posting.google.com> A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x > There is something I found as I thought about these undefineable > numbers; I am no expert in set theory, but this sounds VERY weird to > me: suppose you have a well-ordering of the real numbers (we know that > there exists one) and now consider the set of all these undefinable > real numbers. We know that this set has a least element, so we have > defined the undefinable number... > Can you explain this? Does it have something to do with `s > paradox? Let c be the well-ordering at hand and let x be the c-least element of the set of undefinable[1] real numbers. This is your definition of x, right? But this definition has a parameter in it, namely c. This parameter is undefinable in the same sense that x was, I think. So, it doesn't seem like you've really defined x, since you can't write down what you mean by c. Footnotes: [1] I'm taking for granted that this term makes sense. I'm not really sure. -- Jesse Hughes Time and again, history has shown that people who think their beliefs trump reality lose, and lose badly. Luckily, I don't have to listen to you. -- on reality avoidance === Subject: Re: Axiom of choice: is it wrong? > There is something I found as I thought about these undefineable > numbers; I am no expert in set theory, but this sounds VERY weird to > me: suppose you have a well-ordering of the real numbers (we know that > there exists one) and now consider the set of all these undefinable > real numbers. We know that this set has a least element, so we have > defined the undefinable number... No, you haven't defined anything, unless the well-ordering that you used is itself definable. We know there exists a well-ordering of the reals by invoking the axiom of choice, not because anyone has actually found a way to define such a well-ordering. > Can you explain this? Does it have something to do with `s > paradox? Not particularly. -- Dave Seaman Judge Yohn's mistakes revealed in Mumia Abu-Jamal ruling. === Subject: Re: Axiom of choice: is it wrong? > There is something I found as I thought about these undefineable > numbers; I am no expert in set theory, but this sounds VERY weird to > me: suppose you have a well-ordering of the real numbers (we know that > there exists one) and now consider the set of all these undefinable > real numbers. We know that this set has a least element, so we have > defined the undefinable number... > Can you explain this? Does it have something to do with `s > paradox? This was proposed 70 (or more?) years ago as a proof that there is no well-ordering of the reals. (The other replies show why it is no such thing.) Back then, a response was formulated, showing it has nothing to do with well-ordering the reals.... The natural numbers are well-ordered, right? Each natural number has a definition (actually, many definitions) in English words. But some natural numbers can be defined in fewer words than others (since there are finitely many words, etc...). But consider this: The least natural number not definable in less than twenty English words. That uniquely defines a natural number. And it has less than twenty words. === Subject: Brand new nature law discovery Professor Rudy J Boxman Professor Rudy J Boxman says: all what we do will be recorded by a new nature-law, and played again, later on in time.Very scary, but a fact. Why Albert Einstein didn't discover this, is a riddle! _______________________________________________________________________ The Dutch Professor Rudy J Boxman has done a discovery, what will be the discovery of all centuries.In 1998, while he was reading one of his favourite books , the holy bible itself, he discovered, that all of the events that are or identical event, which took place only a little later in time.He about it and here are some of these facts about this brand new discovery : > Dutch Press. > RUDY J. BOXMAN is proud to announce the existance of a brand new law of nature.The discovery was done a few years ago,but now presented to the world-press. Maybe because of the high speed of our planet earth through the universe,something strange is going on with the present events on our planet.On each original event...an exact copy will follow up later on...on that exact location.It depents on a several conditions, when the copy is going to hit.Water sourches like lakes and canals or oceans,do have a great impact on the time between two similar events.The events...are similar in details and equal to eachother,but the copy event is 80% .Scientific research was done by RUDY J. BOXMAN . This great news will turn this world complete up-side-down.Why?? Everyone on this planet is infected by this new nature law. We are talking about..killings..,disasters of any kind that will copy itself..automaticly.We can't do,whatever we want to do...anylonger!! The RJB ORG. gave it the new name...: BOXMAN-NATURE LAW. Dutch translation... Vertaling ...Dutch translation....... Rudy J. Boxman is verheugd om het bestaan,aan te kondigen van een geheel nieuwe natuurwet.De ontdekking werd al een paar jaar geleden gedaan,maar nu pas bekent gemaakt aan de wereld-pers. Mischien vanwege de hoge snelheid van onze planeet,de Aarde,door het universum..,maar er is iets vreemd aan de hand met de huidige gebeurtenissen op onze planeet. Op elke orginele gebeurtenis op Aarde ontstaat later,een exacte kopie daarvan op precies dezelfde locatie. Het hangt af van verschillende omstandigheden,waneer de kopie..toeslaat. Water-bronnen als meren,kanalen,of oceanen,hebben een grote invloed op de tijd tussen twee gelijke gebeurtenissen,maar de kopie is altijd tachtig procent(80%). Wetenschappelijk onderzoek werd gedaan door de Hengelo-er Rudy J. Boxman. Dit geweldige nieuws zal de complete wereld op zijn kop zetten. De RJB organisatie..gaf het de nieuwe naam...BOXMAN-NATUURWET. Every event in the holy bible, in fact, happens two times, because as a result of this nature-law.So,every killing, war, and in fact every other event in this book happens exactly, two times,on almost the same location, and an other fact is,the copy event is exactly 80 %.So we are having an event which will be an original event and an exactly copy event which will be about 80 % from the original. This is the universal key to understand the entire book of the bible and this is what not only happened centuries ago, but right now as we live today ! That explains why first the Titanic Ship who hit an ice-berg in 1912,( this was an original event) and later on in time..the ship with the name BRITANICA who was an exact copy of the TITANIC ,was going to the bottom of sea in 1916, four years later in time.Both ships left the same harbour in England.And every other event has it's original and copy just as the same as the Titanic and the Britanica.The airplane disaster above Lockerby for example.It has to be a copy of an event that has happened a few years ago ,when a airplane hit the sea near Ireland,under the same circumstances...years before the Lockerby disaster. And there are more evidential events that shows us, that whatever people do..no matter what..,if they are doing something new, what they do will copy itself 80 % exactly just the same.So every action, will has it's original and 80 % copy. Not only visable in the human world, but in the animal world too ! What ever an animal is doing for the first time, an exactly 80 % copy will follow up, a little later in time.And this is going on with every moving atom or molecule. Professor Rudy J Boxman, Holland. === Subject: clock puzzle... , Here's a tough puzzle: Suppose a clock has 3 hands of equal length (hour, minute, second). Consider the triangle connecting the ends of the hands. At what time is the area of the triangle the largest? Enjoy! === Subject: Re: clock puzzle... > Here's a tough puzzle: > Suppose a clock has 3 hands of equal length (hour, minute, second). > Consider the triangle connecting the ends of the hands. > At what time is the area of the triangle the largest? For instance at 4h00min40sec, ie when the triangle is equilateral. Steps: 1) Your problem is equivalent to find out when the product of the (length of the) sides of your triangle is maximal (using Area = abc/(2R)). That product is maximal when the triangle is equilateral: 2) *If the triangle is not equilateral*, let c be the length of one of its sides such that a and b (the two lengths of the other sides) are different. Now we suppose c is fixed. 3) The product a*b is maximal when the area of the triangle (with the side c fixed, ie only a and b can vary) is maximal (using Area = 1/2*a*b*sin(?)). 4) That area (with the side c fixed) is maximal => a = b. 5) Hence if the initial triangle is not equilateral, the product abc can be increased. 6) The solution is the equilateral triangle. -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France === Subject: Re: clock puzzle... > At what time is the area of the triangle the largest? > 6) The solution is the equilateral triangle. Can someone wake me up at equilateral triangle time? Herc === Subject: Re: clock puzzle... AI5Wa.139$Qx6.7025@nnrp1.ozemail.com.au... > At what time is the area of the triangle the largest? > 6) The solution is the equilateral triangle. > Can someone wake me up at equilateral triangle time? > Herc And actually my solution was quite stupid: the problem is trivial (fix one side c and show that the area is maximal => a = b)... === Subject: Re: clock puzzle... > And actually my solution was quite stupid: the problem is trivial (fix one > side c and show that the area is maximal => a = b)... To show that an inscribed triangle maximizes its area when it is equilateral, is quite trivial. Your solution does not convince me however, that there is _a time_ when this occurs, neither does it say anything about what this time actually _is_. -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: Re: clock puzzle... > Your solution does not convince me however, that there is _a time_ when > this occurs, neither does it say anything about what this time actually > _is_. > -- You're right, I forgot to take into account that the heads (?) move continuously!! === Subject: Re: clock puzzle... >For instance at 4h00min40sec, ie when the triangle is equilateral. Not quite. 40 seconds after 4 o'clock, the hour and minute hands have moved on a bit. You have to consider the constraints between the hand positions. -- Richard -- Spam filter: to mail me from a .com/.net site, put my surname in the headers. FreeBSD rules! === Subject: Re: clock puzzle... > , > Here's a tough puzzle: > Suppose a clock has 3 hands of equal length (hour, minute, second). > Consider the triangle connecting the ends of the hands. > At what time is the area of the triangle the largest? > Enjoy! For a 12 hr clock face. Given 3 points (x1, y1), (x2, y2), (x3, y3), the signed area of a triangle is [(x1 y2 - x2 y1) + (x2 y3 - x3 y2) + (x3 y1 - x1 y3)] / 2 With s as seconds since 12 o'clock, the coodinates of the end of the hands are (Sin [Pi s / 30], Cos[Pi s / 30]), (Sin [Pi s /1800], Cos[Pi s /1800]), (Sin[ Pi s / 21600], Cos[Pi s / 21600]) for second, minute and hour hands respectively The signed area of the triangle is therefore (Sin[11 Pi s / 21600] + Sin[59 Pi s / 1800] + Sin[719 Pi s / 21600] ) / 2 differentiating and set to 0 (11 Cos[11 Pi s / 21600] + 708 Cos[59 Pi s / 1800] + 719 Cos[719 Pi s / 21600] ) / 2 = 0 maximum area is at 20949.12338408509465052277227452846 sec after 12 and 22250.87661591490534947722772547154 sec after 12. These are the times 05 hr 49 min 09.12 sec and 06 hr 10 min 50.88 sec Note, the times add up to 12 hr. The angles (degrees) between hands are 120.3363102 119.8357237 119.8279661 24 hr clock face The signed area of the triangle is (Sin[23 Pi s / 43200] + Sin[59 Pi s / 1800] + Sin[1439 Pi s / 43200] ) / 2 and the max area is at 66366.094400066583277 sec after 00 hr. These are the times 05 hr 33 min 53.91 sec 18 hr 26 min 06.09 sec Note, the times add up to 24 hr. These are more equilateral than the 12 hr face. the angles (degrees) between hands being 120.043039607 119.915953327 120.041007066 It seems that there is no true equilateral time === Subject: Re: clock puzzle... >Suppose a clock has 3 hands of equal length (hour, minute, second). >Consider the triangle connecting the ends of the hands. >At what time is the area of the triangle the largest? If there was a time when the triangle was exactly equilateral, that would be it. But there isn't. The positions of the three hands at time t hours can be written as P_H = [ sin(pi t/6), cos(pi t/6),0 ], P_M = [ sin(2 pi t), cos(2 pi t),0 ], P_S = [ sin(120 pi t), cos(120 pi t),0 ] and twice the area is |(P_M - P_H) x (P_S - P_H)| (using the vector cross product), which works out to |f(t)| where f(t) = sin(719 pi t/6) - sin(11 pi t/6) - sin(118 pi t). So we want to maximize or minimize f(t). By symmetry, f(-t) = -f(t), we may as well maximize it. There are lots of critical points (approximately two per minute). Of course we expect the maximum to be near one of the times when the hour and minute hand are separated by 120 degrees, i.e. t is approximately 4(2+3j)/11 where j is an integer, 0<=j<=10. It can be thought of this way. We can write f(t) = 2 sin(11 pi t/12) (cos(1427 pi t/12) - cos(11 pi t/12)) where cos(1427 pi t/12) varies quickly compared to sin or cos of 11 pi t/12. The maxima of 2 sin(11 pi t/12) (1 - cos(11 pi t/12)) are at t = 8(1+3j)/11. The best j should be one that makes cos(1427 pi t/12), i.e. cos(2854 (1 + 3 j) pi/33), very close to 1, or 1427 (1 + 3 j) nearly a multiple of 33. Now 1427 (1 + 3 j) = 8 + 24 j mod 33, which is -1 mod 33 for j = 1. So we look for a solution near t = 32/11. Solving numerically, the maximum is at approximately t = 2.9096004696658838724, where f = 2.5980706142146652690. The time in hr:min:sec is approximately 2:54:34.5616907971819 approximately. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: clock puzzle... >Suppose a clock has 3 hands of equal length (hour, minute, second). >Consider the triangle connecting the ends of the hands. >At what time is the area of the triangle the largest? > If there was a time when the triangle was exactly equilateral, that would > be it. > But there isn't. > The positions of the three hands at time t hours can be written as > P_H = [ sin(pi t/6), cos(pi t/6),0 ], > P_M = [ sin(2 pi t), cos(2 pi t),0 ], > P_S = [ sin(120 pi t), cos(120 pi t),0 ] > and twice the area is |(P_M - P_H) x (P_S - P_H)| (using the vector > cross product), which works out to |f(t)| where > f(t) = sin(719 pi t/6) - sin(11 pi t/6) - sin(118 pi t). > So we want to maximize or minimize f(t). By symmetry, f(-t) = -f(t), > we may as well maximize it. There are lots of critical points > (approximately two per minute). Of course we expect the maximum > to be near one of the times when the hour and minute hand are separated > by 120 degrees, i.e. t is approximately 4(2+3j)/11 where j is an integer, > 0<=j<=10. > It can be thought of this way. We can write > f(t) = 2 sin(11 pi t/12) (cos(1427 pi t/12) - cos(11 pi t/12)) > where cos(1427 pi t/12) varies quickly compared to sin or cos of > 11 pi t/12. The maxima of 2 sin(11 pi t/12) (1 - cos(11 pi t/12)) > are at t = 8(1+3j)/11. The best j should be one that makes > cos(1427 pi t/12), i.e. cos(2854 (1 + 3 j) pi/33), very close to 1, > or 1427 (1 + 3 j) nearly a multiple of 33. Now > 1427 (1 + 3 j) = 8 + 24 j mod 33, which is -1 mod 33 for j = 1. > So we look for a solution near t = 32/11. Solving numerically, > the maximum is at approximately t = 2.9096004696658838724, where > f = 2.5980706142146652690. The time in hr:min:sec is > approximately 2:54:34.5616907971819 approximately. but you already knew that! from rec.puzzles 2 months ago! Herc === Subject: Re: clock puzzle... >> So we look for a solution near t = 32/11. Solving numerically, >> the maximum is at approximately t = 2.9096004696658838724, where >> f = 2.5980706142146652690. The time in hr:min:sec is >> approximately 2:54:34.5616907971819 approximately. >but you already knew that! from rec.puzzles 2 months ago! Glad to see somebody's paying attention. Although this is similar to the rec.puzzles question === optimized is different, and the answer is therefore slightly different: the answer in that thread was approximately 2:54:34.520548. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Common denominations: 1/pi & 1/g Using 1/pi; we find that (such) a pieth is a section of a 360 degree (whole) circle; whose arc is 360/pi = 114.591 degrees, and/or two radians: Since one radian is a section whose arc length is equal to its radius, wouldn't the pieth help in defining, and understanding the function called pi? Where pi is (just) the _ratio_ of a circle's circumference, to its diameter [2r]. Using 1/g: We find that 1 lbf is 1/g of a slug? Since one slug = f/a = 1# secî/foot = w/g = 32# secî/32'; is the constant _unit_ of mass, and/or inertia in the customary U.S. (& British gravitational) system of weights and measures, and is a constant anywhere; on any planet that is similar to Earth or the moon; wouldn't 1/g = 1 lbf, also be a constant? [On the moon 1/g = 1/(5.33' secî/foot).] Nonflamable comments please:-? === Subject: Re: constructing an ovoid from N given points > ... So you've got five degrees of freedom, which means you can expect > to pass a conic through most any set of five points, and that conic is > unique; if it looks like the points lie on an ellipse, they probably do. Else (excluding degenerate cases ), do they lie on a hyperbola of two branches? That raises a question of how the conic section originated. Given : a x^2+ 2 h x y + b y^2 + 2 f x + 2 g y + 1 = 0 ( 5 constants ) can we find the semi-vertical angle of cone and, the angle between cone axis to the cutting plane in terms of these 5 constants ? === Subject: Re: CURVATURE OF EARTH SURFACE PER MILE?? > I know it involves determining the height of a chord through a given > circle, but beyond that I don't know how to go about it. > Earthdia = 7,926(miles) or 41,849,280(feet) > circ = 24,900(miles) or 131,472,000(feet) > radius = 1/2 dia = 3963miles or 20,924,640feet. > What is formula or procedure to find the answer? I'm learning > disabled so please explain it as to a 10 year old. > Chris === Subject: Re: CURVATURE OF EARTH SURFACE PER MILE?? > I know it involves determining the height of a chord through a given > circle ... No, it involves angle turned in relation to distance travelled along surface of arc. If you traveled 24,900 miles, then you turn through 360 degrees full circle. If you traveled 2,4900 miles, then you turn through 36 degrees, but same curvature. Curvature is defined as the angle ( in radian measure ) turned for a given radius of a circle through center of globe. If you turned 360 degrees (2 pi radians ) from equator to north pole, the curvature of earth is (2 pi/ 24,900) reciprocal miles . This is simply the reciprocal of the Radius of curvature of the Earth. === Subject: Das kleine Einmaleins Dedicated to the sanctuarity of Inspiration : http://www.gvdnet.dk/~hagen/faistos.htm http://www.gvdnet.dk/~hagen/einszwei.gif Ole Hagen === Subject: dumb (but new) math joke Q: What's the highest number in group theory A: a belian === Subject: Re: Eigenvalues > Let A be a square complex matrix. Around every element a_{ii} on the > diagonal of the matrix, we draw a circle with radius the sum of the > norms of the other elements on the same row > sum_{jneq i}abs( a_{ij} ). > Such circles are called Gershgorin discs. > Theorem: Every eigenvalue of A lies in one of these Gershgorin discs. > Argh... precisely the exercise... I couldn't hide it !!!!! === Subject: Re: Euler's Proof >Where can I find a description of the gist of Euler's proof equating the two >forms of the zeta function, i.e,, the infinite series of ordinary numbers, >to the product series involving prime numbers? For one place, Conway JB. Functions of One Complex Variable. -- Stephen J. Herschkorn herschko@rutcor.rutgers.edu === Subject: Re: Euler's Proof rpbprb > Where can I find a description of the gist of Euler's proof equating the two > forms of the zeta function, i.e,, the infinite series of ordinary numbers, > to the product series involving prime numbers? A more general statement is proved in Serre's little gem Cours d'Arithm.8etique, or A Course in Arithmetic (Springer), chapter 5 section 3. Watch for typos. Larry === Subject: Re: Euler's Proof > Where can I find a description of the gist of Euler's proof equating the two > forms of the zeta function, i.e,, the infinite series of ordinary numbers, > to the product series involving prime numbers? The proof is very straightforward. Start from the product representation, expand and multiply: zeta(s) = prod_p 1/(1-p^-s) = prod_p (1 + 1/p^s + 1/p^{2s} + 1/p^{3s} + ...) = sum_{n=1}^oo (1/n^s) -mb === Subject: Re: Factorial/Exponential Identity, Infinity >>... >>I have no idea whare Ross is trying to go with this, but I, for one, >>do not wish to follow him there. > That's too bad. > I've gone through my inductive chain of logic to assert a claim, or > proof as they're called, a couple times, and think it's good. I > admit it's a suprising result. If you can show an error I'd be > interested to know of it. > One issue is that of probabilistic distribution over an infinite set. > I say things like half the integers are multiples of two. One way > to consider it is this: select a hundred different integers between > one and a hundred inclusive. Fifty of a hundred, or 1/2, are even > numbers. Thus it's fair to say that as random as a method of > selection could be over a uniform probability distribution over the > numbers from one to a hundred, the probability of selecting an even > number would be 0.5. > It's so for a hundred, it's so for ten million, it's so for all > positive integer multiples of two. Thus, I say that it is true for > all integers, agreeing with the asymptotic density result of number > theory, where the asymptotic density of evens in the integers is one > half. > Of all the infinite bit strings half of them have equal numbers of > ones and zeros. Thus, the probability of selecting one of those at > random from some uniform probability distribution over the strings is > also equal to one half. It's not just a sample of the population > that's considered to determine projections about the population, it's > the entire population. > I'm not quite sure where all those sequences with equal numbers of > ones and zeros are in that set of those sequences, where for any two > sequences a and b where a is having equal numbers of ones and zeros > that ab. That's not true for any two sequences c and d, for > example .100... = 0.011..., 1/2. > That is to say, the sequences with equal numbers of zeros and ones, F, > for flat, are not uniformly distributed among those sequences in their > order. For example, starting from the first value .000..., there > would be infinitely many sequences with a higher valuation than zero > and not being an element of F where each is less than any element of > F. F does not alternate in the reals. That implies that rational and > irrational sequences of F are indefinite neighbors, besides that they > are at some point dense in the reals. > I really wouldn't put this forward if I didn't think it was a correct > line of reasoning, I don't lie to sci.math. That's one reason why > mathematics is so great, there are definite right and wrong answers, > where right and left are irrelevant. Then again, I think the > political right and left are extremist and not representative of the > moderate polity. > Anyways I'm interested in these sequences because I think they display > in their interrelations some characteristics of the reals that are > valuable to know and that have meaning and utility. > Plus, I have Virgil as a punching bag on sci.math, and he has me. I > can probably take a punch, and have some knowledge of nerve junctures > and the mechanics of joint action. But that's mostly machismo. > Anyways I want people to find errors where they exist in my logical > development of the factorial/exponential identity. Virgil makes a > point about how the result of the expression appears to diverge, yet > the same would be true for a horseshoe shaped function, or saddle, > what-have-you. Until it reaches the point of inflection the function > is increasing and increasingly increasing. > So, go through the proof of the identity and verify it or contradict > it. > Ross (I sent this out oonce, but it seems to have gotten lost.) Not quite right. If you apply Stirling's theorem in the form n! ~ sqrt(2*pi*n)*(n/e)^n, you get (2n)!/(4^n*(n!)2) ~ 1/sqrt(pi*n) This means that n! ( (n/2)!^2 2^n) ~ 1/sqrt(pi*n/2) or n! ( (n/2)! 2^n) ~ (n/2)! /sqrt(pi*n/2) Martin Cohen === Subject: Re: Factorial/Exponential Identity, Infinity > [...] > Not quite right. If you apply Stirling's theorem in the form > n! ~ sqrt(2*pi*n)*(n/e)^n, you get > (2n)!/(4^n*(n!)^2) ~ 1/sqrt(pi*n) > Martin Cohen I read some more about Stirling's approximation. Of note is Stirling's approximation which gives an approximate value for n!, gamma(n+1), for large values of n. MathWorld gives Stirling's approximation as n! ~= sqrt(2pi) n^(n+1/2)e^(-n) Consider it as n -> oo, with having n! = (n/2)! 2^n Divide each side of the approximation by the corresponding side of the other equation. 1 ~= ( sqrt(2pi) n^n n^(1/2) ) / ( e^n (n/2)! 2^n ) 1 ~= ( sqrt(n2pi) n^n ) / ( (n/2)! (2e)^n ) lim n->oo ( sqrt(n2pi) n^n ) / ( (n/2)! (2e)^n) ) ~= 1 Does lim n->oo n! / ( (n/2)! 2^n ) = 1? Why or why not? Ross === Subject: Re: Factorial/Exponential Identity, Infinity > >>... >>I have no idea whare Ross is trying to go with this, but I, for one, >>do not wish to follow him there. > That's too bad. I've gone through my inductive chain of logic to assert a claim, or > proof as they're called, a couple times, and think it's good. I > admit it's a suprising result. If you can show an error I'd be > interested to know of it. Ross has claimed that if f(n) = n!/( (n/2)! 2^n) ) then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he reconsider. Ross, consider this, which you may easily verify yourself: f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. Thus f(2n+2) > f(2n) > 1 for every integer n > 2, and f(2n) gets further away from 1 as n increases, and, in fact, diverges towards +oo. === Subject: Re: Factorial/Exponential Identity, Infinity > >>... >>I have no idea whare Ross is trying to go with this, but I, for one, >>do not wish to follow him there. > That's too bad. I've gone through my inductive chain of logic to assert a claim, or > proof as they're called, a couple times, and think it's good. I > admit it's a suprising result. If you can show an error I'd be > interested to know of it. > Ross has claimed that if f(n) = n!/( (n/2)! 2^n) ) > then lim_{n -> +oo} f(n) = 1, and has rejected my suggestion that he > reconsider. > Ross, consider this, which you may easily verify yourself: > f(6) = 15/8 and f(2n+2)/f(2n) = n + 1/2 for all positive integers n. > Thus f(2n+2) > f(2n) > 1 for every integer n > 2, > and f(2n) gets further away from 1 as n increases, > and, in fact, diverges towards +oo. f(n) = n! / ( (n/2)! 2^n ) f(2n+2) = (2n+2)! / ( (n+1)! 2^(2n+2) ) f(2n) = (2n)! / ( n! 2^(2n) ) f(2n+2)/f(2n) = (2n+2)! n! 2^(2n) / (2n)! (n+1)! 