I forgot to mention the important fact that the S's (not super-S) are all equal-area shapes, but are not necessarily identical. (If they were not required to be equal area, we could just take one big circle and add as many TINY circles as we want to to the edge of the big circle.) (sorry about starting a new thread) Original post below: ------------ Define circleness (a concept probably already called something else) as being one of these three definitions: For a (not necessarily?) convex closed-shape, S: alpha(S) = (area a maximum inscribed circle)/(area of S) beta(S) = (area of S)/(area a minimum circumscribed circle) gamma(S) = alpha(S)*beta(S) = (area a maximum inscribed circle)/(area a minimum circumscribed circle) By maximum inscribed circle I mean the largest circle which can be drawn so as to be completely contained within S. By minimum circumscribed circle I mean the smallest circle which can be drawn so that S can be placed completely within it. So, this is really 3 puzzles -- which one puzzle being a result of which definition of circleness we are using. Actually, this is really an infinite number of puzzles, because I am wondering about the solutions for any positive integer n. Question (I am wondering and do not personally know the answer): For a fixed n: What are the shapes, S(1), S(2),..., S(n), such that the product of the circlenesses of these S's, and this product multiplied in-turn by the circleness of super-S (see below), all by one definition of circleness, is maximized? super-S is any simply-connected and topologically circular (ie. no holes) shape formed by arranging S(1), S(2),..., S(n) so that they are all touching and not overlapping. An example: 3 squares with unit-area can be combined as: --- --- ! ! ! --- --- ! ! --- (Hopefully, this comes out okay on your browser.) Each square has alpha(S) = pi/4. And alpha(super-S) = x (do not feel like figuring it out now...) Then the product is: pi^3 *x /64, which I doubt is maximal. Leroy Quet ==== Portfolio of PAF as of 12AUG03: 50 BCE 22.24 $1,112.00 50 BLS 25.58 $1,279.00 1,050 BMY 26.19 $27,499.50 50 DT 15.34 $767.00 51,000 Q 4.05 $206,550.00 8,100 SBC 23.40 $189,540.00 2,100 SGP 16.00 $33,600.00 380 VZ 35.85 $13,623.00 80 WYE 44.29 $3,543.20 realestate land 3APR03 of 3 lots $19,000 realestate land 30JUL03 another lot $11,500 art of lithographs & porcelain JAN-JUN03 for $12,000 Well, today I gained 100 free shares of SBC plus some cash extras. Today, BMY had a 3 point Crossover compared to SBC of May this year where SBC was higher than BMY: I tried very diligently today to have that 3 point spread and I placed the sell 1000 shares BMY at 26.15 and the buy 1100 SBC at 23.10 and as the day progressed I sold BMY at 26.12 and bought the SBC at 23.40. I could not bear the thought of ending the day in cash. So, today was a Crossover day of about 3 points between SBC and BMY and I gained 100 free shares of SBC. And I expect that within the next 3 months that SBC will be more than 3 points above BMY where I make the switch over in the Crossover and gain at least 100 free new shares of BMY. Life, really cannot be any easier than this. whole entire Universe is just one big atom where dots of the electron-dot-cloud are galaxies ==== any compound rotation (initial torque about different axes) resolve to a single rotation about a single axis, just as any two do; these problems can be solved by purely imaginary quaternions, such as pitch/yaw/roll of a craft. try throwing a bottle in the air with two axes o'spin. > As to the main question, consider that any complex rotation can be > broken down into simple rotations about more than one axis. You can add > momentums together and you have one angular momentum vector for the entire > asteroid. --les ducs d'Enron! http://members.tripod.com/~american_almanac ==== >> To envision this question, imagine I have a gyroscope with dozens >> of different circles, or chambers, and I set them in motion, one by >> one, and describe the complex rotation by a number of mathematical >> function. Exactly how many times can I do this before I don't have >> to describe it by any more numbers of mathematical functions? Or is >> there a limit at all? I'm sorry, but you faded there. What does it mean for a gyroscope > to have circles? What does it mean for a gyroscope to have chambers? > Are you talking about adding different gyrscopes with different > rotating wheels? I believe he means a single rotor, mounted inside a gimbal, which is in turn mounted inside another gimbal, and so forth, all turning in different directions simultaneously. He wants to know whether there is some number of gimbals beyond which adding yet another rotating gimbal has no effect on the complexity of the net motion of the central rotor. -- Jeff, in Minneapolis . ==== >any compound rotation (initial torque >about different axes) resolve to a single rotation >about a single axis, just as any two do; >these problems can be solved by purely imaginary quaternions, >such as pitch/yaw/roll of a craft. > try throwing a bottle in the air with two axes >o'spin. This is a common mistake. What is true is that for any two times t1 and t2, the change in orientation of a rigid body from t1 to t2 can be described by a single rotation about a single axis. But it's not true that the evolution of the orientation in time is described by rotations about a fixed axis. Indeed, this won't happen for a free motion when the body is non-symmetric (all three moments of inertia distinct) and the motion is not rotation about one of the principal axes. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== This brings up a question: An irregular asteroid (Toutatis?) that was >visited by the Galileo or another spacecraft was described as having >a chaotic spin, or different spins on different axes. If angular momentum >is a simple vector quantity, how is this possible? Yes, Toutatis. See > The same angular momentum can be obtained by different motions. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Robert, thanks for the clarifications. I have one more question - and apologies, my math is extremely rusty here - I'm fairly sure I knew this stuff at one point: Angular velocity about one axis can be described as a vector, which when referenced to the center of rotation is in the direction of the axis, with magnitude equal to the angular velocity about that axis in a clockwise direction (correct so far?). In the case of an object such as Toutatis, which I understand to be described as rotating at different angular velocities about each of two different axes, I don't think it can be described by a single vector. I'm not sure though. 1. Is this correct? Are two vectors required here? 2. If so, what is the maximum number of vectors required to describe the arbitrary rotations of a three dimensional object? I'm thinking the answer is either one or three, but I don't know how to prove it. Many thanks in advance for any enlightenment Krill ==== > >This brings up a question: An irregular asteroid (Toutatis?) that was >visited by the Galileo or another spacecraft was described as having >a chaotic spin, or different spins on different axes. If angular > momentum >is a simple vector quantity, how is this possible? Yes, Toutatis. See > The same angular momentum can be obtained by different motions. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Robert, > thanks for the clarifications. I have one more question - and apologies, my math is extremely rusty here - > I'm fairly sure I knew this stuff at one point: > Angular velocity about one axis can be described as a vector, which when > referenced to the center of rotation is in the direction of the axis, with > magnitude equal to the angular velocity about that axis in a clockwise > direction (correct so far?). Yes. > In the case of an object such as Toutatis, which I understand to be > described as rotating at different angular velocities about each of two > different axes, I don't think it can be described by a single vector. I'm > not sure though. 1. Is this correct? Are two vectors required here? Exactly what are you hoping to describe? The angular velocity at any particular time is a single vector. However, this changes as time goes on. The angular momentum stays constant (in the absence of torque), but if the object is not symmetric the moments of inertia about different axes are different, so the angular velocity and angular momentum don't have to be in the same direction. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > >> To envision this question, imagine I have a gyroscope with dozens >> of different circles, or chambers, and I set them in motion, one by >> one, and describe the complex rotation by a number of mathematical >> function. Exactly how many times can I do this before I don't have >> to describe it by any more numbers of mathematical functions? Or is >> there a limit at all? I'm sorry, but you faded there. What does it mean for a gyroscope > to have circles? What does it mean for a gyroscope to have chambers? > Are you talking about adding different gyrscopes with different > rotating wheels? > > I believe he means a single rotor, mounted inside a gimbal, > which is in turn mounted inside another gimbal, and so forth, > all turning in different directions simultaneously. He wants > to know whether there is some number of gimbals beyond which > adding yet another rotating gimbal has no effect on the > complexity of the net motion of the central rotor. > > -- Jeff, in Minneapolis > > . Yes, that is exactly what I mean. THANK YOU! So, how far does this go? Does this go only up to 3, which I would assume is the limit in 3 dimensions, or does it go past that? (...Starblade Riven Darksquall...) ==== > >This brings up a question: An irregular asteroid (Toutatis?) that was >visited by the Galileo or another spacecraft was described as having >a chaotic spin, or different spins on different axes. If angular > momentum >is a simple vector quantity, how is this possible? Yes, Toutatis. See > The same angular momentum can be obtained by different motions. Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Robert, > thanks for the clarifications. I have one more question - and apologies, my math is extremely rusty here - > I'm fairly sure I knew this stuff at one point: Angular velocity about one axis can be described as a vector, which when > referenced to the center of rotation is in the direction of the axis, with > magnitude equal to the angular velocity about that axis in a clockwise > direction (correct so far?). > I really don't know... I've looked up the mathematical definition of angular momentm, and I find that sometimes it doesn't point in a direction parallel to its axis. Like if it's a spinning sphere, and I its origin and the regular momentum vector, but I chose a point on the top or bottom, it will be pointing diagonally. At least I'm assuming origin means the actual center rather than simply something having to do with the curvature, which would make a lot more sense if it was. > In the case of an object such as Toutatis, which I understand to be > described as rotating at different angular velocities about each of two > different axes, I don't think it can be described by a single vector. I'm > not sure though. > I think if you define it one way, it can... but if you define it another way it can't. I'm not so sure about this though. > 1. Is this correct? Are two vectors required here? > 2. If so, what is the maximum number of vectors required to describe the > arbitrary rotations of a three dimensional object? > That's what I'm asking! And I think angular momentum is just one vector, but that the complexity of rotations is much more than one vector... and may not be a vector at all! > I'm thinking the answer is either one or three, but I don't know how to > prove it. Many thanks in advance for any enlightenment Krill (...Starblade Riven Darksquall...) ==== must they be the three orthogonal axes? > try throwing a bottle in the air with two axes >o'spin. This is a common mistake. What is true is that for any two > times t1 and t2, the change in orientation of a rigid body from > t1 to t2 can be described by a single rotation about a single > axis. But it's not true that the evolution of the orientation > in time is described by rotations about a fixed axis. Indeed, > this won't happen for a free motion when the body is > non-symmetric (all three moments of inertia distinct) and the motion is > not rotation about one of the principal axes. --Dec.2000 'WAND' Chairman Paul O'Neill, reelected to Board. Newsish? http://www.rand.org/publications/randreview/issues/rr.12.00/ http://members.tripod.com/~american_almanac ==== >A better way of putting it is this: What is the most complex rotation >>function or class of rotation functions you can make? A (pseudo) vector. ... No limit. Its angular momentum is a single vector quantity. This brings up a question: An irregular asteroid (Toutatis?) that was > visited by the Galileo or another spacecraft was described as having > a chaotic spin, or different spins on different axes. If angular momentum > is a simple vector quantity, how is this possible? > -- > -Mike Suppose that a rigid body possesses three orthogonal angular moments of inertia such that I1 < I2 < I3. Any component of rotation about I2 is unstable. [Old Man] ==== >must they be the three orthogonal axes? What do you mean by they? Any massive body has a (not necessarily unique) set of three orthogonal principal axes, defined by eigenvectors of the moment of inertia tensor. When there is a double eigenvalue, you can choose any two orthogonal vectors in the two-dimensional eigenspace for that eigenvalue as principal axes. In the case where the eigenvalues are all distinct, a rotation about a non-principal axis can not be a free motion, i.e. it will require torque to maintain it. This follows from the Euler equations. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== > Let {a(k)} be any sequence of integers where the sum below converges. Let {b(j)} be a sequence such that: > sum{j=0 to oo} b(j) x^j /j! = exp(A_n(x)), > where A_n(x) = sum{j>=2, GCD(j-1,n)=1} a(j) x^j /j!, where this sum is over all integers j, j >= 2, where > n is coprime to (j-1), n = any fixed positive integer. > Then: b(j) is an integer sequence, and (the main result...): > n divides b(n+1). > A generalization: m also divides b(m+1), for ALL positive integers m where every prime which divides m also divides n. (obvious, but noteworthy) Leroy Quet ==== |> Let {a(k)} be any sequence of integers where the sum below converges. |> Let {b(j)} be a sequence such that: |> |> sum{j=0 to oo} b(j) x^j /j! = |> |> exp(A_n(x)), |> |> where A_n(x) = |> |> sum{j>=2, GCD(j-1,n)=1} a(j) x^j /j!, |> |> where this sum is over all integers j, j >= 2, where |> n is coprime to (j-1), n = any fixed positive integer. |> |> Then: |> |> b(j) is an integer sequence, and (the main result...): |> |> n divides b(n+1). | |A generalization: | |m also divides b(m+1), | |for ALL positive integers m where every prime which divides m also divides n. | |(obvious, but noteworthy) I like this one. I don't know how you prove it, though. It doesn't seem all that obvious to me. Perhaps I'm missing something simple. By computing a few of them, it seems that b_k is the sum over all of the ways of partitioning k distinct elements of the product over the elements* of the partition of a_i where i is the number of elements in it. [*There's got to be a more descriptive term than element. I was about to write partitions in the partition but that's worse. The partition corresponds to an equivalence relation, and I mean the individual equivalence classes. There should be a good term for them.] It looks like that's not so hard to prove by induction. The value of the k-th derivative of e^A at x=0 gives b_k. The k-th derivative itself is the sum over the equivalence relations on a k element set of e^A * (d/dx)^{i1}(A) * ... * (d/dx)^{it}(A) where i1,i2,...,it are the numbers of elements in the equivalence classes. Taking the derivative of that term gives us t+1 terms which correspond to the t+1 partitions of a k+1 element set that restrict to the given partition on the first k elements. The term we get by taking the derivative of (d/dx)^{ij}(A) corresponds to adding the k+1-st element to the j-th equivalence class. The term we get by taking the derivative of e^A (which gives A'e^A) corresponds to creating an additional partition to put the k+1-st element into. And of course the value of the ij-th derivative of A at x=0 is a_ij, while A(0)=0 and so e^A(0)=1. It seems that the result is true because the coefficient of each term is separately divisible by m. The partitions of m+1 elements can be transformed by cyclic permutation of the first m elements. This preserves the number of elements in the equivalence classes. So the number of equivalence relations having i1,i2,...,it elements in their equivalence classes is a multiple of m unless at least one of them is fixed by a cyclic permutation of the first m elements. The only ones which are fixed are ones where the equivalence class that contains the m+1-st element is fixed, since the k+1-st element is fixed. So if there are i_j elements in that equivalence class, for it to be fixed there must be a common factor between i_j-1 and m. The orbits of a nontrivial cyclic permutation of m elements have equal order, all dividing m. But since A_n(x) is a sum only over the j where (j-1,n)=1, that amounts to setting a_ij=0 when (j-1,n)>1. So for the nonzero terms, the number of associated equivalence relations is actually divisible by m. I'm curious to know in what terms you've been thinking about it. Keith Ramsay ==== in the korea high school, (high school = as the front step of university, 3 years period) All students are requested to be at the school before 7:30 every morning. and go to the home after 9:00 or 10:00 night. All students have spend great part of time studying in school. i am concerned about attending school time and ending time in your country. ==== To what country are you refering? USA? 3:00pm. Lurch > in the korea high school, (high school = as the front step of university, 3 years period) All students are requested to be at the school before 7:30 every morning. and go to the home after 9:00 or 10:00 night. All students have spend great part of time studying in school. i am concerned about attending school time and ending time in your country. ==== oh...envious USA it's time is equal to korea elementary school time. ==== > in the korea high school, (high school = as the front step of university, 3 years period) All students are requested to be at the school before 7:30 every morning. and go to the home after 9:00 or 10:00 night. All students have spend great part of time studying in school. i am concerned about attending school time and ending time in your country. And end the regular schools hours at 2:30 to 3:30 PM. Most schools have some sort of optional after-school hour events: like sports, tutoring, accelerated classes, or work placement for a few hours after that. But, I don't think many go later than 5-6 pm. But, many American students look for alternatives to University Studies also. Since the cost or American Universoties is incredible to begin with. And many American University curriculums don't even begin to satisify the job requirements that industry needs. ==== European high schools (a.k.a. secondary schools) start at 8am and end somewhere around 3pm (in Ireland and England it's more like 9am to 4pm. It's totally beyond me why someone from Korea (north or south) would have any idea that their school system is superior to anywhere else. Especially when one considers that the rest of the world, the US in particular, has to prevent the Koreans from killing each other. If Bush wins another term as president my Korean friend, you won't need to worry about high school opening hours any longer; there'll be shag all left in your country once he gets to test his new weaponry on you. davidoff404 ==== in the korea high school, (high school = as the front step of university, 3 years period) All students are requested to be at the school before 7:30 every morning. and go to the home after 9:00 or 10:00 night. All students have spend great part of time studying in school. i am concerned about attending school time and ending time in your country. In Britain 9 am to 4 pm (it varies) weekdays only. But there will be homework for the evenings and weekends. Some pupils will choose to attend extra-curricula clubs etc. Are you at school in North Korea or South Korea? Whichever, I hope that you find time to enjoy yourself as well. -- G.C. ==== Hot-girl, please ignore this nastiness! George European high schools (a.k.a. secondary schools) start at 8am and end > somewhere around 3pm (in Ireland and England it's more like 9am to 4pm. It's > totally beyond me why someone from Korea (north or south) would have any > idea that their school system is superior to anywhere else. Especially when > one considers that the rest of the world, the US in particular, has to > prevent the Koreans from killing each other. If Bush wins another term as president my Korean friend, you won't need to > worry about high school opening hours any longer; there'll be shag all left > in your country once he gets to test his new weaponry on you. davidoff404 -- G.C. ==== > So how many of you have designed your own numerals? Sixteen should > cover the most used bases (including the cancerous one based on the > first and third prime). Surely you don't use the terribly designed > Arabic ones? I have my own set brewing myself, just wanted to check > in on all of yours. Sixteen? Given up on the idea of mixed base number systems? Maybe you should re-fill your bong. ==== In sci.math, Thinkit <2db4f40b.0308110544.1fa88aaa@posting.google.com>: > So how many of you have designed your own numerals? Sixteen should > cover the most used bases (including the cancerous one based on the > first and third prime). Surely you don't use the terribly designed > Arabic ones? I have my own set brewing myself, just wanted to check > in on all of yours. Color me slightly puzzled but, apart from esoteric uses such as making up an alien script suitable for movies such as Independence Day or shows such as Star Trek and Doctor Who I fail to see the point in replacing 0-9A-F, the traditional digits for computer hexadecimal work. On the other hand, the digits are decidedly ad hoc, except perhaps for '1'. The Chinese are a little more logical, at least for the first three; 1 horizontal stroke, 2 horizontal strokes, 3 horizontal strokes. The Romans used 1-4 vertical strokes; the V might have been a tally, the X a doubletally. The ancient Greek system was largely alphabetic. I did create for my amusement a 10-digit set where the number of strokes equals the digit, except for 0; a blank just didn't make sense as a digit. :-) However, I now have no idea where I've put it, but 4 was a square, 3 was a triangle, for example. Or perhaps 3 was 3 strokes; I don't remember now. A 15-stroke last digit would probably be a little complicated but not that complicated. However, I'm not familiar with kanjii; the only other glyph I know is woman. I'm not sure how the Arabic numeral set can be construed as badly designed, although I can't say it's well-designed either. For various uses (e.g., Braille) of course it's highly inappropriate, mostly because one can't tell the difference between say 8 and 9, or 8 and 0. But that's handled by using a different font: raised dot patterns. Nor is it clear how 10 is cancerous. At best, 10 is pernicious, because of a biological accident; Nature could have given us the far more logical (numerologically speaking) base of 12, for example. But how do 10's multiply and mutate? :-) Still, it's an intriguing question. -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== > So how many of you have designed your own numerals? Sixteen should > cover the most used bases (including the cancerous one based on the > first and third prime). Surely you don't use the terribly designed > Arabic ones? I have my own set brewing myself, just wanted to check > in on all of yours. Color me slightly puzzled but, apart from esoteric uses > such as making up an alien script suitable for movies > such as Independence Day or shows such as Star Trek and > Doctor Who I fail to see the point in replacing 0-9A-F, > the traditional digits for computer hexadecimal work. Well, there is one possible reason: using the symbols A-F as digits, in programming languages or in algebra, will interfere with the use of the letters A-F as variables; this often necessitates the awkward use of prefixes 0x (etc.) or the suffix h. Now, I think the poster has gone out on a limb by claiming the Arabic numerals are terribly designed; in fact, the Arabic numerals are very convenient for writing (requiring very few strokes). But, admittedly there's doesn't really seem to be anything logical about their design; for example, 6 and 9 are rotations of each other, yet they don't seem to possess any special numerical relationship to warrant this. Also, the digits 1, 4, and 7 are peculiar in that they are drawn using only lines (the other digits involve curves), yet there doesn't seem to be any logical reason for this. There is also a nice property of base 10 that most bases do not have. Now, this may not seem significant, but let me explain it anyhow ... Most anyone who has done some computational number theory has no doubt heard of the right shift algorithm of finding the GCD of two numbers. This algorithm usually applies when the numbers are stored in binary format, but it can also be performed in decimal, or any other base, with an appropriate modification. It's best illustrated with an example: consider finding the GCD of 71 and 1891. Recall the basic fact of GCD's that you can add or subract multiples of one number to the other and this will not change their GCD (i.e., GCD(a,b) = GCD(a,b+na) ). For example, we can subtract a 71 away from 1891 to get GCD(71,1891) = GCD(71,1820) Now, notice that 1820 ends in a zero; we can chop off this zero (i.e., right shifting) and this will not change the GCD (because all we're doing is removing a factor of 10, which is valid because 71 has no common factors with 10), so: GCD(71,1820) = GCD(71,182) At this point, if we recognize that 71 is prime, we can stop here and announce the GCD is 1. Now, suppose the example had been to find GCD(31, 1787). Then we proceed by adding 3 times 31 to 1787 GCD(31,1787) = GCD(31,1880) = GCD(31,188) = GCD(31,47) = 1 As you can see, this method can be used to test if a prime divides a particular number, and it is much quicker to do it this way than to perform a division. The great calculator George Bidder used a variation of this method when factoring 5-digit integers (see p. 273 in http://users.lk.net/~stepanov/mnemo/biddere.html). The special thing about decimal is that in carrying out the right shift method, assuming we have divided out all powers of 2 and 5 from both numbers, at any point we can proceed by adding or subtracting the smaller number or 3 times the smaller number to the larger number and make the larger number end in a 0. In other bases, it is not quite so convenient; in hexadecimal, for instance, we may have to add/subtract the smaller number, or 3, 5, or 7 times that number. In dozenal, we have to add/subtract the smaller number or 5 times that number. So, we see that in this respect, decimal has a slight advantage over the other bases. Now, granted, the advantage is very slim (particularly against dozenal), but it seems worthwhile to point out a good property of base ten since it is so rare to see (I think it's fairly clear that overall dozenal is somewhat superior to decimal). - Brent ==== Let b[1] = c[1] = 1; Let, {b[k]} and {c[k]} be such that, for every integer m >= 2, b[m] = (1/m) sum{k=1 to m} c[GCD(k,m)] and c[m] = sum{k=1 to m} b[GCD(k,m)]. ('GCD' is the Greatest Common Divisor function.) What is the closed-form (ie. well-known non-recursive definition) for the two sequences? This puzzle is too close to an old math-puzzle that I posted once (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=b4be2fd f.0212211642.bf65366%40posting.google.com&rnum=3&prev= ), and the solution much too simple, for me to wait to post the solution. So it is below: | | V | | V | | V | | V | | V | | V b[m] = the number of positive divisors of m. c[m] = the sum of positive divisors of m. Leroy Quet ==== No I disagree. if f(n) = O(g(n)) i.e. 0 <= f(n) <= C g(n), where C is some constant then you can clearly see that g(n) = Omega(f(n)) and so this holds for the lower bounds of g(n) = Theta(f(n)) => C f(n) <= g(n) Although I agree this won't hold for g(n) <= C2 f(n) but don't you think it's correct saying that g(n) = Theta(f(n)) is true? Anderson > A quickie: Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n)) cheers, Anderson >> f(n), g(n), and h(n) can be any positive functions.. by that >> definition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture >> true :) >> cheers, >> -Andre > >> what I actually meant was that if what you said was the case, then >> the following would also hold: >> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants >> => f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. >> f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) >> and this would then mean that: >> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) >> and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + >> g(n) + h(n)) > There is no contradiction between these. If it's Theta, then it's also > Omega. Let's back up a bit. You said >> For three positive functions, I believe it can be graphically shown >> that the minimum of either of these functions will not (and cannot) >> have an asymptotic lower bound of these functions altogether. > To which I replied >> No. It _can be_ true that min{f(n), g(n), h(n)} = >> Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)). > So my statement was that in this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) Now, as you noticed, more than that is true in this case: min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) But that's fine: saying it's Theta(...) means that it's both O(...) > and Omega(...). Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== Hmm.. I guess you're right. But yes, it *may* hold true for: g(n) <= C f(n) as g(n) may be equal to f(n) at some value of C and n. -Andre > No I disagree. if f(n) = O(g(n)) i.e. 0 <= f(n) <= C g(n), where C is some constant then you can clearly see that g(n) = Omega(f(n)) and so this holds for > the lower bounds of > g(n) = Theta(f(n)) => C f(n) <= g(n) Although I agree this won't hold for g(n) <= C2 f(n) but don't you think it's correct saying that g(n) = Theta(f(n)) is true? > Anderson >> A quickie: >> Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n)) >> cheers, >> Anderson > f(n), g(n), and h(n) can be any positive functions.. by that > definition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture > true :) cheers, -Andre >>> what I actually meant was that if what you said was the case, then >>> the following would also hold: >>> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants >>> =>> f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e. >>> f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n) >>> and this would then mean that: >>> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n)) >>> and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + >>> g(n) + h(n)) >> There is no contradiction between these. If it's Theta, then it's also >> Omega. >> Let's back up a bit. You said >>> For three positive functions, I believe it can be graphically shown >>> that the minimum of either of these functions will not (and cannot) >>> have an asymptotic lower bound of these functions altogether. >> To which I replied >>> No. It _can be_ true that min{f(n), g(n), h(n)} = >>> Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)). >> So my statement was that in this case, where f(n) = Theta(g(n)) and >> f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n)) >> Now, as you noticed, more than that is true in this case: >> min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n)) >> But that's fine: saying it's Theta(...) means that it's both O(...) >> and Omega(...). >> Robert Israel israel@math.ubc.ca >> Department of Mathematics http://www.math.ubc.ca/~israel >> University of British Columbia Vancouver, BC, Canada V6T 1Z2 > > ==== > a^3 + 3va^2 - (v^3 + 1) = 0 is also true as an equation within the system. You *have* to consider the entire system, rather than try to pull > pieces out of it, and claim a single piece is the entire thing. the entire P(m). A point to expressing it this way is to show that the m=0 case reduces it to. a^3 -3a^2 = 0 You use two distinct arguments to conclude that a_3 is coprime to f. One is that to show that a_3 is coprime to f at m=0, and to conclude that therefore a_3 is coprime to f at m<>0. You don't use any mathematical argument to do this. By expressing the a's directly, it's clear that your conclusion is invalid. The other way is by claiming that b_1*b_2*a_3, which is an algebraic integer coprime to f, implies that a_3 is coprime to f. This is only necessarily true when b_1 and b_2 are algebraic integers. Furthermore you do this without any mathematical arguments. This is all notwithstanding the numerous claims you are being shown in other threads that none of the a's are coprime to 5. And those problems aren't dependent on on how f^2 divides off. There is no problem with your algebra in dividing the f^2 off the way you do, and I have not any any point argued with it. I agree that your m and f are independent, just most interesting when f is coprime to 3 and m. It's just your conclusions that need proving, in the face of countless alleged disproofs that are straightforward to verify, but so far not refuted. Why won't you at least admit the m dependency your position would > require on how f^2 divides off? Repeatedly you've claimed you're not asserting such a dependency, > which goes against the math. No I won't and no it doesn't. > James Harris Note that your entire Advance Polynomial Factorization paper hinges on the unproven assertion: > Given that P(m) has a factor f^2 that separates off, two of the g's > should have a factor of f which would force two of the a's to have a > factor that is f. > Therefore, that leaves one factor coprime to f This is directly relevant to my arguments. It's in your interests to fix this up properly and show the mathematics before you can expect to make headway. You won't be published with that statement in the paper. Phil Nicholson ==== > a^3 + 3va^2 - (v^3 + 1) = 0 is also true as an equation within the system. You *have* to consider the entire system, rather than try to pull > pieces out of it, and claim a single piece is the entire thing. the entire P(m). That's a false statement, which I can easily show to be false, by putting in some numbers. Using f=1, x=2, u=1, with P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) I have P(m) = 8m^3 - 24m^2 + 24m + 6 - 6m + 1 so P(m) = 8m^3 - 24m^2 + 18m + 7. But a^3 + 3va^2 - (v^3 + 1) = 0 is a^3 + 3(-1+m) -((-1+m)^3+1) = 0. Now then, would you mind explaining how you can use the second expression to get P(m)? > A point to expressing it this way is to show that the m=0 case reduces > it to. > a^3 -3a^2 = 0 You use two distinct arguments to conclude that a_3 is coprime to f. > One is that to show that a_3 is coprime to f at m=0, and to conclude > that therefore a_3 is coprime to f at m<>0. You don't use any mathematical > argument to do this. By expressing the a's directly, it's clear that your > conclusion is invalid. Why? You put in a lot into one paragraph and I think you just hoped to toss out a conclusion which *you* want readers to believe, possibly hoping that you'd not have to elaborate. I want you to explain yourself in detail. What I will do is point out what you're trying to deny, which is that f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) is an expression that represents a complex system that you can't reduce to a particular polynomial. Even a^3 + 3va^2 - (v^3 + 1) is still not a polynomial because you have variable coefficients. What I do is take P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) and use m=0 to find the constant coefficient. Now there's nothing improper or invalid about that, and of course, the constant term of the expression viewed as a polynomial with respect to m, will be given by m=0. That gives me P(0) = u^2 f^2(3x + uf). Now you can also multiply the expression out, group terms with m, and eventually come upon the same expression as your last coefficient, but you can also just use m=0. > The other way is by claiming that b_1*b_2*a_3, which is an algebraic integer > coprime to f, implies that a_3 is coprime to f. This is only necessarily true > when b_1 and b_2 are algebraic integers. Furthermore you do this > without any mathematical arguments. Here's the mathematical argument. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must be coprime to 3, if f is coprime to 3, which leaves b_3 = 3, or a unit multiple of 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 is coprime to f, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now it can be shown that w_1 w_2 is coprime to 3 without regard to m. That is, the w's can now be shown to all be constant with regard to m, so they have the same value no matter what the value of m is, so they are also constant with f coprime to 3. Introducing a_1, a_2, and a_3, that is seen by considering P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) with f=3, as then you have P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u) so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides off as a constant so it is without a dependency on m. Now looking at P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) I can again check at m=0, to see P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u), which forces w_3 to have a factor that is 3. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. > This is all notwithstanding the numerous claims you are being shown in other > threads that none of the a's are coprime to 5. I'm not interested in claims. I'm interested in proofs. > And those problems aren't dependent on on how f^2 divides off. There is no > problem with your algebra in dividing the f^2 off the way you do, and I have not > any any point argued with it. I agree that your m and f are independent, > just most interesting when f is coprime to 3 and m. > It's just your conclusions that need proving, in the face of countless alleged > disproofs that are straightforward to verify, but so far not refuted. I have refuted them. Why won't you at least admit the m dependency your position would > require on how f^2 divides off? Repeatedly you've claimed you're not asserting such a dependency, > which goes against the math. No I won't and no it doesn't. Well then, do you or do you not believe that m=0 is a special case? > Note that your entire Advance Polynomial Factorization paper hinges > on the unproven assertion: Given that P(m) has a factor f^2 that separates off, two of the g's > should have a factor of f which would force two of the a's to have a > factor that is f. > Therefore, that leaves one factor coprime to f This is directly relevant to my arguments. It's in your interests to fix this up > properly and show the mathematics before you can expect to make headway. You > won't be published with that statement in the paper. Phil Nicholson Well, as usual readers, you see a poster working to convince, and in this case it's with something tossed at you out of the blue. In my paper Advanced Polynomial Factorizaton, you have P(m) = g_1 g_2 g_3 g_1 = (a_1 x + uf), g_2 = (a_2 x + uf), g_3 =(a_3 x + uf) and I prove that one of the g's must be coprime to f, which *should* force two of the g's to each have a factor that is f, for reasons explained in detail, but the ring of algebraic integers has this weird, esoteric problem, which means that you're forced out of that ring. James Harris ==== > let p(x)=x^2 + Ax + B in R[x] > if factor ring R[x]/(p(x)) is field , > find basis of field R[x]/(p(x)) on R > ------------------------------------ > i think that basis is {1,x} > but in solution paper, > deg p(x) =2 > a exist such that p(a)=0 > thus {1,a} is basis > ------------------------------- > what do you think about it ? Well, x is an element of R[x], but not of R[x]/(p(x)). Let pi:R[x] - R[x]/(p(x)) given by g(x) -> g(x) + (p(x)) be the canonical projection map. The a they refer to is pi(x) in R[x]/(p(x)). Notice that with a = pi(x), a^2 + Aa + B is equal to pi(x^2 + Ax + B) which is 0 in the field R[x]/(p(x)). This is the sense in which they mean p(a) = 0. It seems in the solution they are representing the field R[x]/(p(x)) with a more notationally convenient representation (such as {ca + d | c and d are in R and a is such that p(a) = 0}). Note also that 1 is not the polynomial 1 in R[x] but rather pi(1). Anyway, thinking of a as pi(x), we see that it is true that {1,a} is linearly independent. Hope that helps, Hugh ==== >Let me revise that a little bit: > > >let p(x)=x^2 + ax + b in R[x] > >First we should note that there are too many a's >in the notation - let's change p to p(x)=x^2 + Ax + B. if factor ring R[x]/(p(x)) is field , find basis of field R[x]/(p(x)) on R ------------------------------------ i think that basis is {1,x} but in solution paper, deg p(x) =2 a exist such that p(a)=0 thus {1,a} is basis ------------------------------- >what do you think about it ? > >_If_ basis of field R[x]/(p(x)) on R means >>a basis for R[x]/(p(x)), regarded as a vector >>space over R then I think you're right and >>the solution paper is obviously wrong because >>{1, a} is not independent. >>There are at least three possibilities: >>(i) I'm wrong >>(ii) the solution paper is wrong >>(iii) my assumption that basis of field R[x]/(p(x)) on R >>means a basis for R[x]/(p(x)), regarded as a vector >>space over R is wrong, the phrase actually means >>something else here. >> > >It's true that {1, x} is a basis (or rather {1 +

, x +

} >is a basis, where

is the ideal generated by p.) But when I read the statement a exist such that p(a)=0 >thus {1,a} is basis I was assuming that a was an element >of R, which is obviously impossible. I suspect you didn't >tell us exactly what the solution paper says: If a is an >element of some _extension_ of R and p(a) = 0 then >{1, a} is a basis as well. > p(a) = 0, then R[x]/(p(x)) is just {0}[*], which isn't a field. So if R[x]/(p(x)) is a field, and p(a) = 0, then a is not in R. [*] Geez, what a headache. It's obviously true, I don't see a proof. I could be talking total nonsense. I guess it's time for me to leave, I hope I'm on the right track and helping. Jon no FLT proof Miller ==== >let p(x)=x^2 + ax + b in R[x] >>but in solution paper, >>deg p(x) =2 >>a exist such that p(a)=0 >>thus {1,a} is basis >> > >I think the author is using the letter a in two different >senses (as a coefficient and as a zero of the polynomial p). >This is very dodgy practice. > > I don't remember the exact quote, but it's something along the lines of of course, the two A's in theorem 37 are represent different constants. Hardy and Wright. Jon Miller ==== >Let me revise that a little bit: >> > > >let p(x)=x^2 + ax + b in R[x] >> >>First we should note that there are too many a's >>in the notation - let's change p to p(x)=x^2 + Ax + B. >> >>if factor ring R[x]/(p(x)) is field , >>find basis of field R[x]/(p(x)) on R >>------------------------------------ >>i think that basis is {1,x} >>but in solution paper, >>deg p(x) =2 >>a exist such that p(a)=0 >>thus {1,a} is basis >>------------------------------- >>what do you think about it ? >> >_If_ basis of field R[x]/(p(x)) on R means >a basis for R[x]/(p(x)), regarded as a vector >space over R then I think you're right and >the solution paper is obviously wrong because >{1, a} is not independent. There are at least three possibilities: (i) I'm wrong (ii) the solution paper is wrong (iii) my assumption that basis of field R[x]/(p(x)) on R >means a basis for R[x]/(p(x)), regarded as a vector >space over R is wrong, the phrase actually means >something else here. > >It's true that {1, x} is a basis (or rather {1 +

, x +

} >>is a basis, where

is the ideal generated by p.) >>But when I read the statement a exist such that p(a)=0 >>thus {1,a} is basis I was assuming that a was an element >>of R, which is obviously impossible. I suspect you didn't >>tell us exactly what the solution paper says: If a is an >>element of some _extension_ of R and p(a) = 0 then >>{1, a} is a basis as well. >p(a) = 0, then R[x]/(p(x)) is just {0}[*], which isn't a field. That's not so. For example (with R = {reals}, say) let p(x) = x^2 - 4, a = 2. Then p(a) = 0 although R[x]/p(x) is certainly not 0. > So if >R[x]/(p(x)) is a field, and p(a) = 0, then a is not in R. _That's_ true (except for the word so) - hence the words obviously impossible above! (At least when R is a PID, in particular when R is a field) R[x]/p(x) is a field if and only if p is irreducible; in the example above p has degree 2, so if it has a root in R it's not irreducible. >[*] Geez, what a headache. It's obviously true, I don't see a proof. I >could be talking total nonsense. I guess it's time for me to leave, I >hope I'm on the right track and helping. Jon no FLT proof Miller ************************ David C. Ullrich ==== I am a computer science major and i was wondering to myself if there is anything that can not be represented in binary format...I mean since any physical object can be represented as a set of numbers representing different attributes and since any number can be represented by a binary sequance then that to me suggest there is nothing that can not be represented in binary format. if the above is true and anything can be represented then is there a mathematical proof for it? thanks in advace Guy ==== > if the above is true and anything can be represented then is there a > mathematical proof for it? sqrt(2) or pi etc can only be approximated, same with any measurement that comes from real life. Is that what you are looking for? TS ==== I renig on my comment. I still do not see a difference, the same end result comes from the optimized way - just faster. In your example, my method would be the same question (-not differnet-) - only, I get the same end result - only faster with less work. So I still ask: Why bother multiplying by 3 and adding 1 when you can just add one and keep the problem simple and quick? Mike Curry Would it not be faster to add +1 to an odd number rather then multiply by > three and add 1? If I add +1 to an odd number instead, I get faster > results... If you sere sitting an exam and were asked to multiply 148263 by 237893, > what would you expect to happen if instead of answering the question > as set, you argued that it would be faster to multiply 100000 by 200000 > and did that instead? :-) -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > The League of Gentlemen ==== I renig on my comment. I still do not see a difference, the same end result > comes from the optimized way - just faster. In your example, my method would be the same question (-not differnet-) - > only, I get the same end result - only faster with less work. So I still ask: > Why bother multiplying by 3 and adding 1 when you can just add one and keep > the problem simple and quick? > **************************************************************************** ********** What problem are you referring to? The Collatz conjecture has a definite statement. If you change this statement, you are no longer talking about the same problem. _________________________________________________________ Eric J. Wingler (wingler@math.ysu.edu) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817 ==== I renig on my comment. I still do not see a difference, the same end result > comes from the optimized way - just faster. How do you *know* it's the same? Your method always produces 1, but the 3x+1 method has never been proved to produce 1 every time. So, until you prove that, you are merely making an unsupported assertion that the answer is the same. You've made exactly zero progress toward the desired goal. In your example, my method would be the same question (-not differnet-) - > only, I get the same end result - only faster with less work. You've missed the whole point. Yes, your method is easy, but it tells us nothing we didn't already know about the natural numbers. That makes it uninteresting. Sorry. So I still ask: > Why bother multiplying by 3 and adding 1 when you can just add one and keep > the problem simple and quick? Why bother? Because it's an interesting question. If we can figure out how to answer it, that might tell us something important (perhaps even useful) about numbers. ==== >I renig on my comment. I still do not see a difference, the same end result >comes from the optimized way - just faster. You seem to be assuming that the end result is always 1. That is not known - the _problem_ is to _prove_ that it always leads to 1. If you can prove that the original procedure _always_ leads to a 1 that will make you famous. >In your example, my method would be the same question (-not differnet-) - >only, I get the same end result - only faster with less work. Look: _assuming_ that both processes both lead to a 1 eventually, here's an even _faster_ way to get the same end result: GIven an integer n, let the next integer in the sequence be 1. >So I still ask: >Why bother multiplying by 3 and adding 1 when you can just add one and keep >the problem simple and quick? You're still totally missing the point to what the problem _is_. >Mike Curry > Would it not be faster to add +1 to an odd number rather then multiply >by >> three and add 1? If I add +1 to an odd number instead, I get faster >> results... >> If you sere sitting an exam and were asked to multiply 148263 by 237893, >> what would you expect to happen if instead of answering the question >> as set, you argued that it would be faster to multiply 100000 by 200000 >> and did that instead? :-) >> -- >> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >> The League of Gentlemen > ************************ David C. Ullrich ==== Ok, I am catching on now... (doh!). If the Collatz Conjecture Optimization is ever proven, it would be interesting to see if this other way could be proven as well: t(x)=x/2 if x is even, t(x)=(x+1)/2 if x is odd Mike Curry I renig on my comment. I still do not see a difference, the same end result >comes from the optimized way - just faster. You seem to be assuming that the end result is always 1. That is not > known - the _problem_ is to _prove_ that it always leads to 1. If you > can prove that the original procedure _always_ leads to a 1 that > will make you famous. In your example, my method would be the same question (-not differnet-) - >only, I get the same end result - only faster with less work. Look: _assuming_ that both processes both lead to a 1 eventually, > here's an even _faster_ way to get the same end result: GIven an integer n, let the next integer in the sequence be 1. So I still ask: >Why bother multiplying by 3 and adding 1 when you can just add one and keep >the problem simple and quick? You're still totally missing the point to what the problem _is_. Mike Curry > Would it not be faster to add +1 to an odd number rather then multiply >by >> three and add 1? If I add +1 to an odd number instead, I get faster >> results... >> If you sere sitting an exam and were asked to multiply 148263 by 237893, >> what would you expect to happen if instead of answering the question >> as set, you argued that it would be faster to multiply 100000 by 200000 >> and did that instead? :-) >> -- >> Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html >> The League of Gentlemen > ************************ David C. Ullrich ==== Ok, I am catching on now... (doh!). If the Collatz Conjecture Optimization Why do you say optimization? This problem is not about optimization at all. The point is not to try to make it easier, the point is simply to show that it is true, as stated. > is ever proven, it would be interesting to see if this other way could be > proven as well: t(x)=x/2 if x is even, > t(x)=(x+1)/2 if x is odd But that is *easy* to prove (and has been proved). That's why I said it is uninteresting as a matter of research for mathematicians. Your t(x) gives a result that is always *less than* x, for every natural number greater than 1. And never zero or negative. It follows that any sequence of such numbers will decrease steadily until it reaches 1. The difficult thing about the Collatz conjecture is that 3x+1 is *greater* than x. Furthermore, the multiplier is 3, which is greater than the divisor for even numbers (2). So, at first glance, it looks like the sequence should blow up, in general. Yet it does not, for any number that has been checked. In other words, nobody knows if the conjecture is true or false. That makes it interesting. ==== > I renig on my comment. I still do not see a difference, the same end result > comes from the optimized way - just faster. In your example, my method would be the same question (-not differnet-) - > only, I get the same end result - only faster with less work. So I still ask: > Why bother multiplying by 3 and adding 1 when you can just add one and keep > the problem simple and quick? As others have pointed out, this is silly. By changing the rules, you're changing the problem. But there is a legitimate optimization you can make. You can remove ALL the factors of 2 in one step instead of one at a time. So instead of the usual sequence for 7: 7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1 you only see the odd numbers: 7 11 17 13 5 1 This is useful if you're only interested in how many branches are on the tree without fundamentally changing the problem. 