2^(2n+2) = (2n+1)(2n+2) / ( (n+1) 4 ) = (2n+1)2/4 = n + 1/2 Yes, n + 1/2 is divergent, showing f(2n+2) infinitely greater than f(2n). That seems to be a counterexample. (2n+2) / 2n = 1 + 1/n lim (1 + 1/n) = 1 That doesn't really help here because f(x)/f(y) != f(x/y), f(1) = 1/ 2 gamma(3/2) = 1/2gamma(1/2) = 1/2sqrt(pi) != n+1/2 Considering (n+1/2) / (n+1), in the limit it equals one. Here's one consideration: find a similar form for some other identity. If Stirling's identity is lim n! e^n / n^n sqrt(n 2pi) = 1, with having f(n) = n! e^n / n^n sqrt(n)sqrt(2pi) Then for n+1 and n f(n+1) = (n+1)! e^(n+1) / (n+1)^(n+1) sqrt(n+1) sqrt(2pi) f(n+1)/f(n) = ( (n+1)! e^(n+1) n^n sqrt(n) sqrt2pi ) / n! e^n (n+1)^(n+1) sqrt(n+1) sqrt(2pi) ) = ( (n+1) e n^n sqrt(n) ) / (n+1)^(n+1) sqrt(n+1) = e n^n sqrt(n) / (n+1)^n sqrt(n+1) That appears to go to zero. That is to say, f(n) is infinitely larger than f(n+1). That is because (n+1)^n >> n^n. I think that to show its convergence there in that way that what you want there is to show that it would go to one. Is that similar in example to the proferred counterexample? I'm more interested in contradictions to the steps of the proof, yet I still have to figure out how to respond to the proferred counterexample, unless you do. Ross === Subject: Re: Factorization lemma, example |My paper does rest fully on the following rather basic lemma: | |Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c |is a factor of the contant term P(0), while r=g-c, so it varies if g |varies. | |That simple lemma not only anchors my paper, and my proof of Fermat's |Last Theorem, it can be used in lots of places. My best guess is that you think saying that g varies means something more than it does. |Now I'll use it with |the roots of | | y^3 + 3y - 2 | |and the three roots are | | y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} | | y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - | | (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} | | y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - | | (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} I haven't checked, but I'll assume this is okay. |where the polynomial I'm interested in is | | P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. | |Letting | | g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, | |I have c=1, so r=g-1. | |(Some may worry about my choice for c, I'll come back to that later.) | |Now using the substitution g=r+1, with x=sqrt(2), with my roots, I |have | | y_1 = r+1 - 1/(r+1) | | y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) | | y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) | |so I have | | y_1 = (r^2 + 2r)/(r+1) | | y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) | | y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) | |which is | | y_1 = (r^2 + 2r)/(r+1) | | y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) | | y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) | |which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* |be an algebraic integer, though it provably is not, which is a |problem. Doesn't it bother you at least a little bit that you always have to have this kind of rabbit-out-of-a-hat step in each argument you write down? What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic integers, and all three of them divide 2 in the algebraic integers: 2 = y_1 (y_1^2 + 3) 2 = y_2 (y_2^2 + 3) 2 = y_3 (y_3^2 + 3) It follows that none of them is coprime with 2. |Now you may worry about picking c=1, when I had 1^{1/3}, but all that |happens with picking one of the others is that r becomes coprime to 2, |which I'll leave as an exercise for the reader. | |So as I've said my work is irrefutable, and you may have notice people |avoiding that lemma which starts my paper, and has a key place on my |website. Because the lemma has almost no content! |Now you may know why and you may understand why I've been as upset as |I have. | |It IS algebra, after all. Sooner or later, you're going to realize I'm not saying your argument isn't persuasive because I think I can get away with it. I'm saying your argument isn't persuasive because it just isn't. I could never see it as convincing no matter how I tried. It would be like trying to believe that the moon is actually made of cheese. Keith Ramsay === Subject: Re: Factorization lemma, example > Some of you may face civil liabilities, maybe worse--angry > parents--when the full story comes out. Very witty, you're good. ;) > School is about to start, and if universities teach the flawed > mathematics they may also face federal authorities for deliberate > fraud. Ouch, school is just about to start, thanks to you. Hats off, buddy! > Yup, some of you are looking at chaos as you scramble to figure out > how to prevent yourself from being legally liable, but I've warned for > some time now. > It's your fault if you ignored me. Guys: you've been warned. James told you how to get out of trouble, but you won't listen. > It is mathematics after all. Oh, is it? As for me, I would call that megalomania, but whatever. :) === Subject: Re: Factorization lemma, example > |My paper does rest fully on the following rather basic lemma: > | > |Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > |is a factor of the contant term P(0), while r=g-c, so it varies if g > |varies. > | > |That simple lemma not only anchors my paper, and my proof of Fermat's > |Last Theorem, it can be used in lots of places. > My best guess is that you think saying that g varies means something more > than it does. Actually no, it IS that simple, which is why I've been amazed at how successful people have been arguing against it. Note you missed something obvious below Keith, which surprised me. > |Now I'll use it with > |the roots of > | > | y^3 + 3y - 2 > | > |and the three roots are > | > | y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} > | > | y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - > | > | (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} > | > | y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - > | > | (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} > I haven't checked, but I'll assume this is okay. It is. > |where the polynomial I'm interested in is > | > | P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. > | > |Letting > | > | g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, > | > |I have c=1, so r=g-1. > | > |(Some may worry about my choice for c, I'll come back to that later.) > | > |Now using the substitution g=r+1, with x=sqrt(2), with my roots, I > |have > | > | y_1 = r+1 - 1/(r+1) > | > | y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) > | > | y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) > | > |so I have > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) > | > | y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) > | > |which is > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) > | > | y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) > | > |which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* > |be an algebraic integer, though it provably is not, which is a > |problem. > Doesn't it bother you at least a little bit that you always have to have > this kind of rabbit-out-of-a-hat step in each argument you write down? Maybe this will help. Remember that at x=sqrt(2) r+1 = (1+sqrt(2))^{1/3}, which is a unit, so 1/(r+1) = 1/(1+sqrt(2))^{1/3} = (-1+sqrt(2))^{1/3}, so I have y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1+sqrt(2))^{1/3} y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1+sqrt(2))^{1/3} so you have r^2/2, which has to reduce, or y_2 and y_3 can't be algebraic integers, when they are. However, that means that r^2 has a factor of 2 that IS 2, which proves that ONLY y_1 is not coprime to 2. > What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic > integers, and all three of them divide 2 in the algebraic integers: > 2 = y_1 (y_1^2 + 3) > 2 = y_2 (y_2^2 + 3) > 2 = y_3 (y_3^2 + 3) > It follows that none of them is coprime with 2. It does not follow. Merely repeating false mathematics does not make it so Keith Ramsay. You have a choice as you've had all along, believe mathematics, or just make things up that suit you. > |Now you may worry about picking c=1, when I had 1^{1/3}, but all that > |happens with picking one of the others is that r becomes coprime to 2, > |which I'll leave as an exercise for the reader. > | > |So as I've said my work is irrefutable, and you may have notice people > |avoiding that lemma which starts my paper, and has a key place on my > |website. > Because the lemma has almost no content! It's basic, as I've said. That actually removes excuses for you and others Keith Ramsay. > |Now you may know why and you may understand why I've been as upset as > |I have. > | > |It IS algebra, after all. > Sooner or later, you're going to realize I'm not saying your argument > isn't persuasive because I think I can get away with it. I'm saying > your argument isn't persuasive because it just isn't. I could never see it > as convincing no matter how I tried. It would be like trying to believe > that the moon is actually made of cheese. > Keith Ramsay However, *basic* algebra refutes you. Mathematics is not for everyone. Some people prefer to make up truth rather than accept mathematical logic. Which is unfortunate. === Subject: Re: Factorization lemma, example > |My paper does rest fully on the following rather basic lemma: > | > |Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > |is a factor of the contant term P(0), while r=g-c, so it varies if g > |varies. > | > |That simple lemma not only anchors my paper, and my proof of Fermat's > |Last Theorem, it can be used in lots of places. > My best guess is that you think saying that g varies means something more > than it does. > |Now I'll use it with > |the roots of > | > | y^3 + 3y - 2 > | > |and the three roots are > | > | y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} > | > | y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - > | > | (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} > | > | y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - > | > | (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} > I haven't checked, but I'll assume this is okay. > |where the polynomial I'm interested in is > | > | P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. > | > |Letting > | > | g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, > | > |I have c=1, so r=g-1. > | > |(Some may worry about my choice for c, I'll come back to that later.) > | > |Now using the substitution g=r+1, with x=sqrt(2), with my roots, I > |have > | > | y_1 = r+1 - 1/(r+1) > | > | y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) > | > | y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) > | > |so I have > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) > | > | y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) > | > |which is > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) > | > | y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) > | > |which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* > |be an algebraic integer, though it provably is not, which is a > |problem. > Doesn't it bother you at least a little bit that you always have to have > this kind of rabbit-out-of-a-hat step in each argument you write down? > What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic > integers, and all three of them divide 2 in the algebraic integers: > 2 = y_1 (y_1^2 + 3) > 2 = y_2 (y_2^2 + 3) > 2 = y_3 (y_3^2 + 3) > It follows that none of them is coprime with 2. Stunningly simple and convincing. may complain that you have not shown the y_i's are not units: see [***] below. Essentially this same argument can be applied to refute Harris's claims in Advanced Polynomial Factorization, regarding the polynomial P(x) = 65*x^3 - 12*x + 1. Specifically, you have provided yet another way to prove that, given a factorization of P(x) of the form (a1*x + 1)*(a2*x + 1)*(a3*x + 1) where the a's are algebraic integers, then EACH of the a's is not coprime to 5. Your argument has the great virtue that it is so simple that even Harris cannot fail to understand it. He has submitted APF to several journals, including one to which it is currently submitted. Under these circumstances, any ethical author would withdraw the submission and would withdraw the paper from his or her website. > |Now you may worry about picking c=1, when I had 1^{1/3}, but all that > |happens with picking one of the others is that r becomes coprime to 2, > |which I'll leave as an exercise for the reader. > | > |So as I've said my work is irrefutable, and you may have notice people > |avoiding that lemma which starts my paper, and has a key place on my > |website. > Because the lemma has almost no content! > |Now you may know why and you may understand why I've been as upset as > |I have. > | > |It IS algebra, after all. > Sooner or later, you're going to realize I'm not saying your argument > isn't persuasive because I think I can get away with it. I'm saying > your argument isn't persuasive because it just isn't. I could never see it > as convincing no matter how I tried. It would be like trying to believe > that the moon is actually made of cheese. > Keith Ramsay [***]: Proof that the y_i's are not units: Let y be any root of y^3 + 3*y - 2 = 0. Then y is an algebraic integer. If y is a unit, then y = 1/t for some algebraic integer t. This means that 1 + 3*t^2 - 2*t^3 = 0, which by a well-known theorem implies that t cannot be an algebraic integer. Hence a contradiction. Hence y is not a unit. Nora B. === Subject: Re: Factorization lemma, example James, go back and play in your sandbox. Your threats to universities to sue them for teaching wrong mathematics is totally bogus. Let me ask you a question (not like you'll reply anyway). I'm a science major as well as a mathematics minor. Are you telling me that I've wasted all this time learning math that isn't even right??? David Moran === Subject: Re: Factorization lemma, example >>|My paper does rest fully on the following rather basic lemma: >>| >>|Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c >>|is a factor of the contant term P(0), while r=g-c, so it varies if g >>|varies. >>| >>|That simple lemma not only anchors my paper, and my proof of Fermat's >>|Last Theorem, it can be used in lots of places. >>My best guess is that you think saying that g varies means something more >>than it does. > Actually no, it IS that simple, which is why I've been amazed at how > successful people have been arguing against it. Note you missed > something obvious below Keith, which surprised me. > >>|Now I'll use it with >>|the roots of >>| >>| y^3 + 3y - 2 >>| >>|and the three roots are >>| >>| y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} >>| >>| y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - >>| >>| (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} >>| >>| y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - >>| >>| (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} >>I haven't checked, but I'll assume this is okay. > It is. >>|where the polynomial I'm interested in is >>| >>| P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. >>| >>|Letting >>| >>| g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, >>| >>|I have c=1, so r=g-1. >>| >>|(Some may worry about my choice for c, I'll come back to that later.) >>| >>|Now using the substitution g=r+1, with x=sqrt(2), with my roots, I >>|have >>| >>| y_1 = r+1 - 1/(r+1) >>| >>| y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) >>| >>| y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) >>| >>|so I have >>| >>| y_1 = (r^2 + 2r)/(r+1) >>| >>| y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) >>| >>| y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) >>| >>|which is >>| >>| y_1 = (r^2 + 2r)/(r+1) >>| >>| y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) >>| >>| y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) >>| >>|which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* >>|be an algebraic integer, though it provably is not, which is a >>|problem. >>Doesn't it bother you at least a little bit that you always have to have >>this kind of rabbit-out-of-a-hat step in each argument you write down? > Maybe this will help. Remember that at x=sqrt(2) > r+1 = (1+sqrt(2))^{1/3}, which is a unit, so > 1/(r+1) = 1/(1+sqrt(2))^{1/3} = (-1+sqrt(2))^{1/3}, so I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1+sqrt(2))^{1/3} > y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1+sqrt(2))^{1/3} > so you have r^2/2, which has to reduce, or y_2 and y_3 can't be > algebraic integers, when they are. No it doesn't. What has to reduce is: [(r^2+2r)(-1-sqrt(-3))-2sqrt(-3)](-1+sqrt(2))^(1/3) / 2 There's nothing that says it has to reduce the way you have proposed. If the numerator equals 2, then it clearly reduces, regardless of the nature of the components. The behavior of the whole does NOT correspond to the behavior of the parts. > However, that means that r^2 has a factor of 2 that IS 2, which proves > that ONLY y_1 is not coprime to 2. >>What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic >>integers, and all three of them divide 2 in the algebraic integers: >> 2 = y_1 (y_1^2 + 3) >> 2 = y_2 (y_2^2 + 3) >> 2 = y_3 (y_3^2 + 3) >>It follows that none of them is coprime with 2. > It does not follow. Merely repeating false mathematics does not make > it so Keith Ramsay. Did you bother to check his work? Did you apply the definition of coprime? > You have a choice as you've had all along, believe mathematics, or > just make things up that suit you. I would rather believe mathematics than make things up as you have. > >>|Now you may worry about picking c=1, when I had 1^{1/3}, but all that >>|happens with picking one of the others is that r becomes coprime to 2, >>|which I'll leave as an exercise for the reader. >>| >>|So as I've said my work is irrefutable, and you may have notice people >>|avoiding that lemma which starts my paper, and has a key place on my >>|website. >>Because the lemma has almost no content! > It's basic, as I've said. That actually removes excuses for you and > others Keith Ramsay. >>|Now you may know why and you may understand why I've been as upset as >>|I have. >>| >>|It IS algebra, after all. >>Sooner or later, you're going to realize I'm not saying your argument >>isn't persuasive because I think I can get away with it. I'm saying >>your argument isn't persuasive because it just isn't. I could never see it >>as convincing no matter how I tried. It would be like trying to believe >>that the moon is actually made of cheese. >>Keith Ramsay > However, *basic* algebra refutes you. Mathematics is not for > everyone. > Some people prefer to make up truth rather than accept mathematical > logic. Pot, Kettle. Kettle, Pot. > Which is unfortunate. > -- === Subject: Re: Factorization lemma, example |My paper does rest fully on the following rather basic lemma: > | > |Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > |is a factor of the contant term P(0), while r=g-c, so it varies if g > |varies. > | > |That simple lemma not only anchors my paper, and my proof of Fermat's > |Last Theorem, it can be used in lots of places. My best guess is that you think saying that g varies means something more > than it does. > Actually no, it IS that simple, which is why I've been amazed at how > successful people have been arguing against it. Note you missed > something obvious below Keith, which surprised me. > > |Now I'll use it with > |the roots of > | > | y^3 + 3y - 2 > | > |and the three roots are > | > | y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} > | > | y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - > | > | (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} > | > | y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - > | > | (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} I haven't checked, but I'll assume this is okay. > It is. > |where the polynomial I'm interested in is > | > | P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. > | > |Letting > | > | g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, > | > |I have c=1, so r=g-1. > | > |(Some may worry about my choice for c, I'll come back to that later.) > | > |Now using the substitution g=r+1, with x=sqrt(2), with my roots, I > |have > | > | y_1 = r+1 - 1/(r+1) > | > | y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) > | > | y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) > | > |so I have > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) > | > | y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) > | > |which is > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) > | > | y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) > | > |which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* > |be an algebraic integer, though it provably is not, which is a > |problem. Doesn't it bother you at least a little bit that you always have to have > this kind of rabbit-out-of-a-hat step in each argument you write down? > Maybe this will help. Remember that at x=sqrt(2) > r+1 = (1+sqrt(2))^{1/3}, which is a unit, so > 1/(r+1) = 1/(1+sqrt(2))^{1/3} = (-1+sqrt(2))^{1/3}, so I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1+sqrt(2))^{1/3} > y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1+sqrt(2))^{1/3} > so you have r^2/2, which has to reduce, or y_2 and y_3 can't be > algebraic integers, when they are. You are making an unjustified assumption here. I believe you are right that r^2/2 is not an algebraic integer, and that implies that (r^2/2 + r) also is not an algebraic integer. Does that, however, imply that the product (r^2/2 + r)*(-1 + sqrt(-3)) is not an algebraic integer? That is, if A is *not* an algebraic integer and B *is* an algebraic integer, does it necessarily follow that A*B is not an algebraic integer? Do I need to provide a counterexample, or would you prefer to find your own? > However, that means that r^2 has a factor of 2 that IS 2, which proves > that ONLY y_1 is not coprime to 2. > What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic > integers, and all three of them divide 2 in the algebraic integers: 2 = y_1 (y_1^2 + 3) > 2 = y_2 (y_2^2 + 3) > 2 = y_3 (y_3^2 + 3) It follows that none of them is coprime with 2. > It does not follow. Merely repeating false mathematics does not make > it so Keith Ramsay. I cannot believe you just totally blow off Keith's elegant, simple argument here with this useless arrogant phrase it does not follow with no justification whatsoever, preferring instead to expound your own incorrect B.S.. Here is what has happened. You have seen several previous proofs now that your claims in Advanced Polynomial Factorization were false, and perhaps we got through to you that your old proof there has a gaping hole. So yesterday you worked on this and came up with a new argument, this r + 1 argument that you have presented here. It sweeps aside all the previous objections that several people have mentioned regarding m = 0 and the degenerate case, etc.. It does not make use of any of that stuff. But within hours of this new argument being proposed, Keith Ramsay showed with incredible clarity that it too had to be wrong. You react by (1) ignoring his very short proof, and (2) producing a new (but equally incorrect) argument intended to show that you were right. You seem to think that anything you write down is a work of genius and should not be questioned, and that anything written down by anyone else is a pack of lies. But, man, look at the record! And if you don't want to look at the record, AT LEAST take an honest look at Ramsay's argument! Actually at this point it's irrelevant. You don't have a shred of credibility left. If you had any speck of honor or self-respect you would immediately withdraw your paper from consideration at whatever journal to which it has been sent. Stop wasting the reviewers' time. > You have a choice as you've had all along, believe mathematics, or > just make things up that suit you. > > |Now you may worry about picking c=1, when I had 1^{1/3}, but all that > |happens with picking one of the others is that r becomes coprime to 2, > |which I'll leave as an exercise for the reader. > | > |So as I've said my work is irrefutable, and you may have notice people > |avoiding that lemma which starts my paper, and has a key place on my > |website. Because the lemma has almost no content! > It's basic, as I've said. That actually removes excuses for you and > others Keith Ramsay. > |Now you may know why and you may understand why I've been as upset as > |I have. > | > |It IS algebra, after all. Sooner or later, you're going to realize I'm not saying your argument > isn't persuasive because I think I can get away with it. I'm saying > your argument isn't persuasive because it just isn't. I could never see it > as convincing no matter how I tried. It would be like trying to believe > that the moon is actually made of cheese. Keith Ramsay > However, *basic* algebra refutes you. Mathematics is not for > everyone. Yes. That is one statement for which you have actually provided acceptable proof. Nora B. > Some people prefer to make up truth rather than accept mathematical > logic. > Which is unfortunate. > === Subject: Re: Factorization lemma, example Got it!!! Relatively minor correction finishes the argument. See below... |My paper does rest fully on the following rather basic lemma: > | > |Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c > |is a factor of the contant term P(0), while r=g-c, so it varies if g > |varies. > | > |That simple lemma not only anchors my paper, and my proof of Fermat's > |Last Theorem, it can be used in lots of places. My best guess is that you think saying that g varies means something more > than it does. > Actually no, it IS that simple, which is why I've been amazed at how > successful people have been arguing against it. Note you missed > something obvious below Keith, which surprised me. > > |Now I'll use it with > |the roots of > | > | y^3 + 3y - 2 > | > |and the three roots are > | > | y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} > | > | y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - > | > | (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} > | > | y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - > | > | (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} I haven't checked, but I'll assume this is okay. > It is. > |where the polynomial I'm interested in is > | > | P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. > | > |Letting > | > | g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, > | > |I have c=1, so r=g-1. > | > |(Some may worry about my choice for c, I'll come back to that later.) > | > |Now using the substitution g=r+1, with x=sqrt(2), with my roots, I > |have > | > | y_1 = r+1 - 1/(r+1) > | > | y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) > | > | y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) > | > |so I have > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) > | > | y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) > | > |which is > | > | y_1 = (r^2 + 2r)/(r+1) > | > | y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) > | > | y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) > | > |which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* > |be an algebraic integer, though it provably is not, which is a > |problem. Doesn't it bother you at least a little bit that you always have to have > this kind of rabbit-out-of-a-hat step in each argument you write down? > Maybe this will help. Remember that at x=sqrt(2) > r+1 = (1+sqrt(2))^{1/3}, which is a unit, so > 1/(r+1) = 1/(1+sqrt(2))^{1/3} = (-1+sqrt(2))^{1/3}, so I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1+sqrt(2))^{1/3} > y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1+sqrt(2))^{1/3} Here I forgot that (-1+sqrt(-3)) and (-1-sqrt(-3)) each have a factor of 2 that is 2. Letting u_1 = (-1+sqrt(-3))/2 and u_2 = (-1-sqrt(-3)/2, I have y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} y_2 = [(r^2 + 2r)u_1 + sqrt(-3)](-1+sqrt(2))^{1/3} y_3 = [(r^2 + 2r)u_2 - sqrt(-3)](-1+sqrt(2))^{1/3} so grouping a bit differently I have y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} y_2 = [2r u_1 + r^2 u_1 + sqrt(-3)](-1+sqrt(2))^{1/3} y_3 = [2r u_2 + r^2 u_2 - sqrt(-3)](-1+sqrt(2))^{1/3} but I can subtract and add 1 internally to get y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} y_2 = [2r u_1 + r^2 u_1 -1 + 1 + sqrt(-3)](-1+sqrt(2))^{1/3} y_3 = [2r u_2 + r^2 u_2 -1 + 1 - sqrt(-3)](-1+sqrt(2))^{1/3} which forces any non unit factors in common with 2 for y_2 to be in common with r^2 u_1 - 1 and similiarly for y_3. BUT I can also *add* and subtract 1 internally, so I have y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} y_2 = [2r u_1 + r^2 u_1 +1 - 1 + sqrt(-3)](-1+sqrt(2))^{1/3} y_3 = [2r u_2 + r^2 u_2 +1 - 1 - sqrt(-3)](-1+sqrt(2))^{1/3} which completes the proof. Notice that this works because *both* 1+sqrt(-3) and 1-sqrt(-3) have a factor of 2 that is 2. > so you have r^2/2, which has to reduce, or y_2 and y_3 can't be > algebraic integers, when they are. > However, that means that r^2 has a factor of 2 that IS 2, which proves > that ONLY y_1 is not coprime to 2. > What is true is that none of y_1, y_2, and y_3 is a unit in the algebraic > integers, and all three of them divide 2 in the algebraic integers: 2 = y_1 (y_1^2 + 3) > 2 = y_2 (y_2^2 + 3) > 2 = y_3 (y_3^2 + 3) It follows that none of them is coprime with 2. > It does not follow. Merely repeating false mathematics does not make > it so Keith Ramsay. Yup. > You have a choice as you've had all along, believe mathematics, or > just make things up that suit you. Yup. > |Now you may worry about picking c=1, when I had 1^{1/3}, but all that > |happens with picking one of the others is that r becomes coprime to 2, > |which I'll leave as an exercise for the reader. > | > |So as I've said my work is irrefutable, and you may have notice people > |avoiding that lemma which starts my paper, and has a key place on my > |website. Because the lemma has almost no content! > It's basic, as I've said. That actually removes excuses for you and > others Keith Ramsay. It is basic. It just took me a while. > |Now you may know why and you may understand why I've been as upset as > |I have. > | > |It IS algebra, after all. Sooner or later, you're going to realize I'm not saying your argument > isn't persuasive because I think I can get away with it. I'm saying > your argument isn't persuasive because it just isn't. I could never see it > as convincing no matter how I tried. It would be like trying to believe > that the moon is actually made of cheese. Keith Ramsay > However, *basic* algebra refutes you. Mathematics is not for > everyone. > Some people prefer to make up truth rather than accept mathematical > logic. > Which is unfortunate. Trust the mathematical logic. === Subject: Re: Factorization lemma, example >Got it!!! Relatively minor correction finishes the argument. I doubt that anyone has noticed that the amazing things you say showing that mathematicians are all wrong _always_ seem to require later corrections. Doesn't matter of course, since you got it right _this_ time. === Subject: Re: Factorization lemma, example > Got it!!! Relatively minor correction finishes the argument. > See below... >|My paper does rest fully on the following rather basic lemma: >| >|Given any factor g of a polynomial P(x), g=r+c, where c=g at x=0, so c >|is a factor of the contant term P(0), while r=g-c, so it varies if g >|varies. >| >|That simple lemma not only anchors my paper, and my proof of Fermat's >|Last Theorem, it can be used in lots of places. >My best guess is that you think saying that g varies means something more >than it does. >>Actually no, it IS that simple, which is why I've been amazed at how >>successful people have been arguing against it. Note you missed >>something obvious below Keith, which surprised me. >> >|Now I'll use it with >|the roots of >| >| y^3 + 3y - 2 >| >|and the three roots are >| >| y_1 = (1+sqrt(2))^{1/3} - 1/(1+sqrt(2))^{1/3} >| >| y_2 = (1+sqrt(2))^{1/3}(-1+sqrt(-3))/2 - >| >| (-1-sqrt(-3))/2(1+sqrt(2))^{1/3} >| >| y_3 = (1+sqrt(2))^{1/3}(-1-sqrt(-3))/2 - >| >| (-1+sqrt(-3))/2(1+sqrt(2))^{1/3} >I haven't checked, but I'll assume this is okay. >>It is. >|where the polynomial I'm interested in is >| >| P(x) = x+1, which has the non-polynomial factor (1+x)^{1/3}. >| >|Letting >| >| g = (1+x)^{1/3}, at P(0), g=1^{1/3}, and choosing the solution 1, >| >|I have c=1, so r=g-1. >| >|(Some may worry about my choice for c, I'll come back to that later.) >| >|Now using the substitution g=r+1, with x=sqrt(2), with my roots, I >|have >| >| y_1 = r+1 - 1/(r+1) >| >| y_2 = (r+1)(-1+sqrt(-3))/2 - (-1-sqrt(-3))/2(r+1) >| >| y_3 = (r+1)(-1-sqrt(-3))/2 - (-1+sqrt(-3))/2(r+1) >| >|so I have >| >| y_1 = (r^2 + 2r)/(r+1) >| >| y_2 = [(r^2 + 2r)(-1+sqrt(-3)) + 2sqrt(-3)]/2(r+1) >| >| y_3 = [(r^2 + 2r)(-1-sqrt(-3)) - 2sqrt(-3)]/2(r+1) >| >|which is >| >| y_1 = (r^2 + 2r)/(r+1) >| >| y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)]/(r+1) >| >| y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)]/(r+1) >| >|which proves that y_2 and y_3 are coprime to 2 because r^2/2 *should* >|be an algebraic integer, though it provably is not, which is a >|problem. >Doesn't it bother you at least a little bit that you always have to have >this kind of rabbit-out-of-a-hat step in each argument you write down? >>Maybe this will help. Remember that at x=sqrt(2) >> r+1 = (1+sqrt(2))^{1/3}, which is a unit, so >> 1/(r+1) = 1/(1+sqrt(2))^{1/3} = (-1+sqrt(2))^{1/3}, so I have >> y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} >> y_2 = [(r^2/2 + r)(-1+sqrt(-3)) + sqrt(-3)](-1+sqrt(2))^{1/3} >> y_3 = [(r^2/2 + r)(-1-sqrt(-3)) - sqrt(-3)](-1+sqrt(2))^{1/3} > Here I forgot that (-1+sqrt(-3)) and (-1-sqrt(-3)) each have a factor > of 2 that is 2. Letting u_1 = (-1+sqrt(-3))/2 and u_2 = > (-1-sqrt(-3)/2, I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > > y_2 = [(r^2 + 2r)u_1 + sqrt(-3)](-1+sqrt(2))^{1/3} > > y_3 = [(r^2 + 2r)u_2 - sqrt(-3)](-1+sqrt(2))^{1/3} > so grouping a bit differently I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > > y_2 = [2r u_1 + r^2 u_1 + sqrt(-3)](-1+sqrt(2))^{1/3} > > y_3 = [2r u_2 + r^2 u_2 - sqrt(-3)](-1+sqrt(2))^{1/3} > but I can subtract and add 1 internally to get > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > > y_2 = [2r u_1 + r^2 u_1 -1 + 1 + sqrt(-3)](-1+sqrt(2))^{1/3} > > y_3 = [2r u_2 + r^2 u_2 -1 + 1 - sqrt(-3)](-1+sqrt(2))^{1/3} > which forces any non unit factors in common with 2 for y_2 to be in > common with > r^2 u_1 - 1 > and similiarly for y_3. > BUT I can also *add* and subtract 1 internally, so I have > y_1 = (r^2 + 2r)(-1+sqrt(2))^{1/3} > > y_2 = [2r u_1 + r^2 u_1 +1 - 1 + sqrt(-3)](-1+sqrt(2))^{1/3} > > y_3 = [2r u_2 + r^2 u_2 +1 - 1 - sqrt(-3)](-1+sqrt(2))^{1/3} > which completes the proof. Except for the minor snag that (-1+sqrt(2))^(1/3) is an algebraic integer factor of 2, with the other factor being 2/(-1+sqrt(2))^(1/3). Specifically: 2/(-1+sqrt(2))^(1/3) is a root of x^6-16x^3-64. y_3 and y_2 are NOT coprime to 2. The common factor is staring you in the face. You missed a potential common factor. Your proof is flawed and cannot be repaired. > Trust the mathematical logic. > Since you may not trust me, here's the algebra: x=2/(-1+sqrt(2))^(1/3) 2 = x(-1+sqrt(2))^(1/3) 8 = x^3 (-1+sqrt(2)) x^3+8 = sqrt(2) x^3 x^6 + 16x^3 + 64 = 2 x^6 0 = x^6 - 16x^3 - 64 x=(-1+sqrt(2))^(1/3) x^3 = -1+sqrt(2) x^3 + 1 = sqrt(2) x^6 + 2x^3 + 1 = 2 x^6 + 2x^3 - 1 = 0 -- === Subject: Re: Factorization lemma, example Visiting Assistant Professor at the University of Montana. [.snip.] >Except for the minor snag that (-1+sqrt(2))^(1/3) is an algebraic >integer factor of 2, with the other factor being 2/(-1+sqrt(2))^(1/3). >Specifically: 2/(-1+sqrt(2))^(1/3) is a root of x^6-16x^3-64. >y_3 and y_2 are NOT coprime to 2. The common factor is staring you in >the face. -1+sqrt(2) is a unit in the ring of algebraic integers: (-1+sqrt(2))(1+sqrt(2))=1; therefore, (-1+sqrt(2))^{1/3} is a unit; therefore, even though it is a common factor of both y_2 and 2, that does not prove that y_2 is not coprime to 2. Your element 2/(-1+sqrt(2))^{1/3} is simply an associate of 2. What exactly is wrong with Keith Ramsay's lucid, clear, short proof that y_1, y_2, and y_3 are all nonunit divisors of 2? They are not units because their minimal polynomial over Q is y^3+3y-2, whose constant term is not 1 or -1. And we have that (y_i)*(y_i^2+3) = y_i^3 + 3y_i = 2, since y_i^3+3y_i-2 = 0 for i=1,2,3. Therefore, each y_i is not a unit, and each is a divisor of 2, therefore, none of them are coprime to 2: For suppose that there exist algebraic integers r and s such that r*y_i + 2s = 1. Then 1 = r*y_i + 2s = r*y_i + (y_i)(y_i^2+3)s = (y_i)(r + (y_i^2+3)s) so y_i would be a unit, which we already know is not the case. Thus, no y_i is coprime to 2 in the ring of all algebraic integers (or, in fact, in any subring thereof which contains the integers and contains y_i, since the cofactor is just y_i^2+3...) ====================================================================== Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A man's capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan ====================================================================== === Subject: Re: Factorization lemma, example > [.snip.] >>Except for the minor snag that (-1+sqrt(2))^(1/3) is an algebraic >>integer factor of 2, with the other factor being 2/(-1+sqrt(2))^(1/3). >>Specifically: 2/(-1+sqrt(2))^(1/3) is a root of x^6-16x^3-64. >>y_3 and y_2 are NOT coprime to 2. The common factor is staring you in >>the face. > -1+sqrt(2) is a unit in the ring of algebraic integers: > (-1+sqrt(2))(1+sqrt(2))=1; therefore, (-1+sqrt(2))^{1/3} is a unit; > therefore, even though it is a common factor of both y_2 and 2, that > does not prove that y_2 is not coprime to 2. > Your element 2/(-1+sqrt(2))^{1/3} is simply an associate of 2. Dang. > What exactly is wrong with Keith Ramsay's lucid, clear, short proof > that y_1, y_2, and y_3 are all nonunit divisors of 2? Other than the fact that JSH doesn't seem to understand it? I thought it was simple and convincing. [rest deleted since it's already posted] -- === Subject: Re: Factorization lemma -- refuted [snip error-ridden, hysterical meanderings] Here is a theorem which directly refutes your claim, and involves only basic algebra. THEOREM: Every root of a monic polynomial with integer coefficients divides the constant term within the ring of algebraic integers. (This means that the quotient obtained by dividing the constant term by any root is an algebraic integer. Hence none of the roots are coprime to the constant.) Let P(x) = x^n+ax^(n-1)+bx^(n-1)+...+px +q be a monic polynomial with integer coefficients. (a,b,..p,q are integers.) Then every root, 'r', of 'P(r)' is an algebraic integer and is a solution of the equation: r^n+ar^(n-1)+br^(n-2)+...+pr+q = 0 Note that each term on the left is an algebraic integer, because it is the product of algebraic integers. (This is a fundamental property of the ring.) Consider the following rearrangement: r^n+ar^(n-1)+br^(n-2)+...+pr = -q then r(r^(n-1)+ar^(n-2)+br^(n-3)+...+p) = -q Now 'r' is an algebraic integer, and the sum inside the outer parentheses is also an algebraic integer and 'q' is an (algebraic) integer. Hence, r(...) = -q is satisfied for every root of 'P' and the quotient: -q/r is an algebraic integer. This applies to every root whether any root, or some roots, or all roots, or no roots are units. Therefore, the constant term is *not* and cannot be coprime to any root. QED This theorem directly contradicts James' fundamental result, and consequently invalidates it. -- 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: Homology and Cohomology concepts Hi folks, a few questions... I've seen a phrase a few times and I'm not getting it. If a form is closed does it imply that, at least locally, it is exact. If so how so. Also, if the kth-homology group tells us about the interesting k-cycles (those that aren't boundaries) then how do I interpret the zeroth homology, group which tells us how many connected components there are, in terms of 'interesting' 0-cycles? Also does anyone know of a good conceptual discussion of cohomology? homology seems so geometric and suddenly i'm trying to picture what a coboundary means geometrically.. Kevin === Subject: Re: Homology and Cohomology concepts > Hi folks, a few questions... > I've seen a phrase a few times and I'm not getting it. If a form > is closed does it imply that, at least locally, it is exact. If so how > so. > Also, if the kth-homology group tells us about the interesting > k-cycles (those that aren't boundaries) then how do I interpret the > zeroth homology, group which tells us how many connected components > there are, in terms of 'interesting' 0-cycles? A 0-chain is a formal linear combination of [singular, or simplicial] 0-simplices, which are just points. If you take two such points in the same path-component of the space, one with a coefficient of 1, the other with a coefficient of -1, you'll find that you can construct a 1-chain by using a path from the one point to the other. Depending on the sign convention you choose, you'll either go from the negative 0-simplex to the positive one, or vice-versa. In homology, the positive versions of each of those 0-chains are equivalent, since their difference is the boundary of that path-chain I mentioned earlier. > Also does anyone know of a good conceptual discussion of > cohomology? homology seems so geometric and suddenly i'm trying to > picture what a coboundary means geometrically.. Cohomology is generated by cochains, which are essentially linear functions from chains to your favorite ring, or group, or sheaf. As a function, a cochain has what might be termed support; that would be the set of chains on which it is nonzero. Imagine, for a moment, a cochain that is supported in a single simplex S. Properly speaking, then, it would be supported on any chain that included S. The coboundary of that cochain would be supported along simplices whose boundary contains S. Much more detail can be developed, but I'd have to dedicate more time than I have available at the moment. > Kevin === Subject: how to construct Jordon form matrix? , I have to ask how to construct Jordon form matrix using Filippov's Three-Step algorithm to decompose matrix such as [0 1 2; 0 0 0; 0 0 0]? I have to confess that I missed the class. And the Jordon Form Section in Gilbert Strang is terribly unreadable. The oversimplified presentation made me lost for several hours. If you don't have to time to show me the Three-Step Filippov, can you give me some online resources with clear presentation from which I can study the algorithm? -Walala === Subject: how to construct Jordon form matrix using Filippov's Three Step Algorithm? , I have to ask how to construct Jordon form matrix using Filippov's Three-Step algorithm to decompose matrix such as [0 1 2; 0 0 0; 0 0 0]? I have to say that Jordon Form Section in Gilbert Strang is terribly unreadable. The oversimplified presentation made me lost for several hours. If you don't have to time to show me the Three-Step Filippov Algorithm, can you give me some online resources with clear presentation from which I can study the algorithm? -Walala === Subject: Re: how to construct Jordon form matrix using Filippov's Three Step Algorithm? > If you don't have to time to show me the Three-Step Filippov Algorithm, > can you give > me some online resources with clear presentation from which I can study > the algorithm? I have no idea what the Three-Step Filippov Algorithm is but I have a little MS Computing the Jordan Canonical Form at http://www.maths.ex.ac.uk/~rjc/courses/teach.html -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: how to evaluate this infinite sum? > , > I want to ask how to evaluate this infinite sum: > Sum[1/(4*n^2-1)^2, n is integer from -inf to inf] Their may be an easier way, but did a partial fraction decomposition. You end up with four terms. If you multiply the two terms with linear denominators together this term will go to zero in the summation. The other two are then easier to evaluate. Darren > -Walala === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? >I want to ask if the Fourier Series Expansion is unique to each function? >i.e., if I have two different functions having exactly the same Fourier >Series Expansion, can I claim that these two functions are actually the >same? Assuming the functions are Lebesgue integrable on the interval in question, they are equal almost everywhere (change a function on a set of measure 0 and you don't affect its Fourier series). >As an example, I am facing with the following two functions: >f(t)=0, for -pi<=t<=0, and f(t)=sin(t), for 0<=tg(t)=-1/2*sin(t), for -pi<=t<=0, and g(t)=1/2*sin(t), for 0<=tI find Fourier Series Expansion coefficients c_n=1/(2*pi)*Integrate[f(t), t >from -pi to pi], then I can write f(t)=Sum[c_n*exp(i*n*t), n from -inf to >inf]. >Surprisingly, the Fourier Series Expansion for the two functions are exactly >the same. No they're not. f(t) - g(t) = 1/2*sin(t). The n=1 coefficients will be different. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? > I want to ask if the Fourier Series Expansion is unique to each function? For continuous bounded functions f on [-Pi, Pi), yes. === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? >> I want to ask if the Fourier Series Expansion is unique to each function? > For continuous bounded functions f on [-Pi, Pi), yes. Even for L^1 functions if one identifies function equal Lebesgue almost everywhere. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? > Can anybody give me a hand on this? In order to prove the unicity, your functions need to be sufficiently regular. Hint: Parseval. -- Julien Santini, CMI Technop.99le de Ch.89teau-Gombert, France === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? >> I want to ask if the Fourier Series Expansion is unique to each function? For continuous bounded functions f on [-Pi, Pi), yes. > Even for L^1 functions if one identifies > function equal Lebesgue almost everywhere. measures, for which you can assert actual uniqueness, and beyond that are distributions, for which the same can be said. But the above is a simple class of functions on [-Pi, Pi) yielding uniqueness in the classical sense that I thought would be the most helpful to the OP as he continues his wild cramming for exams. === Subject: Re: Is Fourier Series expansion unique to each function? It's wierd that two different functions can be proved to be the same? > I want to ask if the Fourier Series Expansion is unique to each function? >> >> For continuous bounded functions f on [-Pi, Pi), yes. >> >> Even for L^1 functions if one identifies >> function equal Lebesgue almost everywhere. >measures, for which you can assert actual uniqueness, and beyond that are >distributions, for which the same can be said. Hell, why stop there? Let P be the space of trigonometric polynomials (just a vector space, no topology). Let P' be the space of all linear functionals on P. Then every element of P' has a unique Fourier series expansion... (and now _every_ trig series has become a Fourier series, so this is as far as we can go. I'm gonna be famous for this, the world's most general notion of Fourier series (on the circle).) >But the above is a simple >class of functions on [-Pi, Pi) yielding uniqueness in the classical sense >that I thought would be the most helpful to the OP as he continues his wild >cramming for exams. === Subject: Is this known? m= (sqrt(((x/2)^2)+1))+x/2 Where x = an integer =>1 then m = Phi and the metal means. Also where 1/m = (the decimal expansion of m) Where each mean starting with Phi where its continued fraction --- cf = [1:1,1,1,1,1,1,1,1...] cf = [2:2,2,2,2,2,2,2,2...] cf = [3:3,3,3,3,3,3,3,3...] cf = [4:4,4,4,4,4,4,4,4...] cf = [5:5,5,5,5,5,5,5,5...] etc. This of course is all known, but is the following? I found that all metal means can also have a proportional reduction of its rectangles like Phi's' golden rectangle. Then with the input of another poster ( Burglund) who found that x can be any value greater than zero that is plugged into this equation will do the same thing and have proportional reducing rectangles as well! Where x can = transcendental or irrational or rational or integer values. Another fact also is if a rational x of n decimal length like as an example x = .141. Then its decimal length (n) = 3 is plugged into the equation the outcome will be m = an irrational and 1/m = m's decimal expansion after (n) decimal length in m and 1/m. This holds true for Phi and the metal means where the match of m = 1/m decimal expansion is from the first decimal place because x = an integer with zero decimal places. Dan === Subject: Is this system of PDEs solvable? I have a system of PDEs: f_x = g_t f_t = (c^2)g_x By S_t I mean the partial derivative of S with respect to t. By S_x I mean the partial derivative of S with respect to x. f= f(x,t) g= g(x,t) c=c(t) I don't have any particular c(t) in mind; just so long as it's NOT constant. How can I find general solutions to the PDEs for f, g and c? I realize that this question is similar to a single PDE I asked about before. I'm hoping that this formulation might be easier. Eugene Shubert http://www.everythingimportant.org/relativity/generalized.htm === Subject: Re: Is this system of PDEs solvable? >I have a system of PDEs: >f_x = g_t >f_t = (c^2)g_x >f= f(x,t) >g= g(x,t) >c=c(t) >I don't have any particular c(t) in mind; just so long as it's NOT constant. >How can I find general solutions to the PDEs for f, g and c? If you don't really care what c(t) is, you can replace the second equation by (f_t/g_x)_x = 0, i.e. f_{xt} g_x = f_t g_{xx}, and define c(t) = sqrt(f_t/g_x) (of course you'll want f_t/g_x > 0, but hopefully that will be true at least in some region). Maple 9 then finds three families of solutions. Two are not of interest because f_t or g_x is 0, but the third may be of some interest although it's not very general: f(x,t) = c1 x + c3 x^2/2 + F1(t), g(x,t) = c2 + c1 t + (c3 t + c4) x where c^2(t) = F1'(t)/(c3 t + c4). Here c1, c2, c3, c4 are arbitrary constants and F1 is an arbitrary function; of course we want F1'(t)/(c3 t + c4) > 0, which would cause a singularity at t=-c4/c3 if c3 <> 0 unless F1'(-c4/c3) = 0. Thus with c3 = 1, c4 = 0, F1(t) = t^2/2 + t^4/2 + t^6/6, we get solutions f(x,t) = c1 x + x^2/2 + t^2/2 + t^4/2 + t^6/6 g(x,t) = c2 + c1 t + x t c(t) = 1+t^2 I also found another interesting family of polynomial solutions: c(t) = c[0] + c[1] t 2 3 2 4 f(x, t) = 2/3 b[2] c[1] x t + 1/2 a[2] c[1] t 2 + 2 c[0] c[1] b[2] x t 2 3 2 + (4/3 c[0] c[1] a[2] + 1/3 b[1] c[1] ) t + a[2] x 2 2 2 + 2 c[0] b[2] x t + (c[0] a[2] + c[0] c[1] b[1]) t 2 + a[1] x + c[0] b[1] t + a[0] 2 4 3 2 g(x, t) = 1/6 b[2] c[1] t + 2/3 c[0] c[1] b[2] t + b[2] x 2 2 + 2 a[2] x t + c[0] b[2] t + b[1] x + a[1] t + b[0] Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: kalman filter issue, question My textbooks solve the Kalman filter for the following system of equations: x(t) = A x(t-1) + w(t-1) z(t) = H x(t) + v(t) However, I am interested in the following system instead: x(t) = A x(t-1) + s(t) z(t) = H x(t) + v(t) where s(t) is noise. Can I identify s(t) as w(t-1) and use the textbook Kalman filter equations? === Subject: Re: kalman filter issue, question > My textbooks solve the Kalman filter for the following system of equations: > x(t) = A x(t-1) + w(t-1) > z(t) = H x(t) + v(t) > However, I am interested in the following system instead: > x(t) = A x(t-1) + s(t) > z(t) = H x(t) + v(t) > where s(t) is noise. > Can I identify s(t) as w(t-1) and use the textbook Kalman filter > equations? The process noise vector w is assumed to be white Gaussian noise, so w(t) versus w(t-1) is not an issue. So proceed with the standard Kalman filter equations. OUP === Subject: kinds of Algebras In high school and community colleges, we can study something called Elementary Algebra, sometimes called Introductory Algebra; later if we want or are required, we can study Intermediate Algebra. Once we study these things, we believe we know what algebra is and what we can do with it. Soon, even before we may finish the Intermediate level, some people and a few instructors make mention of Modern Algebra and Abstract Algebra. And then we may read further that other sorts of Algebras exist. I thought algebra was algebra and that it just has deeper or higher levels, extending from earlier levels, all based from the first, Elementary one. What are the other algebras, and what happens in them? Is math including Algebra, already an abstract thing? Why is there something called Abstract Algebra when algebra already is abstract without calling some particular kind of algebra, Abstract? Are any of those other algebras in reach for someone who currently maintains some Intermediate Algebra knowledge, and who has only faint memories of Calculus (three semesters) with which he struggled many many years ago? G C === Subject: Re: kinds of Algebras > In high school and community colleges, we can study something called > Elementary Algebra, sometimes called Introductory Algebra; later if we want > or are required, we can study Intermediate Algebra. Once we study these > things, we believe we know what algebra is and what we can do with it. Soon, > even before we may finish the Intermediate level, some people and a few > instructors make mention of Modern Algebra and Abstract Algebra. And then > we may read further that other sorts of Algebras exist. > I thought algebra was algebra and that it just has deeper or higher levels, > extending from earlier levels, all based from the first, Elementary one. > What are the other algebras, and what happens in them? Is math including > Algebra, already an abstract thing? Why is there something called Abstract > Algebra when algebra already is abstract without calling some particular kind > of algebra, Abstract? Are any of those other algebras in reach for someone > who currently maintains some Intermediate Algebra knowledge, and who has only > faint memories of Calculus (three semesters) with which he struggled many many > years ago? > G C Elementary algebra abstracts from numbers to variables. Abstract algebra abstracts from variables that represent numbers to variables that represent sets of numbers. Basically, it's an issue of what level the abstraction occurs at. Are we talking about the conventional meaning of + on ordinary numbers, or some new + on a new set of objects? -- === Subject: Re: kinds of Algebras quite some years ago, makes a three-way distinction among high-school algebra (where we study polynomials), college algebra (where we study rings, fields, groups,...) and university algebra (where we study categories). What he does is, he says, high-school algebra. Lee Rudolph === Subject: Laundry problem You have n towels of different sizes placed randomly into a single pile( after doing your laundry). The one move you are allowed to make is to take a stack of towel off the top of the pile and place this stack onto the bottom of the pile. The goal is to get the towels into a pile from the largest sized towel to the smallest sized towel. What is the least number of moves required? === Subject: Re: Laundry problem >You have n towels of different sizes placed randomly into a single pile( >after doing your laundry). The one move you are allowed to make is to take a >stack of towel off the top of the pile and place this stack onto the bottom >of the pile. The goal is to get the towels into a pile from the largest >sized towel to the smallest sized towel. What is the least number of moves >required? If you just place the stack as-is on the bottom of the pile, you're just doing a rotation, e.g. ABCDE -> CDEAB, and you'll never sort the pile that way. So if this is a real problem rather than a trick, perhaps you are allowed to turn the top stack over before putting it at the bottom, e.g. ABCDE -> CDEBA. I don't know the answer in general, but I have some numerical results: n=3: 2 4: 3 5: 5 6: 6 7: 7 8: 8 My guess is the answer is n for n >= 5. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Laundry problem > You have n towels of different sizes placed randomly into a single pile( > after doing your laundry). The one move you are allowed to make is to take a > stack of towel off the top of the pile and place this stack onto the bottom > of the pile. The goal is to get the towels into a pile from the largest > sized towel to the smallest sized towel. What is the least number of moves > required? Hmm, when we bury a stack of towels to the bottom, are we allowed to flip the stack upside-down in the process? If so, then the problem is solvable and is interesting; but, otherwise, we're limited to just cycling through the towels, and there is clearly no solution (except trivially when the towels are already in order, or one move away from being so). === Subject: Re: Laundry problem > You have n towels of different sizes placed randomly into a single pile( > after doing your laundry). The one move you are allowed to make is to take a > stack of towel off the top of the pile and place this stack onto the bottom > of the pile. The goal is to get the towels into a pile from the largest > sized towel to the smallest sized towel. What is the least number of moves > required? This cute problem has been around for thirty years or so. Search for pancake sort. Rick === Subject: Re: Laundry problem > You have n towels of different sizes placed randomly into a single pile( > after doing your laundry). The one move you are allowed to make is to take a > stack of towel off the top of the pile and place this stack onto the bottom > of the pile. The goal is to get the towels into a pile from the largest > sized towel to the smallest sized towel. What is the least number of moves > required? > This cute problem has been around for thirty years or so. Search for > pancake sort. I found this site useful ... http://www.cut-the-knot.org/SimpleGames/Flipper.shtml === Subject: Re: Laundry problem > You have n towels of different sizes placed randomly into a single pile( > after doing your laundry). The one move you are allowed to make is to > take a > stack of towel off the top of the pile and place this stack onto the > bottom > of the pile. The goal is to get the towels into a pile from the largest > sized towel to the smallest sized towel. What is the least number of > moves > required? This cute problem has been around for thirty years or so. Search for > pancake sort. > I found this site useful ... > http://www.cut-the-knot.org/SimpleGames/Flipper.shtml > === Subject: Lines Crossing In A Nonagon: puzzle In: http://mathforum.org/discuss/sci.math/m/525673/525673 # of crossings: path within m-gon, I mentioned the general case from which this puzzle is derived. I also asserted that the idea MIGHT be the basis of an interesting puzzle. So I am posting this here so as to test this assertion. Start with a regular nonagon (9-gon). Draw a path consisting of connected straight line-segments within the nonagon, so that each segment starts/ends on the vertexes of the 9-gon, and where each vertex is visited exactly once, and where the 1st and last segment are connected. (The total path will cross itself several times.) And the path is restricted as to keep it from visiting adjacent vertexes (along the perimeter) consecutively. And I mention again, the path must be closed, ie the first and last segment must be connected. Now the path must be such that there are a total of exactly 13 crossings inside the nonagon. (There are exactly 2 line-segments meeting at each crossing....I am sure, if I drew the path/nonagon correctly when I was coming up with this puzzle.) (The meetings of the line-segments at the vertexes are NOT considered to be crossings.) In what order does the path visit the vertexes? Of course, there are many solutions, given reflections and rotations. But, aside from reflections and rotations, is there a unique solution? I might give my answer in a few days. Leroy Quet === Subject: Re: Lines Crossing In A Nonagon: puzzle http://mathforum.org/discuss/sci.math/m/525673/525673 ># of crossings: path within m-gon, >I mentioned the general case from which this puzzle is derived. >I also asserted that the idea MIGHT be the basis of an interesting puzzle. >So I am posting this here so as to test this assertion. >Start with a regular nonagon (9-gon). >Draw a path consisting of connected straight >line-segments within the nonagon, so that each segment starts/ends on >the vertexes of the 9-gon, and where each vertex is visited exactly >once, and where the 1st and last segment are connected. >(The total path will cross itself several times.) >And the path is restricted as to keep it from visiting adjacent >vertexes (along the perimeter) consecutively. >And I mention again, the path must be closed, ie the first and last >segment must be connected. >Now the path must be such that there are a total of exactly 13 crossings >inside the nonagon. >(There are exactly 2 line-segments meeting at each crossing....I am sure, >if I drew the path/nonagon correctly when I was coming up with this >puzzle.) >(The meetings of the line-segments at the vertexes are NOT considered to >be crossings.) >In what order does the path visit the vertexes? >Of course, there are many solutions, given reflections and rotations. >But, aside from reflections and rotations, is there a unique solution? >I might give my answer in a few days. >Leroy Quet This rule generates the right sequence: Starting at an arbitrary vertex, visit alternately the second and third vertex from the last, when you find that you would be revisiting a vertex, skip to the next empty one. -- Patrick Hamlyn posting from Perth, Western Australia Windsurfing capital of the Southern Hemisphere Moderator: polyforms group (polyforms-subscribe@egroups.com) === Subject: Re: Logical fallacies Robert B. Israel scribbled the following on sci.math: >> The Bible has a fine one. >> If they aren't against me, they're with me >> If they aren't with me, they're against me > These statements are logically equivalent. >> What about those who are neither with you nor against? > He's stating that there aren't any. This may or may not be true, > depending on what is included in they, but it certainly isn't a > logical fallacy. However, both statements allow people to be BOTH with you AND against you, which might not be what the Bible intended. -- /-- Joona Palaste (palaste@cc.helsinki.fi) --------------------------- | Kingpriest of The Flying Lemon Tree G++ FR FW+ M- #108 D+ ADA N+++| | http://www.helsinki.fi/~palaste W++ B OP+ | ----------------------------------------- Finland rules! ------------/ You could take his life and... - Mirja Tolsa === Subject: Re: Lottery Probability > Nevertheless, the local newspaper lists hot numbers in the lottery > (numbers that have occurred most frequently recently) and overdue > numbers (numbers that have occurred least frequently recently). In most lotteries it is actually a *bad* idea to choose numbers claimed to be hot, because the winning amounts are usually shared among all winners. Hot numbers are chosen more often, thus will have more winners and finally result in less money for each individual winner. For the same reason, picking 19 is almost always a bad idea, because it appears in most birthdays. Felix === Subject: Re: Lottery Probability Nevertheless, the local newspaper lists hot numbers in the lottery > (numbers that have occurred most frequently recently) and overdue > numbers (numbers that have occurred least frequently recently). > > In most lotteries it is actually a *bad* idea to choose numbers claimed > to be hot, because the winning amounts are usually shared among all > winners. Hot numbers are chosen more often, thus will have more > winners and finally result in less money for each individual winner. I wouldn't mind splitting a quarter ion dollars with a hundred others. > For the same reason, picking 19 is almost always a bad idea, because > it appears in most birthdays. Once the ticket agent who was scanning my deck of 42 lotto cards accidentally ran one of them twice. She caught the mistake and said I didn't have to pay for it. I said, No, I'll pay for it. If that's the winning number, I don't want to share it with you. Talk about paranoid optimism. > Felix === Subject: Re: Lottery Probability >Sure is. And maybe hurricanes are even one of them. >(Say, maybe the mean annual sea temperature near the Azores >builds up and up and *up* until it can spawn some hurricanes >and cool itself off. ... Well, that *particular* theory >sounds hopelessly bogus, but maybe there's another one.) It's pretty damn close to the theory of El Nino. But that's in the Pacific. I know some people who bet on hurricanes for a living. They don't believe hurricanes have a memory. I don't have any reason to disbelieve them, but I also haven't checked their reasoning. To paraphrase Erdos, If the actuaries believe it, it's good enough for me. Jon Miller === Subject: Re: Lottery Probability >>Sure is. And maybe hurricanes are even one of them. >>(Say, maybe the mean annual sea temperature near the Azores >>builds up and up and *up* until it can spawn some hurricanes >>and cool itself off. ... Well, that *particular* theory >>sounds hopelessly bogus, but maybe there's another one.) >It's pretty damn close to the theory of El Nino. >But that's in the Pacific. There are various cycles on different time scales (including the El Nino cycle) that influence north Atlantic hurricanes. See e.g. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Lottery Probability >>Sure is. And maybe hurricanes are even one of them. >>(Say, maybe the mean annual sea temperature near the Azores >>builds up and up and *up* until it can spawn some hurricanes >>and cool itself off. ... Well, that *particular* theory >>sounds hopelessly bogus, but maybe there's another one.) >It's pretty damn close to the theory of El Nino. Well, the part about the mean annual sea temperature near the Azores influencing Atlantic hurrican production is certainly part of the current theory as I've heard it expounded. But I don't think that theory goes on to say that the hurricanes which follow such an increase in that temperature themselves cause a decrease in that temperature; and ly I find it hard to believe that they could (though maybe it's so), since despite their awesome power they surely must dissipate much less energy in toto than the differential, over several years of increased insolation, of the energy stored in the increasingly warm waters around the Azores--so I said that the theory, as deliberately sabotaged by me, sounds hopelessly bogus (but maybe I'm wrong about that; it's not like I actually have any facts or figures, much less any models to feed them into). Lee Rudolph === Subject: Re: Mapping of integers to reals ~ Cantor's disproof <0307241604010.11311-100000@gandalf.math.ukans.edu> <0307241639250.11752-100000@gandalf.math.ukans.edu> We can look at the definition of g and we can look at the value of > f(k). That is insufficient, forall functions you must look at the definition of f. >If f(k)=0, we know that g(k) will never halt (and so H(k)=1). > If f(k) != 0, then g(k) will halt (and so H(k) = 0). f(k) = the kth decimal place of pi/4, and H(k) = the >nth decimal place of r. So r and pi/4 can't be the >same real numbers, since they differ in position k. >The same proof goes through for every computable real. >No computable real can be equal to r. You used white box inspection on the function g but not the function f, under what conditions does function f apply to a similar disqualification? Forall f? Halting problem is about limitations of knowledge, your proof doesn't go through for every computable function, it goes through for every known computable function. Why? Because by the halting problem we don't know systematically if a function is computable. Your proof is not sufficient as far as I can see to determine that a function does not exist that calculates r. Call a contender function for Halt HH. HH is not known whether it is computable or not, in the sense of the term that HH terminates so far on its test parameters and has given suitable results for H. But H comes across a difficult parameter and hasn't given a result yet! What conclusion can you draw about the halting nature of HH? Nothing. Herc === Subject: Re: Math argument, round and round > No, actually I think all the mathematicians in the world, except > , are being subtly influenced by the qi-waves broadcast > by a transmitter buried in a secret location in Switzerland by a > chapter of the Bavarian Illuminati, and that the CIA know about > this but have been persuaded to keep quiet by the Jesuits who are > ultimately controlled by the alien lizards who have been running the > world for the last 4000 years. I know this *must* be true because > the alternative would be that the elementary mathematical errors > which I think I can see in ' proof are really there, > and that is making a fool of himself by posting the > same old rubbish over and over again on Usenet rather than finding > something useful to do. You forgot the Knights Templar and Saint Germaine. Bob Kolker === Subject: Re: Math argument, round and round > I have a math proof. That math proof highlights an error made by > mathematicians long ago with methods that also lead to a short proof > of Fermat's Last Theorem. Rather than deal with the argument, posters > go to conclusions from the error in their math, and claim I'm wrong > because of their error!!! > That so-called proof of yours is pure bunkum. It is so incoherent, it is > not even wrong. > Bob Kolker Well luckily for me I found an alternate and simple demonstration using only basic algebra, which I posted on math newsgroups last night. Those of you on sci.physics are no longer needed. You didn't turn out to be any use anyway, more's the pity. Yup, it's the smoking gun. The only question now is how long it will take before mathematicians admit the truth. That math arguments are over people. Those of you who wish can continue to spout off but a lot of action should start taking place behind the scenes. === -- #191, ewill3@earthlink.net -- and remember, I'm not (a) being (from) Sirius It's still legal to go .sigless. === Subject: Re: Math argument, round and round >I have a math proof. That math proof highlights an error made by >mathematicians long ago with methods that also lead to a short proof >of Fermat's Last Theorem. Rather than deal with the argument, posters >go to conclusions from the error in their math, and claim I'm wrong >because of their error!!! >>That so-called proof of yours is pure bunkum. It is so incoherent, it is >>not even wrong. > Well luckily for me I found an alternate and simple demonstration > using only basic algebra, which I posted on math newsgroups last > night. > Those of you on sci.physics are no longer needed. You didn't turn out > to be any use anyway, more's the pity. > Yup, it's the smoking gun. The only question now is how long it will > take before mathematicians admit the truth. You mean like the fact that you *consistently* fail to adequately address *all* objections? > That math arguments are over people. Those of you who wish can > continue to spout off but a lot of action should start taking place > behind the scenes. === Subject: Re: Math argument, round and round is this the announcement of the early resolution of your ten-year programme?... congratulation, again. so, if you cannot actually reply to the arguments of others, repeated *cogito etc. ad vomitorium* in different ways, what is the ideal of your vast postage to the Silent Majority on Googol/Usenet? is there a fan-club in the offing, or does it already exist, other than on cranks.net etc. > Well luckily for me I found an alternate and simple demonstration > using only basic algebra, which I posted on math newsgroups last > night. > Those of you on sci.physics are no longer needed. You didn't turn out > to be any use anyway, more's the pity. === Subject: Matrix Multiplication Could someone please name some applications or areas in computer science where matrix multiplication is used/useful? (Like 3D imaging and === Subject: Re: Matrix Multiplication > Could someone please name some applications or areas in computer science > where matrix multiplication is used/useful? (Like 3D imaging and Yes, 3D imaging. Look in the the OpenGL Programming Guide, for example. -- Stephen Montgomery-Smith stephen@math.missouri.edu http://www.math.missouri.edu/~stephen === Subject: Morley's theorem via complex arithmetic? Maybe somebody with Maple can smoke out the answer to this one. Let A,P,Q,R be distinct complex numbers in the upper half plane such that these six numbers are real and positive: A/P^3 R/P^2 A/(Q-A)^3 (R-A)/(Q-A)^2 Q-2P A-3P. Then (2P-Q-R)=s(-P+2Q-R) where s is one of the two roots of ss+s+1=0. It's the same as this little result, for which a geometric proof is rather easy: http://planetmath.org/encyclopedia/CorollaryOfMorleysTheorem.html (where we have identified B with zero) but it would be nice to have a polynomial identity which makes the proof a no-brainer. Better still would be a proof of Morley itself http://planetmath.org/encyclopedia/ProofOfMorleysTheorem.html along the same lines. If anybody comes up with a polynomial identity to prove it, then I will, of course, immortalize him with a credit over at PlanetMath :) Larry === Subject: Re: Morley's theorem via complex arithmetic? this is covered with a fabulous generalization in _Complex Geometry_ by a Chinese guy, an MAA paperback > http://planetmath.org/encyclopedia/CorollaryOfMorleysTheorem.html > (where we have identified B with zero) but it would be nice to have a > polynomial identity which makes the proof a no-brainer. Better still would > be a proof of Morley itself > http://planetmath.org/encyclopedia/ProofOfMorleysTheorem.html > along the same lines. If anybody comes up with a polynomial identity to --les ducs d'Enron! http://members.tripod.com/~american_almanac === Subject: nonlinear system I posted a few weeks back and am still puzzled. I want to confirm my suspicion that the following system has a unique nonzero solution (or find a counterexample). Let A be n by n matrix with positive entries and largest magnitude eigenvalue > 1 . Show that there is exactly one nonzero x=(x1,x2,...,xn) with all entries between 0 and 1 such that: 1 - e^{-(Ax)_1} = x1 1 - e^{-(Ax)_2} = x2 . . . 1 - e^{-(Ax)_n} = xn where (Ax)_k is kth entry of vector Ax, (Ax)_k = A_{k1} x1 + A_{k2} x2 + ... + A_{kn} xn Of course, x = (0,0,...,0) is a trivial solution. For the few examples that I tried, if you pick a vector with entries between 0 and 1 and iterate the left hand side of the above equations on this point you converge to the solution. I was thinking about some kind of fixed point theorem but only know the contraction mapping principle and couldn't see how to get the conditions to apply for general A. Anyhow, it would not rule out another nonzero solution outside the complete subspace of [0,1]^n which would have to exclude zero, if such an approach worked. If that makes any sense. I note that the n=1 case is just P(y)=y, where y=1-x and P is the Poisson prob gen function with mean A>1, which has unique solution in (0,1). Is there a pgf generalization that can be applied here ?? Any suggestions welcome! === Subject: Number theory and provability Regarding the recent thread about the Peano postulates and their potential to establish a rich axiomatic foundation for number theory: Are they really insufficient? I was surprised at that suggestion. If Peano axioms don't do the job then, what currently does do the job? What is the typical way in which mathematicians formally prove theorems about number theory? Or what is the preferred method of formalization? === Subject: Re: Number theory and provability Originator: tchow@lagrange.mit.edu.mit.edu (Timothy Chow) >Regarding the recent thread about the Peano postulates and their >potential to establish a rich axiomatic foundation for number >theory: >Are they really insufficient? I was surprised at that suggestion. The Peano postulates could mean several things. They could refer to Peano's original postulates, which by modern standards are slightly vague. They could refer to what modern mathematicians call first-order Peano Arithmetic, or they could refer to what they call second-order Peano Arithmetic. Neither of these suffices to prove, or even state, everything that mathematicians have proved in number theory, and first-order PA does not even suffice to *characterize* the natural numbers---meaning that there are other things (nonstandard integers) besides the natural numbers that satisfy those axioms. Second-order PA does characterize the natural numbers in this sense, so in that sense they're sufficient, but again there are theorems of number theory that are proved by other mathematical methods that cannot be reduced to proofs within second-order PA. === Subject: Optimization Problems seem to make any progress after looking at them over and over again. I would appreciate if someone could first let me know if I am just missing out on a concept of optimization, and then give me information that would help me relate the rates so that I could use the algorithm for extreme values to solve the problems. 1. Find the maximum area of a rectangle inscribed in the ellipse 16x^2+25y^2=400. This one is probably my fault for not taking math 12 before I did this course so I don't know much about an ellipse. I can't figure out how to relate these rates. I can figure out y=á, but don't see how that relates to the length and width of a triangle. Area=l*w A(l)=? 2. A rectangular beam is cut out of a cylindrical log with diameter 30cm. Given that strength of a beam is jointly proportional to its width and the square of its depth, find the dimensions of the beam with maximum strength. l(_)l h w Strength= k*w*(h)^.5 I don't understand how to relate these. I know I have to use d=30 somewhere, but don't see how because I would have to know area or volume. (I think) 3. Find the dimensions of the right circular cylinder of maximum volume that can be inscribed in a sphere of radius a. Volume of cylinder=pi(r)^2h Volume of sphere=4/3pi(a)^2 (note:how do you write pi on your comp?)ii? Again I have trouble relating these. Thank you for your time. Scott Eliason === Subject: Re: Optimization Problems > seem to make any progress after looking at them over and over again. > I would appreciate if someone could first let me know if I am just > missing out on a concept of optimization, and then give me information > that would help me relate the rates so that I could use the algorithm > for extreme values to solve the problems. > 1. Find the maximum area of a rectangle inscribed in the ellipse > 16x^2+25y^2=400. > This one is probably my fault for not taking math 12 before I did this > course so I don't know much about an ellipse. I can't figure out how > to relate these rates. I can figure out y=.83, but don't see how that > relates to the length and width of a triangle. > Area=l*w > A(l)=? Drawing the diagram will reveal: l=2x, w=2y. > 2. A rectangular beam is cut out of a cylindrical log with diameter > 30cm. Given that strength of a beam is jointly proportional to its > width and the square of its depth, find the dimensions of the beam > with maximum strength. > l(_)l h > w > Strength= k*w*(h)^.5 You've done the sqrt, not square. Strength = k*w*h^2 Look at the diagonal as it relates to l=h and w. > I don't understand how to relate these. I know I have to use d=30 > somewhere, but don't see how because I would have to know area or > volume. (I think) You don't need area or volume. > > 3. Find the dimensions of the right circular cylinder of maximum > volume that can be inscribed in a sphere of radius a. > Volume of cylinder=pi(r)^2h > Volume of sphere=4/3pi(a)^2 > (note:how do you write pi on your comp?)ii? You will have to use relate the radius of the sphere to the radius and height of the cylinder. Again, you are dealing with triangles. Use pi to reperesent pi. > Again I have trouble relating these. > Thank you for your time. > Scott Eliason All of your problems boil down to finding the appropriate right triangle in the problem and applying the Pythagorean Theorem. -- === Subject: Re: Lynds > Does anyone know about this guy? Without any examination of the paper > admittedly, my inexpert eyes tell me that this belongs to the CRANK > department. But then again, I might be wrong. > ------------------------------------------------------------------------- > http://www.quad-net.com/archive/TIME6-25AGB My usual source in such cases www.crank.net has nothing on this guy. But it does look cranky as hell. LH === Subject: Re: Lynds > Does anyone know about this guy? seem to be of mathematical interest. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: Lynds > Does anyone know about this guy? Without any examination of the paper > admittedly, my inexpert eyes tell me that this belongs to the CRANK > department. But then again, I might be wrong. > ------------------------------------------------------------------------- > No matter how small the time interval, or how slowly an object moves > during that interval, it is still in motion and its position is > constantly changing, so it can't have a determined relative position > at any time, whether during an interval, however small, or at an > instant, he said in a statement. If it did, it couldn't be in > motion. This problem (with instantaneous position) has been addressed and solved by the application of limit processes. The writer, who is described as a drop-out, obviously dropped out before he learned calculus. Send to CRANK department! -- 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: planar cayley graphs Hello. I'm interested in Cayley graphs that are essentially planar. Unfortunately, the graph-theory definion of planar is so prevalent, I'm having trouble searching for it. Call an (infinite) graph essentially planar if it can be embedded in the plane with no two nodes closer than some distance, and no edge longer than some distance. So, the Cayley graph of Z x Z, the cross product of the infinite cyclic group with itself, is planar. Also groups like: < a, b | a^4, b^4, abab > or < a, b | a^6, b^2, ababab > I think are essentially planar. To put it another way, a group is essentially planar if the number of elements with shortest word length less than r goes like r^2. There are hyperbolic groups, I think? (Is the fundamental group of a genus two 2-manifold hyperbolic?) It seems plausible to me that hyperbolic groups with zero curvature would be my essentially planar groups, but I've not been able to make much progress understanding what they are. (My reference is Word Processing in Groups by Epstein and Cannon) Could someone here point me to where I could learn what this idea essentially planar is really called? icholas === Subject: pretty simple problem for any reputable mathematician... :) Hi all I have a simple problem but just can't remember how to solve it due to old age and lack of intelligence :) I have a tarp 100' x 100'. I want to use it for a roof of a shelter type structure. I originally planned to make the shelter 100' x 100'. I soon realized that the tarp would not fit due to the center of the structure being 10' high and the outer edges only being 8' high. I thought the answer may be as simple as figuring the partial surface area of a pyramid, but got to thinking maybe calculus was the ticket. So the question is, how big can I make the structure to fully use the tarp stretched taught over the top if there is a two foot drop from center to outer edges. The structure is supposedly square, but could be changed a bit to make this work if necessary. Any takers. I am just as interested in the mathematics to solve this as i am the solution so please include the process in the solution... many thanx in advance for any pointers in solving this problem... === Subject: Re: pretty simple problem for any reputable mathematician... :) >I have a tarp 100' x 100'. I want to use it for a roof of a shelter >type structure. I originally planned to make the shelter 100' x 100'. > I soon realized that the tarp would not fit due to the center of the >structure being 10' high and the outer edges only being 8' high. I >thought the answer may be as simple as figuring the partial surface >area of a pyramid, but got to thinking maybe calculus was the ticket. >So the question is, how big can I make the structure to fully use the >tarp stretched taught over the top if there is a two foot drop from >center to outer edges. If you think about this, you'll see that you _can't_ spread a square tarp taut (note spelling) over that surface -- well, not unless the material is very stretchy, which tarps typically are not. If the roof is a shallow square pyramid, with the base 100 ft by 100 ft and the central peak 2 ft higher than the base, then the tarp must measure a bit more than 100 ft by 100 ft but will not lie completely flat on all parts of the roof. -- Fortunately, I live in the United States of America, where we are gradually coming to understand that nothing we do is ever our fault, especially if it is really stupid. --Dave Barry === Subject: Re: pretty simple problem for any reputable mathematician... :) > Hi all > I have a simple problem but just can't remember how to solve it due to > old age and lack of intelligence :) > I have a tarp 100' x 100'. I want to use it for a roof of a shelter > type structure. I originally planned to make the shelter 100' x 100'. > I soon realized that the tarp would not fit due to the center of the > structure being 10' high and the outer edges only being 8' high. I > thought the answer may be as simple as figuring the partial surface > area of a pyramid, but got to thinking maybe calculus was the ticket. > So the question is, how big can I make the structure to fully use the > tarp stretched taught over the top if there is a two foot drop from > center to outer edges. The structure is supposedly square, but could > be changed a bit to make this work if necessary. > Any takers. I am just as interested in the mathematics to solve this > as i am the solution so please include the process in the solution... > many thanx in advance for any pointers in solving this problem... Look at a cross section of the roof through the center of the structure. You'll have a triangle with height 2' and two sides sloping down of length 50'. Looking at only the left half gives you a right triangle with height 2', hypotenuse 50'. a^2 + b^2 = c^2. -- === Subject: Re: pretty simple problem for any reputable mathematician... :) >I have a tarp 100' x 100'. I want to use it for a roof of a shelter >type structure. I originally planned to make the shelter 100' x 100'. > I soon realized that the tarp would not fit due to the center of the >structure being 10' high and the outer edges only being 8' high. I >thought the answer may be as simple as figuring the partial surface >area of a pyramid, but got to thinking maybe calculus was the ticket. So the question is, how big can I make the structure to fully use the >tarp stretched taught over the top if there is a two foot drop from >center to outer edges. > If you think about this, you'll see that you _can't_ spread a square > tarp taut (note spelling) over that surface -- well, not unless the > material is very stretchy, which tarps typically are not. > If the roof is a shallow square pyramid, with the base 100 ft by > 100 ft and the central peak 2 ft higher than the base, then the tarp > must measure a bit more than 100 ft by 100 ft but will not lie > completely flat on all parts of the roof. Hi Stan Thanx for the reply. It does make sense that there will be a bit of excess tarp but I am hoping it will be very minimal and hopefully a solution will present itself. Thanx for the headz up. === Subject: Re: Pure math, physics, and FLT Lawrence E. McKnight > [snip... > Interesting. One of the very earliest Western mathematicians, >Euclid, proved around 300 BC that the sqrt(2) is irrational. >It is a model proof by contradiction which is still one of the >first things a mathematician learns. Euclid never claimed that >sqrt(2) does not exist. Perhaps you are thinking of the Southern >politicians in the 20th century who refused to accept that pi is >irrational, and wanted to legislate that it equals 22/7. >Mathematicians objected. However I do not believe the construction >workers lobbied very hard against it. > Southern? I thought it was Indiana. I believe that Indiana wanted to make pi equal to 3. === Subject: Re: Pure math, physics, and FLT > Lawrence E. McKnight politicians in the 20th century who refused to accept that pi is > irrational, and wanted to legislate that it equals 22/7. >> Southern? I thought it was Indiana. > I believe that Indiana wanted to make pi equal to 3. None of the above, though exactly what the 's authors *did* intend is a little hard to work out. Look it up at and decide for yourself. === Subject: Re: Pure math, physics, and FLT >> I believe that Indiana wanted to make pi equal to 3. > None of the above, though exactly what the 's authors *did* intend > is a little hard to work out. Look it up at >Today many mathematicians engage in what they call pure math. >However, if pure math had been the rule for mathematics, would >humanity even know of irrational numbers like sqrt(2)? Irrational >square roots come up in construction. Construction drove a lot of >early mathematical research, like working out pi. You know, for >building things with circles. >Now I've extended mathematics yet again by considering non-polynomial >factors of polynomials. Go ahead, take a look in all of established >mathematics where you see polynomials, and notice all there is about >polynomial factors, but what about non-polynomial factors? Uh, right. The fact that you're unaware of something means that it simply doesn't exist. Got it. >It's like if pure math mathematicians had refused to consider >irrational numbers, like sqrt(2), and were busily occupying themselves >with integer factors. >Mathematicians were pushed not only into using irrational >numbers--notice the name--but into using imaginary numbers, now also >called complex numbers. >Intriguingly, factoring polynomials of degree 3 and higher pushed >mathematicians there, but those of you in physics also know that >complex numbers have their role in helping humanity understand the >real world. >Throughout history mathematicians have been pushed and they push back. > Numbers that weren't perfect integers were called irrational. When >the square root of negative numbers were needed mathematicians fought >back calling them imaginary. >Um, mathematicians don't have a great record here people. _Mathematicians_ don't have a great record. But you do. Nice one. >Now I've pushed with the factoring of polynomials into non-polynomial >factors. >Factoring polynomials into polynomial factors is like factoring >integers into integer factors. >Like x^2 + 3x + 2 = (x+2)(x+1), with x=3, you have 20=5(4). > 3^2 + 3(3) + 2 = 20 >Why am I pushing this issue? Because mathematicians, pushing back >against a new concept, are fighting a short proof of Fermat's Last >Theorem that uses factoring of polynomials into non-polynomial >factors. >That's like mathematicians fighting use of sqrt(2), but construction >workers disagreed. Oh for heaven's sake. Exactly what mathematicians fought the use of sqrt(2)? (I mean in the last, say, 2000 years.) And exactly what construction workers disagreed? >Or like mathematicians fighting sqrt(-1), saying it was imaginary, but >physics now disagrees. And exactly who fought sqrt(-1)? >Now they fight non-polynomial factorization, and I'm stuck. >What gives? >The problem is pure math which means that mathematicians don't have >to go down paths they don't like. If previous mathematicians were >into pure math then humanity might not even have sqrt(2), oh, but >those construction workers wouldn't have gone along. >So now I need people *outside* of mathematics to push. Make them >follow mathematical logic instead of doing what they like, and show >that not only construction workers can make mathematicians follow the >math. Power to the people. Right on, man. > === Subject: Re: Pure math, physics, and FLT >> Interesting. One of the very earliest Western mathematicians, >> Euclid, proved around 300 BC that the sqrt(2) is irrational. >I don't think the proof is originally due to Euclid, is it? Boyer and Merzbach, _A History of Mathematics_: Considering the fame of the author and his best seller, remarkably little is known of Euclid's life. So obscure was his life that no birthplace is associated with his name. ... There is no new discovery attributed to him, but he was noted for expository skill. This is the key to the success of his greatest work, the Elements. It was ly a textbook and by no means the first one. We know of at least three earlier such elements, including that by Hippocrates of Chios; but there is no trace of these, nor of other potential rivals from ancient times. The Elements of Euclid so far outdistanced competitors that it alone survived. Keith Ramsay === Subject: Re: Pure math, physics, and FLT your personal Saviour! Build a shrine! Start a Church! Run through > the streets screaming, is God! is God! > is God! ... but I just finished building my shrine to |-|erc, the Truman! Are you implying that I have been worshipping false idols? Harris forgive me! I'm just joking... I'm still faithful to Archimedes Plutonium. The universe is obviously one big Plutonium atom. Surely anyone can see that. What is it about math and physics that brings out the nutcases, anyway? Sometimes I wonder why I read these groups. Scary thought. -Mike === Subject: Re: Pure math, physics, and FLT your personal Saviour! Build a shrine! Start a Church! Run through > the streets screaming, is God! is God! > is God! > ... but I just finished building my shrine to |-|erc, the Truman! Are you > implying that I have been worshipping false idols? Harris forgive me! > I'm just joking... I'm still faithful to Archimedes Plutonium. The > universe is obviously one big Plutonium atom. Surely anyone can see that. > What is it about math and physics that brings out the nutcases, anyway? > Sometimes I wonder why I read these groups. Scary thought. I think it's simply because those are subjects which are good at capturing interest; however, I think there's fundamentally two ways people deal with something they are interested in, but that is hard to understand. 