22_7 34_11 52_17 26 40_13 20 10 16_5 8 4 2 1 becomes _7 _11 _17 _13 _5 1 The even numbers are still accounted for, I'm just not showing them. 7 is still 5 braches away from 1. And as far as this being an optimization, that depends on how you're doing the math. I use this when I represent binary numbers as text strings. Since the number of factors of two is equal to the number of contiguous least significant 0s, my perl program drops all the 0s in one fell swoop: 101 : 5 10000 : 16 (drop all four 0s in one step) 1 : 1 But keep in mind that you can't go around claiming that the stopping time for 7 is 5, because the usual meaning of stopping time includes the iterations of x/2. Although some like to use the rules x/2 and (3x+1)/2. Why I don't know. But it means their stopping time statistics are not apllicable to the proper Collatz Problem. For example, someone spoke of locating all the numbers of stopping time 7. He has a nice little chart for figuring this out that works real nice, but it's worthless to me because the sequence (3x+1)/2 x/2 x/2 x/2 x/2 x/2 x/2 would be stopping time 8 under the proper rules. and it's not just that the count is different, when he says ALL numbers, he's including the sequences (3x+1)/2 (3x+1)/2 x/2 x/2 x/2 x/2 x/2 (3x+1)/2 (3x+1)/2 (3x+1)/2 x/2 x/2 x/2 x/2 which would have stopping times of 9 and 10 under the proper rules. So he's comparing apples to oranges. In my example of removing all factors of 2 in one step, I don't speak about stopping time for that reason. I use the term ORDER, where ORDER is the count of odd numbers in the sequence (or the number of branches away from 1). I say that 7 is ORDER 5. With this definition, it makes no difference how I remove the factors of 2. Whether I remove them all at once or one at a time, the ORDER doesn't change. That's why it is a legitimate optimization. Your proposal changes the order of 7 from 5 to 1 and is, therefore, worthless. Mike Curry Would it not be faster to add +1 to an odd number rather then multiply > by > three and add 1? If I add +1 to an odd number instead, I get faster > results... If you sere sitting an exam and were asked to multiply 148263 by 237893, > what would you expect to happen if instead of answering the question > as set, you argued that it would be faster to multiply 100000 by 200000 > and did that instead? :-) -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html > The League of Gentlemen ==== , > I renig on my comment. I still do not see a difference, the same end result > comes from the optimized way - just faster. In your example, my method would be the same question (-not differnet-) - > only, I get the same end result - only faster with less work. So I still ask: > Why bother multiplying by 3 and adding 1 when you can just add one and keep > the problem simple and quick? Because that makes it an interesting and unsolved problem. ==== , > Ok, I am catching on now... (doh!). If the Collatz Conjecture Optimization > is ever proven, it would be interesting to see if this other way could be > proven as well: t(x)=x/2 if x is even, > t(x)=(x+1)/2 if x is odd Shouldn't take anyone reading sci.math more than five seconds. ==== Define circleness (a concept probably already called something else) as being one of these three definitions: For a (not necessarily?) convex closed-shape, S: alpha(S) = (area a maximum inscribed circle)/(area of S) beta(S) = (area of S)/(area a minimum circumscribed circle) gamma(S) = alpha(S)*beta(S) = (area a maximum inscribed circle)/(area a minimum circumscribed circle) By maximum inscribed circle I mean the largest circle which can be drawn so as to be completely contained within S. By minimum circumscribed circle I mean the smallest circle which can be drawn so that S can be placed completely within it. So, this is really 3 puzzles -- which one puzzle being a result of which definition of circleness we are using. Actually, this is really an infinite number of puzzles, because I am wondering about the solutions for any positive integer n. Question (I am wondering and do not personally know the answer): For a fixed n: What are the shapes, S(1), S(2),..., S(n), such that the product of the circlenesses of these S's, and this product multiplied in-turn by the circleness of super-S (see below), all by one definition of circleness, is maximized? super-S is any simply-connected and topologically circular (ie. no holes) shape formed by arranging S(1), S(2),..., S(n) so that they are all touching and not overlapping. An example: 3 squares with unit-area can be combined as: --- --- ! ! ! --- --- ! ! --- (Hopefully, this comes out okay on your browser.) Each square has alpha(S) = pi/4. And alpha(super-S) = x (do not feel like figuring it out now...) Then the product is: pi^3 *x /64, which I doubt is maximal. Leroy Quet Abuse-Reports-To: durward at hillsboro.net to report improper postings ==== > Define circleness A circle is a polygon that has been stripped of it's epaulets. ==== In sci.math, dustbird : > Define circleness A circle is a polygon that has been stripped of it's epaulets. > A polygon is a circle with a few rough edges. :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. >>And that's false. I am pretty sure that Dale produced explicit common >>factors; but in any case, your claim here is certainly false, since >>their product is not coprime to 65. > I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit > factors in common with 5 in the ring of algebraic integers. > And he proves (below) that they do. At this point you have to do more than point at your own proof. You must find the error in his. >Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >>any elements of R. If a and b are coprime to c, then a*b is coprime to >>c. >>Proof. We use the characterization of coprime valid for commutative >>rings with 1: a and b are coprime in R if and only if there exist x >>and y in R such that ax+by = 1. > By that definition only *one* of the a's is coprime to 5, but none of > them has a factor in common with 5 either. Do you even understand what he's saying? The application would be: If a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, which means a_1*a_2*a_3 is coprime to 5. The ring of algebraic integers is really screwed up. No, your understanding of them is. For those who don't understand, consider that in the ring of evens, > which does not have 1, you can't use that definition of coprime that > Arturo Magidin gives, though it is, interestingly enough, true that in > fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is > not in the ring. Why do you make unnecessary side trips to bizarre rings? However, rather than use dueling definitions or argue about > definitions I can simply switch to saying that 2 does not share > non-unit factors in the ring of evens with 6. >>Since a and c are coprime by assumption, there exist n and m in R such >>that an+cm = 1. Since b and c are coprime by assumption, there exist r >>and s in R such that br+cs = 1. >>Multiplying both together, we have >>1 = (an+cm)(br+cs) >> = abrn + acns + cbmr + c^2*ms >> = ab(rn) + c(ans + bmr + cms). >>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and >>ab*x + c*y = 1. Therefore, ab and y are coprime. QED >>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY >>non-unit factors in common with 5 in the ring of algebraic >>integers. Then, by the lemma, neither does a1*a_2; and applying the >>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which >>clearly has 5 as a nonunit common factor with 5. This contradicts the >>assumption that none of a_1, a_2, a_3 have common non-unit factors >>with 5 in the ring of algebraic integers. Therefore, your assertion is >>false. > Well by your definition of coprime NONE of the a's have a factor in > common with 5, in the ring of algebraic integers, and you cannot prove > that any of them do. No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. This shows that there is a mistake in your work since you are claiming the product of three algebraic integers coprime to 5 *is* 65. Now if you don't want to call that coprime fine. It doesn't change > the situation. What I can do is show that with a very quick argument using basic > algebra as I've done. You keep posting the same text as if it hasn't been riddled with holes. How about dealing with people's counter-arguments directly instead of posting the same thing over and over? [usual proof deleted] I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where -1 and 1 are the > only members that are both a unit, i.e. factor of 1, and an integer, > where no non-unit member is a factor of any two integers that are > coprime. How does the Ring of Objects differ from the algebraic integers? Are you sure your definition is consistent? In which ring are you referring to the two integers being coprime? -- Will Twentyman ==== >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. >>And that's false. I am pretty sure that Dale produced explicit common >>factors; but in any case, your claim here is certainly false, since >>their product is not coprime to 65. > I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit > factors in common with 5 in the ring of algebraic integers. > And he proves (below) that they do. At this point you have to do more > than point at your own proof. You must find the error in his. He uses a special definition of coprime, and assumes that at least two of the three numbers is coprime. However, by his definition of coprime, only one of the numbers is coprime. >Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >>any elements of R. If a and b are coprime to c, then a*b is coprime to >>c. >>Proof. We use the characterization of coprime valid for commutative >>rings with 1: a and b are coprime in R if and only if there exist x >>and y in R such that ax+by = 1. > By that definition only *one* of the a's is coprime to 5, but none of > them has a factor in common with 5 either. Do you even understand what he's saying? The application would be: If > a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, > a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, > which means a_1*a_2*a_3 is coprime to 5. By the definition he gave only ONE of the a's is coprime to 5. The others don't have non unit factors in common with 5, in the ring of algebraic integers, but each have a factor that is sqrt(5) in a higher ring. The ring of algebraic integers is really screwed up. No, your understanding of them is. Nope. The ring of algebraic integers is really screwed up. For those who don't understand, consider that in the ring of evens, > which does not have 1, you can't use that definition of coprime that > Arturo Magidin gives, though it is, interestingly enough, true that in > fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is > not in the ring. Why do you make unnecessary side trips to bizarre rings? Arturo Magidin uses a special definition of coprime which depends on certain assumptions which fail with the ring of algebraic integers, and I highlight for readers how that definition can also fail in another ring. However, rather than use dueling definitions or argue about > definitions I can simply switch to saying that 2 does not share > non-unit factors in the ring of evens with 6. >>Since a and c are coprime by assumption, there exist n and m in R such >>that an+cm = 1. Since b and c are coprime by assumption, there exist r >>and s in R such that br+cs = 1. >>Multiplying both together, we have >>1 = (an+cm)(br+cs) >> = abrn + acns + cbmr + c^2*ms >> = ab(rn) + c(ans + bmr + cms). >>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and >>ab*x + c*y = 1. Therefore, ab and y are coprime. QED >>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY >>non-unit factors in common with 5 in the ring of algebraic >>integers. Then, by the lemma, neither does a1*a_2; and applying the >>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which >>clearly has 5 as a nonunit common factor with 5. This contradicts the >>assumption that none of a_1, a_2, a_3 have common non-unit factors >>with 5 in the ring of algebraic integers. Therefore, your assertion is >>false. > Well by your definition of coprime NONE of the a's have a factor in > common with 5, in the ring of algebraic integers, and you cannot prove > that any of them do. No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then > their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. > This shows that there is a mistake in your work since you are claiming > the product of three algebraic integers coprime to 5 *is* 65. Only one of them is coprime to 5, and the other two should have some non unit factor in common with 5 but don't because the ring of algebraic integers is flawed. Now if you don't want to call that coprime fine. It doesn't change > the situation. What I can do is show that with a very quick argument using basic > algebra as I've done. You keep posting the same text as if it hasn't been riddled with holes. > How about dealing with people's counter-arguments directly instead of > posting the same thing over and over? [usual proof deleted] Deleting out a mathematical argument does not make it go away. I've found the Ring of Objects which includes the ring of algebraic > integers, and does not have this problem, as the b's are all included > in it. The Ring of Objects is the set of all numbers where -1 and 1 are the > only members that are both a unit, i.e. factor of 1, and an integer, > where no non-unit member is a factor of any two integers that are > coprime. How does the Ring of Objects differ from the algebraic integers? Are you sure your definition is consistent? In which ring are you referring to the two integers being coprime? The Ring of Objects is a higher ring than the ring of algebraic integers, which doesn't have its problems. The two integers are coprime in the Ring of Objects. The definition removes the possibility of the fundamental contradiction which distinguishes a field from a ring, which is that in a field coprimeness does not exist, while in the ring of integers, it does. So fields are at odds with the ring of integers. I've simply highlighted the crucial difference between them. James Harris ==== This post is classic Harris, marred only by the incoherent mumblings of others. I remove those mumblings to reveal James' undiluted efflux: ####### > He uses a special definition of coprime, and assumes that at least two > of the three numbers is coprime. However, by his definition of coprime, only one of the numbers is > coprime. ####### > By the definition he gave only ONE of the a's is coprime to 5. The others don't have non unit factors in common with 5, in the ring > of algebraic integers, but each have a factor that is sqrt(5) in a > higher ring. ####### > Nope. The ring of algebraic integers is really screwed up. ####### > Arturo Magidin uses a special definition of coprime which depends on > certain assumptions which fail with the ring of algebraic integers, > and I highlight for readers how that definition can also fail in > another ring. ####### > Only one of them is coprime to 5, and the other two should have some > non unit factor in common with 5 but don't because the ring of > algebraic integers is flawed. ####### > Deleting out a mathematical argument does not make it go away. ####### > The Ring of Objects is a higher ring than the ring of algebraic > integers, which doesn't have its problems. The two integers are coprime in the Ring of Objects. The definition removes the possibility of the fundamental > contradiction which distinguishes a field from a ring, which is that > in a field coprimeness does not exist, while in the ring of integers, > it does. So fields are at odds with the ring of integers. I've simply highlighted the crucial difference between them. > > James Harris ####### ==== > >>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >>factors in common with 5 in the ring of algebraic integers. And that's false. I am pretty sure that Dale produced explicit common >factors; but in any case, your claim here is certainly false, since >their product is not coprime to 65. >> I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit >> factors in common with 5 in the ring of algebraic integers. >> And he proves (below) that they do. At this point you have to do more >> than point at your own proof. You must find the error in his. He uses a special definition of coprime, and assumes that at least two >of the three numbers is coprime. No, I use the standard definition of coprime: two elements a and b are coprime in R if and only if the ideals (a) and (b) are coprime, if and only if there is no prime ideal of R which contains both (a) and (b). It is a THEOREM that in a commutative ring with 1, two elements a and b are coprime if and only if there exist x and y in R such that ax+by=1. What is the definition YOU are using? The definition you are using is a and b are coprime if and only if they have no non-unit common factors. In the ring of all algebraic integers, the two are EQUIVALENT. However, your definition is WEAKER than the correct meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime under your incorrect usage, but they are not coprime under correct usage. >However, by his definition of coprime, only one of the numbers is >coprime. Coprime to ->what<-? And prove it. [.snip.] >Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >any elements of R. If a and b are coprime to c, then a*b is coprime to >c. Proof. We use the characterization of coprime valid for commutative >rings with 1: a and b are coprime in R if and only if there exist x >and y in R such that ax+by = 1. >> By that definition only *one* of the a's is coprime to 5, but none of >> them has a factor in common with 5 either. >> Do you even understand what he's saying? The application would be: If >> a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, >> a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, >> which means a_1*a_2*a_3 is coprime to 5. By the definition he gave only ONE of the a's is coprime to 5. Prove it. >The others don't have non unit factors in common with 5, in the ring >of algebraic integers, but each have a factor that is sqrt(5) in a >higher ring. Which doesn't prove anything. If the others don't have non-unit factors in common with 5 in the ring of all algebraic integers, then they would be coprime under the correct definition as well. [.snip.] >> For those who don't understand, consider that in the ring of evens, >> which does not have 1, you can't use that definition of coprime that >> Arturo Magidin gives, though it is, interestingly enough, true that in >> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is >> not in the ring. >> Why do you make unnecessary side trips to bizarre rings? Arturo Magidin uses a special definition of coprime which depends on >certain assumptions which fail with the ring of algebraic integers, I use the STANDARD definition; it is you who uses a SPECIAL one. Please list, explicitly: 1. The special definition; 2. The one you use; 3. The certain assumptions on which my supposedly special definition depends. 4. A proof that they fail in the ring of algebraic integers. It is not enough to say the ring is really screwed up, because that is supposed to be your conclusion, not your hypothesis. [.snip.] ====================================================================== 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 ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >>>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >>>factors in common with 5 in the ring of algebraic integers. >>And that's false. I am pretty sure that Dale produced explicit common >>factors; but in any case, your claim here is certainly false, since >>their product is not coprime to 65. I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit >factors in common with 5 in the ring of algebraic integers. >>And he proves (below) that they do. At this point you have to do more >>than point at your own proof. You must find the error in his. He uses a special definition of coprime, and assumes that at least two > of the three numbers is coprime. What definition of coprime would you suggest that he use? What definition of coprime are you using? > However, by his definition of coprime, only one of the numbers is > coprime. I'll address this below. >>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >>any elements of R. If a and b are coprime to c, then a*b is coprime to >>c. >>Proof. We use the characterization of coprime valid for commutative >>rings with 1: a and b are coprime in R if and only if there exist x >>and y in R such that ax+by = 1. >By that definition only *one* of the a's is coprime to 5, but none of >them has a factor in common with 5 either. >>Do you even understand what he's saying? The application would be: If >>a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, >>a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, >>which means a_1*a_2*a_3 is coprime to 5. By the definition he gave only ONE of the a's is coprime to 5. The others don't have non unit factors in common with 5, in the ring > of algebraic integers, but each have a factor that is sqrt(5) in a > higher ring. Which ring is that? If you mean your Ring of Objects, please explain why sqrt(5) is a factor there and not in the algebraic integers. >For those who don't understand, consider that in the ring of evens, >which does not have 1, you can't use that definition of coprime that >Arturo Magidin gives, though it is, interestingly enough, true that in >fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is >not in the ring. >>Why do you make unnecessary side trips to bizarre rings? Arturo Magidin uses a special definition of coprime which depends on > certain assumptions which fail with the ring of algebraic integers, > and I highlight for readers how that definition can also fail in > another ring. What are the assumptions that fail with the ring of algebraic integers? What definition *should* he be using if this one is not appropriate? The question becomes: what do you mean by coprime, and why does your definition differ from his in the algebraic integers? >However, rather than use dueling definitions or argue about >definitions I can simply switch to saying that 2 does not share >non-unit factors in the ring of evens with 6. >Since a and c are coprime by assumption, there exist n and m in R such >>that an+cm = 1. Since b and c are coprime by assumption, there exist r >>and s in R such that br+cs = 1. >>Multiplying both together, we have >>1 = (an+cm)(br+cs) >> = abrn + acns + cbmr + c^2*ms >> = ab(rn) + c(ans + bmr + cms). >>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and >>ab*x + c*y = 1. Therefore, ab and y are coprime. QED >>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY >>non-unit factors in common with 5 in the ring of algebraic >>integers. Then, by the lemma, neither does a1*a_2; and applying the >>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which >>clearly has 5 as a nonunit common factor with 5. This contradicts the >>assumption that none of a_1, a_2, a_3 have common non-unit factors >>with 5 in the ring of algebraic integers. Therefore, your assertion is >>false. >Well by your definition of coprime NONE of the a's have a factor in >common with 5, in the ring of algebraic integers, and you cannot prove >that any of them do. >>No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then >>their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. >>This shows that there is a mistake in your work since you are claiming >>the product of three algebraic integers coprime to 5 *is* 65. > Only one of them is coprime to 5, and the other two should have some > non unit factor in common with 5 but don't because the ring of > algebraic integers is flawed. James, should be don't is the same as don't. What you are saying could be rephrased as They are all coprime to 5, but I want 2 of them not to be. If I'm not understanding what you are saying here, please explain it. Note: I'm assuming that no non-unit factor in common with 5 is equivalent to coprime to 5 in the algebraic integers. If you disagree with that, please state why. >Now if you don't want to call that coprime fine. It doesn't change >the situation. What I can do is show that with a very quick argument using basic >algebra as I've done. >>You keep posting the same text as if it hasn't been riddled with holes. >> How about dealing with people's counter-arguments directly instead of >>posting the same thing over and over? >>[usual proof deleted] Deleting out a mathematical argument does not make it go away. No, but you haven't said anything new in a while, and it has nothing to do with what I was talking about. >I've found the Ring of Objects which includes the ring of algebraic >integers, and does not have this problem, as the b's are all included >in it. The Ring of Objects is the set of all numbers where -1 and 1 are the >only members that are both a unit, i.e. factor of 1, and an integer, >where no non-unit member is a factor of any two integers that are >coprime. >>How does the Ring of Objects differ from the algebraic integers? >>Are you sure your definition is consistent? >>In which ring are you referring to the two integers being coprime? > The Ring of Objects is a higher ring than the ring of algebraic > integers, which doesn't have its problems. Aside from the algebraic integers not doing what you think they should do, what problems do they have? The two integers are coprime in the Ring of Objects. Ok, so depending on what is in the Ring of Objects, 2 and 3 may or may not be coprime... how would I check this? > The definition removes the possibility of the fundamental > contradiction which distinguishes a field from a ring, which is that > in a field coprimeness does not exist, while in the ring of integers, > it does. Coprimeness exists in a ring, it's just not very interesting. > So fields are at odds with the ring of integers. Considering the ring of integers is *not* a field, this should not be surprising. > I've simply highlighted the crucial difference between them. You mean the whole point of this months of discussion has been about the fact that the ring of integers is not a field? Am I missing something? > James Harris -- Will Twentyman ==== >> >>The problem is that neither a_1, a_2, nor a_3 have ANY non-unit >>factors in common with 5 in the ring of algebraic integers. >>>>And that's false. I am pretty sure that Dale produced explicit common >>>factors; but in any case, your claim here is certainly false, since >>>their product is not coprime to 65. >>I've proven it true that neither a_1, a_2 nor a_3 have ANY non-unit >>factors in common with 5 in the ring of algebraic integers. And he proves (below) that they do. At this point you have to do more >than point at your own proof. You must find the error in his. >> He uses a special definition of coprime, and assumes that at least two >> of the three numbers is coprime. What definition of coprime would you suggest that he use? What >definition of coprime are you using? I believe that James incorrectly believes that a is coprime to b means a has no nonunit common factors with b. That definition is equivalent to the standard one in many rings (e.g., in the ring of integers, in the ring of all algebraic integers, in any PID), but not in others. The interesting point, of course, is that the definition James is using, if it were not equivalent to the usual one, would be useless for his purposes. James needs the coprime property in order to remove factors in a congruence, but that step is not valid if you only have have no nonunit common factors. You need coprime in the usual sense, with its equivalent statement that there are x and y such that ax+by=1. (If the two are equivalent in the ring in question, of course, there is no problem; but they are not always equivalent). For instance, in Z[sqrt(-5)], the two statements are not equivalent. 2 and 1+sqrt(-5) have no common factors other than units, but they are not coprime; we have that (1+sqrt(-5))*(1-sqrt(-5)) = 0 (mod 2) in Z[sqrt(-5)], but we cannot conclude from the fact that 2 and 1+sqrt(-5) have no factors in common that 1-sqrt(-5)=0 (mod 2); it clearly does not (all multiples of 2 are of the form 2a+2b*sqrt(-5), a and b integers). Rather, the usual argument is that if a*b=0 (mod c) and a and c are coprime, then we know that c|ab. Since there exist elements x,y such that ax+cy = 1, multiplying by b we have abx + cby = b, and since c divides abx and cby, it divides b; thus b=0 (mod c) is a valid conclusion. If you want another example that James likes, as he points out (his conclusion is correct, although his reasoning is incorrect), 2 and 6 are coprime in 2Z: the ideal (2) consists of all integers divisible by 4, and is maximal; (6) consists of all integers divisible by 12; so (6) is contained in (2). The only ideal that contains both (2) and (6) is (2); and (2) is not a prime ideal: 2*2 lies in (2), but 2 does not. Thus, there is no prime ideal of 2Z containing both (2) and (6), so 2 and 6 are coprime (they are not, however, comaximal; the two conditions are equivalent in a ring with 1, assuming the Axiom of Choice). However, even though 2 and 6 are coprime, we cannot cancel in congruence in this ring. 12=0 (mod 2), since 12=2*6. However, if we write 12=2*6=0 (mod 2) (in 2Z) and argue that 6 is coprime to 2, and therefore we can cancel from the congruence, we would get that 2=0 (mod 2) (in 2Z), and that statement is false: there is no element k in 2Z such that 2=0+2k. So, in the end, the definition which James is using is not the one he should be using for his purposes, unless he can prove that in the ring he is working on it is equivalent to the usual one. Never mind that there is no ring of objects as he has defined it. [.snip.] >Which ring is that? If you mean your Ring of Objects, please explain >why sqrt(5) is a factor there and not in the algebraic integers. There is no Ring of Objects according to his definition. Even allowing for the possibility that number means complex number or algebraic number, there is no unique ring determined by the conditions he gives, yet he assumes there is one and only one such ring. [.snip.] ====================================================================== 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 ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== of the three numbers is coprime. No, I use the standard definition of coprime: two elements a and b are > coprime in R if and only if the ideals (a) and (b) are coprime, if and > only if there is no prime ideal of R which contains both (a) and (b). It is a THEOREM that in a commutative ring with 1, two elements a and > b are coprime if and only if there exist x and y in R such that > ax+by=1. That's fascinating. Why don't you give the theorem, as it must be in error for the ring of algebraic integers. And yes, I would like to see that theorem as I'd be fascinated to see some evidence of how deep the weird algebraic integer ring problem has gone in terms of what mathematicians believe to be true. Still my guess is that it depends on assertions about units, where you base your belief, once again, on assuming that a number that is not a unit in the ring of algebraic integers must be a non-unit factor in the ring of algebraic integers. The problem is fascinating indeed, and easily leads to confusion, if you aren't careful to be *very* careful with your definitions and assumptions. That is, you need to follow the math, not what you *think* is the truth. > What is the definition YOU are using? I've given the following definition for coprime in the proper ring, which is the ring of objects: Two objects are coprime if they don't share a non-unit factor e.g. 2 is coprime to 3, while 2 and 4 share 2 as a non-unit factor. > The definition you are using is a and b are coprime if and only if > they have no non-unit common factors. Yeah, that's basically it. > In the ring of all algebraic integers, the two are > EQUIVALENT. However, your definition is WEAKER than the correct > meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime > under your incorrect usage, but they are not coprime under correct usage. They are coprime in the ring Z[sqrt(-5)]. It's strange to me that you'd miss something so obvious, as you could also consider something like the ring of evens, where 2 is coprime to 6, in the ring of evens. Apparently, you wish to use the knowledge that in a *higher* ring 1+sqrt(-5) and 2 share sqrt(2) as a factor, for the lesser ring. But just as 2 is a factor of 6 in the ring of integers, but is NOT in the ring of evens, it simply is the case that in the ring Z[sqrt(-5)] 1+sqrt(-5) and 2 are coprime, by the logical definition. What appears to have happened is that mathematicians considered various rings, and tried to find some general way to handle them, rather than simply finding the object ring. It'd be like someone fiddling around with the ring of evens, rather than just integers. Sometimes they put in 3, other times they put in 7 with integers, when they really just need to put in 1, and be done. >However, by his definition of coprime, only one of the numbers is >coprime. Coprime to ->what<-? And prove it. One of the a's is coprime to f, and in particular 5, with f=sqrt(5), m=1. The proof follows. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) proving that w_1 w_2 must be coprime to 3, if f is coprime to 3, which leaves b_3 = 3, or a unit multiple of 3. Essentially objections to how f^2 divides off now come down to claiming that the w's are functions of m, but consider that w_1 w_2 = 1, when m=0, if f is coprime to 3. But that was an arbitrary choice, so let f=3. Now it can be shown that w_1 w_2 is coprime to 3 without regard to m. That is, the w's can now be shown to all be constant with regard to m, so they have the same value no matter what the value of m is, so they are also constant with f coprime to 3. Introducing a_1, a_2, and a_3, that is seen by considering P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) with f=3, as then you have P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u) so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides off as a constant so it is without a dependency on m. Now looking at P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) I can again check at m=0, to see P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u), which forces w_3 = 3, or a unit multiple of 3. Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, f=sqrt(2), but that's a special case as generally they are not, which shows a problem with the ring of algebraic integers. > [.snip.] Lemma. Let R be the ring of all algebraic integers, and let a, b, c be >any elements of R. If a and b are coprime to c, then a*b is coprime to >c. Proof. We use the characterization of coprime valid for commutative >rings with 1: a and b are coprime in R if and only if there exist x >and y in R such that ax+by = 1. >> By that definition only *one* of the a's is coprime to 5, but none of >> them has a factor in common with 5 either. >> Do you even understand what he's saying? The application would be: If >> a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, >> a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, >> which means a_1*a_2*a_3 is coprime to 5. By the definition he gave only ONE of the a's is coprime to 5. Prove it. I've given the proof above. One of the a's is coprime by that definition, while the others do not have non-unit factors in common with 5 in the ring. And in fact your example with Z[sqrt(-5)] outlines how, as similarly, in that ring 1+sqrt(-5) is, by a *logical* definition*, coprime to 2. However, by the definition you've been using, it would be that 1+sqrt(-5) and 2 do not share non-unit factors in the ring Z[sqrt(-5)]. >The others don't have non unit factors in common with 5, in the ring >of algebraic integers, but each have a factor that is sqrt(5) in a >higher ring. Which doesn't prove anything. If the others don't have non-unit > factors in common with 5 in the ring of all algebraic integers, then > they would be coprime under the correct definition as well. Actually, if the ring weren't flawed they *would* have non-unit factors in common with 5. > [.snip.] > For those who don't understand, consider that in the ring of evens, >> which does not have 1, you can't use that definition of coprime that >> Arturo Magidin gives, though it is, interestingly enough, true that in >> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is >> not in the ring. >> Why do you make unnecessary side trips to bizarre rings? Arturo Magidin uses a special definition of coprime which depends on >certain assumptions which fail with the ring of algebraic integers, I use the STANDARD definition; it is you who uses a SPECIAL > one. Please list, explicitly: 1. The special definition; I quote from above. It is a THEOREM that in a commutative ring with 1, two elements a and b are coprime if and only if there exist x and y in R such that ax+by=1. > 2. The one you use; I've given the following definition for coprime in the proper ring, which is the ring of objects: Two objects are coprime if they don't share a non-unit factor e.g. 2 is coprime to 3, while 2 and 4 share 2 as a non-unit factor. > 3. The certain assumptions on which my supposedly special definition > depends. My guess is that you assume that all numbers that *should* be factors in the ring of algebraic integers are units, when in fact the ring is flawed in that certain numbers which *should* be units aren't in the ring. These numbers are multiplicative inverses of other numbers that *are* in the ring of algebraic integers, but are not units, though they are coprime to all non-unit algebraic integers. The ring of algebraic integers is flawed in a fascinating way. > 4. A proof that they fail in the ring of algebraic integers. Well I just explained it, but I'll repeat that the problem is that certain numbers which *should* be units in the ring, are not, and their multiplicative inverses aren't even in the ring, though they should be. > It is not enough to say the ring is really screwed up, because that > is supposed to be your conclusion, not your hypothesis. I've proven it repeatedly. Rather than work to get to the truth, I've seen posters work to convince, presumably to protect their *beliefs* about mathematics as a discipline. Their social preoccupation has been of interest to me. My fear is that many mathematicians actually are participants in what can be considered to be a dogmatic relgion, where rather than challenging what they're taught, math students learn to accept from authority. However, science is about challenging what is known to make certain that it fits with reality, not with giving your trust to particular human beings. Trust, but verify, as Ronald Reagan said, in one of the best quotes ever. James Harris ==== >He uses a special definition of coprime, and assumes that at least two >>of the three numbers is coprime. >> No, I use the standard definition of coprime: two elements a and b are >> coprime in R if and only if the ideals (a) and (b) are coprime, if and >> only if there is no prime ideal of R which contains both (a) and (b). >> It is a THEOREM that in a commutative ring with 1, two elements a and >> b are coprime if and only if there exist x and y in R such that >> ax+by=1. That's fascinating. Why don't you give the theorem, as it must be in >error for the ring of algebraic integers. And yes, I would like to see that theorem as I'd be fascinated to see >some evidence of how deep the weird algebraic integer ring problem has >gone in terms of what mathematicians believe to be true. Still my guess is that it depends on assertions about units, where you >base your belief, once again, on assuming that a number that is not a >unit in the ring of algebraic integers must be a non-unit factor in >the ring of algebraic integers. You are speaking nonsense. A non unit factor of ->what<-? If you again mean, a non-unit factor of itself, then there is no belief, there is only mathematical truth: by definition, x is a factor of y in the commutative ring R if and only there exists an element z of R such that x*z=y. If a is an algebraic integer which is not a unit in the ring of all algebraic integers, then since 1 is an algebraic integer and a*1=a, it follows, necessarily, that a is a non-unit factor of itself in the ring of all algebraic integers. I'll include the definitions, since I doubt you know them. DEF. Let R be a ring. A subset I of R is an ideal of R if and only if: (i) I is nonempty; (ii) If x and y are in I, then so is x-y. (iii) If x is in I and r is any element of R, then rx is in I and xr is in I. Note that if the ring is commutative, it is enough to check that rx is in I. Lemma 0. If I is an ideal of R, then 0 is in I; and if a,b are in I, then a+b is in I. Proof. I is nonempty, so there is an a in I. Since 0 is in R, 0*a=0 lies in I by (iii). So 0 is in I. If a,b are in I, then -b=0-b lies in I by (ii). Since a,-b lie in I, then a-(-b)=a+b lies in I by (ii). QED. DEF. Let R be a commutative ring, and let a be an element of R. The principal ideal generated by a is the collection { ra : r in R}. It is denoted by (a). Lemma 1. Let R be a commutative ring. Then (a) is an ideal of R. Proof. Since R is a ring, it contains a 0. Therefore, 0*a=0 is in (a), so (a) is nonempty. If ra and sa are in (a), then ra-sa=(r-s)a, which by definition is in I. Finally, if ra is in (a), and s is any element of R, then s*(ra) = (sr)a, which lies in (a). QED DEF. Let R be a ring. An ideal I of R is said to be a prime ideal if and only if it is not equal to R, and for any two elements x and y of R, if x*y is in I, then either x is in I or y is in I. DEF. Let R be a commutative ring. Two ideals I and J of R are said to be coprime (in R) if and only if there is no prime ideal P of R which contains both I and J. DEF. Let R be a commutative ring. Two elements a and b of R are said to be coprime (in R) if and only if the principal ideals (a) and (b) are coprime in R. DEF. Let R be a commutative ring. An ideal I of R is said to be a maximal ideal if and only if I is not equal to R, and there is no ideal J of R which properly contains I and is properly contained in R. DEF. Let R be a commutative ring. Two ideals I and J of R are said to be comaximal if and only if there does not exist a maximal ideal of R which contains both I and J. Lemma 2. Let I be an ideal of R, and let a be an element of R not in I. Then the set { x + ra : x in I, r in R} is an ideal of R, and moreover, it is the smallest ideal of R which contains I and a. Proof. Let S denote the set {x+ra : x in I, r in R}. First, since I is an ideal, it is nonempty. Let x in I. Then x+0*a is in S, so S is nonempty. Assume that b,c are in S. Then b=x+ra for some x in I, r in R; and c=y+sa for some y in I, s in R. Then b-c = (x+ra) - (y+sa) = (x-y) + (r-s)a. Since I is an ideal, x-y is in I; and r-s is in R. So this is an element of S; thus, if b and c are in S, then so is b-c. Finally, let y be any element of S and let r in R. Then we may write y=x+sa with x in I and s in R. Then r*y = r*(x+sa) = (r*x) + r*sa = (rx) + (rs)*a. Since I is an ideal, rx is in I; so this is an element of S. Therefore, S is an ideal. Now assume that J is any ideal which contains I and contains a. We prove that it contains S. For let x in I be arbitrary and let r in R. We want to show that x+ra is in J. Since J is an ideal, and a is in J, it follows that ra is in J. Since J is an ideal, and ra is in J, it follows that 0=0*ra is in J. Since 0 and ra are in J, it follows that -ra = 0-ra is in J. Since x is in J and -ra is in J, it follows that x-(-ra)=x+ra is in J. This proves that J contains S, and proves the Lemma. QED. THEOREM 3. Let R be a commutative ring with 1. If I is a maximal ideal of R, and a is any element of R not in I, then there exists an element x in I and an element r in R such that x+ra=1. Proof. Let S={x+ra : x in I, r in R}. This set is an ideal, and contains I. It also contains a, since a=0+1a. Therefore, the set properly contains I. Since I is maximal, the only ideal which properly contains I is R itself, so S=R. Since 1 is in R, we conclude that 1 is in S, and therefore, 1 = x+ra for some x in I and r in R, as claimed. QED. THEOREM 4. Let R be a commutative ring with 1. If I is a maximal ideal of R, then it is also a prime ideal of R. Proof. Assume not. Then there exist elements a,b of R, none of them in I, such that ab lies in I. Since a is not in I, by Theorem 3 there exist x in I and r in R such that 1 = x+ra. Multiplying by b we have b = b*1 = b(x+ra) = bx + bra = (bx) + r(ba). Now, x is in I, hence so is bx; likewise, ba is in I, hence so in r*ba. Therefore, bx+r(ba) lies in I. But that means that b is in I, contradicting the choice of a and b. Therefore, I must be a prime ideal. QED THEOREM 5. (Zorn's Lemma) If C is a nonempty partially ordered set with the property that any chain is bounded above, then C has maximal elements. Proof omitted. Known to be equivalent to the Axiom of Choice. Lemma 6. Let R be a commutative ring with 1, and let I be an ideal. If I contains 1, then I=R. Proof. We know I is contained in R. To prove the reverse inclusion, let x in R. We need to show it is in I. Since 1 in I and x in R, x*1=x must lie in I, so x is in I. QED. THEOREM 7. Let R be a ring with 1. If I is any ideal of R and is not equal to R, then there exists a maximal ideal M of R which contains I. Proof. Let S= { J ideal of R: J is not equal to R and J contains I}. Order S by inclusion. Then I is in S, so S is nonempty. Let C be a chain in S, that is, C is contained in S and C={J_k}, with the property that for any indices m,n, either J_m is contained in J_n or J_n is contained in J_m. If C is empty, I is an upper bound for C. Assume C is nonempty. Let J be the set theoretic union of C. Then J contains I, since each element of J contains I. So it is nonempty. If a,b in J, then a in J_k and b in J_n for some k and n. Either J_k is contained in J_n (and so J_n contains both a and b) or J_n contained in J_k (so J_k contains both a and b). So assume without loss of generality that J_n contains both a and b. Since J_n is an element of S, it is an ideal, so a-b is in J_k, and so a-b is in J. Finally, if a in J, and r in R, then a in J_k for some k, and since this is an ideal, ra is in J_k, so ra is in J. Therefore, J is an ideal of R which contains S. J contains all elements of C, so it is a bound for C. We just need to show that it lies in S; the only way in which it would not lie in S is if it violated the condition J is not equal to R. But that would mean that J=R, so 1 is an element of J. Since J is the set-theoretic union of the J_k, so 1 must be in some J_k. But that would imply, by Lemma 6, that J_k=R; this contradicts the fact that J_k is an element of S. The contradiction arose from assuming that R=J, so J is not equal to R, so J lies in S. In conclusion, S satisfies the hypothesis of Zorn's Lemma, and has maximal elements. Let M be a maximal element of S. That means that: (1) M is an ideal of R; (2) M is not equal to R; and (3) M contains I [all this by virtue of being in S]. Finally, it also means that there does not exist any ideal J of R, different from R, which contains I, contains M, and is not equal to M [by virtue of being maximal in S]. We need only show that M is maximal to prove the theorem. But if M is not maximal, then there exists an ideal K of R, different from R, which properly contains M. But then it would necessarily contain I, and so be an element of S larger than M. This contradicts the maximality of M in S, and so we conclude that M is a maximal ideal that contains I. QED THEOREM 8. Let R be a commutative ring with 1. Two ideals I and J are coprime if and only if they are comaximal. Proof. If I and J are coprime, and M is a maximal ideal, then M cannot contain both I and J, for then M would also be a prime ideal containing both I and J, which is impossible. So I and J are comaximal. Conversely, suppose that I and J are comaximal. If P is a prime ideal containing both I and J, then by Theorem 7 there exists a maximal ideal M containing P, and therefore containing I and J, contradicting comaximality. So I and J are coprime. QED THEOREM 9. Let R be a commutative ring with 1, and let a and b be two elements of R. Then a and b are coprime in R if and only if there exist x and y in R such that ax+by=1. Proof. First, assume that there exists x and y in R such that ax+by=1. Note that since R has a 1, a lies in (a) (being equal to 1*a), and b lies in (b). So if P is any ideal containing both (a) and (b), then it contains both a and b as elements. Since a is in P, then ax must be in P; since b is in P, then b*y is in P. Since ax and by are both in P, we must have that ax+by is in P. But 1=ax+by, so 1 is in P. Therefore, P=R. In particular, there is no ideal other than R which contains both (a) and (b); therefore, there is no maximal ideal which contains both (a) and (b); therefore, by definition, (a) and (b) are comaximal. By Theorem 8, they are also coprime. Therefore, a and b are coprime in R. Conversely, assume that a and b are coprime in R. That means that (a) and (b) are coprime, hence comaximal by Theorem 8. Let I = {ra + sb : r,s in R}. We claim that this is an ideal of R which contains both (a) and (b). Indeed, it is an ideal, by the same argument as Lemma 2: it is equal to the set of all x+sb with x in (a) and s in R. It contains (a) because if ra is in (a), then ra = ra+0*b lies in I; and it contains (b) because if sb is in (b), then sb = 0*a+sb which is in I. Therefore, I is an ideal which contains both (a) and (b). If I is not equal to R, then by Theorem 7 there exists a maximal ideal M of R which contains I; therefore, M contains (a) and (b). This contradicts the comaximality of (a) and (b). The contradiction arises from assuming that I is not equal to R, so we must conclude that I=R. Therefore, 1 is an element of R, so 1 is in {ra+sb : r,s in R}. Therefore, there exist x and y in R such that 1 = xa+yb, as claimed. QED Done. >The problem is fascinating indeed, and easily leads to confusion, if >you aren't careful to be *very* careful with your definitions and >assumptions. Which you are not. >That is, you need to follow the math, not what you *think* is the >truth. > What is the definition YOU are using? I've given the following definition for coprime in the proper ring, >which is the ring of objects: There is no ring of objects. Your definition does not have a referent, as I have pointed out many times. This definition is in general weaker than the usual one, but is equivalent to the usual one in the ring of all algebraic integers. DEF. Let R be a commutative ring, a and b elements of R. We say that a is a factor of b (in R) if and only if there exists c in R such that ac=b. DEF. Let R be a commutative ring with 1, and let u in R. Then u is a unit in R if and only if there exists v in R such that u*v=1. DEF. Let R be a commutative ring, a,b,c elements of R. We say that c is a common factor of a and b (in R) if and only if c is a factor of a in R and c is also a factor of b in R. THEOREM. Let R be a ring commutative ring with 1, and let a and b in R. If a and b are coprime in R, then any common factor of a and b in R is a unit. Proof. Since a and b are coprime in R, there exist x,y in R such that ax+by=1. Let c in R be a common factor of a and b. Then there exists d in R such that cd=a; and there exists e in R such that ce=b. Therefore, cdx=ax, and cey=by. Therefore, 1 = ax+by = cdx + cey = c(dx+ey). Since R is a ring, dx and ey are in R, and so is dx+ey. Therefore, dx+ey is an element of R which multiplied by c yields 1. Therefore, c is a unit in R. QED EXAMPLE. A commutative ring with 1 R in which two elements have no common factor other than units, but are not coprime. Proof. Let R = Z[sqrt(-5)], a = 2, b= 1+sqrt(-5). First, we show that a and b are not coprime. For assume that we had x,y in R such that ax+by = 1. Write x = r+s*sqrt(-5), y = t+u*sqrt(-5) with r,s,t,u integers. Then 1 = 2x+(1+sqrt(-5))y = 2r + 2s*sqrt(-5) + t + u*sqrt(-5) -5u +t*sqrt(-5) = (2r + t - 5u) + (2s+u+t)*sqrt(-5). Therefore, 2r+t-5u = 1 2s+u+t = 0 Substituting t=-2s-u, we have 1 = 2r -2s -u -5u = 2r - 2s - 6u. But since r,s,u are integers, that means that 2r-2s-6u is an even integer, and therefore cannot equal 1. Therefore, a and b are not coprime in R. To prove that they have no common factors, consider the function N:Z[sqrt(-5)]->Z given by N(a+b*sqrt(-5)) = a^2 + 5b^2, where a and b are integers. Then it is easy to verify by direct calculation that for all x,y in Z[sqrt(-5)] N(xy) = N(x)N(y). Assume that x is a factor of y in Z[sqrt(-5)]. Then there exists z in Z[sqrt(-5)] such that xz=y. Therefore, N(y) = N(xz) = N(x)N(z). Therefore, the integer N(y) is a product of the integers N(x) and N(z), so N(x) is a factor of N(y) IN THE INTEGERS. That is: if x is a factor of y in Z[sqrt(-5)], then N(x) is a factor of N(y) in Z. So, assume that a+b*sqrt(-5) is a common factor of 2 and 1+sqrt(-5). Then N(a+bsqrt(-5)) = a^2 + 5b^2 must be a factor, IN THE INTEGERS, of N(2) = 4 and of N(1+sqrt(-5)) = 1+1*5 = 6. The only possibilities are that N(a+b*sqrt(-5)) = 2 or 1 (note N(x) is always nonzero). But 2 = a^2+5b^2 has no solution with a and b integers, so we must have N(a+b*sqrt(-5)) = 1. The only solutions to 1 = N(a+b*sqrt(-5)) = a^2 + 5b^2 are b=0, a=1; or b=0, a=-1. So either a+b*sqrt(-5)=1, hence is a unit; or a+b*sqrt(-5)=-1, hence is a unit. This proves that (1) 2 and 1+sqrt(-5) are not coprime in Z[sqrt(-5)]; but (2) 2 and 1+sqrt(-5) have no non unit common factors in Z[sqrt(-5)]. This proves that in general the two notions are not equivalent; since coprime implies no common factors other than units, but the converse does not hold, no common factors other than units is a weaker proposition than coprime, in general. The only thing that would remain is to prove that the two notions are equivalent in the ring of all algebraic integers. This is a deep theorem due to Dedekind and Kronecker, and uses the finiteness of the class number. It has been sketched before, but here are the broad strokes: First, let R be the ring of all algebraic integers. For any a and b in R, there exists c in R such that (a,b) = {xa + yb : x, y in R} = (c) To prove this, let A be the ring of integers of Q(a,b). Then there exists a positive integer k such that (a,b)^k = (d) for some d in A (finiteness of the class number). Let A' be the ring of integers of Q(a,b,d^{1/k}), where d^{1/k} is any of the complex k-th roots of d. In A', we have (a,b)^k = (d) = (d^{1/k})^k, so by unique factorization of ideals in rings of integers, we have equal, so letting c=d^{1/k} gives c. Let a and b be any two algebraic integers, and assume that a and b have no common non-unit factors in the ring of all algebraic integers. Let c be an algebraic integer such that (a,b)=(c). Then c is a common factor of a and b: for a is in (a,b), and so a is in (c)={rc: r in R}; that means, there exists r in R such that rc=a. Likewise, b is in (a,b), hence in (c), and so there exists s in R such that cs=b. So c is a common factor of both a and b. Since a and b have no common factors other than units, that means that c is a unit in the ring of algebraic integers. That means that there exists v in R such that vc=1. But c is in the ideal (c), and v is in R, so vc=1 is in (c). Since (c) contains 1, it follows that (c)=R. Therefore, (a,b) = {xa+yb : x,y in R} = (c) = R, so 1 is in (a,b). Therefore, there exist x and y in R such that ax+by =1. Therefore, a and b are coprime in R, as claimed. [.rest deleted.] >> In the ring of all algebraic integers, the two are >> EQUIVALENT. However, your definition is WEAKER than the correct >> meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime >> under your incorrect usage, but they are not coprime under correct usage. They are coprime in the ring Z[sqrt(-5)]. Not under the correct usage. >It's strange to me that you'd miss something so obvious, as you could >also consider something like the ring of evens, where 2 is coprime to >6, in the ring of evens. They are coprime under the usual definition, and under your bastardization thereof. However, in the ring Z[sqrt(-5)], they are coprime under your bastardization, but not under the correct usage. [.snip.] >>However, by his definition of coprime, only one of the numbers is >>coprime. >> Coprime to ->what<-? >> And prove it. One of the a's is coprime to f, and in particular 5, with f=sqrt(5), >m=1. The proof follows. Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). > You have no u in your assertions before; what is u? >Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) You have not proven that b1, b2, b3, u, w1, w2, w3 are algebraic integers. >where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m), and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the b's must equal 0, which means WHEN m = 0. > P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3) which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf) WHEN m = 0. >proving that w_1 w_2 must be coprime to 3 YOu have not proven that w1*w2 is an algebraic integer, so this statement is unsupported right now. >if f is coprime to 3, which >leaves >b_3 = 3, or a unit multiple of 3. WHEN m = 0. You have, however, not proven that, under my definition, they are coprime. >Essentially objections to how f^2 divides off now come down to >claiming that the w's are functions of m, but consider that w_1 w_2 = >1, when m=0, if f is coprime to 3. This is nonsense and is based on your misunderstanding of obejctions. >But that was an arbitrary choice, so let f=3. WHEN m = 0. >Now it can be shown that w_1 w_2 is coprime to 3 without regard to m. Please do so, using the correct definition of coprime, since you claim that under my definition you will do so. Be sure to prove that w_1*w_2 is an algebraic integer while you do so. >That is, the w's can now be shown to all be constant with regard to m, Please do so. >so they have the same value no matter what the value of m is, so they >are also constant >with f coprime to 3. Please provide a complete proof. See above for what a real proof looks like. >Introducing a_1, a_2, and a_3, that is seen by considering P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) You have not proven that a1, a2, a3 are algebraic integers. >with f=3, as then you have P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u) so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides >off as a constant so it is without a dependency on m. when f=3. You have not shown that a_1/3, a_2/3, a_3/3 are algebraic integers. >Now looking at P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) I can again check at m=0, to see P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u), which forces w_3 = 3, or a unit multiple of 3. When m=0. >Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf) where you'll notice that the b's are algebraic integers with m=1, I don't notice that; I assume, by the way, that you screwed up above > a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 you meant > b_1 = a_1/3, b_2 = a_2/3, b_3 = a_3/3. >f=sqrt(2), but that's a special case as generally they are not, which >shows a problem with the ring of algebraic integers. YOu have analized a number of special cases, and said a lot of things, most of which are not supported. In addition, you CLAIMED that you would prove that, under my definition (i.e., the standard definition), the ai would be shown to be coprime to 5 for the polynomial in question, 65x^3 - 12x + 1. I do not see the values of r_i and s_i such that r_i*a_i + s_i*5 = 1, which is the definition you claimed you were going to use. So all you did was waste space and my time. [.snip.] >>By the definition he gave only ONE of the a's is coprime to 5. >> Prove it. I've given the proof above. One of the a's is coprime by that >definition, I do not see the elements r and s that make r*a+s5 = 1. That's what the definition says, after all. PROVE IT. >while the others do not have non-unit factors in common >with 5 in the ring. You have not proven anything. >And in fact your example with Z[sqrt(-5)] outlines how, as similarly, >in that ring 1+sqrt(-5) is, by a *logical* definition*, coprime to 2. The STANDARD DEFINITION gives that 1+sqrt(-5) and 2 are NOT coprime in Z[sqrt(-5)]. I've proven it. >However, by the definition you've been using, it would be that >1+sqrt(-5) and 2 do not share non-unit factors in the ring >Z[sqrt(-5)]. That is true regardless of whether you use my definition of coprime (i.e., the correct one), or your definition of coprime (i.e., the bastardization). The point is that this property does not imply coprimality, using the correct definition. >>The others don't have non unit factors in common with 5, in the ring >>of algebraic integers, but each have a factor that is sqrt(5) in a >>higher ring. >> Which doesn't prove anything. If the others don't have non-unit >> factors in common with 5 in the ring of all algebraic integers, then >> they would be coprime under the correct definition as well. Actually, if the ring weren't flawed they *would* have non-unit >factors in common with 5. You are assuming your conclusion. That's called a ciruclar argument. >>Arturo Magidin uses a special definition of coprime which depends on >>certain assumptions which fail with the ring of algebraic integers, >> I use the STANDARD definition; it is you who uses a SPECIAL >> one. >> Please list, explicitly: >> 1. The special definition; I quote from above. It is a THEOREM that in a commutative ring with 1, two elements a and > b are coprime if and only if there exist x and y in R such that > ax+by=1. You see the big capitalized word THEOREM? You did not give the special definition I supposedly use. >> 2. The one you use; I've given the following definition for coprime in the proper ring, >which is the ring of objects: There is no ring of objects. Your definition is incoherent as written, and has no referent when fixed. >Two objects are coprime if they don't share a non-unit factor e.g. 2 >is coprime to 3, while 2 and 4 share 2 as a non-unit factor. This is not the correct definition of coprime. > 3. The certain assumptions on which my supposedly special definition >> depends. My guess You ->guess<-? But you made a very definitive ASSERTION above. Are you trying to say that when you asserted that I was depending on some assertions, you did not KNOW that to be the case, you were merely ->guessing<-? Is that how you work? [rhetorical question: yes, it is how you work] >is that you assume that all numbers that *should* be factors >in the ring of algebraic integers are units, when in fact the ring is >flawed in that certain numbers which *should* be units aren't in the >ring. All this should really means numbers which I WOULD REALLY, REALLY, REALLY like to have the properties I want so my proof wouldn't be nonsense. [.snip.] >These numbers are multiplicative inverses of other numbers that *are* >in the ring of algebraic integers, but are not units, though they are >coprime to all non-unit algebraic integers. If the multiplicative inverses are NOT algebraic integers, then you cannot talk about them being coprime to algebraic integers in the ring of algebraic integers. If they are multiplicative inverses of something in your ring, then they are units, and therefore coprime to verbiage. >The ring of algebraic integers is flawed in a fascinating way. So you claim. >> 4. A proof that they fail in the ring of algebraic integers. Well I just explained it, but I'll repeat that the problem is that >certain numbers which *should* be units in the ring, are not, and >their multiplicative inverses aren't even in the ring, though they >should be. Translation: numbers that you ->fervently wish<- were there are not, and that's a problem. For you. >> It is not enough to say the ring is really screwed up, because that >> is supposed to be your conclusion, not your hypothesis. I've proven it repeatedly. You've given a lot of verbiage, but you have proven nothing. [.snip.] ====================================================================== 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 ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > ...I'd be fascinated to see... and > ...The problem is fascinating indeed, ... and > ...is flawed in a fascinating way... James. Enough with the fascinating. Please drop the Yup., Um., odd, amazing, etc. as well. If your vocabulary needs work, try remedial English. [snip] > Well I just explained it, but I'll repeat that the problem is that > certain numbers which *should* be units in the ring, are not, and > their multiplicative inverses aren't even in the ring, though they > should be. Fine. Name one. Give us a number which *should* be a unit in the ring of algebraic integers, but which is not. Just one. Pretty please! For God's sake man, name just ONE numeric quantity which supports your claim! -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com ==== I'm working on an optimization problem where my objective function is the sum of two functions: a strictly convex function and a strictly quasiconvex function. Both functions are continuous and differentiable. In general, the sum of quasiconvex functions is not necessarily quasiconvex, but are there any conditions where the sum turns out to be quasiconvex. Particularly in the case of adding a convex function and quasiconvex function, under what circumstances could I claim the sum to be quasiconvex? Any pointers/references in this regard would be most appreciated. Karthik ==== > I'm working on an optimization problem where my objective function is > the sum of two functions: a strictly convex function and a strictly > quasiconvex function. Both functions are continuous and > differentiable. In general, the sum of quasiconvex functions is not necessarily > quasiconvex, but are there any conditions where the sum turns out to > be quasiconvex. Particularly in the case of adding a convex function > and quasiconvex function, under what circumstances could I claim the > sum to be quasiconvex? Please define quasiconvex and strictly quasiconvex. ==== > ...start with a dodecagon (12-gon) [...] make a path that moves > from vertex to vertex [...] visiting each vertex exactly one time. > And the path returns to the starting-point. But in this puzzle, > consecutive vertexes MAY be connected by a segment. So, I give a list of nonnegative integers below. As the path is drawn > (as opposed to after the path is completed), the n_th segment crosses > a(n) previously drawn segments, where a(n) is the n_th term of the > integer-list. The path starts at 12. And the first segment goes from 12 to 8. > The list {a(n)}: 0, 0, 1, 0, 2, 2, 0, 3, 2, 2, 5, 2 > ... > 17 of the 10! paths that start off {12, 8} match that crossing- > count sequence, so solutions seem fairly rare - about 1 per 213000 > paths - but on the other hand, there are perhaps O((n-3)!) possible > crossing-count sequences, so 17 could instead be an unusually high > prevalence, and this case no more rare than thousands of others. > I plan to look at this more and find out, next weekend. > -jiw > ... > I wonder which sequences (if any), for the 12-gon, produce one > solution, but are interesting. > If you, or anyone, happens to find such a sequence, then it would be > interesting to post it to sci.math and rec.puzzles as a challenge for > us all. > ... For the 12-gon there are 893597 different sequences that begin 12,8, > and only 315027 of them belong to unique paths; I don't know if any > of those paths are interesting. :) For example, the > sequential-crossing-counts list 0 0 0 1 1 1 1 0 6 7 8 2 belongs to > a unique path. On the other hand, the list 0 0 0 0 1 2 2 3 4 4 5 4 > belongs to 164 different paths, and no list belongs to more than that. paths account for 80% of all sequences. Sequences belonging to 17 > paths (like your original sequence, 0 0 1 0 2 2 0 3 2 2 5 2) are > relatively rare, around 1 in 300. I think my estimate of O((n-3)!) possible crossing-count sequences > was high and that O((n-4)!) is much closer. If I have a chance, > I'll modify my program to look at all paths starting at 12, rather > than only those that start at 12 and proceed to 8. That may have > a somewhat less arbitrary complexity. Here is a slightly-revised extract from http://pat7.com/jp/counts-r512 : > Occ.Count #Occ.Count Extension An example count sequence > 1 315027 315027 0 0 0 1 1 1 1 0 6 7 8 2 > 2 179162 358324 0 0 0 1 1 1 1 0 6 0 7 7 > 3 102670 308010 0 0 0 1 1 1 1 0 0 0 1 7 > 4 71484 285936 0 0 0 1 1 1 0 0 0 1 1 1 > The first column is how many paths a pattern belongs to. > The second column is how many patterns belong to that number of > paths. The third column is the product of the first two, or > sum of products when the first column is a range. The last > group of columns shows an example of a count sequence that > belongs to the number of paths given in column 1. > (In the file, there is also an Eg# column that shows program > code numbers of the example count sequences. Also, the file > shows similar data for 5, 6, ... 11-gons as well as 12-gons.) > -jiw Interesting. So a puzzle (which *I* could NEVER solve, most probably) for rec.puzzles and sci.math readers is to find the solution for the path found by James' computer, the *unique* path which follow the rules of the original puzzle and has the crossing-count sequence: 0 0 0 1 1 1 1 0 6 7 8 2 ;) (I have no idea personally if this path is at all interesting from a puzzle view-point..) Leroy Quet ==== I'm trying to figure out this problem, but I can't find any information (in my book or on the internet) that's related. I'm asked to solve the following system using the elimination method dx/dt = 2x - y dy/dt = -x + 2y I appreciate any help thanks ..Boris.. ==== L1+L2 d(x+y)/dt=x+y L1-L2 d(x-y)/dt=3(x-y) you get x+y and x-y terminated > I'm trying to figure out this problem, but I can't find any > information (in my book or on the internet) that's related. I'm asked to solve the following system using the elimination method dx/dt = 2x - y > dy/dt = -x + 2y I appreciate any help thanks ..Boris.. ==== > L1+L2 d(x+y)/dt=x+y L1-L2 d(x-y)/dt=3(x-y) you get x+y and x-y terminated That's a nice way to do it. It may not be what the OP meant by the elimination method. The OP may have meant, write the first equation in the form y = stuff involving only x and t, then differentiate with respect to t to get dy/dt = other stuff involving only x and t, then in the second of the original equations replace y and dy/dt by the respective stuffs involving only x and t. This procedure eliminates y and leaves a 2nd order equation for x, which OP has presumably already learned how to solve. > I'm trying to figure out this problem, but I can't find any > information (in my book or on the internet) that's related. I'm asked to solve the following system using the elimination method dx/dt = 2x - y > dy/dt = -x + 2y -- ==== L1+L2 d(x+y)/dt=x+y L1-L2 d(x-y)/dt=3(x-y) you get x+y and x-y terminated That's a nice way to do it. It may not be what the OP meant by > the elimination method. The OP may have meant, write the first > equation in the form y = stuff involving only x and t, then differentiate with respect to t to get dy/dt = other stuff involving only x and t, then in the second of the original equations replace y and dy/dt > by the respective stuffs involving only x and t. This procedure > eliminates y and leaves a 2nd order equation for x, which OP has > presumably already learned how to solve. I've been refreshing myself a bit on ODEs lately. Some texts define a differential operator D, whereby The OP's first equation dx/dt = 2x - y can be written Dx = 2x - y or (D-2)x + y = 0 The second equation dy/dt = -x + 2y similarly would be written x + (D-2)y = 0 The resulting system (D-2)x + y = 0 x + (D-2)y = 0 can be readily solved by systematic elimination for x and y. The results will be the characteristic equations of second order linear equations expressed as a linear second order differential operator. Essentially the same method described by Myerson above. I fear I may get pulverized by PDE in the Fall after my long math haitus, so I thought I'd introduce myself to the group. Andrew > I'm trying to figure out this problem, but I can't find any > information (in my book or on the internet) that's related. I'm asked to solve the following system using the elimination method dx/dt = 2x - y > dy/dt = -x + 2y ==== > Point is ASU, USA mathematicians have unified all of the known (and > previously unknown) primary laws (equations) of physics for my four > dimensional space time physiker friends, by simple multiplication > and division of the nine (9) primary universal base unit values. > In short, the physiker boneheads could not unify the foundation of > all their published theories, so the ASU, USA mathematicians did. > -- The 9 Primary Universal Base Unit Values... 1) mass... (hc/G)^1/2 = 5.4563026(39) x 10^-8 kg > 2) length... (hG/c^3)^1/2 = 4.0507625(38) x 10^-35 m > 3) time... (hG/c^5)^1/2 = 1.3511889(41) x 10^-43 s > 4) current... e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A > 5) temperature... b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K > 6) substance... M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol > 7) intensity... (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd > 8) Al Einstein solid angle = 6.8517999(97) x 10^1 sr > 9) Max Planck plane angle = 1.3703599(97) x 10^2 rad multiplied and divided by themselves, up to > and including their fourth (4th) power yield... The Primary Universal Physical Constants 001) irradiance constant i. = 4.5211591(52) x 10-122 s^3/kg > 002) radiance constant i.= 3.0978078(26) x 10-120 s^3-sr/kg > 003) radiant volume constant = 1.3554094(15) x 10^-113 m-s^2/kg > 004) measurement volume = 6.6467654(65) x 10^-104 m^3 > 005) graviton volume constant = 1.2181812(31) x 10^-96 m^3/kg > 006) luminous efficacy = 3.7229937(53) x 10^-96 cd-sr-s^3/kg-m^2 > 007) electric current volume = 1.3838190(49) x 10^-93 m^2/A > 008) luminous energy = 1.8257115(55) x 10^-86 cd-sr-s > 009) electric charge volume = 4.1485851(42) x 10^-85 m^3/A-s > 010) molar volume = 4.0027765(33) x 10^-77 m^3/kmol > 011) moment of inertia = 8.9530708(38) x 10^-77 kg-m^2 > 012) graviton fluidity = 1.0031235(26) x 10^-70 m-s/kg > 013) measurement area = 1.6408677(14) x 10^-69 m^2 > 014) electric moment = 6.4900363(91) x 10^-54 A-s-m > 015) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2 > 016) magnetic moment = 1.9456639(62) x 10^-45 A-m^2 > 017) luminous intensity = 1.9720204(06) x 10^-45 cd > 018) graviton frequency i. = 1.3511889(41) x 10^-43 s > 019) luminous flux = 1.3511889(41) x 10^-43 cd-sr > 020) graviton moment = 2.2102186(33) x 10^-42 kg-m > 021) inductance constant = 3.4877980(18) x 10^-39 kg-m^2/A^2-s^2 > 022) absorption-emission = 2.4763819(58) x 10^-36 s/kg > 023) graviton wavelength = 4.0507625(38) x 10^-35 m > 024) Planck constant = 6.6260687(65) x 10^-34 kg-m^2/s > 025) relative expansion = 2.8154241(58) x 10^-33 /K > 026) electric resistivity = 1.0456155(41) x 10^-30 kg-m^3/A^2-s^3 > 027) unified substance = 1.6605387(31) x 10^-27 kmol > 028) kinematic viscosity = 1.2143880(58) x 10^-26 m^2/s > 029) inverse electric current = 8.4334589(42) x 10^-25 /A > 030) Boltzmann constant= 1.3806502(93) x 10^-23 kg-m^2/s^2-K > 031) thermal resistance = 9.7866124(96) x 10^-21 s^3-K/kg-m^2 > 032) graviton molality = 3.0433405(93) x 10^-20 kmol/kg > 033) elementary charge = 1.6021764(62) x 10^-19 A-s > 034) primary radiation = 5.9552136(16) x 10^-17 kg-m^4/s^3 > 035) specific heat = 2.5303770(42) x 10^-16 m^2/s^2-K > 036) magnetic flux q. = 2.0678336(42) x 10^-15 kg-m^2-sr/A-s^2-rad > 037) magnetic flux = 4.1356672(77) x 10^-15 kg-m^2/A-s^2 > 038) electric permittivity = 1.2922425(96) x 10^-13 A^2-s^4/kg-m^3 > 039) magnetic exposure = 2.9363775(58) x 10^-12 A-s/kg > 040) electric constant = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3 > 041) magnetic pole strength = 4.8032041(96) x 10^-11 A-m > 042) Newton constant = 6.6723641(43) x 10^-11 m^3/kg-s^2 > 043) density of states = 2.0392015(07) x 10^-10 s^2/kg-m^2 > 044) S-B primary constant = 1.3897143(30) x 10^-9 kg/s^3-K^4 > 045) radiant distribution constant = 3.335640952... x 10^-9 s/m > 046) graviton mass constant = 5.4563026(39) x 10^-8 kg > 047) molar Planck constant = 3.9903126(87) x 10^-7 kg-m^2/s-kmol > 048) magnetic constant = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr > 049) electric conductance = 3.8740458(43) x 10^-5 A^2-s^3/kg-m^2 > 050) conductance q. = 7.7480916(72) x 10^-5 A^2-s^3-rad/kg-m^2-sr > 051) magnetic permeability = 8.6102258(15) x 10^-5 kg-m/A^2-s^2 > 052) fine-structure constant = 7.2973525(36) x 10^-3 /rad > 053) second radiation constant = 1.4387752(29) x 10^-2 m-K > 054) dielectric constant = 1.4594705(05) x 10^-2 /sr > 055) gravitational momentum = 1.6357583(80) x 10^1 kg-m/s > 056) relative permeability = 6.8517999(97) x 10^1 sr > 057) inverse fine-structure = 1.3703599(97) x 10^2 rad > 058) impedance of vacuum = 3.767303134? x 10^2 kg-m^2/A^2-s^3-sr > 059) molar gas constant = 8.3144720(88) x 10^3 kg-m^2/s^2-kmol-K > 060) spin angle constant = 9.3894326(23) x 10^3 sr-rad > 061) i. conductance q. = 1.2906403(83) x 10^4 kg-m^2-sr/A^2-s^3-rad > 062) von Klitzing constant = 2.5812807(61) x 10^4 kg-m^2/A^2-s^3 > 063) inverse gravitational mass = 1.8327429(14) x 10^7 /kg > 064) Faraday constant = 9.6485341(30) x 10^7 A-s/kmol > 065) speed of light in vacuum = 2.99792458 x 10^8 m/s > 066) graviton energy constant = 4.9038802(52) x 10^9 kg-m^2/s^2 > 067) Josephson primary = 2.4179894(88) x 10^14 A-s^2/kg-m^2 > 068) Josephson q. = 4.8359789(67) x 10^14 A-s^2-rad/kg-m^2-sr > 069) electric displacement = 3.9552465(66) x 10^15 A-s/m > 070) absorbed dose = 8.987551787? x 10^16 m^2/s^2 > 071) luminous density = 2.7467669(26) x 10^17 cd-sr-s/m^3 > 072) gravity displacement = 4.4930470(15) x 10^18 kg-s/m^2 > 073) molar mass constant = 3.2858629(17) x 10^19 kg/kmol > 074) magnetic potential = 1.0209601(87) x 10^20 kg-m/A-s^2 > 075) thermal conductance = 1.0218040(21) x 10^20 kg-m^2/s^3-K > 076) electric current constant = 1.1857530(90) x 10^24 A > 077) luminance constant = 1.2018155(94) x 10^24 cd/m^2 > 078) luminous flux density = 8.2346000(82) x 10^25 cd-sr/m^2 > 079) Avogadro constant = 6.0221419(79) x 10^26 /kmol > 080) gravitational field = 1.3469816(08) x 10^27 kg/m > 081) electric potential = 3.0607616(38) x 10^28 kg-m^2/A-s^3 > 082) electric conductivity = 9.5637446(19) x 10^29 A^2-s^3/kg-m^3 > 083) Celcius temperature = 3.5518626(82) x 10^32 K > 084) graviton wave number = 2.4686709(99) x 10^34 /m > 085) mass flow rate constant = 4.0381492(72) x 10^35 kg/s > 086) molar energy = 2.9531863(13) x 10^36 kg-m^2/s^2-kmol > 087) surface concentration = 1.0119881(80) x 10^42 kmol/m^2 > 088) graviton frequency = 7.4008894(66) x 10^42 /s > 089) superforce constant = 1.2106066(96) x 10^44 kg-m/s^2 > 090) luminous intensity i. = 5.0709414(41) x 10^44 /cd > 091) angular velocity = 1.0141882(87) x 10^45 rad/s > 092) electric flux density = 9.7642024(91) x 10^49 A-s/m^2 > 093) radiant intensity = 5.2968673(52) x 10^50 kg-m^2/s^3-sr > 094) graviton field strength = 2.2187308(44) x 10^51 m/s^2 > 095) superpower constant = 3.6293075(70) x 10^52 kg-m^2/s^3 > 096) magnetic flux density = 2.5204148(03) x 10^54 kg/A-s^2 > 097) thermal conductivity = 2.5224979(53) x 10^54 kg-m/s^3-K > 098) magnetic field strength = 2.9272342(65) x 10^58 A/m > 099) absorbed dose rate = 6.6515877(33) x 10^59 m^2/s^3 > 100) graviton surface density = 3.3252544(33) x 10^61 kg/m^2 > 101) electric field strength = 7.5560134(90) x 10^62 kg-m/A-s^3 > 102) measurement area i. = 6.0943364(99) x 10^68 /m^2 > 103) dynamic viscosity = 9.9688619(97) x 10^69 kg/m-s > 104) molar concentration = 2.4982658(70) x 10^76 kmol/m^3 > 105) surface tension constant= 2.9885896(41) x 10^78 kg/s^2 > 106) electric charge density = 2.4104603(51) x 10^84 A-s/m^3 > 107) angular acceleration = 7.5058954(07) x 10^87 rad/s^2 > 108) thermal transfer = 6.2272175(40) x 10^88 kg/s^3-K > 109) electric current density = 7.2263783(36) x 10^92 A/m^2 > 110) luminous efficacy i. = 2.6860104(16) x 10^95 kg-m^2/cd-sr-s^3 > 111) graviton density constant = 8.2089591(81) x 10^95 kg/m^3 > 112) measurement density = 1.5044911(77) x 10^103 /m^3 > 113) radiant density constant = 7.3778445(74) x 10^112 kg/m-s^2 > 114) radiance constant = 3.2280892(03) x 10^119 kg/s^3-sr > 115) irradiance constant = 2.2118221(59) x 10^121 kg/s^3 measured quantities, and 9^9 (387,420,489) more > (less of course the 115 knowns above). The yielded > unknowns are future laws of physics (discoveries), > which also have been verified by lab measurement. Weighing the Graviton Primary Equations... E=(hc^5/G)^1/2 > 4.9038802 x 10^9 kg-m^2/s^2 = [(6.6260687 x 10^-34 kg-m^2/s) > (2.4216061 x 10^42 m^5/s^5)/(6.6723641 x 10^-11 m^3/kg-s^2)]^1/2 > 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 > [rsu 3.9 x 10^-8] E=c^5/Gv > 4.9038802 x 10^9 kg-m^2/s^2 = (2.4216061 x 10^42 m^5/s^5)/ > (6.6723641 x 10^-11 m^3/kg-s^2)(7.4008894 x 10^42 /s) > 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 > [rsu 3.9 x 10^-8] E=hc^3/Gm > 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s) > (2.6944002 x 10^25 m^3/s^3)/(6.6723641 x 10^-11 m^3/kg-s^2) > (5.4563026 x 10^-8 kg) > 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 > [rsu 3.9 x 10^-8] E=h(Gd)^1/2 > 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s) > [(6.6723641 x 10^-11 m^3/kg-s^2)(8.2089591 x 10^95 kg/m^3)]^1/2 > 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2 > [rsu 3.9 x 10^-8] G=c^3/mv > 6.6723641 x 10^-11 m^3/kg-s^2 = (2.6944002 x 10^25 m^3/s^3)/ > (5.4563026 x 10^-8 kg)(7.4008894 x 10^42 /s) > 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 > [rsu 3.9 x 10^-8] G =wc^4/E > 6.6723641 x 10^-11 m^3/kg-s^2 = (4.0507625 x 10^-35 m) > (8.0776087 x 10^33 m^4/s^4)/(4.9038802 x 10^9 kg-m^2/s^2) > 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 > [rsu 3.9 x 10^-8] G=c^4/F > 6.6723641 x 10^-11 m^3/kg-s^2 = (8.0776087 x 10^33 m^4/s^4)/ > (1.2106066 x 10^44 kg-m/s^2) > 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 > [rsu 3.9 x 10^-8] G=c^5/P > 6.6723641 x 10^-11 m^3/kg-s^2 = (2.4216061 x 10^42 m^5/s^5)/ > (3.6293075 x 10^52 kg-m^2/s^3) > 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2 > [rsu 3.9 x 10^-8] E is graviton energy constant, h is Planck constant, > c is speed of light in vacuum, G is Newton constant, > v is graviton frequency, m is graviton mass constant, > d is graviton density constant, w is graviton wavelength, > F is superforce constant, P is superpower constant, > e is elementary charge, b is second radiation constant, > M is molar mass constant, i. is inverse, q. is quantum http://physics.nist.gov/cuu/Constants/ > [UPDATED 1998 CODATA-NIST VALUES] If the math works, what's wrong with today's physics? Have they steered themselves down a blind alley? Is Energy really energy, or are all values for E problematic? There is a similar post on sci.physics.electromag at: http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8&group=sci.physics.ele ctromag&selm=30067f9b.0308100956.495d52df%40posting.google.com where an equation for Energy does something similar to what is here. The reference link at bottom goes to a page where similar values for each component of the Axiomatic Equation are worked out. I'm posting the original below, should the links not work:-) --------------------------------------------------------------------- ENERGY'S AXIOMATIC EQUATIONS? 1. Energy is the basic common denominator of known physics, as eloquently expressed by Einstein's famous equation: E = mc^2 2. Energy can also be expressed (DeBroglie) as: E = hc/L , where h = Planck's constant, c = light velocity, L = photon lambda, so that we can say: E = hc/L = mc^2 3. Now, if light is an electromagnetic phenomenon, the we can also say (per Hyperphysics: http://hyperphysics.phy-astr.gsu.edu/hbase/waves/emwv.html -and- http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html), per Maxwell's equation: Em/Bm = c , and c = 1/(eo mo)^1/2 so that magnetic value times lightspeed: Bm * c = Em, its electric value equivalent. The question then becomes, can this equation be raised to the level of Energy? 4. A way to do this is to multiply Em by lightspeed, so that: Em * c = (Bm)c^2, which says that an electron accelerated to lightspeed is energy, which if so gives us: Em * c = (Bm)c^2 = E 5. Combining the above into a continuous equation gives us: Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^2 6. However, this is a still incomplete equation, taking E = mc^2, we then have: Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^2 = mc^2 = E * * * Now, this last continuous equation gives us a way to interrelate electric field, magnetic field, Planck's constant, photon lambda, and mass, all as Energy. This looks like a rather unifying equation for these forces, except gravity is still missing. The way to solve this (and here we are treading on new and dangerous ground) is to give mass a gravity and magnetic value. If we set mass as m = 1, to represent one hydrogen atom, which is an incomplete value, since from it is missing a gravitational constant, which is represented here as negative g = 5.9x10^-39, a negative remainder from how the atom is formed, so that mass is now expressed as: m (gravity/kilograms) = (1-g) (as per Gravity Force Coupling at: http://hyperphysics.phy-astr.gsu.edu/hbase/forces/couple.html which is not the same as Newtonian G) Likewise, if we set mass for magnetic, then the internal atomic positive and negative magnetic charges equal zero, m = 0, except for the atom's magnetic remainder, which here is set as a positive value at m = (0+Bm). Therefore, m (magnetic potential) = Bm (This can further be illustrated with E/c^2 = m, which is now E/c^2 = Bm, which translates into E = (Bm)c^2, as per #4 above.) 