1) You can spend years of study and work learning the subject, learning from texts, peers, and lecturers. 2) You can decide that you will *not* study and work learning the subject, and instead develop your own idea of what is should be about, neglecting the work of thousands of intellegent, capable men and women that came before you, thus re-inventing the wheel at best, or riding around on a triangle tired bicycle at worst. As you can see, there are many on the physics newsgroups who are quite proud of their triangular bikes. -- William C. Hogg === Subject: Re: Quantum phase > , > which is meant (most probably) the phase of its wavefunction. It is often > said to be unphysical, but far from being so when it is not constant and > varies along with the real part of the wavefunction. Also I think it > principal reason why interference behaves the strange way it does. Could > you comment please and say in which cases it is known this phase plays a > major role? Quantum phases is an extremely interesting area of physics. Essentially the phase of the wavefunction may be ignored in most cases since one can (almost always) regard the phase as being freely specifiable. However, if you're dealing with a system which undergoes an adiabatic change, the phase can have observable consequences. This is the famous Berry's phase, and you're probably best off going to Sakurai's Modern Quantum Mechanics for the best overview of it at a beginning grad-student level. I also quite like the treatment of it given in chapter 1 of Nakahara's Geometry, Topology and Physics. phil === Subject: Re: Quantum phase > , > which is meant (most probably) the phase of its wavefunction. It is often > said to be unphysical, but far from being so when it is not constant and > varies along with the real part of the wavefunction. Also I think it > principal reason why interference behaves the strange way it does. Could > you comment please and say in which cases it is known this phase plays a > major role? > Also, and this is why I follow-up to sci.math, I would much appreciate a > knowledgable expert to list most famous theorems concerning the phase from > the quantum point of view. > I think this question is often overlooked or very poorly addressed (cf. > google for a host of examples) but that its importance is major in a vast > realm of quantum phenomena. > Best regards. Dirac derived the elementary magnetic monopole charge, g, by noting that only the phase of the electron wave function changes under translation in a magnetic potential. In Gaussian units, g = n (hbar*c / 2e) ~ (137n / 2) e n = 1, 2, 3, ... Never falsified by observation, but also never observed. See: http://www.iw.net/~jakoepke/index.html [Old Man] === Subject: Re: Quantum phase > , > which is meant (most probably) the phase of its wavefunction. It is often > said to be unphysical, but far from being so when it is not constant and > varies along with the real part of the wavefunction. Also I think it > principal reason why interference behaves the strange way it does. Could > you comment please and say in which cases it is known this phase plays a > major role? > Also, and this is why I follow-up to sci.math, I would much appreciate a > knowledgable expert to list most famous theorems concerning the phase from > the quantum point of view. > I think this question is often overlooked or very poorly addressed (cf. > google for a host of examples) but that its importance is major in a vast > realm of quantum phenomena. > Best regards. Dirac derived the elementary magnetic monopole charge, g, by noting that only the phase of the electron wave function changes under translation in a magnetic potential. In Gaussian units, g = n (hbar*c / 2e) ~ (137n / 2) e n = 1, 2, 3, ... Never falsified by observation, but also never observed. See: http://www.iw.net/~jakoepke/index.html [Old Man] === Subject: Re: Quantum phase > , which is meant (most probably) the phase of its wavefunction. It is often > said to be unphysical, but far from being so when it is not constant and > varies along with the real part of the wavefunction. Also I think it > principal reason why interference behaves the strange way it does. Could > you comment please and say in which cases it is known this phase plays a > major role? Also, and this is why I follow-up to sci.math, I would much appreciate a > knowledgable expert to list most famous theorems concerning the phase from > the quantum point of view. I think this question is often overlooked or very poorly addressed (cf. > google for a host of examples) but that its importance is major in a vast > realm of quantum phenomena. Best regards. > Dirac derived the elementary magnetic monopole charge, g, by > noting that only the phase of the electron wave function changes > under translation in a magnetic potential. In Gaussian units, > g = n (hbar*c / 2e) ~ (137n / 2) e n = 1, 2, 3, ... > Never falsified by observation, but also never observed. See: > http://www.iw.net/~jakoepke/index.html > [Old Man] Data point on Google censorship: original post has disappeared from there! A very eclectic censorship this is. === Subject: Re: Quantum phase > Data point on Google censorship: original post has disappeared from there! > A very eclectic censorship this is. I don't think it's Google doing it. It could be, but I'm doubting it. I suspect the problem is connectivity. The OP message didn't make it to Google, because the OP's IP didn't send it properly. Or some server between the OP and Google didn't send it on. The result is, people in the other direction from the OP's IP got the message and replied to it, but Google didn't see it. Alternatively, it could be the OP had the no-archive thing set. Socks === Subject: question about book -Basic mathmatics for the physical sciences- Anyone got this? I guess i am being a bit thick but I cannot seem to find the answers to the end of chapter E questions. Are they actually provided? TIA Ed Walsh === Subject: Re: Quick terminology question >(I tried to submit this posting earlier, but I believe something was wrong >with my news server or something.) Is it axiomize? Because of some reason I thought it was axiomatize. > So far there have been at least three replies to your previous post. > One saying no, it's axiomize, and two in favor of axiomatize (both > citing Google hit counts as evidence - great minds and all that.) The OED gives only the following words based on the term AXIOM: AXIOMATIC AXIOMATICAL AXIOMATICALLY AXIOMATIZE AXIOMATIZATION It does not give AXIOMIZE. Some of my favorite OED entries are those based on the word CUSP. These include: CUSPATE: Shaped like a cusp. CUSPATED: Furnished with a cusp or cusp. CUSPED: (1) Having a cusp or cusps. (2) Of the form of a cusp. CUSPID: A cusp. CUSPIDAL: (1) Belonging to the apex (of a cone). (2) Having, relating to, or of the nature of, a cusp. CUSPIDATE: Having a cusp or sharp point. CUSPIDATED: Having a cusp or cusps. CUSPIDATION: Decorating with cusps, cusping. CUSPING: A formation consisting of cusps, cusp-work. -- Clive Tooth http://www.clivetooth.dk === Subject: *Re: Number-Theory/Calculus Theorem Is the below properly and literally analytical number-theory? Or is it more number-theoretical analysis (not to be confused with numerical analysis)? :) Leroy Below is repost of recent message of mine (which I do not think properly propagated through all of sci.math). --- This is obvious, perhaps, but it is still interesting to me, for it combines number-theory and calculus (as do MANY other well-known theorems in other ways). Let f(x) be any real -> real analytic (about x = 0) function, where f(0) = 0, and all of f's derivatives at x = 0 are integers. (..and where f's and exp(f(x))'s Taylor-series expansions {about x = 0} converge.) For simplicity, let a[j] = the j_th derivative of f(x) at x = 0, and let b[j] = the j_th derivative of exp(f(x)) at x = 0. Now, not only are all a[j]'s integers, but so are the b[j]'s. And also... For p = any prime, p divides (b[p+1] - b[p]*a[1] - a[p+1]) - Examples: If g = exp(sin(x)) or exp(arctan(x)), and a[j] = the j_th deivative of g at x = 0, then: p divides (a[p+1] - a[p]). If g(x) = e^(x*e^(x*e^(x*e^...))) = exp(x*g(x)), then: p divides (a[p+1] - 2*a[p]). (perhaps) All of the above seems to be just a meaningless curiosity. Leroy Quet === Subject: Re: *Re: Number-Theory/Calculus Theorem >Let f(x) be any real -> real analytic (about x = 0) function, where >f(0) = 0, and all of f's derivatives at x = 0 are integers. >(..and where f's and exp(f(x))'s Taylor-series expansions {about x = >0} converge.) If f is analytic at 0, so is exp(f), and their Taylor series at 0 converge in some open disk D in the complex plane centred at 0. >For simplicity, let >a[j] = the j_th derivative of f(x) at x = 0, >and let >b[j] = the j_th derivative of exp(f(x)) at x = 0. >Now, not only are all a[j]'s integers, but so are the b[j]'s. Faa di Bruno's formula gives the derivatives of g(f(x)) in terms of the derivatives of g at f(x) and the derivatives of f at x. In this case g = exp, the derivatives of g at f(0) = 0 are all 1. Your result above follows from the fact that the coefficients in Faa di Bruno's formula are all integers. This is easily seen by induction. >And also... >For p = any prime, >p divides (b[p+1] - b[p]*a[1] - a[p+1]) Interesting. It looks to me like (if f_n and g_n are the n'th derivatives of f and g respectively, and p is prime) D^p(g o f) mod p = (g_1 o f) f_p + (g_p o f) f_1^p D^(p+1)(g o f) mod p = (g_1 o f) f_(p+1) + (g_2 o f) f_1 f_p + (g_(p+1) o f) f_1^(p+1) i.e. all coefficients in Faa di Bruno's formula are divisible by p except for these terms where the coefficients are congruent to 1 mod p. The second of these easily follows from the first, of course. Let's see: the terms in Faa di Bruno's formula for D^p(g o f) correspond to partitions of p, where the partition with k_1 1's, k_2 2's ,..., k_p p's (k_1 + 2 k_2 + ... + p k_p = p) corresponds to term p!/(k_1! 1!^(k_1) k_2! 2!^k_2 ... k_p! p!^(k_p)) (g_(k_1+...+k_p) o f) f_1^k_1 ... f_p^k_p It's easy to see that if p is prime the only ways the factor of p in the numerator can be cancelled by a p in the denominator are from the partitions (k_1,...,k_p) = (p,0,...,0) and (0,...,0,1), corresponding to the two terms (g_p o f) f_1^p and (g_1 o f) f_p respectively. Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: refined: how to prove that 2 relations are equivalent? (for any x,y) (T1(x,y) <-- (x=y)) > (for any x,y) (T1(x,y) <-- G(x,y)) > (for any x,y,z) (T1(x,y) <-- G(x,z) & T1(z,y)) for all x, T1(x,x) for all x,y, (G(x,y) -> T1(x,y) for all x,y,z, (G(x,z) & T1(z,y) -> T1(x,y)) > It is easy to see that there are many T1s that satisfy the sentences. > Let's constrain T1 to satisfy the following minimality requirement: T1 > consist of the smallest set of facts that makes the sentences true. Would you clarify what 'facts' are? > Likewise, let T2 be a minimal relation satisfying Likewise, what's the 'facts'? > (for any x,y) (T2(x,y) <-- (x=y)) > (for any x,y) (T2(x,y) <-- G(x,y)) > (for any x,y,z) (T2(x,y) <-- T2(x,z) & T2(z,y)) for all x, T2(x,x) for all x,y, (G(x,y) -> T2(x,y) for all x,y,z, (G(x,z) & T2(z,y) -> T2(x,y) > Prove that T1 and T2 are equivalent. You're asking to prove for all x,y, (T1(x,y) <-> T2(x,y)) ? What's the topic being considered? It looks like T could be the reflexive transitive closure of a partial order G. Answer: T = intersection of all binary relations satisfying those three properties. Indeed, the properties are intersection or conjunction closed and the great intersection or conjuction of those properties would be minimal, in fact smallest, relation satisfying the properties. That T is the smallest instead of just minimal is now easy to see and hence as both T1 and T2 are 'minimal', they are both equivalent to T. > (BTW, the book Foundation of Databases by Abiteboul et al, p.275 has > arrows oriented my way;-) Computerists and mathematicians speak such different languages that each speculate the other is an alien. However mathematicians have been here centuries before any computerists arrived, so the computerists are the aliens. ;-) === Subject: Re: refined: how to prove that 2 relations are equivalent? > My apologies, as I indeed copied the problem incorrectly. Here it is: > Let T1 be a relation that satisfies the following sentences: > (for any x,y) (T1(x,y) <-- (x=y)) > (for any x,y) (T1(x,y) <-- G(x,y)) > (for any x,y,z) (T1(x,y) <-- G(x,z) & T1(z,y)) > It is easy to see that there are many T1s that satisfy the senetences. > Let's constrain T1 to satisfy the following minimality requirement: T1 > consist of the smallest set of facts that makes the sentences true. > Likewise, let T2 be a minimal relation satisfying > (for any x,y) (T2(x,y) <-- (x=y)) > (for any x,y) (T2(x,y) <-- G(x,y)) > (for any x,y,z) (T2(x,y) <-- T2(x,z) & T2(z,y)) > Prove that T1 and T2 are equivalent. > (BTW, the book Foundation of Databases by Abiteboul et al, p.275 has > arrows oriented my way;-) Well, it is common in logic programming as well, but probably a bit uncomfortable to read for most mathematicans. The description of T2 says, that T2 is the smallest reflexive and transitive relation, that contains G. The idea is, to give an explicit construction of a relation, that turns out to be equal to both T1 and T2. Consider the relation T3, defined as follows: T3(x,y) <=> there is a sequence x_1,...,x_n such that x_i = x_(i+1) or G(x_i,x_(i+1)) Now you can prove the following points: (1) T3 is contained in T1 (2) T3 satisfies the requirements for T1. Hence T1 is contained in T3 by minimality of T1, (3) If R is any reflexive and transitive relation containing G, then R contains T3. In particular T3 is contained in T2. (4) T3 is reflexive and transitive and contains G. Hence T2 contains T3 by minimality of T2. Marc === Subject: Re: refined: how to prove that 2 relations are equivalent? >> Let T1 be a relation that satisfies the following sentences: >> (for any x,y) (T1(x,y) <-- (x=y)) >> (for any x,y) (T1(x,y) <-- G(x,y)) >> (for any x,y,z) (T1(x,y) <-- G(x,z) & T1(z,y)) >for all x, T1(x,x) >for all x,y, (G(x,y) -> T1(x,y) >for all x,y,z, (G(x,z) & T1(z,y) -> T1(x,y)) >> It is easy to see that there are many T1s that satisfy the sentences. >> Let's constrain T1 to satisfy the following minimality requirement: T1 >> consist of the smallest set of facts that makes the sentences true. >Would you clarify what 'facts' are? He means just that T1 should be the smallest relation that satisfies the above conditions. >> Likewise, let T2 be a minimal relation satisfying >Likewise, what's the 'facts'? >> (for any x,y) (T2(x,y) <-- (x=y)) >> (for any x,y) (T2(x,y) <-- G(x,y)) >> (for any x,y,z) (T2(x,y) <-- T2(x,z) & T2(z,y)) >for all x, T2(x,x) >for all x,y, (G(x,y) -> T2(x,y) >for all x,y,z, (G(x,z) & T2(z,y) -> T2(x,y) >> Prove that T1 and T2 are equivalent. >You're asking to prove > for all x,y, (T1(x,y) <-> T2(x,y)) ? >What's the topic being considered? It looks like T could be the > reflexive transitive closure of a partial order G. >Answer: T = intersection of all binary relations satisfying > those three properties. >Indeed, the properties are intersection or conjunction closed and > the great intersection or conjuction of those properties >would be minimal, in fact smallest, relation satisfying the properties. >That T is the smallest instead of just minimal is now easy to see and >hence as both T1 and T2 are 'minimal', they are both equivalent to T. >> (BTW, the book Foundation of Databases by Abiteboul et al, p.275 has >> arrows oriented my way;-) >Computerists and mathematicians speak such different languages that each >speculate the other is an alien. However mathematicians have been here >centuries before any computerists arrived, so the computerists are the >aliens. ;-) === Subject: Re: refined: how to prove that 2 relations are equivalent? >My apologies, as I indeed copied the problem incorrectly. Here it is: >Let T1 be a relation that satisfies the following sentences: >(for any x,y) (T1(x,y) <-- (x=y)) >(for any x,y) (T1(x,y) <-- G(x,y)) >(for any x,y,z) (T1(x,y) <-- G(x,z) & T1(z,y)) > It is easy to see that there are many T1s that satisfy the senetences. >Let's constrain T1 to satisfy the following minimality requirement: T1 >consist of the smallest set of facts that makes the sentences true. >Likewise, let T2 be a minimal relation satisfying >(for any x,y) (T2(x,y) <-- (x=y)) >(for any x,y) (T2(x,y) <-- G(x,y)) >(for any x,y,z) (T2(x,y) <-- T2(x,z) & T2(z,y)) >Prove that T1 and T2 are equivalent. Well, if someone hadn't already given an explanation I'd suggest you think about it a little first. (Since as of yesterday you hadn't even _started_ to think about the question - you can't do that until you've at least _read_ the question correctly!) >(BTW, the book Foundation of Databases by Abiteboul et al, p.275 has >arrows oriented my way;-) >>(for any x,y) (T1(x,y) <-- (x=y)) >>(for any x,y) (T1(x,y) <-- G(x,y)) >>[i] (for any x,y) (T1(x,y) <-- G(x,y) & T1(x,y)) >(for any x,y) (T2(x,y) <-- (x=y)) >>(for any x,y) (T2(x,y) <-- G(x,y)) >>[ii](for any x,y) (T2(x,y) <-- T2(x,y) & T2(x,y)) >> Are you sure you've copied the problem correctly? It doesn't >> make any sense - for example [i] says nothing at all, it's >> true for _any_ T1; similarly [ii] says nothing. === Subject: Re: refined: how to prove that 2 relations are equivalent? > Let T1 be a relation that satisfies the following sentences: (for any x,y) (T1(x,y) <-- (x=y)) > (for any x,y) (T1(x,y) <-- G(x,y)) > (for any x,y,z) (T1(x,y) <-- G(x,z) & T1(z,y)) (for any x,y) (T2(x,y) <-- G(x,y)) > (for any x,y,z) (T2(x,y) <-- T2(x,z) & T2(z,y)) Prove that T1 and T2 are equivalent. The description of T2 says, that T2 is the smallest reflexive and transitive > relation, that contains G. > The idea is, to give an explicit construction of a relation, that turns out > to be equal to both T1 and T2. > Consider the relation T3, defined as follows: > T3(x,y) <=> there is a sequence x_1,...,x_n such that > x_i = x_(i+1) or G(x_i,x_(i+1)) So we need natural numbers! (Which means the proof is not purely logical, right?) Alternatively, I might suggest defining T3(x,y,n) <=> there is a sequence x_1,...,x_n such that G(x_i,x_(i+1)) where x_1,...,x_n aren't necessarily distinct, and T3(x,y) = U T3(x,y,n) > Now you can prove the following points: Here I'm confused, so that I reversed order. > (2) T3 satisfies the requirements for T1. For any n T3(x,y,n) is contained in T1 (by induction?) Union ot T3(x,y,n) is contained in T1 too (by some set axiom?) > (1) T3 is contained in T1 Stuck here as I don't quite understand the difference between (1) and (2). Satisfying the requirements means that you describe a set by some constraints. I understand that when you have wider set of constraints, the set would be smaller. So I have a method of proving of (2) but don't have a method of proving (1) because proving set containment is done via checking constraint satisfiability. Thank you for the input! > Hence T1 is contained in T3 by minimality of T1, > (3) If R is any reflexive and transitive relation containing G, then > R contains T3. In particular T3 is contained in T2. > (4) T3 is reflexive and transitive and contains G. > Hence T2 contains T3 by minimality of T2. > Marc === Subject: Re: ??'s about Metric space topology X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS In , on === >Subject: ??'s about Metric space topology Actually, your questions are not specific to metric spaces. >First, what is the difference between connectedness and >continuity? One applies to topological spaces, the other to functions. >Doesn't connectedness imply continuity? No, it's an apples and oranges thing. >Second, what is the significance of the concept compactness, other >than the fact that it is necessary for a lot of important and >interesting theorems? Well, that's about it: the notions of compact and paracompact turn out to be tremendously usefull. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any 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 not reply to spamtrap@library.lspace.org === Subject: Re: ??'s about Metric space topology > http://www.math.buffalo.edu/~sww/classes/COMPACT/COMPACT1.html > where the guy talks about it being all about holes. > Don't know about the validity of what he's saying but if he's right > then it might add something to understanding. About compactness: the idea of holes does not talk to me: many > fractal sets are full of holes, if they are not scattered > altogether. What is intuitively important to me is etc. etc....... thanks anyway. I notice that the author of the website I highlighted is a professor of mathematics and has written a book on topolgy: http://www.acsu.buffalo.edu/~sww/BATH.html So this should lend a little credence to the the website on compactness and the idea of holes. -- Posted via Mailgate.ORG Server - http://www.Mailgate.ORG === Subject: Set Theory (paradox?) I think I've found a paradox, but I'm far from being a professional. So, please take this with a grain of salt. --- In one of Raymond Smullyan's books he explains the limited abstraction principle by saying, Given any property and given any set A, there exists the set of all elements of the set A that have the property. Now, consider a set A of all things (elements) that are in only one set. That is, these elements occur in only one set in the entire world. Certainly, there is no problem up to this point. However, let's now apply the limited abstraction principle. We will create a set B and put into it all the elements in set A which have the property of only existing in one set. Obviously, there is a contradiction here, because the elements of A can only be found in A,and so can't be put into B. Also, if we leave set B empty then we have not successfully completed the limited abstraction principle. Does that qualify as a paradox? ..,__........_______________.................... | | Raheman Velji / / . . |__/ _, |_ ,. ,_ _, ,_ /,. || | / | | |/ |||/ | | | / / / /| | / _/_|/| |//|||_|/| | / _/_// |/ _ ............ blochee@yahoo.ca .........../|..... | === Subject: Re: Set Theory (paradox?) Visiting Assistant Professor at the University of Montana. >I think I've found a paradox, but I'm far from being a professional. >So, please take this with a grain of salt. >--- >In one of Raymond Smullyan's books he explains the limited >abstraction principle by saying, > Given any property and given any set A, there exists the set of >all elements of the set A that have the property. >Now, consider a set A of all things (elements) that are in only one >set. That is, these elements occur in only one set in the entire >world. Certainly, there is no problem up to this point. I disagree. I see no reason to assume that this set exists, unless you are bounding it as a subset of some set we already know exists. > However, >let's now apply the limited abstraction principle. We will create a >set B and put into it all the elements in set A which have the >property of only existing in one set. Obviously, there is a >contradiction here, because the elements of A can only be found in >A,and so can't be put into B. Again, there is nothing obvious here. Assuming for the sake of argument that you have a set A, then you are defining B to be the set B = {x in A : x occurs in exactly one set}. You know that B is a subset of A; so every element of B is an element of A. If B were a PROPER subset of A (that is, if B were a set which is DIFFERENT from A), then that would cause problems. So the best you can deduce now is that B=A; that is, the 'limited abstraction principle' does not produce a new set when applied to A. > Also, if we leave set B empty then we >have not successfully completed the limited abstraction principle. Incorrect as well. There is absolutely nothing wrong with the limited abstraction principle yielding an empty set. For example, given ANY set A, the set B = {x in A : x is not equal to x} is the empty set, and is obtained by using the limited abstraction principle to the set A and the property x is not equal to x. >Does that qualify as a paradox? No. ====================================================================== = === Subject: Re: Set Theory (paradox?) Visiting Assistant Professor at the University of Montana. >I think I've found a paradox, but I'm far from being a professional. >So, please take this with a grain of salt. >--- >In one of Raymond Smullyan's books he explains the limited >abstraction principle by saying, > Given any property and given any set A, there exists the set of >all elements of the set A that have the property. >Now, consider a set A of all things (elements) that are in only one >set. That is, these elements occur in only one set in the entire >world. Certainly, there is no problem up to this point. I disagree. I see no reason to assume that this set exists, unless you are bounding it as a subset of some set we already know exists. > However, >let's now apply the limited abstraction principle. We will create a >set B and put into it all the elements in set A which have the >property of only existing in one set. Obviously, there is a >contradiction here, because the elements of A can only be found in >A,and so can't be put into B. Again, there is nothing obvious here. Assuming for the sake of argument that you have a set A, then you are defining B to be the set B = {x in A : x occurs in exactly one set}. You know that B is a subset of A; so every element of B is an element of A. If B were a nonempty and PROPER subset of A (that is, if B were a set which is DIFFERENT from A and contains at least one element), then that would cause problems. So the best you can deduce now is that either B is empty, or B=A; that is, the 'limited abstraction principle' does not produce a new set when applied to A. In fact, you may deduce that B=A, since x is in A if and only if it satisfies the property given. > Also, if we leave set B empty then we >have not successfully completed the limited abstraction principle. Incorrect as well. There is absolutely nothing wrong with the limited abstraction principle yielding an empty set. For example, given ANY set A, the set B = {x in A : x is not equal to x} is the empty set, and is obtained by using the limited abstraction principle to the set A and the property x is not equal to x. >Does that qualify as a paradox? No. ====================================================================== = === Subject: Re: Set Theory (paradox?) > I think I've found a paradox, but I'm far from being a professional. > So, please take this with a grain of salt. > --- > In one of Raymond Smullyan's books he explains the limited > abstraction principle by saying, > Given any property and given any set A, there exists the set of > all elements of the set A that have the property. > Now, consider a set A of all things (elements) that are in only one > set. That is, these elements occur in only one set in the entire > world. Certainly, there is no problem up to this point. However, > let's now apply the limited abstraction principle. We will create a > set B and put into it all the elements in set A which have the > property of only existing in one set. Obviously, there is a > contradiction here, because the elements of A can only be found in > A,and so can't be put into B. Also, if we leave set B empty then we > have not successfully completed the limited abstraction principle. > Does that qualify as a paradox? Nope. A = {}, B={}. Consider any element e. If e=1, then {1,2} and {1} contain e. Otherwise, {e} and {e,1} contain e. As a result, the set of all elements that are in exactly one set is {}= the empty set. > ..,__........_______________.................... > | | Raheman Velji / / . . > |__/ _, |_ ,. ,_ _, ,_ /,. || > | / | | |/ |||/ | | | / / / /| | > / _/_|/| |//|||_|/| | / _/_// |/ _ > ............ blochee@yahoo.ca .........../|..... > | -- === Subject: Re: Set Theory (paradox?) > -----Original Message----- > Conversation: Set Theory (paradox?) === > Subject: Set Theory (paradox?) > Now, consider a set A of all things (elements) that are in only one > set. That is, these elements occur in only one set in the entire > world. Certainly, there is no problem up to this point. However, > let's now apply the limited abstraction principle. We will create a > set B and put into it all the elements in set A which have the > property of only existing in one set. Obviously, there is a > contradiction here, because the elements of A can only be found in > A,and so can't be put into B. Also, if we leave set B empty then we > have not successfully completed the limited abstraction principle. > Does that qualify as a paradox? No. You have simply concocted an artful description of the empty set. A = {x | there is exactly one S (x in S) } B = {x in A | there is exactly one S (x in S) } Notice that B = A. Because A and B are the same set, they are one set, not two. So there is absolutely nothing contradictory about the existance of a thing that belongs to one and only one set yet belongs to both A and B On the otherhand, the set C = A / {A} is not equal to A and every element of A belongs to both A and C. Therefore, there can't actually be any elements in A. So A is the empty set. Note: The way you describe it, A is not officially a set. In order to make it a set, you would need to start with some bounding set, say V, and write A = {x in V | there is exactly one S (x in S) } === Subject: sigma(n)=sigma(n+1)=sigma(n+2) Is there a natural number n such that : sigma(n)=sigma(n+1)=sigma(n+2) ? (where sigma() is the sum of divisors ) === Subject: Re: sigma(n)=sigma(n+1)=sigma(n+2) > Is there a natural number n such that : > sigma(n)=sigma(n+1)=sigma(n+2) ? I don't know. Problem B13 of Guy, Unsolved Problems in Number Theory, says Sierpinski has asked if sigma(n) = sigma(n + 1) infinitely often. Hunsucker, Nebb & Stearns ... found 113 solutions 14, 206, 957, 1334, 1364, 1634, 2685, 2974, 4364, ... less than 10^7. The paper is in Math. Student 41 (1973) 285-289. There's also a paper of Guy & Shanks in the Fib. Quart. 12 (1974) 299. Maybe type the sequence above into Sloane's Online Encyclopedia of Integer Sequences and see what comes out. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Solution of a Quadratic Recurrence Equation Hello everybody. As part of my research on poulation genetics, I encountered the following quadratic recurrence equation: f(t+1) = f(t) (1-c f(t)) where c is a constant and f(0) is 0.05. Is there any solution for this equation? F. Habibzadeh, M.D. === Subject: Re: Solution of a Quadratic Recurrence Equation > Hello everybody. As part of my research on poulation genetics, I > encountered the following quadratic recurrence equation: > f(t+1) = f(t) (1-c f(t)) > where c is a constant and f(0) is 0.05. Is there any solution for > this equation? Let g(t) = c f(t). Then g(t+1) = g(t) (1 - g(t)) with g(0) = 0.05 c. This is a special case (r = 1) of the famous logistic map: g(t+1) = r g(t) (1 - g(t)). There are explicit solutions known for r = -2, 2, and 4, but not for r = 1: see http://mathworld.wolfram.com/LogisticMap.html For r = 1, g(t) -> 0 for 0 <= g(0) <= 1 (0 < c <= 20), and g(t) diverges for any other initial condition. -- | Jim Ferry | Center for Simulation | +------------------------------------+ of Advanced Rockets | | http://www.uiuc.edu/ph/www/jferry/ +------------------------+ | jferry@[delete_this]uiuc.edu | University of Illinois | === Subject: Re: Solution of a Quadratic Recurrence Equation > Hello everybody. As part of my research on poulation genetics, I > encountered the following quadratic recurrence equation: > f(t+1) = f(t) (1-c f(t)) > where c is a constant and f(0) is 0.05. Is there any solution for > this equation? I know nothing about non-linear recurrence relations, but would make two observations on yours: (1) By writing g(t) = cf(t) you get read of the c: g(t+1) = g(t) (1 - g(t)). (2) This can be written g(t+1) - g(t) = -g(t)^2. I think one probably gets a good idea of the behaviour as t gets large by considering the corresponding differential equation dg/dt = -g^2 ie dt/dg = -g^{-2} with solution t = -a + g^{-1} ie g(t) = 1/(t+a). I think it would fairly easy to show from this that f(t) ~ 1/ct as t -> infty. Perhaps one could even get a more accurate estimate this way. -- Timothy Murphy e-mail: tim@birdsnest.maths.tcd.ie tel: +353-86-233 6090 s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland === Subject: Re: Solution of a Quadratic Recurrence Equation > g(t+1) - g(t) = -g(t)^2. >I think one probably gets a good idea of the behaviour as t gets large >by considering the corresponding differential equation > dg/dt = -g^2 ... >I think it would fairly easy to show from this that f(t) ~ 1/ct >as t -> infty. This is true if 0 < g(0) < 1: if g(0) < 0 or g(0) > 1 it's easy to see that g(t) -> -infinity as t -> infinity. If 0 < g(0) < 1, then 0 < g(t) < 1/4 for t >= 1, and of course g(t) is decreasing. Let h(t) = 1/g(t). Then h(t+1) = h(t) + 1 + 1/(h(t)-1). Of course h(t+1) > h(t)+1, so h(t) > h(0) + t. For any epsilon > 0, take N so 1/(N-1) < epsilon. Now h(N) > N, so for t > N we have h(t+1) < h(t) + 1 + epsilon, and h(t) < (1+epsilon) t for sufficiently large t. Thus h(t) = (1 + o(t)) t, implying g(t) = (1 + o(t))/t and f(t) = (1 + o(t))/(c t) as t -> infinity. I think a bit more analysis will show that h(t) = t + ln(t) + O(1) as t -> infinity, so g(t) = 1/t - ln(t)/t^2 + O(1/t^2). Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: stuck proving simple(?) tautology >and it's confusing. Do you mean > I / G / (G/T1) ? But it's not confusing for this reason - the fact > that A or B and C means A or (B and C) > is absolutely standard; no more confusing > than the fact that 1 + 2*3 means 1 + (2*3). > I was thinking along the same lines until browsing Disjunctive Normal Form > page at Mathworld. They do use parentesis! The only explanation that I can > think of right now why parenthesis are essential is that / and / are dual > operations in logic so that analogy with multiplication and addition really > fails. Some logical systems supplement their basic formation rules for wffs with conventions of convenience that rank the connectives in terms of binding strength, e.g. & binds more strongly than v, v more strongly than -> and so forth, so that parenthetical clutter can be avoided. So p & q -> r can mean (p & q) -> r, etc. E.J. Lemmon does this in _Beginning Logic_. But I don't agree with David Ullrich that this is absolutely standard. Some texts would even disallow p v q v r because the standard formation rules require v to be explicitly binary. Cheers, Karl === Subject: Style of proofs To me, it seems that mathematical theorems are so often proven using a lot of obscure symbols instead of simple techniques. Take, for example, that for all composite n, (2^n)-1 is composite. My proof is the obvious one, using the binary representation of (2^n)-1 as 111...111 (with n 1's), and showing that this number is equal to 10...010...010......01 times a shorter 111...111. Once I saw a complicated-looking proof of this theorem, and was told that the complicated-looking proof was the same as the one I mentioned here. Take another example: http://www.wikipedia.org/wiki/Proofs_of_Fermat%27s_little_theorem Guess which proof on this page is mine? Where can I find simple proofs of theorems? === Subject: Re: Style of proofs : To me, it seems that mathematical theorems are so often proven using a : lot of obscure symbols instead of simple techniques. : Take, for example, that for all composite n, (2^n)-1 is composite. My : proof is the obvious one, using the binary representation of (2^n)-1 : as 111...111 (with n 1's), and showing that this number is equal to : 10...010...010......01 times a shorter 111...111. Once I saw a : complicated-looking proof of this theorem, and was told that the : complicated-looking proof was the same as the one I mentioned here. It seems that obvious is in the eye of the beholder. To me, it is obvious that x-1 is a polynomial factor of x^r-1, hence 2^m-1 is a factor of 2^n-1 if m is a factor of n. Of course, in a sense, this proof is the same as the one you gave, but it is clearer to me. Also, sometimes the goal is not to give the simplest proof, but to give one that: - uses natural steps rather than slick steps whose motivations are not clear, or; - illustrates the use of a particular technique, that may be used to prove other statements, or; - can be generalized to give a proof of a deeper result. Ted === Subject: Re: Taking Calculus without taking precalcu X-Cise: tanbanso@iinet.net.au X-CompuServe-Customer: Yes X-Coriate: admin@interspeed.co.nz X-Ecrate: tanandtanlawyers.com X-Punge: Micro$oft X-Sanguinate: themvsguy@email.com X-Terminate: SPA(GIS) X-Tinguish: Mark Griffith X-Treme: C&C,DWS at 11:08 PM, Will Self said: >With great respect for Wade and his many excellent posts, I must >disagree on this one. Precalculus is basically college algebra + >trigonometry + sometimes an introduction to transcendental functions. They used to cover transcendental functions in a HS course called Advanced Algebra. If the OP took such a class then he should be able to do quite nicely without precalc, unless he needs the review. Most of the HS course work these days is review of material that the students should have mastered in an earlier grade. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Any 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 not reply to spamtrap@library.lspace.org === Subject: Re: Technical document preparation (lecture notes) > Greetings everyone, > Does anyone know of a company which provides a service of converting > handwritten lecture notes into an electronic format (preferably PDF or > TeX), I have tried searching the web for some time without success. Your best bet would undoubtedly be to corner a grad student and ask him/her to type up the notes for you. Most grad students in math/physics are quite handy with LaTeX and could do a really good job for probably not very much money. How big is the project? === Subject: Re: Technical document preparation (lecture notes) > Greetings everyone, > Does anyone know of a company which provides a service of converting > handwritten lecture notes into an electronic format (preferably PDF or > TeX), I have tried searching the web for some time without success. > Also, has anyone used such a service before? What is going rate of > payment for such work and how high a quality can I expect from them? > Ryan Should have said earlier: I've done things like this in the past for lecturers who've wanted their notes typeset and they've always been happy with the result. If you'd like to discuss getting me to typeset your notes, === Subject: Re: The basic idea behind my great forthcoming proof > david_lawrence_petry@yahoo.com (David Petry) says... >>in the position of telling me that I am an imbecile, >>crackpot, and _untermenschen_ and that I am unworthy >>of being a member of your elite community if I reject >>your fantasy world. That's wrong and dangerous. That's rich coming from Petry, who advocates that that mathematics dealing with infinity should be banned and mathematical inquiry not corresponding to his personal notion of computation should be suppressed. Anyway mathematicians hardly form an elite community --- mathematics is a topic of free inquiry and anyone can join in. Perhaps Petry (and the rest of us) might find it more valuable if he ceased his paranoid polemics and actually did some maths for a change. > It seems to me that you are on the one who is > calling the vast majority of mathematicians crackpots, > imbeciles, and religious fanatics. You go around > saying that they are living in a fantasy world, that > they have dangerous and even immoral ideas. And yet > you are resentful that they aren't more tolerant > of you and your opinions. Don't you see a little bit > of the pot calling the kettle black going on here? So true. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html His mind has been corrupted by colours, sounds and shapes. The League of Gentlemen === Subject: Re: The basic idea behind my great forthcoming proof > David Petry is saying: > 1)Cantor completely changed the way mathematicians thought. That's an exaggeration. All of applied mathematics, and virtually all of undergraduate mathematics, and even most of pure mathematics, would make sense to someone who knew nothing of Cantor. > 2)In fact, he introduced quasi-religious beliefs into mathematics that have > corrupted the subject to this day. Certainly the foundations of mathematics have been corrupted. === Subject: Re: The basic idea behind my great forthcoming proof > And if he's not disputing the > existence of infinite sets, what assumption needed for the proof > |N| < |R| (or |X| < |P(X)|) is controversial to David? If you start with a well defined list of well defined real numbers (i.e. given integers m and n, you can compute a digit which is the n'th digit of the m'th real number), then you can diagonalize over that list to get a new well defined real number. No mathematician would ever find that counter- intuitive or controversial, as far as I know. But it does not lead to the conclusion that |N| < |R|. Cantor claimed it makes sense to talk about lists and numbers which are not well defined (as far as I know, he wasn't aware he was making that claim). He needed that assumption to prove that |N|<|R|. So my claim is that Cantor introduced the idea that mathematics must study not only well defined things, but also things which are not well defined. I believe that idea was never needed before Cantor. That's what mathematicians of his time found to be so counter-intuitive, even if they weren't able to pinpoint the problem. My claim is that mathematics could get along just fine without Cantor's claim. === Subject: Re: The basic idea behind my great forthcoming proof >> David Petry is saying: >> 1)Cantor completely changed the way mathematicians thought. > That's an exaggeration. > All of applied mathematics, and virtually all of > undergraduate mathematics, and even most of pure > mathematics, would make sense to someone who knew > nothing of Cantor. >> 2)In fact, he introduced quasi-religious beliefs into mathematics that have >> corrupted the subject to this day. > Certainly the foundations of mathematics have > been corrupted. How so? Here are two statements about sets. Which of these do you think is corrupted? 1. For each set X, there is an injection from X into its power set P(X), but there is no injection from P(X) into X. 2. There is a bijection between the real numbers R and P(N), the power set of the naturals. Notice that the terms infinite, greater than, less than, and countable are not mentioned anywhere. Is it merely the words that you are objecting to, or is it the underlying concepts? Is it the definitions of the terms injection and bijection? -- === Subject: Re: The basic idea behind my great forthcoming proof :> And if he's not disputing the :> existence of infinite sets, what assumption needed for the proof :> |N| < |R| (or |X| < |P(X)|) is controversial to David? : If you start with a well defined list of : well defined real numbers (i.e. given integers : m and n, you can compute a digit which is : the n'th digit of the m'th real number), : then you can diagonalize over that list to : get a new well defined real number. : No mathematician would ever find that counter- : intuitive or controversial, as far as I know. : But it does not lead to the conclusion that : |N| < |R|. Why does it not lead to the conclusion that |N| < |R|? If you give me a well defined list of well defined real numbers, then using the diagonal argument I can find a well defined real number that is not in your list. Stephen === Subject: Re: The problem of number partition Given a partition into i parts, largest part j, > replace every part of size j with a part of size j - 1, > and introduce a new part who size is the number of parts > you used to have of size j; you get a partition > into i + 1 parts, largest part j - 1. > Take N=7, i=3, j=3. The partitions (1,3,3) and (2,2,3) > are both mapped to (1,2,2,2) by your mapping. Yes, you and Robert Israel are quite right - good thing, too, as Robert found a counterexample to the original conjecture. -- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for email) === Subject: Re: The problem of number partition > > ... > My question is: > If i ... > Sorry, it should be P(N,i,j)<=P(N,i+1, j-1). > ... >> In fact, it seems to me P(12, 4, 4) = 3 while P(12, 5, 3) = 2, >> providing a counterexample to the conjecture. > >No, it's not a counter example. In fact, it is easy to obtain >P(N,i,j)=P(N,j,i). > Huh? Your conjecture was P(N,i,j) <= P(N,i+1,j-1) for i Take N=12, i=j=4. Since 4<4+1, your conjecture would say > P(12,4,4) <= P(12,5,3). But in fact P(12,4,4) = 3 (the possible > partitions being 1+3+4+4, 2+2+4+4 and 2+3+4+4) and P(12,5,3) = 2 > (the possible partitions being 1+2+3+3+3 and 2+2+2+3+3). > How is that not a counterexample? Sorry, there is a mistake in the condition. It should be If ij. can we prove P(N,i,j)Sorry, there is a mistake in the condition. It should be >If ij. can we prove P(N,i,j) The following question has to do with a paper by Anandan and Aharonov, > PRL, 65, 14, 1990. > Does anyone know if the quotient group > U(2)/(U(1) X U(1)) > is a normal subgroup of > U(3)/(U(2) X U(1)) ? Are these groups? -- === Subject: Re: Three-signed arithmetic : T space There are formal methods of logic for the analysis of new mathematical systems. Firstly, one can say that the - in existing systems is either UNARY or BINARY. Thus -5 is a UNARY use of the minus. 10 - 5 is a BINARY use. These are standard terms. Secondly, the unary form often IMPLIES a function. A function takes an ARGUMENT and delivers a RESULT. So argument 5 with the negation function gives -5. The binary form always is a function, because you can see TWO arguments shrinking into ONE result. Functions have properties: distributivity - can one do such as in conventional math when 5(4+3) = 5*4 + 5*3 ? commutativity - can one swap arguments 5*4 = 4*5 ? associativity - can groups be swapped (5+4)+3 = 5+(4+3) ? Finally is there an IDENTITY FUNCTION. For example, with factorials there are TWO: 1!=1 2!=2 === Subject: Re: Three-signed arithmetic : T space Hi. Yes to all of your questions. The associative, distrubutive and commutative properties do work for Y. I'm glad you're staying on with me here. I have laid this out pretty well in the messages above this point. I am sorry if my communication or this format is not efective. Unary and Binary usages of the sign symbols are acceptable. I would stress that there is no 10 in T. 10 is a magnitude. It should be +10, -10, or *10 in T. as 10 is +10 in R I think you will understand this. Think of the signs as you would think of them for real numbers, except that there is one more of them. The star (*) operator is symmetrical to the + operator in reals. for sign. Here are all the examples for a magnitude x: -(-x) = +x. -(+x) = *x. -(*x) = -x. +(-x) = *x. +(+x) = -x. +(*x) = +x. *(-x) = -x. *(+x) = +x. *(*x) = *x. There is a well founded pattern here. The phenomena is rotational in nature. The following operation is an example of summation or superposition: * 1 + 2 Just as - 1 + 2 in reals is. It's fairly easy to say that * 1 * 2 = * 3 just as + 1 + 2 = + 3 and - 1 - 2 = - 3. T is a number system like R, just with an extra sign. The above sums are graphically shown on the equivalent of the number line, which in T has three extensions from the origin rather than two. Here are some more examples of summation: + 1 * 2 - 3 + 2 - 2, * 1 + 1 - 6 + 5 * 3 * 2, + 1.34 * 2.45 * 4.5 - 1.23. In resolving sums I have discovered that they do not necessarily reduce to one single magnitude! This is a result of the conception of cancellation and superposition. If you attempt to resolve -1 + 2 to be message). Cancellation only occurs when we have a component in every sign within the sum (-x+x*x = 0). This is where the dimentionality that I have mentioned is exhibited. For a three-signed space sums of two signs do not resolve to a single magnitude and this has led to the use of the term Y-space since I defined T-space to be a single magnitude with three signs. In general sums are in Y, not in T. For two elements y1 and y2 in Y their sum can be reperesented as y1 * y2. I am sorry if this is annoying because of the standard use of star for products in reals, but symbollically in Y it is effective as sum. Suppose y1 = - 3 + 2 and y2 = - 1 * 3. y1 * y2 = ( - 3 + 2 ) *( - 1 * 3 ) = - 3 + 2 - 1 * 3 = - 4 + 2 * 3 = - 2 * 1 - 2 + 2 * 2 = - 2 * 1. The - 2 + 2 * 2 at the end cancels out to zero. Products are much easier. Just distribute the terms as for reals: (y1)(y2) = ( - 3 + 2 )( - 1 * 3 ) = (-3)(-1) * (-3)(*3) * (+2)(-1) * (+2)(*3) = + 3 - 9 * 2 + 6 = + 7 - 7. The identity function; The factorial example is a little confusing to me. *1 is the identity value. (*1)(*1) = *1. if we take an element y1 in Y and want (y1)(y1') = *1 then we must find something like a complex conjugate: (+1*1)(-1*1) = *1+1-1*1 = *1. I have worked out division for specific examples. The method uses linear equations by supposing that the answer is -a+b*c for a given problem. Too much to print here. I think you are like me in that it is much easier to create my own math than it is to read someone else's. I really do appreciate your questions and hope that you can see it. If not, make it up yourself and I am almost certain you and I will speak the same language. There are choices to make along the way but the choices must yield consistency. > There are formal methods of logic for the analysis of new mathematical > systems. > Firstly, one can say that the - in existing systems is either UNARY > or BINARY. > Thus -5 is a UNARY use of the minus. 10 - 5 is a BINARY use. These are > standard terms. > Secondly, the unary form often IMPLIES a function. A function takes an > ARGUMENT and delivers a RESULT. So argument 5 with the negation > function gives -5. > The binary form always is a function, because you can see TWO > arguments shrinking into ONE result. > Functions have properties: > distributivity - can one do such as in conventional math when > 5(4+3) = 5*4 + 5*3 ? Yes! > commutativity - can one swap arguments 5*4 = 4*5 ? Yes! > associativity - can groups be swapped (5+4)+3 = 5+(4+3) ? Yes! > Finally is there an IDENTITY FUNCTION. > For example, with factorials there are TWO: > 1!=1 > 2!=2 > === Subject: Topology Grill Can Someone explain this Grill? I am starting on Claus Berge Topological spaces . I already encountered a couple of misprints. He uses only this - a grill meets all of A sub i . Where we have a family of sets A. joe === Subject: Re: Topology Grill > Can Someone explain this Grill? > I am starting on Claus Berge Topological spaces . I already encountered a > couple of misprints. He uses only this - a grill meets all of A sub i . > Where we have a family of sets A. > joe Does that mean that if G is a grill, G's intersection with A sub i is not empty for every A sub i in A? === Subject: Re: Topology Grill I thought meet was unclear. it would have been clearer if it done via notation joe > Can Someone explain this Grill? > I am starting on Claus Berge Topological spaces . I already encountered a > couple of misprints. He uses only this - a grill meets all of A sub i . > Where we have a family of sets A. joe Does that mean that if G is a grill, G's intersection with A sub i is not > empty for every A sub i in A? === Subject: Turn off your satelite nasa anyone seen Popular Science May issue? Features the pulsed laser spy satelites that penetrate clouds, only they use it to constantly bombard me with heathen dictators since April 2002, every second I've been bombarded with a laser to sound attack using a dozen retards just designed to me off. 100,000 people in Townville Australia all tune into my constant interrogation from wake up to drop asleep, constant torture 2 years. turn it off nasa, you stole my life and destroyed any life i may have. 2 years of thought deprivation completely unjustified. you all targeted then attacked me theres a ing idiot making spastic deranged comments oh look at this he just said, all through typing this. all my neighbours can hear it, my left ear is mostly deaf because I wear headphones all day. US military plays with advanced torture from a distance tech like nothing else, I'm their ing prototype. Herc === Subject: Re: Turn off your satelite nasa Wrong newsgroup. You should communicate with NASA directly. -- 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: Turn off your satelite nasa > anyone seen Popular Science May issue? Features the > pulsed laser spy satelites that penetrate clouds, > only they use it to constantly bombard me with heathen > dictators since April 2002, every second I've been bombarded > with a laser to sound attack using a dozen retards just designed > to me off. Your mental planet is in conjunction with Uranus. -- Dr. Smartass BAAWA Knight of Heckling -- a.a. #1939 And the knowledge that they fear Is a weapon to be used against them. --Rush, The Weapon === Subject: TV for nothin and your actors for free haven't seen any replies with the answer next to the phone number yet! its not that difficult, this whole town of 100,000 people are all in on it. prefix with 61 7 reply here next to the number if anyone confirms or denies it > 47482160 > 47788360 > 47792822 > 47290642 > 47254486 > 47855847 > 47491445 > 47230018 > 47785779 they called it TRUEman, about one man all of media and government ganged up on, i posted in 2000 i'm getting spied on by govt. and I can't reach a girl who looks like laurie holden, Laurie Holden then costars Jim Carrey in a romance...... aus.tv know it! I'm from Townsville and YOU ARE the Truman! http://tinyurl.com/iky5 I was in Townsville over the weekend, and I heard him. Very spooky! http://tinyurl.com/iky8 >phone someone in Townsville, half of you must know someone there, >every day I go out people say THERES THE TRUMAN I'm in Townsville. We're sick of you. http://tinyurl.com/iky9 http://tinyurl.com/iky4 You rule Truman! worth a phone call to give me my life back? Herc === Subject: UMD-proof Hello... Im searching for proof for the fact that if Banach space has UMD-p for some p between ]1,infty[ then it has UMD-p for every p in ]0,infty[. I know already the proof by Burkholder based on the geometrical characterization of UMD-spaces, but I would like to know also the earlier proof by Pisier. Every text says that it is found in Maureys Systeme de HAar in Seminaires MAurey-Schwartz 1974-5, but unhappily I have so poor franch that I dont understand it. Could somebody tell me where I could find this proof in english or deutche? === Subject: uncertainty analysis I am having some difficulty doing an uncertainty analysis of some experimental data. 1. Is the uncertainty of a sum of independant measurements of some quantity expressed as follows: U(x1+x2+....xn)= sqrt((dx1)squared+(dx2)squared+....(dxn)squared) where dx1,dx2.. are the uncertainties in each measurement. or is it as follows: U(x1+x2+....+xn)= dx1+dx2+....dxn I am confused between two different definitions in two different places. 