7. So, in taking the energy continuous equation to its conclusion, using the new values for mass, we get: Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^2 = (1-g)c^2 = E which now represents an Axiomatic Equation for Energy including electric field, magnetic field, light speed, Planck's constant, photon lambda, mass, and gravity. (Please note the gravity constant is not Newton's Gravity, for it is a proton to proton value, which needs to be converted further to become Newton's G.) This axiomatic equation, with conversions and SI Base Units values (meters, kilograms, time), was developed more extensively at: http://www.humancafe.com/cgi-bin/discus/show.cgi?84/108.html I am posting this equation here to solicit reasonable feedback. Is energy universal? Can it be expressed as per the 7 equations listed above? If not, where lies the error? Is is not all Energy? The conclusion (speculative at this point) from how this axiomatic equation is written is that electromagnetic energy and gravity become inversely proportional to the energy environment within which they are measured. This may mean, a possibility only, that gravity density around Pluto will be greater than that of Mercury, which would then mean that in using the Orbital Equation: v^2*R = GM, yields for us a mass reading that may be inaccurate, since we had always assumed G to be constant. Of course, if it is not constant, then the mass values must represent something else, which may mean that Pluto is perhaps really a dirty water ice-ball planet, whereas Mercury is nearly all metallic. The other ramification from this equation is that if photon lambda is totally canceled out, gravity g reaches its maximum value (from 0 to 1) of g = 1, which means total gravity takes over, mass ceases to exist, light ceases to exist, all is left is gravity... like a black hole. So these equations may be important, if true, since it would give us a new perspective on universal gravity (not equally constant) and provide a way to solve for various components of the equation in relation to each other. Happy computing! -Ivan ==== I want to program some econometric functions in C/C++, VB or SQL for an in-house application. I did see some free source code in Fortran but none in the any of the above languages. In particular, I am looking at Time Series Analysis, Forecasting, GARCH etc. as applied in the finance industry. Does anyone know of a good place from where I can get this - for free or a nominal fee (Source Code or Libraries that can be used in custom applications). Jay ==== That is the method I have used, but there is a flaw of twisting. (Correct me if I m wrong), the normal vector (principle normal) of a curve always point to the concave side. When the curve changes concavity, the normal vector reverses direction, and so does the binormal. For the other question, I was refering to equations like x*sin(x)+y*cos(x)=2*z^2 If you restrict x, y, z between -5 to 5, the graph is a forking tube. I m very interested in exploring such equations and their classifications. Take a smooth curve G, parametrized by r(t) = (x(t),y(t),z(t)). These > vectors can be defined easily: r'(t) = (x'(t),y'(t),z'(t)): the velocity vector T(t) = r'(t) / | r'(t) | : unit tangent vector T'(t) = |r'(t)|^2 K(t) : K = curvature vector N(t) = K(t)/|K(t)| : Principal normal to curve B(t) = T(t) x N(t) : Binormal to curve Then, take a small radius rho, and generate normal circles to the > curve G: C(t, rho, theta) = r(t) > + rho*cos(theta)*N(t) > + rho*sin(theta)*B(t) > If you choose rho sufficiently small (I think smaller than min(1/|K(t)|) > will do the trick), then you'll get a nice embedding of the tube. The > circles form radial 'ribs' along the tube, and tracing out the curves > C(t,rho,theta) for fixed rho & theta, but with varying t, draws lines > that run 'parallel' [kind of parallel, but not really], to the axis of > the tube. I haven't seen anyone in my math department with explicit interest in > problems like these, or courses that explore this area. Are they more > of a computer science guy's interest? > I don't know. The formulas seem kind of trivial, so that may be why the > apparent lack of curiosity you're seeing. Dale. ==== TWISTING: (Correct me if I m wrong), the normal vector (principle normal) of a curve always point to the concave side. When the curve changes concavity, the normal vector reverses direction, and so does the binormal. This ROTATION: If you transform the parametric equation of a circle in the xy plane with the z-axis rotation matrix, and varying theta from 0 to 2pi, you get the equation of a donut ring. There are other ways too, such as using spherical coordinates. I would very much like mathematical equations instead of computer calculations of the normal, binormal, and using them as a basis. For example: x*sin(x)+y*cos(x)=2*z^2 If you restrict x, y, z between -5 to 5, the graph is a forking tube. I m very interested in exploring such equations and their classifications. > I think you have hit upon the basic method. For example, to construct a > helical tube you might use the following equations: x=a cos(t) > y=a sin(t) > z=b t where a and b are constants and t varies to give you different points on the > helix. Then, as you said, you would compute the tangent vector and draw a > circle in the plane perpendicular to the tangent. I am not sure what you > meant by rotating circles or problems (twisting). If you are interested in computer graphics there are news groups for that. > One idea then is to just draw lots of spheres centered on points along the > helix. It might help others here to help you if you told us why you want to > describe a tube. > I haven't seen anyone in my math department with explicit interest in > problems like these, or courses that explore this area. Are they more > of a computer science guy's interest? ==== > How can I describe a tube (imagine an out of control water hose) in > R3? Or examples of equations that resemble tube like structures? > Preferably no twisting nor self-intersections. An approach that deals with notions of twisting is that of the framed curve. Based on another message of yours, I think you are familiar with the Frenet frame (T,N,B). There are other options for frames; in fact, any time-varying orthonormal basis can be used as a frame. Probably you want to restrict to those for which the first vector is T; so you want to consider frames of the form (T(t),e1(t),T(t)xe1(t)). (t is the curve parameter; x is the cross product.) Off the top of my head, I can't think of any obvious framing that will eliminate twisting; but you can measure twisting by how e1(t) varies, and if you can reduce that then you should be OK. Kevin. <3c65f87.0308080830.c3dd6be@posting.google.com> <3c65f87.0308091410.24e9c1d8@posting.google.com> ==== >> Why is 'Braveheart' worth mentioning in this context? >Both Saving Private Ryan and Braveheart are in genres that have > James, > If you had the *slightest* knowledge of *anything* about William Wallace > or Scottish history you would not describe it as a historical battle > epic, but more as quite a fun film, if you totally disregard any facts > but look on it as an amusing sort of Hollywood romp. Ha. Yeah, they did sort of tinker with the details a bit. Guess that's how it has to be when you only have 128 pages or so to work with. Ed Howdershelt - Abintra Press Science Fiction and Semi-Fiction http://abintrapress.tripod.com ==== (x-x/2)(x-x/2+1)... (x-2)(x-1) (x+1)(x+2)...(x+x/2-1)(x+x/2) > ------------------------------------------------------------ > x^x Now it should be pretty obvious that the limit of that expression > evaluates to one. There is (x+n)(x-n) in the numerator and x^2 in the > denominator for n=1 to x/2. > That's ridiculous. To get an expression that equals one there what is necessary is II ( (x+n) (x-n) + n^2 ) That actually equals II x^2. The expression (x+n)(x-n) is greater than (x+n+1)(x-n-1) for positive x and n. Ross ==== lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 False. Stirling's approximation for n! is actually a transformation of a formula that's only true in the limit. That's why some use Gosper's approximation for n!, it's more accurate for some values of n, I have also seen mentioned Burnside's formula, and a Lanczos approximation, and Godfrey's improved Lanczos approximation. Here are some links with information about factorial approximations: http://mathforum.org/library/drmath/view/60692.html http://www.google.com/search?q=factorial+approximation http://www.google.com/search?q=%22Gamma+function%22+approximation Stirling's identity is: lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1 An approximation is: n! ~= n^n sqrt(n2Pi) / e^(n-(1/12)n), n >> 1 The value of n factorial is approximately equal to n to the n'th power times the square root of n times 2Pi divided by e to the n'th power, for n much greater than one. What that says is that for given large finite values of n, that the result of the approximation of n! is near n!. The benefit of using the other approximations is that they are nearer n!. This thread isn't concerned about four trillion factorial or eight hundred and thirty seven thousand factorial, it's about the result of expressions as the variable diverges to infinity, n >> 4000000000000. Particularly, at this point it's concerned with setting the value of one variable of the gamma function to an expression of the other variable which diverges. lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 Here I'll call that the Reformulated Gamma expression, RG(x). RG(1) is: lim n->oo ((2n)!) / (n! n^n) = 1 If that is not true, then it it is perhaps possible to validly compare it with known results of the value of n! or the other terms in the variable expresion and to get a known untrue result. That would definitely help establish the truth value of that identity. As we have previously discussed in this thread, there is a known result for an expression with n!, (n/2)!, 2^n, and sqrt(n). Stirling's formula has terms of n!, n^n, e^n, and sqrt(n). The Duplication Formula for the Gamma function gives an expression for Gamma(2z) in terms of 2^(2z-1/2), Gamma(z), and Gamma(z+1/2). Stirling's formula has n! and n^n on opposite sides of the divisor, where RG(1) has them on the same side of the divisor. Stirling's formula: 1: lim n->oo ( n! e^n ) / ( n^n sqrt(n) sqrt(2P)i ) = 1 The expression f(x) = lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 for x=1: 2: lim n->oo (2n!) / ( n! n^n) = 1 The common terms of those two expressions are n! and n^n. Put n! alone in the numerator for 1 and 2: 3: lim n->oo n! / ( n^n e^(-n) sqrt(n) sqrt (2Pi) ) = 1 4: lim n->oo n! / ((2n)! n^(-n)) = 1 Divide 3 by 4: 5: lim n->oo ( (2n)! e^n ) / ( n^(2n) sqrt(n) sqrt(2Pi) ) = 1 Put n^n alone in the numerator for 1 and 2: 6: lim n->oo n^n / (( n! e^n)/(sqrt(n) sqrt(2Pi)) 7: lim n->oo n^n / ( (2n)! / n! ) = 1 Divide 6 by 7: 8: lim n->oo ( (2n)! sqrt(n) sqrt(2Pi) ) / ( n! n! e^n ) = 1 So we have two equations, 5 and 8, each is the result of reducing common expressions from a product of the two equations 1 and 2. That's about how the l.h.s of each equation is equal to the same value, one. How about gathering the (2n)! terms of 5 and 8 and attempting to divide those? That's probably not very informative. 9: lim n->oo (2n)! / ( n^(2n) sqrt(n) sqrt(2Pi) e^(-n) ) = 1 10: lim n->oo (2n)! / (( n! n! e^n )/(sqrt(n) sqrt(2Pi))) = 1 Divide 9 by 10. 11: lim n->oo n! n! e^n e^n / ( n^(2n) n 2Pi) = 1 That is equivalent to: 12: lim n->oo n! ^2 e^2n / (n^(2n) n 2Pi) = 1 Take the square root of both sides. 1: lim n->oo n! e^n / (n^n sqrt(n) sqrt(2Pi)) = 1 Well, that's just the same thing as Stirling's formula. I think it was derived from Stirling's formula and f(x) for x=1 without simply multiplying Stirling by f(x)/f(x), except in a way it is as f(x)/f(x) = 1 / 1 = 1. Please explain your reasoning. I'm not trying to be obtuse here, I'm trying to be not obtuse here, if it's invalid I want to know and why. That's because it has the same, true meaning. Why do I say lim n->oo ( (1+1/x)n! ) / ( n! n(n/x) ) = 1? That's not asking why it is so, it is asking why I say it is so. I've explained why I think it is so. I think you should address that. Do you have some reason why you say it is not so? You said False and Stirling's formula. That's not an explanation. I use False sometimes, normally in protracted conversations where an explanation is already evident, it's fun as hell, but it's not an explanation, just the assertion of the existence of an explanation. It is so that 2 n! =/= (2n)!, for n =/=1. So anyways, what's your explanation? That is to say, for the benefit of people who don't know what your explanation is, present it. I think it would be great if RG(1) was true. I think it is, for the reasons that I have provided, namely that I think it is a valid use of Euler's Gamma function and that it appears as shown here to not contradict Stirling's formula. While that may be true, I'm not going to ignore valid objections. So, if you say it's false, then explain for our benefit why. I don't mind if you reiterate your explanation and answer the questions of others about it, I think it would be good. Ross ==== lim n->oo ((1+1/x)n!) / (n! n^(n/x)) Set x=1/n: lim n->oo ((1+n)n)! / (n! n^(n^2)) lim n->oo (n^2+n)! / (n! n^(n^2) ) Set x=n: lim n->oo ((1+1/n)n)! / (n! n^(1) ) lim n->oo (n+1)! / (n! n) lim n->oo (n+1) / (n) The result is equal to one. Set x=n/2: lim ((1+2/n)n!) / (n! n^(2)) lim (n+2)! / (n! n^2) lim (n+1)(n+2) / n^2 lim (n^2 + 3n + 2) / n^2 lim (n^2/n^2) + (3n/n^2) + (2/x^2) That appears to start diverging, and also appears to be equal to one. Set x = 2n: lim ((1+ 1/2n)n)! / (n! n^(1/2)) lim (n+1/2)! / (n! n^(1/2)) lim Gamma(n+3/2) / (Gamma(n+1) n^(1/2)) That appears to diverge. Set x=1: lim (2n)! / (n! n^n) lim (n+1)(n+2)...(2n) / ( n^n) lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n!/n^n That might also appear to evaluate to one as all but the first term which evaluates to one evaluate to equal zero, depending on what a, b, ... equal. The value of a might be sum n, which is greater than n, with b being the value of the sum of the product of each pair of N, I wonder if that is n^2. If that were so on and so forth for the other terms then the value of the expression would be n. If a, b, ... were each less than n, n^2, etcetera, that would not be so. lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n! / n^n If a is sum n, then a is (n+1)(n/2) = n^2/2 + n/2, and an^(n-1) is (n^(n+1) + n^n)/2, and that divided by n^n is n/2+1/2. The limit of the n'th term, n! / n^n, is zero. Let's see, consider a finite case of (x+1)(x+2)...(x+4). (x^2 + 3x + 2)(x+3)(x+4) (x^3 + 6x^2+11x + 6)(x+4) (x^4 + 10x^3 + 35x^2 + 50x + 24) Yep, it appears a = sum n for n= 1 to 4, 5*2, (n+1)/(n/2). The value of b is the sum of each product of each pair of {1, 2, 3, 4), the value of c would be the sum of each product of each triple of 1, 2, 3, 4, etcetera, d is the product 1, 2, 3, 4. Normally in the polynomial the coefficient a is on the first term, this starts with a on the second. 1*2 + 1*3 + 1*4 +2*3 + 2*4 + 3*4 = 35 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50 1*2*3*4 = 24 Let's see, 35 = 1*(2+3+4) + 2*(3+4) + 3*4, the sum for i from 1 to n-1 of the product of i and the sum of n from i+1 to n. The third term c is 1*(2*3 + 2*4 + 3*4) + 2*(3*4) ..., a slightly more convoluted expression where more terms would be necessary to display it by example. So, it appears that the expression diverges. Now I want to figure out what it diverges to, symbolically. Then you could divide the divergent expression by that and get a convergent expression. Set x=1/2: lim ((1+2)n)! / (n! n^(2n)) lim (3n)! / (n! n^(2n)) More of the same. lim (n+1)(n+2)...(3n) / (n^(2n) lim n^2n/n^2n + a n^(2n-1) / n^(2n) + b n^(2n-2) / n^(2n) + ... If a is (2n+1)n = 2n^n + n, then the second term would be (n^(2n+1)+ n^2n) / n^2n = n+1, where in the above it was (n+1)/2. What are the closed forms for the coefficients of the polynomial of (x+1)(x+2)...(x+n)? They are {1, sum n, ..., n!), what is the form of each of the intervening coefficients? Going back, set x=1/n: lim n->oo ((1+n)n)! / (n! n^(n^2)) lim n->oo (n^2+n)! / (n! n^(n^2) ) lim (n+1)(n+2)...(n^2+n) / n^(n^2) lim (n^(n^2))/(n^(n^2)) + ((n^2+n+1)(n^2+n)/2) (n^(n^2-1)) / (n^(n^2)) + ... lim 1 + (n^4 + 2n^3 + n^2 + n) (n^(n^2-1)) / 2(n^(n^2)) + ... lim 1 + (n^(n^2+3) + 2n^(n^2+2) + n^(n^2+1) + n^(n^2))/2(n^(n^2)) + ... lim 1 + (n^3)/2 + n^2 + n/2 + n/2 + 1/2 + ... So there appear to be five or seven cases, where x is a constant f(x)=c, x is a function of a constant times n, f(x)=cn, for c=1, 01, and x is a function of a constant times the reciprocal of n, f(x)=c/n, for c=1, 01. For f(x)= lim n->oo ((1+1/x)n)! / (n! n^(n/x)) = 1, that is true when x=cn, 0> lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 >> False. > lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1 So, use Stirling's formula to obtain the asymptotics (at least the dominant term) of ((1+1/x)n)!/(n! n^(n/x)). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen <3c6b9c1e.0308091559.485216f4@posting.google.com> <3c6b9c1e.0308100457.73ff5939@posting.google.com> <3c6b9c1e.0308101544.48f969ec@posting.google.com> <3c6b9c1e.0308111623.38637b13@posting.google.com> <3c6b9c1e.0308112243.4fb4e3bc@posting.google.com> ==== : :That appears to start diverging, and also appears to be equal to one. : and then later. : :Bollocks, indeed. : :Ross : Speaks for itself. ==== I want to figure out what are the coefficients. These are 0! = 1, ? = (n+1)(n/2), and basically for each an expression of the sum of the product of each triple, quadruple, quintuple, etcetera, of {1, 2, 3, 4, ..., n}, which would be the same as for {0, 1, 2, ..., n}. HMF: An infinite series whose general term is a rational function of the index may always be reduced to a finite series of psi and polygamma functions. Let's see, f(x) = II_i=1^n (n+x) I go to MathWorld and read about polynomial. http://mathworld.wolfram.com/Polynomial.html The function is a univariate polynomial with constant coefficients. There are n many monomial summands. The order or degree of f is n. Exchanging the coefficients of a univariate polynomial end-to-end produces a polynomial whose roots are the reciprocals of the 1/x_i of the original roots x_i. The roots x_i of f are -1, -2, -3, ..., their reciprocals 1/x_i are -1, -1/2, -1/3, .... Consider Vieta's Formulas. http://mathworld.wolfram.com/VietasFormulas.html I noticed the s1 = r1 + r2 + r3 + r4, it is similar to the 1 + 2 + 3 + 4, and so on for the following values of s. Consider Stirling Numbers of the First Kind. The Stirling Number generating functions are defined in terms of a falling factorial, where f is defined in terms of a rising factorial, almost. The rising factorial or Pochhammer is x(x+1)(x+2)...(x+n-1), f is Pochhammer(x,n) * 2. The function f is easily defined as a falling factorial (2n)(2n-1)...(n+1), that is (2n)_n. What is to be done here is to determine the symbolic form of the coefficients of the n_1 monomials of the polynomial, (n+1)(x+2)...(n+n), to sum them together to get another polynomial. The first coefficient is 1, the second (n+1)(n/2), and the last n!. There are thus the n-2 other coefficients that require determination. The coefficients form another polynomial with each i'th coefficient multiplied by n^(n+1-i). Their sum is a univariate polynomial of n. I'm trying to figure out what its degree would be, as the number of monomials is a function of the variable, and the coefficients are each a function of the variable. Call the polynomial g(n). This polynomial is for the case of RG(1), the expression lim n->oo (2n!) / (n! n^n) is under consideration. lim n->oo (2n)! / (n! n^n g(n)) = 1 Basically g(n) = (n+1)(n+2)...(n+n) / n^n = RG(1). In that sense g(n) for RG(x) is RG(x), and when x=cn, 0oo ((1+1/x)n!)/(n! n^(n/x)), it equals one for certain values of x without explicitly multiplying it by its reciprocal. Also, it should be possible to determine forms for g(n). Ross ==== I cannot think of a counterexample to the following question. >If f is an automorphism of a finite group G such that, for every g in G, >the elements g and f(g) are conjugate, this does not necessarily >imply that f is inner, does it? You can have counterexamples to statements (at least to false statements), but what do you mean by a counterexample to a question? OK, so you are looking for an example of a finite group G with an outer automorphism f such that, for every g in G, the elements g and f(g) are conjugate. I do not any such example. Please let me know if you find one! Derek Holt. ==== NASA's > hypervelocity guns. > Roughly, they use a gunpowder charge to compress hydrogen which propels the > projectile to hypervelocity. > Frank Yes, I am beginning to see all energy transformations as either obeying the > IdealGasLaws which are linear in math form and the EM laws which are > sinusoidal and have cancelling of energies. The linear transformation allows > the exceeding of barriers such as sound barrier. The sinusoidal form such as > slingshot, crossbow, fusion tokamaks have so much cancelling of energy > that they seldom can ever perform as well as the linear transformation. > this must be the first time anyone has directly compared a crossbow to a tokamak. amazing > The most common linear transformation of energy is explosion and burning > of fuel and of course air pressure such as airguns. Another linear > transformation > is Fission because the neutrons obey the IdealGasLaws. We can bifurcate all energy transformations as either direct and linear or > as indirect, cancelling, and sinusoidal. What I am anxious to prove in the above is that barriers exist such as the > Sound Barrier. And airguns can exceed the Sound Barrier but can bow and arrow > exceed the Sound Barrier or can a slingshot (another form of EM) > ever exceed the Sound Barrier. But the most important of these questions > is whether Fusion can ever exceed 2/3 breakeven and still be under **control** > There is no such thing as the 'sound barrier.' Raz ==== Consider the sequence where, for m = any positive integer, a(m) = the number of integers j, 1 <= j <= m, where floor(m/j) divides m. Is there an easy direct way to calculate the sequence? The best way I can come up with is: a(m) = sigma(m) - sum{k|m} floor(m/(1+k)), where sigma is the sum of the positive divisors of m. (I guess my means of calculating this sequence is good enough, if it is a true means to calculate it.) Leroy Quet ==== > What I've endeavored to do to prevent math discussions that go on > endlessly as issues are debated all over the map is focus in on > particular points of contention. I'd thought I'd removed all room for debate on the issue of whether or > not, given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) one of the a's must be coprime to f, when f is coprime to 3, without > regard to m, by focusing on P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) with f=3, as I'd *said* that then each of the w's must equal 3^{1/3}, > but a poster pointed out that in fact I get another factor of 3 that > divides off, and it's indeterminate as to how it divides off. That > poster continued thinking that shot down my arguments, and presented > their beliefs, which would be that the w's vary with m. However, I'll destroy that position using f=3 by considering P(m). Returning to the full expression with f=3, I have P(m) = 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3) = (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u). At issue has been a claim by several posters, who've spent *months* > arguing with me that f divides off as a variable of m, such that the > a's can have varying factors in common with f, dependent on its value. > They make this argument because I can use m=0 to show that when f is > coprime to 3, one of the a's does not share non-unit factors in common > with f. They claim the m=0 is a special case as they wish for factors > of f to move around in the factorization dependent upon whether or not > it factors over rationals. Yup, if you've been wondering about all the arguing and all the posts, > a lot of it has been about that one point of contention, which I'm > going to shoot down here. Now then, if these other posters were correct, then a variable factor > in common with 3 should divide off through each of the a's, but > looking at P(m), 3 itself must divide through AS A CONSTANT without > regard to the value of m. That's because P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) proves that *each* of the a's must have a factor that is 3 without > regard to m, so m=0 doesn't matter. > What you have been claiming is about the form of the factorization of P(m)/f^2. In terms of the factorization that you gave above, > P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3), you want to show (direct quote from your text above): each of the w's must equal 3^{1/3}. What you REALLY need to show is that the analog of this is true for f = prime > 3, and for some reason you think it will be sufficient to show it only for f = 3. But even with f = 3 you have a problem. What you have shown here is not about P(m)/f^2. You have instead considered P(m)/f^3. The reasons are (1) you know that I have already shown that for f = 3, the w's are indeterminate in the factorization of P(m)/f^2, and (2) you have lost track of what you were doing. Bottom line: what you have done here regarding P(m)/f^3 for f = 3 is not what you set out to do, and it does not imply what you wanted to do either. I actually don't disagree with the result you noted for the a_i's above, but it is in the so-what category: it is not what you need. You want to focus on the factorization of P(m)/f^2. Second bottom line: my examples showing that there are infinitely many choices for the w's when you are considering what you really intended - P(m)/f^2 - are all still completely valid. What you have done here is irrelevant to that also. The implication of these examples is: you CAN'T do what you wanted to do when f = 3. Third bottom line: What you REALLY need is a demonstration that what you (incorrectly) think holds for f = 3, also holds for f = 5, 7, 11, 13, .... This you have not even touched. As I noted in the other thread, you need to prove something about the trees deep in the forest (f = prime > 3). You CANNOT (and very definitely have not!) done that by considering the underbrush out on the edge of the forest (f = 3). Reasoning that a pattern on the boundary must extend into the interior takes proving also. And what you have done here doesn't even establish the pattern out on the boundary (f = 3). Fourth and final bottom line: After all this work, you are back where you started: no proof at all that the pattern of factorization you have observed for m = 0 extends to m <> 0. Nora B. > BUT, these posters believe that the a's have a *variable* function of > f, and my having f=3 wouldn't change that fact, so if true, it'd force > some of the a's, depending on the value of m, to not have a factor > that is 3. [propaganda deleted] > I await attempts at further debate from posters, but I warn them that > my job is to focus on the issues, and take away their ability to > endlessly argue, expecially through distraction. > James Harris ==== > |BUT, these posters believe that the a's have a *variable* function of > |f, and my having f=3 wouldn't change that fact, so if true, it'd force > |some of the a's, depending on the value of m, to not have a factor > |that is 3. You *say* having f=3 wouldn't change it, but why? The rest of > the posting doesn't provide an argument. Doesn't it bother you > to be stuck repeating such a claim without any way to show > that it's actually true besides telling us how weird it would be > if it weren't? Keith Ramsay Posters repeatedly delete out the math in replies. Here the pertinent expressions are P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) where if f divides off the a's as a function of m, then you get a > contradiction, as in fact, it's clear that f divides off as a constant > independent of m. Of course, here f=3, but what's amazing is that posters would try to > convince anyone that having f change its value to being coprime to 3, > could *introduce* a dependency on m. And that explains why they delete out pertinent expressions as they > want to convince readers. Or does it? I'm curious about what readers think. Why do YOU think posters like Keith Ramsay keep deleting out the math? There is a much more general question. Why does anyone read these James Harris threads at all? However, I will address your first questions as a casual reader. The removal of the constantly repeated algebra is a good idea since there must be hundreds of repetitions of the stuff at hand. Same old stuff over and over again. Never the slightest attempt to clarify. Just the same tired stuff endlessly. Now to my more general question. In my case, I'm learning from the various posters who have posted refutations to your proof. These posters mostly explain very well their steps and provide definitions and other useful information. They are more focused on the mathematics than you are. They seem to have an excellent grasp on the subject. Indeed, they are rminding me of material I studied 30 years ago and never used. Chuck -- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net ==== I'm curious about what readers think. Why do YOU think posters like Keith Ramsay keep deleting out the math? I think if YOU deleted all the math, we'd be better off. ==== > What I've endeavored to do to prevent math discussions that go on > endlessly as issues are debated all over the map is focus in on > particular points of contention. I'd thought I'd removed all room for debate on the issue of whether or > not, given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf) one of the a's must be coprime to f, when f is coprime to 3, without > regard to m, by focusing on P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3) with f=3, as I'd *said* that then each of the w's must equal 3^{1/3}, > but a poster pointed out that in fact I get another factor of 3 that > divides off, and it's indeterminate as to how it divides off. That > poster continued thinking that shot down my arguments, and presented > their beliefs, which would be that the w's vary with m. That is, I made a mistake, which I've acknowledged. Intriguingly, Nora Baron appears to be trying to convince you the reader of something else. See below... > However, I'll destroy that position using f=3 by considering P(m). Returning to the full expression with f=3, I have P(m) = 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3) = (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u). At issue has been a claim by several posters, who've spent *months* > arguing with me that f divides off as a variable of m, such that the > a's can have varying factors in common with f, dependent on its value. > They make this argument because I can use m=0 