2. How do i find the uncertainty in the slope of a line given that: each of the points has an uncertainty in the x value each of the points has an uncertainty in the y value It is possible to do it graphically but I need an analytical method for this. There is a function called linest in excel which gives regression statistics for a line including uncertainty in slope and intercept but it doesnt consider uncertainties in individual measurements and only looks at the scatter of the data. Shyam. === Subject: Re: V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Solution for R > V = (1+sqrt(R/F)) / (1-sqrt(R/F)) > Why can't I seem to figure out how to solve for R? Multiply both sides by [1 - sqrt(R/F)] and then solve for sqrt(R/F). That will give you sqrt(R/F) = (V-1) / (V+1) Square both sides and solve for R. If I did it right, you should come up with R = F [(V-1)/(V+1)]^2 which is the well known, but grossly erroneous, W4NTI voltage / frequency / resistance formula (He got kicked out of engineering school because of his weird theory). I'm surprised that anyone other than Dan is still using that formula. === Subject: V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Solve for R? V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Why can't I seem to figure out how to solve for R? === Subject: Re: V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Solve for R? > V = (1+sqrt(R/F)) / (1-sqrt(R/F)) > Why can't I seem to figure out how to solve for R? It depends on what you've tried. -- === Subject: Re: V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Solve for R? > V = (1+sqrt(R/F)) / (1-sqrt(R/F)) > Why can't I seem to figure out how to solve for R? Let x = sqrt(R/F). Then V = (1 + x)/(1 - x) (1 - x)V = (1 + x) V - Vx = 1 + x V - 1 = Vx + x V - 1 = x(V+1) x = (V-1)/(V+1) sqrt(R/F) = (V-1)/(V+1) R/F = [(V-1)/(V+1)]^2 R = F*(V-1)^2 / (V+1)^2 === Subject: Re: V = (1+sqrt(R/F)) / (1-sqrt(R/F)) Solve for R? In message , >V = (1+sqrt(R/F)) / (1-sqrt(R/F)) >Why can't I seem to figure out how to solve for R? Multiply both sides by (1-sqrt(R/F)). Then simplify. Mike -- M.J.Powell === Subject: Re: vector problem > i have a vector (a line) impacting on a surface > and i am looking for the reflection. I know the > normal to this surface and the reflected vector > will basically lie in the same plane made by > the incident vector and the normal, at the same > angle to the normal but on the other side, so > to speak. > Let I be in the incident vector, N the unit normal vector, R the reflection > vector. A simple vector projection argument shows > R = I - (2)N, > where <,> is the dot product. (I'm pretty sure this is what Dale has, but > he had funny words in his post.) surely so. Notice that I in the above expression is in the opposite direction to the path of the incident vector, which is traveling towards the surface === Subject: Re: which miserably boring nonrigorous applied math course should i take? Wow, you're not kidding--the SMU department appears to be *amazingly* > highly staffed with applied mathematicians, and the courses taught > reflect that. >> i'm not sure how you knew what school i go to...but this was excellent >> advice. > Hey! How *did* Lee know what school he goes to? > Even his NNTP-Posting-Host header isn't so helpful, since it was a Verizon > DSL address, and not a school computer. His Path header doesn't seem very > useful, since he posted via Google. (So why *does* he have an > NNTP-Posting-Host header if he didn't post to an NNTP server?) So, how > did Lee do it? It seems he's not going to tell us. Boo hoo. Curiosity is really killing me here. === Subject: Re: which miserably boring nonrigorous applied math course should i take? >> >> Hey! How *did* Lee know what school he goes to? >> >> Even his NNTP-Posting-Host header isn't so helpful, since it was a Verizon >> DSL address, and not a school computer. His Path header doesn't seem very >> useful, since he posted via Google. (So why *does* he have an >> NNTP-Posting-Host header if he didn't post to an NNTP server?) So, how >> did Lee do it? >It seems he's not going to tell us. Boo hoo. Curiosity is really killing >me here. Google is one's friend (evidently Three Dog Night was wrong). Lee Rudolph === Subject: Re: which miserably boring nonrigorous applied math course should i take? <871xweny2a.fsf@phiwumbda.localnet> A8EwTYfhf*u~,Eu,tf6$HN*MY&)u0G =N' x<%)/s=GZ_BD2Qz9m=S%4v^I+>T|'1{w70ZY=ih,=)kMY_}?{%)x0)];K~@J6m5.EN?>Zh Xh;Y V|',x(js'Jfq02joVpj|#x > Google is one's friend (evidently Three Dog Night was wrong). If every number aside from one has at least two friends, then we can continue to rely on Three Dog Night's keen insight into arithmetic sociology. -- Jesse Hughes Basically there are two angry groups. I am a harsh force of one. Against me is a society of mathematicians. So far it's been a draw. -- JSH gives another display of keen insight. === Subject: Re: which miserably boring nonrigorous applied math course should i take? Do you mean to tell me that Jeramiah wasn't a Bull frog! > Wow, you're not kidding--the SMU department appears to be *amazingly* > highly staffed with applied mathematicians, and the courses taught > reflect that. >> i'm not sure how you knew what school i go to...but this was excellent >> advice. Hey! How *did* Lee know what school he goes to? Even his NNTP-Posting-Host header isn't so helpful, since it was a Verizon > DSL address, and not a school computer. His Path header doesn't seem very > useful, since he posted via Google. (So why *does* he have an > NNTP-Posting-Host header if he didn't post to an NNTP server?) So, how > did Lee do it? > It seems he's not going to tell us. Boo hoo. Curiosity is really killing > me here. == Subject: Re: 4 of 6 combinations oh oops... I just realized that I read your problem incorrectly. Please excuse my carelessness. BJ MacNevin > If I need to work out a 4-digit code using numbers 1 to 6, am I right > in saying there are only 6^4 (i.e. 1296) different combinations? Then > if any of the six numbers (1 to 6) can be used more than once, i.e. up > to four times, would the formula for this be 6^4^6? > Thank you for your help. === Subject: Re: 4 of 6 combinations If I understant what you are saying, then I think the basic code has only 6^4 possibilities. Position 1 has 6 possibilties Position 2 has 6 possibilities Position 3 has 6 possibilities Position 4 has 6 possibilities. = 6^4 = 1 296 possible codes But if each number can be used only once, then your number of possibilities decreases by one each time: Position 1 has 6 possibilties Position 2 has 5 possibilities Position 3 has 4 possibilities Position 4 has 3 possibilities. = 6*5*4*3 = 360 possible codes Well, at least I think that makes sense. -BJ MacNevin > If I need to work out a 4-digit code using numbers 1 to 6, am I right > in saying there are only 6^4 (i.e. 1296) different combinations? Then > if any of the six numbers (1 to 6) can be used more than once, i.e. up > to four times, would the formula for this be 6^4^6? > Thank you for your help. === Subject: inverse function f(x) = x^4 + x^3 + 1 for 0<=x<=2 g(x) = inverse of f (x) F(x) = f(2g(x)) Find F(3). I can't find an expression for the inverse of f(x). === === Subject: re: inverse function update This problem is copied from an exercise in a textbook: f(x) = x^4 + x^3 + 1 for 0<=x<=2 g(x) = inverse of f (x) F(x) = f(2g(x)) Find F(3). Correct answer is 88/7. === Subject: Re: inverse function > f(x) = x^4 + x^3 + 1 for 0<=x<=2 > g(x) = inverse of f (x) > F(x) = f(2g(x)) > Find F(3). > I can't find an expression for the inverse of f(x). Hint1: You don't need to find an expression for the inverse of f(x). Hint2: You will need to find the root of 3 = x^4 + x^3 + 1 which lies between 0 and 2. (But you can probably do that by inspection.) David === === Subject: re: inverse function the answer is 88/7. How can I get this answer? === === Subject: Re: inverse function > the answer is 88/7. > How can I get this answer? Either you stated the problem incorrectly in the original post, or the answer above is not correct. To remind the casual reader, the original question posted was as follows: > f(x) = x^4 + x^3 + 1 for 0<=x<=2 > g(x) = inverse of f (x) > F(x) = f(2g(x)) > Find F(3). The key to this problem, as has been hinted elsewhere, is that if f(a) = 3 then g(3) = a because f and g are inverse functions. You should be able to guess the value of a where 0<=a<=2 for which f(a) = a^4 + a^3 + 1 = 3 Once you have guessed that value for a, you will know g(3). Then you will be able to work out the above problem and find F(3), but I don't think you will get 88/7. === http://www.thinkspot.net/sheila/ http://www.k12groups.org/ === === Subject: Re: math > : The base of a rectangular aquarium tank measures 4.3 meters by 5.7 > : meters with a depth of 9.4 meters. When filled to a depth one foot > : below the top, the tank holds about 58,800 gallons. One week after the > : tank was filled, the water level has dropped about eight inches. Andy > : estimates that about 57,500 gallons of water are now in the tank. > A bunch of archaic and ambiguous units! Archaic?? If you mean old, fine, but normally 'archaic' means no longer in use, a meaning that doesn't apply to any of the units used above, all of which are in current use. (Otoh, you're Canadian: perhaps these units lack currency there.) http://math.wustl.edu/~msh210/ Likely your mail's by mistake been deleted. === === Subject: Addressing math standards What, if any, procedures have been put in place by your principal or others in your building to ensure that you cover all math standards each year? === === Subject: Re: Technology and Math Use PowerPoint, or computer programs... or do what most do... show the students how a calculator can assist them with math, but not do their homework for them! I'm your school has the overhead projector for the graphing calculator. I know at the college level we use Derive to graph things in 3D....Derive is a CAS (computer algebra system). === Subject: Re: finding fractions between fractions poster, > I understand there are many ways to find fractions between > fractions. I am attempting to teach my class that one way to do it > is to add the two numerators and add the two denominators and the > resulting fraction will be in between the two original fractions. > EX: 2/3 and 4/5 are the two fractions and the question is find a > fraction between these. I know that the value of 6/8 (or 2+4/3+5) > is in between these fractions. My question is, WHY? > -- > This is called the 'mediant' of two fractions. > Here's a visual proof of why the mediant is between the two > terms: http://mathpages.com/home/kmath055.htm Wow! Cool site. It's interesting to find a property of fractions for which all the fractions in an equivalence class for a given ratio do NOT have the same property. === === === Subject: Calculating Grades I need some help in calulating my grades for the semester. My grades are basically broken down into 3 groups. Tests (which are 70%), Assignments (which are 20%) and Class Participation (which is 10%) is what determines the final grade for the class. I need to know how to do that. Is there a type of formula I can use to determine this. I would like to be able to put the formula(s) in Excel and then enter my grades from my tests, and assignments, etc and be able to figure out Jason === Subject: Re: 3-d integration routine is needed For the future reference, I am now using Dcuhre.f written by genz, it works quite good. But I also have f90 code of cubpack which i havent tried it yet. > Thank you for all your suggestions, I am going to work on these suggestions > and try to come up with a solution. By the way my integration limits are all > scalar which makes the problem even simpler. > google but I could not find anything. Do you have any suggestions? Any > reliable routine which involves c/c++/fortran is good enough for me. === Subject: free incomplete Cholesky code? Can anyone recommend free source code (C or Fortran is fine) for incomplete Cholesky (IC) decomposition (e.g. for preconditioning conjugate gradient) of a general Hermitian sparse matrix in compressed-row (or compressed-column) format? Ideally, it should support complex-Hermitian matrices, not just real-symmetric. I searched around on Google for a while, and I found some Fortran 77 code on SLATEC for real matrices that I could potentially adapt, but I was hoping to see if I have any other alternatives. Cordially, Steven G. son === Subject: consulting wanted Researchers at UCLA are looking for coding help in projects about computation and estimation for some financial markets. The project involves computation of an equilibrium using a fixed point method, and estimation of underlying parameters from the data. Experiences in these areas will be a plus. Langages can be either of Matlab, Gauss, C++, or your favorite. Short-term rather intensive work. Competitive compensation. If interested, please send inquires to === Subject: please remove the previous post programming help. Will you kindly remove this message (and this message) from the posting. Thank you, === Subject: Re: FFTW and polynomial Multiplicaion {more questions} > But shouldn't it not matter, because my 'in' coefficient vector is all real > with complex part (in[i][1]) as all zeros. ?? No, because the transform of a real array is not (in general) real. > And further-more I have incorporated squaring the complex-part in my program > with no success... > [...] > void square_FftwVector(fftw_complex *arr, int len){ > int i=0; > for(i;i arr[i][0]=arr[i][0]*arr[i][0]; > arr[i][1]=arr[i][1]*arr[i][1]; > } This is not how you square a complex number. double re = arr[i][0], im = arr[i][1]; arr[i][0] = re * re - im * im; arr[i][1] = 2.0 * re * im; You need to go back and read some basic introductions to complex arithmetic and the use of FFTs for convolutions. === Subject: Re: FFTW and polynomial Multiplicaion {more questions} > I have made the change as pointed out by Steven, but I am > still lost about the output I am getting. I have scoured > the web and but invain > [..] > for(i=0;i < N; i++) > out[i][0]= out[i][0]*out[i][0]; /*the O(n) step of FFT */ > } > ``out'' is a complex array, but you are only squaring the real part. But shouldn't it not matter, because my 'in' coefficient vector is all real with complex part (in[i][1]) as all zeros. ?? And further-more I have incorporated squaring the complex-part in my program with no success... I am pasting the complete source below, followed by the output, Pls let me know how to proceed and thanks for your patience. #include #include void print_FftwVector(fftw_complex *arr, int len){ int i=0; for(i;i I have made the change as pointed out by Steven, but I am > still lost about the output I am getting. I have scoured > the web and but invain > [..] > for(i=0;i < N; i++) > out[i][0]= out[i][0]*out[i][0]; /*the O(n) step of FFT */ > } ``out'' is a complex array, but you are only squaring the real part. Cheers, Matteo Frigo === Subject: FFTW and polynomial Multiplicaion {more questions} I have made the change as pointed out by Steven, but I am still lost about the output I am getting. I have scoured the web and but invain I am working on squaring a polynomial, say x^7 + x^6 + x^5 .... + 1 { n = 8 } so the 'in' array is (imaginary part being all zeros) (1,1,1,1,1,1,1,1,0,0,0,0,0,0,0) and we expect the output coefficient vector = (1,2,3,4,5,6,7,8,7,6,5,4,3,2,1) crux of the code. ---------------------------------------------------------- int deg=7, n=deg+1 , N = 2*deg+1; fftw_complex *in = fftw_malloc(N* sizeof (fftw_complex)); fftw_complex *out = fftw_malloc(N* sizeof (fftw_complex)); fftw_plan p1, p2; for(int i=0;i < N; i++) { if(i < n) { in[i][0] = 1 ; } else { in[i][0] = 0 ; } in[i][1]= 0; } p1 = fftw_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE); fftw_execute(p1); for(i=0;i < N; i++) out[i][0]= out[i][0]*out[i][0]; /*the O(n) step of FFT */ } p = fftw_plan_dft_1d(N,out,out, FFTW_BACKWARD, FFTW_ESTIMATE); fftw_execute(p); print (out); ---------------------------------------------------------- now I realize that the above output is is not normalized so the output should be N * (1,2,3,4,5,6,7,8,7,6,5,4,3,2,1) but I get (67.5, 63.75, 63.75, and all remaining coefficients are 63.75) can any one help me get on the track? === Subject: Re: Discrete Chebyshev: best algorithm? ha scritto nel > nag_linf_fit is part of the NAG C Library - you can buy it. > (see http://www.nag.co.uk/numeric/CL/manual/pdf/E02/e02gcc.pdf ) > It seems to use a modified version of ACM algo 495, for which the > Fortran source is available at > http://www.netlib.org/toms/495 Thank you very much, I saw it but I though it was a demonstrative incomplete piece of code (I don't speak fortran :)) > If neither efficiency nor readability is an issue for your application > you could try to throw the Fortran source into f2c and see what > comes out ... I have compiled in g77 and linked in gcc and it seems to work! > But what is the best one? Is Borrodale' and Phillips' still > preferable? In my simulations I see that it depends on the number od points; for few points (10,100...) 495 is faster, while for many point (10000...) Seidel's one is faster. Clearly I'm not sure if the implementations I am using are good ones, but I hope so :) Thank you very much m > Thank you very much > m > === Subject: help with boundary condition for schroedinger equation I need to have transparent boundary conditions because i need to propagate for a long time. I tried to implement Hadley's BC [IEEE Journ. Quant. El., v. 28, January 1992, p 363-370], but i can obtain good results. Maybe this is due to my lack in programming, because the algorithm, i believe is straightforward. If anyone has already implemented such an algorithm and can help me... welcome. Thank you very much. === Subject: Re: Supreme : I'm in my second week of my first year of university and hoping for a : little help with my analysis homework as I am finding it quite a : struggle. The question is: : Let A and B be bounded, nonempty sets of real numbers. : Is it always true that sup(AB) = supAsupB, where AB = {xy: x belongs : to A, y belongs to B}? Try for exemple A={-1}, B={-1,1} Then A={-2,1}, B={-2,1} -- Pierre Verpeaux === Subject: Re: Simplifying logical proposition > I would like to know if anyone could help me understanding how Maple did the > following simplification (using bsimp): Proposition: (not p or q) and ((q and not r) or (r and not q)) > Simplified proposition: r and not p and not q > My Maple 9 bsimp gives > (r and not p and not q) or (q and not r) > I don't think bsimp has been changed in years. What Maple are you using? > Something must have changed since you both got different and *wrong* > results. > I have done the problem by hand, and I stand by the answer that I got with > Maple. Agrees with tht truth table also that I posted. === Subject: Re: How to get a structured irregular mesh in FDTD? > This sounds like a photonic crystal. Did you search the literature on > the simulation of photonic crystals? > Yeah! > Do you have an good idea? No, sorry. As I didn't know which field your problems comes from, I didn't know whether you had this idea allready. Bye, Christof === Subject: Re: How to get a structured irregular mesh in FDTD? > I want to get the EM field in a cubic with several non-perfect metal > spheres in it with FDTD. > This sounds like a photonic crystal. Did you search the literature on > the simulation of photonic crystals? > Bye, > Christof Yeah! Do you have an good idea? === Subject: Re: How to get a structured irregular mesh in FDTD? > I want to get the EM field in a cubic with several non-perfect metal > spheres in it with FDTD. This sounds like a photonic crystal. Did you search the literature on the simulation of photonic crystals? Bye, Christof === Subject: How to get a structured irregular mesh in FDTD? I want to get the EM field in a cubic with several non-perfect metal spheres in it with FDTD. I try to apply a structured irregular mesh to accurately model the spheres, but i do not know how to get such a mesh. After searching several databases, I found that there are few several papers dealling with such a mesh. Does this mean that this method can not get a good result? Who can tell me how to get such a mesh? And there is another problem, that is, why the papers always model perfect metal,not non-perfect metal. Thank you for your help. === Subject: Re: Need help with nleq1 This is minipack c double precision function dpmpar(i) c c where c c i is an integer input variable set to 1, 2, or 3 which c selects the desired machine parameter. If the machine has c t base b digits and its smallest and largest exponents are c emin and emax, respectively, then these parameters are c c dpmpar(1) = b**(1 - t), the machine precision, c c dpmpar(2) = b**(emin - 1), the smallest magnitude, c c dpmpar(3) = b**emax*(1 - b**(-t)), the largest magnitude. c c c Machine constants for IEEE machines. c data dmach(1) /2.22044604926d-16/ data dmach(2) /2.22507385852d-308/ data dmach(3) /1.79769313485d+308/ This is in nleq1.f C C DOUBLE-PRECISION MACHINE CONSTANTS C C D1MACH( 1) = B**(EMIN-1), THE SMALLEST POSITIVE MAGNITUDE. C C D1MACH( 2) = B**EMAX*(1 - B**(-T)), THE LARGEST MAGNITUDE. C C D1MACH( 3) = B**(-T), THE SMALLEST RELATIVE SPACING. C C D1MACH( 4) = B**(1-T), THE LARGEST RELATIVE SPACING. C C D1MACH( 5) = LOG10(B) I am quite new to computing and I am sorry if I mislead you with my messages. I used minipack recently and I was sure that it is working quite well so I change the last subroutine accordingly. > In minipack I find this > [...] > data dmach(1) /2.22044604926d-16/ > data dmach(2) /2.22507385852d-308/ > data dmach(3) /1.79769313485d+308/ > Did you modify the last subroutine in nleq1.f accordingly? > And I change nleq a little bit. It looks like it is working for me now. > I am _very_ curious, what this little bit means precisely, especially > since you called it a compiler issue before. > By being more specific you could help others, who didn't resolve the > problem yet. > Alois === Subject: Re: Need help with nleq1 > In minipack I find this > [...] > data dmach(1) /2.22044604926d-16/ > data dmach(2) /2.22507385852d-308/ > data dmach(3) /1.79769313485d+308/ Did you modify the last subroutine in nleq1.f accordingly? > And I change nleq a little bit. It looks like it is working for me now. I am _very_ curious, what this little bit means precisely, especially since you called it a compiler issue before. By being more specific you could help others, who didn't resolve the problem yet. Alois === Subject: Re: Need help with nleq1 In minipack I find this c double precision function dpmpar(i) c c where c c i is an integer input variable set to 1, 2, or 3 which c selects the desired machine parameter. If the machine has c t base b digits and its smallest and largest exponents are c emin and emax, respectively, then these parameters are c c dpmpar(1) = b**(1 - t), the machine precision, c c dpmpar(2) = b**(emin - 1), the smallest magnitude, c c dpmpar(3) = b**emax*(1 - b**(-t)), the largest magnitude. c c c Machine constants for IEEE machines. c data dmach(1) /2.22044604926d-16/ data dmach(2) /2.22507385852d-308/ data dmach(3) /1.79769313485d+308/ And I change nleq a little bit. It looks like it is working for me now. > Problem was related to compiler and now it is solved thanks for everyone. > since that is a quite straightforward program: > Could you give any details? > Alois === Subject: Re: Need help with nleq1 > Problem was related to compiler and now it is solved thanks for everyone. since that is a quite straightforward program: Could you give any details? Alois === Subject: Re: Need help with nleq1 Problem was related to compiler and now it is solved thanks for everyone. > I need to learn how to use this routine to solve some nonlinear equations > but first I wanted to test it and could not make it work. > The code below is actually from maine1.f which is included in that routine > as an example but I only change FCN. > The answer for the equations is supposed to be 1,2,3 but I could not make it > work. > Please give me some advice, if you made this work before. > INTEGER NN > INTEGER N,I,N1 > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(NN) > EXTERNAL NLEQ1E, SECOND > REAL STIME,ETIME,CPTIME > N = 2 > 10 IF(N.LE.NN)THEN > RTOL = 1.0D-5 > N1 = N+1 > DO 20 I=1,N > X(I)=DBLE(I)/DBLE(N1) > 20 CONTINUE > CALL SECOND(STIME) > CALL NLEQ1E(N,X,RTOL,IERR) > CALL SECOND(ETIME) > CPTIME = ETIME-STIME > IF (CPTIME.NE.0.0) THEN > 1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), > $ /) > WRITE(*,1001)CPTIME > ENDIF > N = N+1 > GOTO 10 > ENDIF > write(*,*) x(1) , x(2) , x(3) > END > C > SUBROUTINE FCN(N,X,FVEC,IFLAG) > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 > RETURN > END === Subject: Re: Need help with nleq1 > did you look at the last subroutine d1mach in nleq1? > There some machine-specific constants are set. The default settings > are for Motorola processors. If you use some Intel processors, you > need to modify the subroutine. > This problem is mentioned in the comments right at the start of nleq1. > Good luck > Alois > -- > Alois Steindl, Tel.: +43 (1) 58801 / 32558 > Inst. for Mechanics II, Fax.: +43 (1) 58801 / 32598 > Vienna University of Technology, > A-1040 Wiedner Hauptstr. 8-10 === Subject: Re: Need help with nleq1 did you look at the last subroutine d1mach in nleq1? There some machine-specific constants are set. The default settings are for Motorola processors. If you use some Intel processors, you need to modify the subroutine. This problem is mentioned in the comments right at the start of nleq1. Good luck Alois === Alois Steindl, Tel.: +43 (1) 58801 / 32558 Inst. for Mechanics II, Fax.: +43 (1) 58801 / 32598 Vienna University of Technology, A-1040 Wiedner Hauptstr. 8-10 === Subject: Re: Need help with nleq1 Hi again. I still get singular jacoian with this simple fcn. INTEGER N DOUBLE PRECISION RTOL INTEGER IERR DOUBLE PRECISION X(N) EXTERNAL NLEQ1E RTOL = 1.0D-5 x(1)=1.d1 x(2)=2.d2 x(3)=3.d1 CALL NLEQ1E(N,X,RTOL,IERR) write(*,*) x(1) , x(2) , x(3), IERR END SUBROUTINE FCN(N,X,FVEC,IFLAG) INTEGER N DOUBLE PRECISION X(N),FVEC(N) FVEC(1) = X(1)-1.d0 FVEC(2) = x(2)-2.d0 FVEC(3) = x(3)-3.d0 RETURN END === Subject: Re: Need help with nleq1 I made a mistake here, ignore this and the previous message. I am going to try a few things and come back. > With this one, although my initial values are exact answer to the eqns, I > get singular jacobian error. Other than that, is this a proper way to drive > the routine? > INTEGER N > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(N) > EXTERNAL NLEQ1E > REAL STIME,ETIME,CPTIME > RTOL = 1.0D-5 > x(1)=1.d0 > x(2)=2.d0 > x(3)=3.d0 > CALL NLEQ1E(N,X,RTOL,IERR) > write(*,*) x(1) , x(2) , x(3) > END > SUBROUTINE FCN(N,X,FVEC,IFLAG) > INTEGER N > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 > RETURN > END Spellucci equations >but first I wanted to test it and could not make it work. >The code below is actually from maine1.f which is included in that > routine >as an example but I only change FCN. >The answer for the equations is supposed to be 1,2,3 but I could not > make it >work. >Please give me some advice, if you made this work before. > INTEGER NN > INTEGER N,I,N1 > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(NN) > EXTERNAL NLEQ1E, SECOND > REAL STIME,ETIME,CPTIME > N = 2 > !!!!!!!!ERROR your n=3 >10 IF(N.LE.NN)THEN > !!!!!!!!!! why this? you have only _one_ type of fcn > RTOL = 1.0D-5 > N1 = N+1 > DO 20 I=1,N > X(I)=DBLE(I)/DBLE(N1) >20 CONTINUE > !!!!!!!! ERROR you had n=2 hence x(3) uninitialized > CALL SECOND(STIME) > CALL NLEQ1E(N,X,RTOL,IERR) > CALL SECOND(ETIME) > CPTIME = ETIME-STIME > IF (CPTIME.NE.0.0) THEN >1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), > $ /) > WRITE(*,1001)CPTIME > ENDIF > N = N+1 > GOTO 10 > !!!!!!!! this makes no sense here (they had sevral testcases within > _one_ > fcn) > ENDIF > write(*,*) x(1) , x(2) , x(3) > END >C > SUBROUTINE FCN(N,X,FVEC,IFLAG) > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 > RETURN > END === Subject: Re: Need help with nleq1 With this one, although my initial values are exact answer to the eqns, I get singular jacobian error. Other than that, is this a proper way to drive the routine? INTEGER N DOUBLE PRECISION RTOL INTEGER IERR DOUBLE PRECISION X(N) EXTERNAL NLEQ1E REAL STIME,ETIME,CPTIME RTOL = 1.0D-5 x(1)=1.d0 x(2)=2.d0 x(3)=3.d0 CALL NLEQ1E(N,X,RTOL,IERR) write(*,*) x(1) , x(2) , x(3) END SUBROUTINE FCN(N,X,FVEC,IFLAG) INTEGER N DOUBLE PRECISION X(N),FVEC(N) FVEC(1) = x(2)*x(3)-3*x(1) FVEC(2) = x(2)*x(3)/x(1)-6.0 FVEC(3) = x(3)*x(1)-3.0 RETURN END >I need to learn how to use this routine to solve some nonlinear equations >but first I wanted to test it and could not make it work. >The code below is actually from maine1.f which is included in that routine >as an example but I only change FCN. >The answer for the equations is supposed to be 1,2,3 but I could not make it >work. >Please give me some advice, if you made this work before. > INTEGER NN > INTEGER N,I,N1 > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(NN) > EXTERNAL NLEQ1E, SECOND > REAL STIME,ETIME,CPTIME > N = 2 > !!!!!!!!ERROR your n=3 >10 IF(N.LE.NN)THEN > !!!!!!!!!! why this? you have only _one_ type of fcn > RTOL = 1.0D-5 > N1 = N+1 > DO 20 I=1,N > X(I)=DBLE(I)/DBLE(N1) >20 CONTINUE > !!!!!!!! ERROR you had n=2 hence x(3) uninitialized > CALL SECOND(STIME) > CALL NLEQ1E(N,X,RTOL,IERR) > CALL SECOND(ETIME) > CPTIME = ETIME-STIME > IF (CPTIME.NE.0.0) THEN >1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), > $ /) > WRITE(*,1001)CPTIME > ENDIF > N = N+1 > GOTO 10 > !!!!!!!! this makes no sense here (they had sevral testcases within _one_ > fcn) > ENDIF > write(*,*) x(1) , x(2) , x(3) > END >C > SUBROUTINE FCN(N,X,FVEC,IFLAG) > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 > RETURN > END === Subject: Re: Need help with nleq1 Hi Thank you for your attention. As far as I understood, maine1.f is not a good example to show how to use this routine. Do you have any other example ? >I need to learn how to use this routine to solve some nonlinear equations >but first I wanted to test it and could not make it work. >The code below is actually from maine1.f which is included in that routine >as an example but I only change FCN. >The answer for the equations is supposed to be 1,2,3 but I could not make it >work. >Please give me