to show that when f is > coprime to 3, one of the a's does not share non-unit factors in common > with f. They claim the m=0 is a special case as they wish for factors > of f to move around in the factorization dependent upon whether or not > it factors over rationals. Yup, if you've been wondering about all the arguing and all the posts, > a lot of it has been about that one point of contention, which I'm > going to shoot down here. Now then, if these other posters were correct, then a variable factor > in common with 3 should divide off through each of the a's, but > looking at P(m), 3 itself must divide through AS A CONSTANT without > regard to the value of m. That's because P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) proves that *each* of the a's must have a factor that is 3 without > regard to m, so m=0 doesn't matter. What you have been claiming is about the form of the > factorization of P(m)/f^2. In terms of the factorization > that you gave above, P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3), you want to show (direct quote from your text above): each of the w's must equal 3^{1/3}. See what I mean? The poster Nora Baron posted as if in fact I didn't myself acknowledge the mistake above!!! In fact, two of the w's are coprime to 3, while the remaining one has a factor that is 3. > What you REALLY need to show is that the analog of this is > true for f = prime > 3, and for some reason you think it > will be sufficient to show it only for f = 3. But even with > f = 3 you have a problem. What you have shown here is not > about P(m)/f^2. You have instead considered P(m)/f^3. > The reasons are (1) you know that I have already shown that > for f = 3, the w's are indeterminate in the factorization of > P(m)/f^2, and (2) you have lost track of what you were doing. But they are not indeterminate, which can be seen by considering P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) as that's the only way the 3^3 can divide off when f=3. There is NO other way, so it's wrong for Nora Baron to claim it's indeterminate. > Bottom line: what you have done here regarding P(m)/f^3 for > f = 3 is not what you set out to do, and it does not imply > what you wanted to do either. I actually don't disagree > with the result you noted for the a_i's above, but it is in > the so-what category: it is not what you need. You want to > focus on the factorization of P(m)/f^2. Actually, it blows away your assertions that f^2 divides off in different ways dependent on the value of m. Look at P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 = (a_1/f x + u)(a_2/f x + u)(a_3 x + uf) where here now, if f is coprime to 3, that *last* f is blocked by a_3. Now I've proven that factorization, while Nora Baron is working to convince you that the f's divide out differently dependent on m. But when f=3 that's easily shown to be just wrong. So that leaves Nora Baron trying to convince you that the equations check to see if f is coprime to 3. Why would the poster fight so hard? Because accepting the truth means accepting that there's a problem with the ring of algebraic integers. > Second bottom line: my examples showing that there are infinitely > many choices for the w's when you are considering what you > really intended - P(m)/f^2 - are all still completely valid. > What you have done here is irrelevant to that also. The > implication of these examples is: you CAN'T do what you > wanted to do when f = 3. Third bottom line: What you REALLY need is a demonstration > that what you (incorrectly) think holds for f = 3, also holds for > f = 5, 7, 11, 13, .... This you have not even touched. As > I noted in the other thread, you need to prove something > about the trees deep in the forest (f = prime > 3). You > CANNOT (and very definitely have not!) done that by considering > the underbrush out on the edge of the forest (f = 3). Reasoning > that a pattern on the boundary must extend into the interior > takes proving also. And what you have done here doesn't even > establish the pattern out on the boundary (f = 3). Actually, what I demonstrate with f=3 is that f^2 does NOT divide off as dependent on m. The f^2 divides off one way from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > Fourth and final bottom line: After all this work, you are back where > you started: no proof at all that the pattern of factorization you > have observed for m = 0 extends to m <> 0. > Nora B. > BUT, these posters believe that the a's have a *variable* function of > f, and my having f=3 wouldn't change that fact, so if true, it'd force > some of the a's, depending on the value of m, to not have a factor > that is 3. [propaganda deleted] And notice how much effort the poster Nora Baron goes to in order to try and convince you the reader. Here you're told there was propaganda but it was deleted, so if you want to know what was deleted, you have to go back to the earlier post and find it. > I await attempts at further debate from posters, but I warn them that > my job is to focus on the issues, and take away their ability to > endlessly argue, expecially through distraction. But they're still trying, as they've made it their job to hide the truth, and thereby hide the embarrassment to mathematicians. James Harris ==== |Posters repeatedly delete out the math in replies. Let me remind you that you were the one who kept saying that the postings we were writing to each other should keep getting shorter and more to the point. That's not possible if I keep requoting the same stuff of yours over and over. |Here the pertinent expressions are | | P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - | | (-1+m 3^2 )x u^2 + u^3 = | | (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) | |where if f divides off the a's as a function of m, then you get a |contradiction, as in fact, it's clear that f divides off as a constant |independent of m. A contradiction is when the same premisses logically imply both A and A is false. This is not a contradiction. This is just a situation that looks wrong to you. |Of course, here f=3, but what's amazing is that posters would try to |convince anyone that having f change its value to being coprime to 3, |could *introduce* a dependency on m. You find the possibility amazing. This is apparently all you have to go on. There are plenty of mathematical claims where it would be amazing if they were false, but nobody knows how to prove them yet. For instance, nobody can prove yet that the decimal expansion of pi has infinitely many 3's in it. The idea that they should appear on average every tenth digit out to the tens of billions of places and then disappear would be amazing. That's not a proof. It's not the case that every function of m and f which is independent of m for f having common factors with 3 is also independent of m for other values of f. So to show that it doesn't happen in this case, you'd have to tell us something specific about these common divisors that makes it not just amazing to you but not actually possible. |And that explains why they delete out pertinent expressions as they |want to convince readers. | |Or does it? | |I'm curious about what readers think. | |Why do YOU think posters like Keith Ramsay keep deleting out the math? It seems to me that you present a rather insulting picture of the lurkers. You describe them as if they were such mindless sheep that they can't read what you're saying and understand it, unless I keep requoting it to them for you. Not just a few times either, but over and over again. Just who do you think they are? Even supposing they were intimidated somehow, don't you suppose right, there really is a contradiction in the definition of algebraic integer, how can we convince the world? I think it's safe to say you're failing to attract support like that because you do NOT have a clear and convincing case. Even supposing for the sake of argument that we're all terribly confused and you're actually right in your final conclusions, can't you see that it would be unclear that you're correct? Keith Ramsay ==== [snip bottomless pit of demented arguments] > ... they've made it their job to hide the > truth, and thereby hide the embarrassment to mathematicians. Speaking of hiding things, you forgot to post one of the numbers you discovered which support your claim that the ring of algebraic integers is flawed. You claim that there are numbers which *should be* algebraic integers, but which have been *left out* of the ring of algebraic integers. Name one. Just one will do. (You wouldn't be trying to 'hide the truth, and thereby hide the embarrassment' would you?) -- 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 ==== > |Posters repeatedly delete out the math in replies. Let me remind you that you were the one who kept saying that > the postings we were writing to each other should keep getting > shorter and more to the point. That's not possible if I keep > requoting the same stuff of yours over and over. Yes it is. Your conclusion does not follow logically. Given a mathematical argument there are typically points not in dispute, so you can delete those out until you reach the *last* point you don't disagree with, which you should leave in for context, and then you consider the *next* point you do disagree with, and state why you believe it doesn't follow logically. Now that would typically lead to replies shorter than the entire math argument, unless possibly the error took a really long time to explain. Remember, a math proof begins with a truth and proceeds by logical steps to a conclusion which then must be true. If you think someone has a false claim of discovering a proof, then all you need do is either show their argument does not begin with a truth, or show a break in the logical chain. > |Here the pertinent expressions are > | > | P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - > | > | (-1+m 3^2 )x u^2 + u^3 = > | > | (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u) > | > |where if f divides off the a's as a function of m, then you get a > |contradiction, as in fact, it's clear that f divides off as a constant > |independent of m. A contradiction is when the same premisses logically imply > both A and A is false. This is not a contradiction. This is > just a situation that looks wrong to you. Actually your claim here is just wrong Keith Ramsay. The question is whether or not f^2 divides off as a function of m, which would suppose that it varies dependent on m. However, I show that when f=3, not only f^2 but an additional factor of f, divides off, showing that it divides off as a constant. Maybe this will help you see P(m)/3^3 = (a_1/w_1(m) x + u)(a_2/w_2(m) x + u)(a_3/w_3(m) x + u) where I've introduced w's as a function of m, where w_1(m) w_2(m) w_3(m) = 3^3. However, *any* variation of the w's that leaves any of them without a factor that is 3, will force non-unit factors in common with 3 into P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 which would be a contradiction. > |Of course, here f=3, but what's amazing is that posters would try to > |convince anyone that having f change its value to being coprime to 3, > |could *introduce* a dependency on m. You find the possibility amazing. This is apparently all you have > to go on. There are plenty of mathematical claims where it would be > amazing if they were false, but nobody knows how to prove them yet. > For instance, nobody can prove yet that the decimal expansion of pi has > infinitely many 3's in it. The idea that they should appear on average > every tenth digit out to the tens of billions of places and then disappear > would be amazing. That's not a proof. So are you then claiming that you consider it possible that factors of f divide off with a dependency on m? > It's not the case that every function of m and f which is independent > of m for f having common factors with 3 is also independent of m > for other values of f. So to show that it doesn't happen in this case, > you'd have to tell us something specific about these common divisors > that makes it not just amazing to you but not actually possible. It's simple enough, as P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) where the focus is on m, and the question is how f^2 divides off. Now it might help to see the underlying P(m) without so many variables, so let x=1, u=1, f=7, so you have P(m) = 7^2((m^3 7^4 - 3m^2 7^2 + 3m) - 3(-1+m 7^2 ) + 7), which is P(m) = 49(2401 m^3 - 147 m^2 + 3m + 3 - 147m + 7), which is P(m) = 49(2401 m^3 - 147 m^2 -144 m + 10). Now the assertion by posters is that the 49 might divide off of P(m) as a function of m, because they attack my use of P(0), which in this case gives P(0) = 49(7) as a special case. And yes readers, you *can* stick in numbers to help yourself understand, and then you may also have the dubious pleasure of catching posters, who have advanced math training, like from Berkeley University's Ph.D program, lying to you. What I use is the non-polynomial factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf), which in this case is P(m) = (a_1 + 7)(a_2 + 7)(a_3 + 7) = 49(2401 m^3 - 147 m^2 -144 m + 10) with a special expression I call an uber-polynomial, which again is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f). So, yes, posters have been working to convince you that m=0 is a special case and that they can *tell* you how 49 divides off of P(m) = 49(2401 m^3 - 147 m^2 -144 m + 10) where they claim reducibility over rationals constrains how 49 divides through (a_1 + 7)(a_2 + 7)(a_3 + 7). The reality is that these methods are something beyond what mathematicians have used before, and with their own techniques, they can't say much about how 49 divides off of P(m), and they can't constrain it as a variable dependent on m. However, they fight the math because they must not believe in mathematics itself as a body of truths. Instead it is a *social* function to them, possibly just a way to pay the bills. So they fight letting society know of an error that has sat in mathematics for over a hundred years. They are working to convince you the reader. The reality is that these people have no way to force 49 to fit into any particular factorization given 49(2401 m^3 - 147 m^2 -144 m + 10) because they don't have advanced factorization techniques. > |And that explains why they delete out pertinent expressions as they > |want to convince readers. > | > |Or does it? > | > |I'm curious about what readers think. > | > |Why do YOU think posters like Keith Ramsay keep deleting out the math? It seems to me that you present a rather insulting picture of the > lurkers. You describe them as if they were such mindless sheep that > they can't read what you're saying and understand it, unless I keep > requoting it to them for you. Not just a few times either, but over and > over again. Just who do you think they are? My audience is a very particular one. > Even supposing they were intimidated somehow, don't you suppose > right, there really is a contradiction in the definition of algebraic > integer, how can we convince the world? I think it's safe to say > you're failing to attract support like that because you do NOT have > a clear and convincing case. Even supposing for the sake of > argument that we're all terribly confused and you're actually right > in your final conclusions, can't you see that it would be unclear > that you're correct? Keith Ramsay Grow up. James Harris ==== > There is a much more general question. Why does anyone read these James > Harris threads at all? > Good question. Maybe to see if he's ever going to bet back on his meds!? ==== James, Looking over recent threads I don't seem to be able to find where you posted one of the numbers which you claim *should be* in the ring of algebraic integers, but which is *left out*. It's not conceivable that you have overlooked this, since it is the fundamental basis for your claim the ring of algebraic integers is flawed, and the motivation for your notion of a ring of objects, whatever that is. Please take advantage of this wonderful opportunity to support your claim and post one of these numeric values (there must be millions of them). Of course, if you have something to hide, that's different --dare I say fascinating? -- 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 ==== >> There is a much more general question. Why does anyone read these James >> Harris threads at all? > Good question. Maybe to see if he's ever going to bet back on his meds!? My main reason is for the comedy. It usually happens like this: People spend weeks and weeks (or months and months) trying to pound XYZ into James' head. He yells and screams about liars, dueling proofs, military to punish mathematicians, etc. Then one day (and this is the delicious part) he suddenly exclaims, Oh no! I've just realized that XYZ may be true! If it is, I'll have to retract my proof! I need to think about this for a while. Then comes a period of anywhere from a few hours to a few days of James getting drunk, singing to the walls, struggling with despair and self-doubt, etc. Finally he emerges with a cry of victory: Eureka! XYZ is true, but it doesn't matter! I've just realized that Z is an uber-super-duper-poly-want-a-cracker-nomial, so my proof is right after all! Now we need to launch a Congressional investigation into why mathematicians have conspired to keep uber-super-duper-poly-want-a-cracker-nomials secret from me! And the whole mess starts all over again. You just can't buy entertainment like that. -- Wayne Brown | When your tail's in a crack, you improvise fwbrown@bellsouth.net | if you're good enough. Otherwise you give | your pelt to the trapper. e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock ==== >> Has this unusual piece been shot down yet? >> http://www.geocities.com/dharwadker/ >> It doesn't display properly for me in Netscape, Opera, Mozilla >> or Konqueror .... > > I can see it in Netscape Communicator 4.8. I can see it in Nutscrape 4.78 (indeed in all of these) > but from Section II onward visual gibberish starts to appear. Unfortunately I am unqualified to tell you if it is still mathematical gibberish even when it isn't visual gibberish. -- G.C. ==== > I can see it in Nutscrape 4.78 (indeed in all of these) > but from Section II onward visual gibberish starts to appear. The problem lies in the use of the Symbol font. There is a semi-official script http://hutchinson.belmont.ma.us/tth/Xfonts.fix which you can apply if you are using X-windows, after which the document becomes readable, although not, in my case, comprehensible. -- Timothy Murphy tel: +353-86-233 6090 ==== > http://www.geocities.com/dharwadker/ ... > (Paul) Let me know if you manage to get through all the jargon and see whether it > is a valid proof. As far as I can see, there is some wishful thinking in the paper. N is somewhere in the set {4,5,6}, true. There is a Steiner triple (n+1,2n,6n) for n=4 but not for n=5 or n=6, true. But how does that relate to colorations of planar graphs? It's none too clear even how he uses planarity or colorations. He does know all about the Witt pattern (5,8,24); but he seems to be too sanguine about its significance. The algebra about split extensions of groups seems to be okay; but most of those extensions are trivial, i.e. just direct products. LH ==== Please see http://www.tln.net/~reriker/fermat.html for the details. ==== >Please see http://www.tln.net/~reriker/fermat.html for the details. You start by taking a^q + b^q = c^q and transforming it into a pythagorean triple by writing (a^{q/2})^2 + (b^{q/2})^2 = (c^{q/2})^2 and then proceed to use the well known characterization of rational pythagorean triples. Now, tell me, if q is not a multiple of 2, then how do you guarantee that a^{q/2}, b^{q/2}, and c^{q/2} are rational numbers? For instance, say q=7. Then you need to take the 7th power of the square root of an integer. What makes you think that a, b, and c are perfect squares to begin with? ====================================================================== It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== > What is your defintion of program then? A formal definition would be: A Base of Computing (aka programming language) is a mapping from a recursively enumerable set onto the set of recursive functions. A program, then, would be any element in the domain of that mapping (Base of Computing.) But to capture the spirit of a program, we need to be less abstract and less formal, and just point out that: 1. A program is composed of constructs (whose semantics are) such as variable assignment, branching and conditionalized execution. 2. Some programs do not terminate on some inputs. A single expression isn't really a program. (Multiple expressions that allow recursion can be.) The problem is that it is too limited. You seem to be inputting and outputting only expressions (couched in the syntax of a LISP program.) That is, the program that you generate is limited to those that consist of a single expression, limiting your capabilities tremendously. It is not general program synthesis. Furthermore, I still don't see where you actually do anything with the expression, other than to input it and then output it again with LISP syntax surrounding it ö but I would like to learn more. I think that if you can generate programs for the 5 problems (specs) below, then that would go a long ways towards explaining, and demonstrating, how general your system is. The problem here is that I happened to give an example (determining if one number is a factor of another) for which there is in fact a LISP expression. In an attempt to determine if what you have in mind is very general (as general as the Predicate Calculus, which I use), here are some useful program specifications that I describe in my papers and the corresponding Predicate Calculus wff. (See my papers for the axioms and formal definitions of Number Theory. The 8 rules are universal for all domains. I will copy all of that here if people want, though it is all in my papers.) What would be the LISP expression for each? Again, my papers are at: http://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/20021008.1/1 and http://www.arxiv.org/html/cs.lo/0003071 1. List all prime numbers between 1000 and 1005. PRIME(x)^BETW(1000,x,1005) 2. Is a given number prime? PRIME(I) 3. Determine the largest proper factor of a given number. FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A) 4. Determine the largest prime factor of a given number. FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A) 5. Determine the smallest prime factor of a given number. FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x) If you can express these as LISP expressions, then we have something to talk about. Otherwise, you have a very limited system that is really just dealing with expressions as opposed to full-fledged programs, and you can't express, much less generate, hardly anything. > You do not want a program generator, you want a computerized human programmer. With all due respect, I think you're missing the point. 1. We can easily imagine a situation in which there is a need for a person to see the programs and subjectively choose his favorite. But this is not the point. 2. The system needs to CONSIDER multiple algorithms when generating the program, to differentiate it from the mere translation from one programming language into another. In the latter case, the user has to conceptualize the algorithm, then express it in one language to be translated into another language. In the former (which my system does), there is no algorithm being expressed by the user, and the system constructs (considers, outputs, whatever) the algorithms, thus producing multiple programs using different algorithms. For example, in the factor checking example, my system constructs 2 algorithms, which I have copied directly from my papers below. The English lines are added by me to explain what the program is doing. The other lines are the input (spec) entered by the user and the output (program) generated by the system: User Input: FAC(I,J) English: Is one given number a factor of another ? Program 1 English: Search for an integer A such that A*I=J. for A=1:1:J set B=A*I if B=J write (true) write (false) Program 2 English: Check the remainder of J/I. set A=rem(J,I) if A=0 write (true) write (false) > What I am doing is showing > that in fact, it is not much different from other systems, that *you* do not > accept as input specification. I would. My system is different from your system in that I can express and generate a wide range of programs, whereas you have yet to demonstrate that a single LISP expression can express that much. Here is your chance. Express the above 5 requirements. (My system is different from other systems in that they give no examples at all or are just programming language translators.) > As I said above, it qualifies as a programming language. No, Predicate Calculus has no assignment, branching or conditionalized execution. It can't get into loops. It is non-procedural. Again, how would you specify what the program is supposed to do? Show that a LISP expression can represent the above 5 problems, and then we have something to discuss. > While authors of bogus papers refuse to answer criticisms such as why > it never sees more than one algorithm, I will gladly respond to any > and all criticisms of my paper and the system that it describes. I am glad you are open to criticism. Please address mine concerning 1/ the > fact that PC qualifies as a programming language as much as Lisp does, and 2/ > the pointlessmess of generating many algorithms when a computer can generate > the best one from the start. See above for both 1 and 2. But first show how a LISP expression can represent the above 5 useful programming tasks. Charlie Volkstorf Cambridge, MA > Sam ==== > A single expression isn't really a program. Then (1 + 1) is not, contrary to what you said in your previous post. > The problem is that it is too limited. > You seem to be inputting and outputting only expressions (couched in > the syntax of a LISP program.) That is, the program that you > generate is limited to those that consist of a single expression, > limiting your capabilities tremendously. It is not general program > synthesis. I did not say it was. You asked me to write a program that generates a program that checks if a number is a factor of another. I did (actually I generalized that a bit, since it checks the truth of any expression). > Furthermore, I still don't see where you actually do anything with the > expression, other than to input it and then output it again with LISP > syntax surrounding it but I would like to learn more. I check for its validity. That's what needed to achieve what you asked. Note it in 30 seconds. My intent was to prove that Lisp an be used to generate programs. > The problem here is that I happened to give an example (determining if > one number is a factor of another) for which there is in fact a LISP > expression. Lisp is Turing complete. Thus anything you can specify using a Turing machine, you can specify with Lisp. > In an attempt to determine if what you have in mind is > very general (as general as the Predicate Calculus, which I use), here > are some useful program specifications that I describe in my papers. > What would be the LISP expression for each? Note that this would be more appropriate in comp.lang.lisp. I will be giving ways to *specify* your jobs. The actual implementation is irrelevant (albeit trivial in these cases), just as Predicate Calculus is only a syntax, whose implementation *is* procedural (partly given in your paper, in the form of APC, which is very akin to pseudo code). > 1. List all prime numbers between 1000 and 1005. > PRIME(x)^BETW(1000,x,1005) (format t ~s (mapcar 'prime' (list))) Where list contains all the elements. (prime x) evaluates to x if x is prime, to nil otherwise. > 2. Is a given number prime? > PRIME(I) (primep x) > 3. Determine the largest proper factor of a given number. > FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A) (max (proper-factors x)) > 4. Determine the largest prime factor of a given number. > FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A) (max (prime-factors x)) > 5. Determine the smallest prime factor of a given number. > FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x) (min (prime-factors x)) > If you can express these as LISP expressions, then we have something > to talk about. Otherwise, you have a very limited system that is > really