some advice, if you made this work before. > INTEGER NN > INTEGER N,I,N1 > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(NN) > EXTERNAL NLEQ1E, SECOND > REAL STIME,ETIME,CPTIME > N = 2 > !!!!!!!!ERROR your n=3 >10 IF(N.LE.NN)THEN > !!!!!!!!!! why this? you have only _one_ type of fcn > RTOL = 1.0D-5 > N1 = N+1 > DO 20 I=1,N > X(I)=DBLE(I)/DBLE(N1) >20 CONTINUE > !!!!!!!! ERROR you had n=2 hence x(3) uninitialized > CALL SECOND(STIME) > CALL NLEQ1E(N,X,RTOL,IERR) > CALL SECOND(ETIME) > CPTIME = ETIME-STIME > IF (CPTIME.NE.0.0) THEN >1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), > $ /) > WRITE(*,1001)CPTIME > ENDIF > N = N+1 > GOTO 10 > !!!!!!!! this makes no sense here (they had sevral testcases within _one_ > fcn) > ENDIF > write(*,*) x(1) , x(2) , x(3) > END >C > SUBROUTINE FCN(N,X,FVEC,IFLAG) > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 > RETURN > END === Subject: Re: Need help with nleq1 I need to learn how to use this routine to solve some nonlinear equations >but first I wanted to test it and could not make it work. >The code below is actually from maine1.f which is included in that routine >as an example but I only change FCN. >The answer for the equations is supposed to be 1,2,3 but I could not make it >work. >Please give me some advice, if you made this work before. INTEGER NN > INTEGER N,I,N1 > DOUBLE PRECISION RTOL > INTEGER IERR > DOUBLE PRECISION X(NN) > EXTERNAL NLEQ1E, SECOND > REAL STIME,ETIME,CPTIME > N = 2 !!!!!!!!ERROR your n=3 >10 IF(N.LE.NN)THEN !!!!!!!!!! why this? you have only _one_ type of fcn > RTOL = 1.0D-5 > N1 = N+1 > DO 20 I=1,N > X(I)=DBLE(I)/DBLE(N1) >20 CONTINUE !!!!!!!! ERROR you had n=2 hence x(3) uninitialized > CALL SECOND(STIME) > CALL NLEQ1E(N,X,RTOL,IERR) > CALL SECOND(ETIME) > CPTIME = ETIME-STIME > IF (CPTIME.NE.0.0) THEN >1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), > $ /) > WRITE(*,1001)CPTIME > ENDIF > N = N+1 > GOTO 10 !!!!!!!! this makes no sense here (they had sevral testcases within _one_ fcn) > ENDIF > write(*,*) x(1) , x(2) , x(3) > END C SUBROUTINE FCN(N,X,FVEC,IFLAG) > DOUBLE PRECISION X(N),FVEC(N) > FVEC(1) = x(2)*x(3)-3*x(1) > FVEC(2) = x(2)*x(3)/x(1)-6.0 > FVEC(3) = x(3)*x(1)-3.0 RETURN > END === Subject: Need help with nleq1 I need to learn how to use this routine to solve some nonlinear equations but first I wanted to test it and could not make it work. The code below is actually from maine1.f which is included in that routine as an example but I only change FCN. The answer for the equations is supposed to be 1,2,3 but I could not make it work. Please give me some advice, if you made this work before. INTEGER NN INTEGER N,I,N1 DOUBLE PRECISION RTOL INTEGER IERR DOUBLE PRECISION X(NN) EXTERNAL NLEQ1E, SECOND REAL STIME,ETIME,CPTIME N = 2 10 IF(N.LE.NN)THEN RTOL = 1.0D-5 N1 = N+1 DO 20 I=1,N X(I)=DBLE(I)/DBLE(N1) 20 CONTINUE CALL SECOND(STIME) CALL NLEQ1E(N,X,RTOL,IERR) CALL SECOND(ETIME) CPTIME = ETIME-STIME IF (CPTIME.NE.0.0) THEN 1001 FORMAT(//,1X,'Time ','used ','=',F9.3,1X,'Sec',//,66('*'), $ /) WRITE(*,1001)CPTIME ENDIF N = N+1 GOTO 10 ENDIF write(*,*) x(1) , x(2) , x(3) END C SUBROUTINE FCN(N,X,FVEC,IFLAG) DOUBLE PRECISION X(N),FVEC(N) FVEC(1) = x(2)*x(3)-3*x(1) FVEC(2) = x(2)*x(3)/x(1)-6.0 FVEC(3) = x(3)*x(1)-3.0 RETURN END === Subject: Re: FFTW input data size > I was reading the FFTW-3.01 manual and I says that FFTW can handle the input data of arbitrary size.... Does that mean that > The number of coefficients in my coefficient vector can be arbitrary, and > the size of each coefficient is > bounded by single/long-double ? Yes. The length of the input vector is arbitrary (modulo memory constraints). The precision of each vector element is determined by the C implementation (float, double, long double). > Is there any way to work with coefficients with arbitrary sizes?? Not in FFTW. That being said, if you are really desperate, FFTW contains an arbitrary precision FFT routine in its test program. The routine is undocumented, dirty, very slow, and 1D only. See libbench2/mp.c. Cheers, Matteo Frigo === Subject: FFTW input data size I was reading the FFTW-3.01 manual and I says that FFTW can handle the input data of arbitrary size.... Does that mean that The number of coefficients in my coefficient vector can be arbitrary, and the size of each coefficient is bounded by single/long-double ? Is there any way to work with coefficients with arbitrary sizes?? Manish === Subject: asking for a consultant A research group at UCLA look for a consultant for a project which involves computation of equilibrium and estimation in financial markets. Some ability and interests in economics and finance will be necessary. A knowledge of matlab or gauss or c++ or equivalent will be needed. A short and intensive work. Please contact === Subject: Calculating Video Plus codes by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9GFknp12417; I would be interested to learn whether you got an answer to your post in August 03 re: how to calculate Video Plus codes. My pressing need is to learn how to pre-program those Rugby World Cup games which are being televised by ITV2, so that I can avoid missing these (particularly those which start a bit too early for an old codger such as myself. Kind regards, Hugh Kenshole (A retired engineer (Canadian Space Agency, etc.) who found simulation so much easier to use, when conducting risk analysis on complex redundant systems, than the more rigorous analytical methods!) === Subject: Re: calculating log of the ratio of determinants > I have to calculate the log ratio of two symmetric p.d 500x500 > determinants S1 and S2, the determinants are too large and will not > fit into a double variable. > Since I cannot calculate the determinants exactly I can think of the > following ways to do it > 1] > calculate S2^{-1}S1, find out its eigenvalues and add up the sum of > their logs > 2] > calculate sum of the logs of the ev's of S1, find out sum of the ev's > of S2 and subtract > Is there any reason to prefer one over the another? >Here is what I would do if I were calculating the thing. Assuming the >determinants are positive ;-) I would modify the LU algorithm (or >whatever else you are using) to keep track of the log of the determinant >rather than the determinant itself. It seems to me that if the determinant >is so large as to exceed 10^308 (the IEEE double-length max, roughly >speaking) the matrix must be ill-conditioned. Instead of working this hard, unless much accuracy is needed, keep the product of the terms in the form m*2^n. This way, not much additional work is needed. === This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin@stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 === Subject: Re: calculating log of the ratio of determinants > It seems to me that if the determinant is so large as to > exceed 10^308 (the IEEE double-length max, roughly > speaking) the matrix must be ill-conditioned. No. A 309x309 diagonal matrix with all entries 10 has determinant 10^309 but is perfectly conditioned. It is *very* common that determinants of large matrices either overflow or underflow. === Subject: Re: calculating log of the ratio of determinants > I have to calculate the log ratio of two symmetric p.d 500x500 > determinants S1 and S2, the determinants are too large and will not > fit into a double variable. > Since I cannot calculate the determinants exactly I can think of the > following ways to do it > 1] > calculate S2^{-1}S1, find out its eigenvalues and add up the sum of > their logs > 2] > calculate sum of the logs of the ev's of S1, find out sum of the ev's > of S2 and subtract > Is there any reason to prefer one over the another? Here is what I would do if I were calculating the thing. Assuming the determinants are positive ;-) I would modify the LU algorithm (or whatever else you are using) to keep track of the log of the determinant rather than the determinant itself. It seems to me that if the determinant is so large as to exceed 10^308 (the IEEE double-length max, roughly speaking) the matrix must be ill-conditioned. You might then want to look at high-precision techniques a la David Smith (Computing in Science and Engineering, earlier this year). But I would try first to accumulate the log rather than the multiply the factors, as I reduced the matrix to LU form. === ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ === Subject: Re: Matrix Maximization May be coding the assignment problem directly in a modeling language is simpler: set d 'days' /monday,tuesday,wednesday,thursday,friday/; set m 'movies' /a,b,c,d,e,f,g/; table hp(d,m) 'happy points' A B C D E F G Monday 37 29 22 22 11 3 14 Tuesday 45 29 27 21 17 14 26 Wednesday 51 39 40 21 20 11 20 Thursday 46 22 28 15 18 16 20 Friday 29 19 12 9 7 6 10 ; positive variables x(d,m) 'assignment variables'; variable z 'objective variable'; equations obj 'objective function (to be maximimized)' assign1(d) 'on each day show exactly one movie' assign2(m) 'each movie can be shown at most one times' ; obj.. z =e= sum((d,m),hp(d,m)*x(d,m)); assign1(d).. sum(m, x(d,m)) =e= 1; assign2(m).. sum(d, x(d,m)) =l= 1; model assignment /obj,assign1,assign2/; solve assignment using lp maximizing z; display x.l; I get a solution with objective 153: ---- 32 VARIABLE x.L assignment variables a b c d g monday 1.000 tuesday 1.000 wednesday 1.000 thursday 1.000 friday 1.000 This model is soo small it can be solved with the free student/demo system. ---------------------------------------------------------------- erwin@gams.com, http://www.gams.com/~erwin ---------------------------------------------------------------- > Since there are only 360360 choices, the simplest method is an > exhaustive search. Here is a program using the old fashioned > GWBASIC interpreter: > 10 DIM A(14),B(14),C(14),D(14),E(14) > 20 FOR I=0 TO 14: READ A(I),B(I),C(I),D(I),E(I): NEXT I > 30 MM=0 > 40 FOR A=0 TO 14 > 50 FOR B=0 TO 14: IF B=A THEN 140 > 60 FOR C=0 TO 14: IF C=B OR C=A THEN 130 > 70 FOR D=0 TO 14: IF D=C OR D=B OR D=A THEN 120 > 80 FOR E=0 TO 14: IF E=D OR E=C OR E=B OR E=A THEN 110 > 90 M=A(A)+B(B)+C(C)+D(D)+E(E) > 100 IF MM<=M THEN MM=M: PRINT M,A;B;C;D;E > 110 NEXT E > 120 NEXT D > 130 NEXT C > 140 NEXT B > 150 NEXT A > 1000 DATA 37, 45, 51, 46, 29 > 1010 DATA 29, 29, 39, 22, 19 > 1020 DATA 22, 27, 40, 28, 12 > 1030 DATA 22, 21, 21, 15, 9 > 1040 DATA 11, 17, 20, 18, 7 > 1050 DATA 3, 14, 11, 16, 6 > 1060 DATA 14, 26, 20, 20, 10 > 1070 DATA 0, 0, 0, 0, 0 > 1080 DATA 0, 0, 0, 0, 0 > 1090 DATA 0, 0, 0, 0, 0 > 1100 DATA 0, 0, 0, 0, 0 > 1110 DATA 0, 0, 0, 0, 0 > 1120 DATA 0, 0, 0, 0, 0 > 1130 DATA 0, 0, 0, 0, 0 > 1140 DATA 0, 0, 0, 0, 0 > which produces output: > 128 0 1 2 3 4 > 131 0 1 2 3 6 > 133 0 1 2 4 3 > 134 0 1 2 4 6 > 135 0 1 2 6 3 > 135 0 3 1 2 6 > 135 0 3 2 4 1 > 137 0 3 2 6 1 > 139 0 6 1 2 3 > 140 0 6 2 4 1 > 141 1 0 2 4 3 > 142 1 0 2 4 6 > 143 1 0 2 6 3 > 143 1 3 2 0 4 > 146 1 3 2 0 6 > 150 1 6 2 0 3 > 153 3 6 2 0 1 > these are: D G C A B > with values: 11 26 40 46 19 === Subject: Re: Matrix Maximization by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9H7JXv09779; Oops, I got the last line wrong, it should be: with values: 22 26 40 46 19 === Subject: Re: Matrix Maximization by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9H6obM07651; Since there are only 360360 choices, the simplest method is an exhaustive search. Here is a program using the old fashioned GWBASIC interpreter: 10 DIM A(14),B(14),C(14),D(14),E(14) 20 FOR I=0 TO 14: READ A(I),B(I),C(I),D(I),E(I): NEXT I 30 MM=0 40 FOR A=0 TO 14 50 FOR B=0 TO 14: IF B=A THEN 140 60 FOR C=0 TO 14: IF C=B OR C=A THEN 130 70 FOR D=0 TO 14: IF D=C OR D=B OR D=A THEN 120 80 FOR E=0 TO 14: IF E=D OR E=C OR E=B OR E=A THEN 110 90 M=A(A)+B(B)+C(C)+D(D)+E(E) 100 IF MM<=M THEN MM=M: PRINT M,A;B;C;D;E 110 NEXT E 120 NEXT D 130 NEXT C 140 NEXT B 150 NEXT A 1000 DATA 37, 45, 51, 46, 29 1010 DATA 29, 29, 39, 22, 19 1020 DATA 22, 27, 40, 28, 12 1030 DATA 22, 21, 21, 15, 9 1040 DATA 11, 17, 20, 18, 7 1050 DATA 3, 14, 11, 16, 6 1060 DATA 14, 26, 20, 20, 10 1070 DATA 0, 0, 0, 0, 0 1080 DATA 0, 0, 0, 0, 0 1090 DATA 0, 0, 0, 0, 0 1100 DATA 0, 0, 0, 0, 0 1110 DATA 0, 0, 0, 0, 0 1120 DATA 0, 0, 0, 0, 0 1130 DATA 0, 0, 0, 0, 0 1140 DATA 0, 0, 0, 0, 0 which produces output: 128 0 1 2 3 4 131 0 1 2 3 6 133 0 1 2 4 3 134 0 1 2 4 6 135 0 1 2 6 3 135 0 3 1 2 6 135 0 3 2 4 1 137 0 3 2 6 1 139 0 6 1 2 3 140 0 6 2 4 1 141 1 0 2 4 3 142 1 0 2 4 6 143 1 0 2 6 3 143 1 3 2 0 4 146 1 3 2 0 6 150 1 6 2 0 3 153 3 6 2 0 1 these are: D G C A B with values: 11 26 40 46 19 === Subject: Re: roots of quadratic matrix equation? >The matrix square root has 2^n solutions and not 2, but is not as difficult as > Maybe I don't understand the question but matrices can have more > square roots than that! For example the identity matrix is the square > of every matrix A^(-1) B A where B = diag(1,-1) and A in GL_2(R) is > arbitrary. Yes, of course. (I saw at least 2^n solutions, but didn't check whether there were more...) === Subject: Re: roots of quadratic matrix equation? Distribution: inet You right. AB = (BA)^T in case of symmetric matrices. :-) >Because all you matrices are symmetric you can use AB = BA for any >symmetric A and B of size nxn. > this is false. counter example > A= [3 0 ; 0 1 ]; B=[2 -1 ; -1 2]; > AB= [ 6 -3 ; -1 2 ]; > BA= [ 6 -1 ; -3 2 ]; > sorry > === Subject: Re: roots of quadratic matrix equation? > Because all you matrices are symmetric you can use AB = BA for any > symmetric A and B of size nxn. There are simple 2 x 2 counterexamples to this assertion. > Thus formally you have two solutions the same way you have > solutions for > n = 1: > Q = 0.5*B*A^(-1) +(-) ( 0.25*B*B - C*A^(-1) )^(1/2) > But in this > (case you have a problem of finding the square root of matrix > (P*P = M => P = M^(1/2) ) which is as difficult as an initial problem. The matrix square root has 2^n solutions and not 2, but is not as difficult as the original problem. A spectral factorization reduces it to n ordinary square roots === Subject: Re: roots of quadratic matrix equation? Distribution: inet >Because all you matrices are symmetric you can use AB = BA for any >symmetric A and B of size nxn. this is false. counter example A= [3 0 ; 0 1 ]; B=[2 -1 ; -1 2]; AB= [ 6 -3 ; -1 2 ]; BA= [ 6 -1 ; -3 2 ]; sorry === Subject: Re: roots of quadratic matrix equation? Distribution: inet Because all you matrices are symmetric you can use AB = BA for any symmetric A and B of size nxn. Thus formally you have two solutions the same way you have solutions for n = 1: Q = 0.5*B*A^(-1) +(-) ( 0.25*B*B - C*A^(-1) )^(1/2) But in this case you have a problem of finding the square root of matrix (P*P = M => P = M^(1/2) ) which is as difficult as an initial problem. So may be using numerical methods is a better idea. Andrey > Is there a formula for the roots of the following quadratic matrix > equation? > QAQ - BQ + C = 0 > where A, B, C and Q are all symmetric nxn matrices, n>1? > thanks === Subject: Re: light bulbs > I've been doing some reading and it seems that conduction of electricity > in metals is actually a fairly complicated process. Inside a light bulb, > conduction through the filament is further complicated by the presence of > the electron cloud. If you know of numerical work that simulates electrical > conduction, especially in light bulbs, please let me know. > I have no practical application for this. I just want to know how it is done. Post your request to the newsgroup: sci.engr.lighting > Ignorantly, > Allan Adler > ara@zurich.ai.mit.edu === Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. === Subject: light bulbs I've been doing some reading and it seems that conduction of electricity in metals is actually a fairly complicated process. Inside a light bulb, conduction through the filament is further complicated by the presence of the electron cloud. If you know of numerical work that simulates electrical conduction, especially in light bulbs, please let me know. I have no practical application for this. I just want to know how it is done. Ignorantly, Allan Adler ara@zurich.ai.mit.edu **** * * * Disclaimer: I am a guest and *not* a member of the MIT Artificial * * Intelligence Lab. My actions and comments do not reflect * * in any way on MIT. Moreover, I am nowhere near the Boston * * metropolitan area. * * * **** === Subject: Re: Are there any storage efficient preconditioners? > We may have a misunderstanding here. When I said: > Take for instance a lower triangular matrix. > I meant: consider as an example. Not consider as a solution to your > problem. > But doesn't this need the complete Cholesky decomposition to precede > the back substitution step, and wouldn't it be much slower then a > preconditioned conjugate gradient solution? > So it may be beside the point for me to remark that with a lower > triangular matrix there is no back substitution. And a lower triangular > matrix is its own LU decomposition (Cholesky does not apply to > unsymmetric matrices). > My point was that even though you compute something that is M^{-1}b, > this does not mean that you actually form M^{-1}. The action of this > inverse can be computed using just M. Look at the algorithm. > So, with an ILU, you double the storage (or maybe not: ILU(0) only needs > the pivots stored, the rest of the elements are the same as of the > matrix), and no inverses needed. > Clear now? > V. Many thanks - yes I understand now. === Subject: Re: Are there any storage efficient preconditioners? We may have a misunderstanding here. When I said: > Take for instance a lower triangular matrix. I meant: consider as an example. Not consider as a solution to your problem. > But doesn't this need the complete Cholesky decomposition to precede > the back substitution step, and wouldn't it be much slower then a > preconditioned conjugate gradient solution? So it may be beside the point for me to remark that with a lower triangular matrix there is no back substitution. And a lower triangular matrix is its own LU decomposition (Cholesky does not apply to unsymmetric matrices). My point was that even though you compute something that is M^{-1}b, this does not mean that you actually form M^{-1}. The action of this inverse can be computed using just M. Look at the algorithm. So, with an ILU, you double the storage (or maybe not: ILU(0) only needs the pivots stored, the rest of the elements are the same as of the matrix), and no inverses needed. Clear now? V. === === Subject: Re: Are there any storage efficient preconditioners? > However, although the > preconditioner matrix M may not need much storage, the inverse of M > (which is needed in the CG algorithm) always seems to require a very > large storage. > The inverse is *not* needed. You just need to be able to solve a system > with M. That's very different. > Take for instance a lower triangular matrix. You can easily solve a > system with that (solve the first variable, then the second, ... ) which > gives you the same result as application its inverse. Yet you never > computed this inverse, in fact, you didn't do any matrix computations at > all, you immeidately generated the desired result. > V. But doesn't this need the complete Cholesky decomposition to precede the back substitution step, and wouldn't it be much slower then a preconditioned conjugate gradient solution? === Subject: Re: Are there any storage efficient preconditioners? > However, although the > preconditioner matrix M may not need much storage, the inverse of M > (which is needed in the CG algorithm) always seems to require a very > large storage. The inverse is *not* needed. You just need to be able to solve a system with M. That's very different. Take for instance a lower triangular matrix. You can easily solve a system with that (solve the first variable, then the second, ... ) which gives you the same result as application its inverse. Yet you never computed this inverse, in fact, you didn't do any matrix computations at all, you immeidately generated the desired result. V. === === Subject: Are there any storage efficient preconditioners? I'd like to ask if anyone could suggest a preconditioner in which the storage for the inverse of the preconditioner matrix M, is no larger than the original A matrix? I have a large sparse matrix that I'm solving using Conjugate Gradient and a Jacobi preconditioner. The matrix is real, definite and banded such that I need just 3 vectors for storage. I'd like to speed up the solution with a better preconditioner, and I've looked at several including Incomplete Cholesky and SSOR. However, although the preconditioner matrix M may not need much storage, the inverse of M (which is needed in the CG algorithm) always seems to require a very large storage. === Subject: Re: an inequality to prove by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id h9HCjVT28452; >t(0)=x; t(n)=1-n*t(n-1); number in parenthesis is subscript >define R=t(20) >show R(1-e^(-1))<1/21 Hi Tracy, a hint: prove by induction that, for n even, t(n) = n! * (sum (k=0,...,n) ((-1)^k / k!) - 1/e ). Best wishes Torsten. === Subject: Re: an inequality to prove >t(0)=x; t(n)=1-n*t(n-1); number in parenthesis is subscript >define R=t(20) >show R(1-e^(-1))<1/21 Not knowing what x is, it is impossible to say what R is. If you want help with your homework, you should at least reproduce it correctly. === Subject: an inequality to prove t(0)=x; t(n)=1-n*t(n-1); number in parenthesis is subscript define R=t(20) show R(1-e^(-1))<1/21 === Subject: Re: Robust Parameter Estimation Hi I would like to know what Algorithm you use to solve this problem by nonlinear least-squares. Algorithm, you only provide (u_i,y_i) and the nonlinear function. Alien+ > I am trying to implement a robust estimation algorithm instead of nonlinear > least squares minimization: > My nonlinear least-squares cost if of the form: > J(e) = sum_{i=1}^{N} [y_i - f(u_i, x)]^2, (Eq. 1) > where f= f(u,x) is the model output for an input u and x is the vector (in > R^p) of model parameters. > The variable y_i is the measured output for an input u_i at time i. > I have an algorithm to solve the above LS problem given a nonlinear function > f(u,x) and the set { (u_i,y_i), i = 1,...,N }. > I want to use the same algorithm with minimal modifications to solve the > following robust minimization problem (which I've seen in a book): > sum_{i=1}^{N} psi(y_i - f(u_i,x)) grad_{x_k}(f) = 0, (similar to normal > equations) (Eq. 2) > k = 1, ..., p, > where psi(x) is the Huber function: > psi(x) = max(-k, min(x,k)), > and grad_{x_k)(f) is the gradient vector of partial derivaties: df/dx(k), k > = 1,...,p. > My problem is that I don't know how to pose the robust problem, Eq. 2, > (solution of the above normal equations) in a similar form to the least > squares minimizations problem, Eq. 1. > I suspect it is NOT simply > J_r(e) = sum_{i=1}^{N} psi(y_i - f(u_i, x))^2. === Subject: Re: Robust Parameter Estimation > I am trying to implement a robust estimation algorithm instead of nonlinear > least squares minimization: > My nonlinear least-squares cost if of the form: > J(e) = sum_{i=1}^{N} [y_i - f(u_i, x)]^2, (Eq. 1) > where f= f(u,x) is the model output for an input u and x is the vector (in > R^p) of model parameters. > The variable y_i is the measured output for an input u_i at time i. > I have an algorithm to solve the above LS problem given a nonlinear function > f(u,x) and the set { (u_i,y_i), i = 1,...,N }. > I want to use the same algorithm with minimal modifications to solve the > following robust minimization problem (which I've seen in a book): > sum_{i=1}^{N} psi(y_i - f(u_i,x)) grad_{x_k}(f) = 0, (similar to normal > equations) (Eq. 2) > k = 1, ..., p, > where psi(x) is the Huber function: > psi(x) = max(-k, min(x,k)), > and grad_{x_k)(f) is the gradient vector of partial derivaties: df/dx(k), k > = 1,...,p. > My problem is that I don't know how to pose the robust problem, Eq. 2, > (solution of the above normal equations) in a similar form to the least > squares minimizations problem, Eq. 1. > I suspect it is NOT simply > J_r(e) = sum_{i=1}^{N} psi(y_i - f(u_i, x))^2. It is J_r(e) = sum_{i=1}^{N} phi(y_i - f(u_i, x)), where phi is the integral of psi. Setting the gradient to zero gives the above conditions. === Subject: Robust Parameter Estimation I am trying to implement a robust estimation algorithm instead of nonlinear least squares minimization: My nonlinear least-squares cost if of the form: J(e) = sum_{i=1}^{N} [y_i - f(u_i, x)]^2, (Eq. 1) where f= f(u,x) is the model output for an input u and x is the vector (in R^p) of model parameters. The variable y_i is the measured output for an input u_i at time i. I have an algorithm to solve the above LS problem given a nonlinear function f(u,x) and the set { (u_i,y_i), i = 1,...,N }. I want to use the same algorithm with minimal modifications to solve the following robust minimization problem (which I've seen in a book): sum_{i=1}^{N} psi(y_i - f(u_i,x)) grad_{x_k}(f) = 0, (similar to normal equations) (Eq. 2) k = 1, ..., p, where psi(x) is the Huber function: psi(x) = max(-k, min(x,k)), and grad_{x_k)(f) is the gradient vector of partial derivaties: df/dx(k), k = 1,...,p. My problem is that I don't know how to pose the robust problem, Eq. 2, (solution of the above normal equations) in a similar form to the least squares minimizations problem, Eq. 1. I suspect it is NOT simply J_r(e) = sum_{i=1}^{N} psi(y_i - f(u_i, x))^2. === Subject: Boole's Rule, ..., Weddle's Rule? Closed Newton-Cotes Rules: n=2 TrapezoidalRule n=3 Simpson's Rule n=4 Simpson's 3/8 Rule n=5 Boole's Rule (sometimes Milne's Rule) n=6 ??? n=7 Weddle's Rule Is there a name attached to the n=6 case? === Subject: Re: Boole's Rule, ..., Weddle's Rule? >n=7 Weddle's Rule I should say, sometimes I've seen this name used for the n=7 closed Newton-Cotes rule and sometimes, as at: http://mathworld.wolfram.com/Newton-CotesFormulas.html for a similar closed 7-point rule but one that has less unwieldy coefficients. === Subject: Re: Boole's Rule, ..., Weddle's Rule? >n=7 Weddle's Rule > I should say, sometimes I've seen this name used for the n=7 closed > Newton-Cotes rule closed Newton-Cotes is the name for the *whole family* of interpolatory quadrature rules with equidistant arguments containing the endpoints. As far as I know, the n=6 rule has no name. The reason is, I think, that it is less efficient than the n=7 rule so no one proposed to actually use it. === Subject: Jacobian elliptic function -- complex argument (u) There is a sine series solution for sn(u|m) listed in Abramowitz & Stegun, p. 575, 16.23.1. Is the series legitimate if the argument u is complex? There is the mention of the special case of complex arguments on same page 16.21.1-->16.21.4. Is that the *only* way to compute the complex case of u? Many thanks. === Subject: Re: num ana. problems > t(0)=x;t(n)=1-n*t(n-1); R(x)=t(20); (the number in parenthesis of t is subscript) > how to show that R(1-e^(-1))<=1/21. ============== [CORRECTED] You have (-1)^k *t(k)/k! - (-1)^{k-1} *t(k-1)/(k-1)! = (-1)^k/k! , k=1,2,...,20. By summing above equalities, one finds t(20)/20! - t(0) = S(20)- 1 where S(n):=SUM_{k=0 to k=n}(-1)^k/k! , that is R(x)= (20!)(x -1 +S(20)) .Therefore (1) R(1-e^{-1}) = (S(20) - e^{-1})*(20!) . Let a>0 .According to Taylor theorem with remainder, there exists c in (-a,0) such that e^{-a} =SUM_{k=0 to k=20}(-1)^k *a^k/k! - a^{21}*e^c/(21!) . For a=1 , n=20 you find e^{-1}=S(20)- e^{c_1}/(21!) , c_1 in (-1,0) . Therefore (2) e^{-1}/(21!) < S(20)-e^{-1}=e^{c_1}/(21!) < 1/(21!) . In conclusion, from (1)-(2) we have 1/e < R(1-e^{-1}) < 1 . === Subject: Re: num ana. problems > t(0)=x;t(n)=1-n*t(n-1); R(x)=t(20); (the number in parenthesis of t is subscript) > how to show that R(1-e^(-1))<=1/21. > ============== > [CORRECTED-I] > You have > (-1)^k *t(k)/k! - (-1)^{k-1} *t(k-1)/(k-1)! = (-1)^k/k! , k=1,2,...,20. > By summing above equalities, one finds > t(20)/20! - t(0) = S(20)- 1 where S(n):=SUM_{k=0 to k=n}(-1)^k/k! , > that is R(x)= (20!)(x -1 +S(20)) .Therefore > (1) R(1-e^{-1}) = (S(20) - e^{-1})*(20!) . > Let a>0 .According to Taylor theorem with remainder, > there exists c in (-a,0) such that > e^{-a} =SUM_{k=0 to k=20}(-1)^k *a^k/k! - a^{21}*e^c/(21!) . > For a=1 , n=20 you find > e^{-1}=S(20)- e^{c_1}/(21!) , c_1 in (-1,0) . > Therefore > (2) e^{-1}/(21!) < S(20)-e^{-1}=e^{c_1}/(21!) < 1/(21!) . > In conclusion, from (1)-(2) we have 1/e < R(1-e^{-1}) < 1 . =============== [CORRECTED-II] You have (-1)^k *t(k)/k! - (-1)^{k-1} *t(k-1)/(k-1)! = (-1)^k/k! , k=1,2,...,20. By summing above equalities, one finds t(20)/20! - t(0) = S(20)- 1 where S(n):=SUM_{k=0 to k=n}(-1)^k/k! , that is R(x)= (20!)(x -1 +S(20)) .Therefore (1) R(1-e^{-1}) = (S(20) - e^{-1})*(20!) . Let a>0 .According to Taylor theorem with remainder, there exists c in (-a,0) such that e^{-a} =SUM_{k=0 to k=20}(-1)^k *a^k/k! - a^{21}*e^c/(21!) . For a=1 , n=20 you find e^{-1}=S(20)- e^{c_1}/(21!) , c_1 in (-1,0) . Therefore (2) e^{-1}/(21!) < S(20)-e^{-1}=e^{c_1}/(21!) < 1/(21!) . In conclusion, from (1)-(2) we have 1/(21*e) < R(1-e^{-1}) < 1/21 . ========Alex= === Subject: Re: num ana. problems > t(0)=x;t(n)=1-n*t(n-1); R(x)=t(20); (the number in parenthesis of t is subscript) > how to show that R(1-e^(-1))<=1/21. =========== Nice problem ! You have (-1)^k *t(k)/k! - (-1)^{k-1} *t(k-1)/(k-1)! = (-1)^k , k=1,2,...,20. By summing above equalities, one finds t(20)/20! - t(0) = 0 , that is R(x)= t(20)=t(0)*20! =x*(20!) Therefore R(1-e^{-1}) = (1-e^{-1})*(20!) Let x<0 ; Try to use that there exists c in (x,0) such that e^x =SUM_{k=0 to k=20}x^k/k! + x^{21}*e^c/(21!) (=Taylor,MacLaurin formula,with remainder) . For x=-1 you find (1-e^{-1})*(20!) = SUM_{k=1 to k=20}(-1)^{k-1}20!/k! + e^c/21 with -1