just dealing with expressions as opposed to full-fledged > programs, and you can't express, much less generate, hardly anything. Refer to my comment above. Lisp is Turing complete. > 2. The system needs to CONSIDER multiple algorithms when generating > the program, to differentiate it from the mere translation from one > programming language into another. Again, you are applying the human way of solving problems to a computer. A well-designed system needs not differentiate [the actual program] from the mere translation from a programming language into another. It does not have the notion of such difference. It is designed to take a certain input that specifies a task to be done and return a program that performs the specified task. >> As I said above, it qualifies as a programming language. No, Predicate Calculus has no assignment, branching or conditionalized > execution. It can't get into loops. It is non-procedural. It still qualifies as a programming language, albeit not a Turing complete one (just as a Maple-like language without the procedural constructs would be) Sam -- Fear is the path to the dark side. Fear leads to anger, anger leads to hatred, hatred leads to suffering. I sense much fear in you. ==== A single expression isn't really a program. Then (1 + 1) is not, contrary to what you said in your previous post. Single expressions are a proper subset of programs. You can talk about a program that consists of a single expression, but that is not programming in general. A system that only creates single expressions is not much of a Program Synthesis system because very few programs consist of just one expression. It is very limited. (And I'm not so sure that your system does even that.) See my 5 examples below. > Furthermore, I still don't see where you actually do anything with the > expression, other than to input it and then output it again with LISP > syntax surrounding it but I would like to learn more. I check for its validity. That's what needed to achieve what you asked. So you just have a program syntax checker? > Note that I am not saying that my program is a general program generator. And I am saying that in fact it is not. So we agree? Well, by Program Synthesis, I mean a general program generator. > Lisp is Turing complete. Thus anything you can specify using a Turing machine, > you can specify with Lisp. With a single expression? Without sepcifying or knowing the algorithm that will be used? Program Synthesis means that the system constructs the algorithms. > 1. List all prime numbers between 1000 and 1005. > PRIME(x)^BETW(1000,x,1005) (format t ~s (mapcar 'prime' (list))) Where list contains all the elements. You don't even refer to 1000 and 1005! How did list get to contain the prime numbers between 1000 and 1005? > (prime x) evaluates to x if x is prime, to nil otherwise. So prime is a predicate supplied in LISP? > 2. Is a given number prime? > PRIME(I) (primep x) So primep is a predicate supplied by LISP? What's the difference between prime and primep? > 3. Determine the largest proper factor of a given number. > FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A) (max (proper-factors x)) So proper-factors is a construct supplied by LISP? > 4. Determine the largest prime factor of a given number. > FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A) (max (prime-factors x)) So prime-factors is a primitive supplied by LISP? > 5. Determine the smallest prime factor of a given number. > FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x) (min (prime-factors x)) > 2. The system needs to CONSIDER multiple algorithms when generating > the program, to differentiate it from the mere translation from one > programming language into another. Again, you are applying the human way of solving problems to a computer. So? Anything a computer can do a human can do. You are disqualifying every possible program. > A well-designed system needs not differentiate [the actual program] from the > mere translation from a programming language into another. It does not have > the notion of such difference. It doesn't. I am saying that if it is not a mere translator from one programming language to another, then multiple programs (using different algorithms) would be generated from one program requirement. >> As I said above, it qualifies as a programming language. No, Predicate Calculus has no assignment, branching or conditionalized > execution. It can't get into loops. It is non-procedural. It still qualifies as a programming language, albeit not a Turing complete one That is a self-contradiction. By definition, a programming language is Turing complete. The point is that it doesn't use programming language comnstructs and is non-procedural. What better way to be able to specify what a program is supposed to do? SUMMARY Are prime, primep, proper-factor and prime-factor all primitives supplied by LISP? If not, then how can you use them? If so, then at some point you will come across a program requirement that is not a primitive supplied by LISP. What do you do then? You seem to be merely declaring that each requirement is a primitive supplied by LISP and displaying trivial expressions using those primitives. I don't see any significance to that at all. You are not describing the general case where the programming requirement is not a LISP primitive. What would you do then? It sounds like you are talking about a programming system in which the merely compiles them into one program. That's just a compiler and not non-procedural input. Charlie Volkstorf Cambridge, MA > Sam ==== >> Furthermore, I still don't see where you actually do anything with the >> expression, other than to input it and then output it again with LISP >> syntax surrounding it but I would like to learn more. >> I check for its validity. That's what needed to achieve what you asked. So you just have a program syntax checker? I don't check the syntax. I check the truth of the expression. >> 1. List all prime numbers between 1000 and 1005. >> PRIME(x)^BETW(1000,x,1005) >> (format t ~s (mapcar 'prime' (list))) >> Where list contains all the elements. You don't even refer to 1000 and 1005! How did list get to contain > the prime numbers between 1000 and 1005? I said I'm not giving the whole implementation. This is not about Lisp programming, this is about specifying requirements. It's trivial to construct a list that contains all integers between 1000 and 1005. >> (prime x) evaluates to x if x is prime, to nil otherwise. So prime is a predicate supplied in LISP? Irrelevant. If it's not supplied, supply it. (indeed it is not part of the CL standard). That applies to the following functions too. If you're arguing that PC is better because it provides a prime checking predicate by default, then you're agreeing that PC is merely a higher level language. Less important, prime is *not* a predicate. A predicate returns either t or nil (true or false). (prime x) returns x if x is prim, nil otherwise. primep (that follows) is a predicate. >> Again, you are applying the human way of solving problems to a computer. So? Anything a computer can do a human can do. You are disqualifying > every possible program. You misunderstood my critic. I'm not saying it's not possible to do so (though it would be hard), I'm saying it's not the most efficient way. >> A well-designed system needs not differentiate [the actual program] from >> the mere translation from a programming language into another. It does not >> have the notion of such difference. It doesn't. I am saying that if it is not a mere translator from one > programming language to another, then multiple programs (using > different algorithms) would be generated from one program requirement. Show it! That is, write an implementation of your system. For now, you only have a paper that describes your ideal system. It does not describe how to actually implement that system. Lisp has implementations. > By definition, a programming language is Turing complete. That is false. A programming language needs not be Turing complete. A programming language is a language that allows you to specify a task to be done by the computer. It needs not be able to specify *any* task the computer can do. > The point is that it doesn't use programming > language comnstructs and is non-procedural. Simply because you provided (in your design, since there is no implementation) enough high level feaures so that we do not need branching to solve the problems you are addressing. Note that it does not provide features tro adress other kinds of problems. Thus the following: Using PC, please provide a non procedural construct that verifies if a given set with 2 laws is a vector space. PC does not provide any facility to verify if a given object is a member of a given set, neither does it provide a way to construct objects or laws, so I doubt this is even adressable. Note that this is not even program generation. If you manage to do that, go on to the following: Using PC, provide a program that generates 2 algorithms that verify if a given subset of a vector space, with the vector space laws, is a sub-vector space. And last, provide a program, using PC, that generates algorithms to check if a given file on a Unix system can be opened by the current user. > SUMMARY [summary snipped] I have adressed these above in this post. > It sounds like you are talking about a programming system in which the > merely compiles them into one program. That's just a compiler and not > non-procedural input. That is so, to a certain extent. But PC is not different, apart from the fact that it is not Turing complete and does provide higher level functions not defined in the Common Lisp standard (but that could be part of a math library). PC is a mathematical non-procedural language. The sole fact that it is purely mathematical makes it unsuitable for program generation outside the context of mathematics. Sam -- Creativity can be a social contribution, but only in so far as society is free to use the results. - Richard Stallman ==== >It is often said that the Lorentz transformation can be derived from >the homogeneity and isotropy of space alone. I'm looking for a >concrete counterexample to this claim. They can't be. Homogeneity (isn't isotropy redundant with homogeneity?) and the assumption of no special frame (principle of relativity) gets you, for a boost in x, x' = a(x - vt) y' = y z' = z t' = bt + cx These are Galilean if a=1, b=1, c=0. To derive the Lorentz transforms you can suppose when the origin of two frames S and S' coincide a flash of light expands out in a sphere. Throw in the additional postulate of the invariance of the speed of light and you get equations in the two frames describing the sphere, (1) x^2 + y^2 + z^2 = c^2 t^2 (2) x'^2 + y'^2 + z'^2 = c^2 t'^2 Note that equation (2) explicitly has the speed of light unchanged and the light shell spherical in S'. Insert the primed quantities defined above into equation (2), rearrange terms, and find which values of a, b, c are needed to reproduce equation (1). The Lorentz transforms can be derived from the principle of relativity and Maxwell's equations. That is, since no experiment seems able to identify an aether or aether frame, rather than taking Lorentz's course of assuming the aether to interact with matter such that it remains undetectable, take Einstein's course and assume there really is no special frame, that the principle of relativity applies to electromagnetism as well as to mechanics. Maxwell's equations remain invariant under Lorentz transformations, so you can do an exercise similar to the above but with more difficult math. Either way, you need something more than just homogeneity. Maybe what you've often seen said just didn't make it clear that Maxwell's equations were the additional input. -- A good plan executed right now is far better than a perfect plan executed next week. -Gen. George S. Patton ==== >>It is often said that the Lorentz transformation can be derived from >>the homogeneity and isotropy of space alone. I'm looking for a >>concrete counterexample to this claim. > They can't be. Homogeneity (isn't isotropy redundant with homogeneity?) No. Homogeneity means the manifold in question looks the same everywhere; isotropy means it looks the same no matter which direction one looks. An anisotropic manifold can be homogeneous, as long as it looks the same everywhere (i.e. some special direction is everywhere the same). E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2 is homogeneous but not isotropic. Of course a simple change of coordinates can make it isotropic -- that is not always the case: Minkowski spacetime is homogeneous but is not isotropic in 4-d, only in its spatial 3-d slices. A manifold that is isotropic everywhere is necessarily homogeneous. > and the assumption of no special frame (principle of relativity) gets you, > for a boost in x, > x' = a(x - vt) > y' = y > z' = z > t' = bt + cx You can further constrain a,b,c by requiring these transforms form a group (i.e. that the composition of two transforms of this type results in another transform of this type). One finds there are three possibilities: 1) the Galileo group 2) the Lorentz group 3) the Euclid group (3) is unphysical (x,y,z,t all look the same, and the composition of large-enough boosts along +x results in a boost along -x). > These are Galilean if a=1, b=1, c=0. To derive the Lorentz transforms you > can suppose when the origin of two frames S and S' coincide a flash of > light expands out in a sphere. Throw in the additional postulate of the > invariance of the speed of light [...] The Lorentz transforms can be derived from the principle of relativity and > Maxwell's equations. [...] Given the constraints imposed by the group property, one merely needs to select from the three possible groups. A postulate like there is an upper bound on speeds suffices, but one needs a measurement to establish that this upper bound is the speed of light. A postulate like pion beams exist serves double duty (and is manifestly true).... > Either way, you need something more than just homogeneity. Yes. Tom Roberts tjroberts@lucent.com ==== >It is often said that the Lorentz transformation can be derived from >the homogeneity and isotropy of space alone. I'm looking for a >concrete counterexample to this claim. >> They can't be. Homogeneity (isn't isotropy redundant with homogeneity?) No. Homogeneity means the manifold in question looks the same >everywhere; isotropy means it looks the same no matter which direction >one looks. An anisotropic manifold can be homogeneous, as long as it >looks the same everywhere (i.e. some special direction is everywhere >the same). E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2 > is homogeneous but not isotropic. Of course a simple change > of coordinates can make it isotropic -- that is not always > the case: Minkowski spacetime is homogeneous but is not > isotropic in 4-d, only in its spatial 3-d slices. A manifold that is isotropic everywhere is necessarily homogeneous. Ehh... I guess I'd buy that, as long as homogeneity is defined as looks the same everywhere along one particular direction. I was thinking of homogeneity as looks the same everywhere, no qualifiers. >> and the assumption of no special frame (principle of relativity) gets you, >> for a boost in x, >> x' = a(x - vt) >> y' = y >> z' = z >> t' = bt + cx You can further constrain a,b,c by requiring these transforms form a >group (i.e. that the composition of two transforms of this type results >in another transform of this type). One finds there are three possibilities: > 1) the Galileo group > 2) the Lorentz group > 3) the Euclid group I'm not familiar with this argument. Could you clue me in? -- A good plan executed right now is far better than a perfect plan executed next week. -Gen. George S. Patton ==== >>It is often said that the Lorentz transformation can be derived from >>the homogeneity and isotropy of space alone. I'm looking for a >>concrete counterexample to this claim. > They can't be. Homogeneity (isn't isotropy redundant with homogeneity?) I tend to agree. To me a point having the same properties as every other point defines homogeneity. What does homogeneity mean without isotropy? > No. Homogeneity means the manifold in question looks the same > everywhere; isotropy means it looks the same no matter which direction > one looks. An anisotropic manifold can be homogeneous, as long as it > looks the same everywhere (i.e. some special direction is everywhere > the same). I understand the distinction you're making technically and wonder if a concrete illustration of an anisotropic, homogeneous manifold exists. > E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2 > is homogeneous but not isotropic. Please pardon my ignorance. I had differential geometry from Ted Frankel. We never covered any practical examples. What are the mathematical definitions of homogeneity and isotropy in this case? Please explain the essential features of this geometry and the special direction. Why are 3 dimensions required? Why doesn't your example work with just ds^2 = 2dx^2 + 3dy^2 ? Eugene Shubert http://www.everythingimportant.org ==== Perfectly Innocent skrev i melding >>It is often said that the Lorentz transformation can be derived from >>the homogeneity and isotropy of space alone. I'm looking for a >>concrete counterexample to this claim. > They can't be. Homogeneity (isn't isotropy redundant with homogeneity?) I tend to agree. To me a point having the same properties as every > other point defines homogeneity. What does homogeneity mean without > isotropy? No. Homogeneity means the manifold in question looks the same > everywhere; isotropy means it looks the same no matter which direction > one looks. An anisotropic manifold can be homogeneous, as long as it > looks the same everywhere (i.e. some special direction is everywhere > the same). I understand the distinction you're making technically and wonder if a > concrete illustration of an anisotropic, homogeneous manifold exists. Many crystalline substances (e.g. quartz) are homogenous, but not isotropic. The em-properties, including the anisotropy, are the same everywhere in the crystal. Think of it. Wouldn't it be rather meaningless to say that a crystal which macroscopically is the same everywhere isn't homogenous? Paul ==== What was I thinking? The cylinder SxR is obviously homogeneous but not isotropic! Eugene Shubert http://www.everythingimportant.org/relativity/simultaneity.htm ==== > To me a point having the same properties as every > other point defines homogeneity. What does homogeneity mean without > isotropy? As applied to fields on a manifold, homogeneity means the field is independent of position, and isotropy means the field is independent of orientation. No nonzero vector field can be isotropic except possibly at isolated points, but can certainly be homogeneous if it is constant everywhere. For the manifold itself, it is common to say the manifold is isotropic or homogeneous iff the metric is. For instance, a 3-d spherically-symmetric manifold is isotropic around its center, but unless it is isometric to E^3 it is not homogeneous, and is not isotropic at any other location. The surface of a sphere S^2 (with the usual metric) is both homogeneous and isotropic. Note I use orientation in the more general sense than the binary choice of orientation for orientable manifolds. I believe there is a technical term for this, but forget what it is. > I understand the distinction you're making technically and wonder if a > concrete illustration of an anisotropic, homogeneous manifold exists. Minkowski spacetime in 4 dimensions. Timelike orientations are DIFFERENT from spacelike orientations, yet the metric is independent of position. The surface of a cylinder RxS (with the usual metric) is isotropic and homogeneous by the above definition. Topologically it certainly has a special orientation, but isotropy is a geometrical property, not a topological one. >> E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2 >> is homogeneous but not isotropic. I misspoke. The manifold and its metric are indeed isotropic, but the COORDINATES are not. Oops. This manifold is isometric to E^3 (up to topological considerations which aren't mentioned). Tom Roberts tjroberts@lucent.com ==== > Be careful with what you mean by one-to-one. Goedel's method is a > method for setting up a one-to-one correspondence between the > expressions in the calculus and a _certain subset_ GC of the integers. In particular, not every integer is a Goedel number. Consider the number > 100 = 2^2*5^2. Looking at this prime factorization it becomes > immediatelly obvious that the number 100 cannot be a Goedel codon, since > powers of the prime 3 are skipped, TOGETHER with the fact that powers of > 5 appear. This is by definition not allowed. On the other hand, given G in GC, you can easily decompose G into its > prime number factorization and assuming the decomposition follows proper > codification rules (i.e. no numbers like 100) therefrom derive a unique > sequence of exponents of the first n primes, which will lead you back to > a sequence of fundamental codons, which can be used to recover the > calculus expression uniquely. The above paragraph can be used slightly modified to prove what you > want. > number we get to a unique sequence of exponents and, given a codification, > to a statement in PM. My question is: how is it proved that there exists a > complete codification between statements in PM and these exponent > (basically natural numbers). In other word how does Godel prove that all > statements in PM can be mapped into the natural numbers? Intuitively what is > the Godel number of the class of all classes? > The essense is that the statements are countable. The Godel numbering shows that any sequence of alphabet symbols is countable. Are you implying that formula may exist that are not sequences of alphabet symbols? Herc ==== > Be careful with what you mean by one-to-one. Goedel's method is a > method for setting up a one-to-one correspondence between the > expressions in the calculus and a _certain subset_ GC of the integers. In particular, not every integer is a Goedel number. Consider the number > 100 = 2^2*5^2. Looking at this prime factorization it becomes > immediatelly obvious that the number 100 cannot be a Goedel codon, since > powers of the prime 3 are skipped, TOGETHER with the fact that powers of > 5 appear. This is by definition not allowed. On the other hand, given G in GC, you can easily decompose G into its > prime number factorization and assuming the decomposition follows proper > codification rules (i.e. no numbers like 100) therefrom derive a unique > sequence of exponents of the first n primes, which will lead you back to > a sequence of fundamental codons, which can be used to recover the > calculus expression uniquely. The above paragraph can be used slightly modified to prove what you > want. a > number we get to a unique sequence of exponents and, given a codification, > to a statement in PM. My question is: how is it proved that there exists a > complete codification between statements in PM and these exponent > (basically natural numbers). In other word how does Godel prove that all > statements in PM can be mapped into the natural numbers? Intuitively what is > the Godel number of the class of all classes? > The essense is that the statements are countable. The Godel numbering shows > that any sequence of alphabet symbols is countable. Are you implying that > formula may exist that are not sequences of alphabet symbols? Herc > What I am asking is: how is it proved that the symbols from which these statements are derived is countable. For example, these symbols cannot include irrational numbers. Taking it a step further does this mean that there are no statements about irrational numbers in PM? ==== >number we get to a unique sequence of exponents and, given a codification, >to a statement in PM. My question is: how is it proved that there exists a >complete codification between statements in PM and these exponent >(basically natural numbers). In other word how does Godel prove that all >statements in PM can be mapped into the natural numbers? Given a specification of the language PM (in particular the alphabet it uses), this is very easy. > Intuitively what is >the Godel number of the class of all classes? It is not mathematical objects, but their descriptions in the language, that have Godel numbers. The same object could have many different descriptions in the language, and each will have a different Godel number. On the other hand, some objects (e.g. almost all real numbers) can not be completely described in the language. I don't know how you'd describe the class of all classes in PM. If you were encoding English, with letters corresponding to their ASCII codes, you could give the class of all classes the Godel number 2^116 * 3^104 * 5^101 * ... * 89^115, where 116, 104, 101, ... are the ASCII codes of t, h, e, .... Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== architechture. Unlike JSH, there is nothing groundbreaking or earthshattering about it. Are there any journals that look for papers on the GR specifically (like Fibonnaci Quarterly for example) or any other journals that would look for a paper on this topic. ==== 8.9 x 14.4 ? ==== Out of curiosity. Why on earth do sci.math'ers - which otherwise seems to be quite intelligent people - even consider replying to James Harris' posts? First of all, he is the crank fatale of our century. Second, he is probably the kind of crank who rolls on the floor laughing his ass off each time someone replies. It appears that replying to Harris is like watching a gold fish in a bowl. You can't stop, yet you know it's not going anywhere but around and around, ad infinitum. ==== > Out of curiosity. Why on earth do sci.math'ers - which otherwise seems > to be quite intelligent people - even consider replying to James > Harris' posts? Another Harris basher. What a surprise. > First of all, he is the crank fatale of our century. Second, he is > probably the kind of crank who rolls on the floor laughing his ass off > each time someone replies. That would make him a troll, not a crank. If you must insult him, at least get your insults right. > It appears that replying to Harris is like watching a gold fish in a > bowl. You can't stop, yet you know it's not going anywhere but around > and around, ad infinitum. It does seem to be going around and around, doesn't it? And here's why: *** James Harris has proved Fermat's Last Theorem *** but people are either unwilling or unable to accept it. I admit that I personally cannot understand his proof: it's beyond my comprehension. James says it should be easy for everyone to understand, but (please forgive my impudence, James, in offering you some constructive criticism) perhaps he has failed to take into account that he is so much smarter than the rest of us that what seems easy to him may not be so easy for us. The reason that his threads go around and around is that people keep having the same misunderstandings, and he keeps trying to enlighten them with the truth . . . but to no avail. I wonder whether people are as thick as they seem, or if they have some sort of mental block against the mere possibility of a simple proof of FLT (at least by a non-mathematician), or, and this is what I fear, if they *do* understand that it's correct, but are attempting to stifle it out of some perverted form of self-interest. But maybe I'm just being paranoid. The reaction to James could be explained by sheer stupidity. There's not necessarily a conspiracy. I myself, until recently, did not accept Harris's proof because . . . (drumroll please) . . . I did not understand it. For example, take his use of objects. I can sense that what he's doing here is profound and radical. In contrast to the rampant abstraction of conventional mathematics, James is working with concrete things, things that have an actual existence, not just some member of a set, which was constructed under the ZFC axioms, with such and so operators defined to have these various properties, yadda yadda yadda. James doesn't need all that crap! He just gets to the core of the matter in one leap! I wish I could follow him. I don't even understand what is and is not an object. But unlike certain abusive Harris harrassers, I have the guts to admit that the fault lies with me for not understanding. How cowardly to blame James for one's incomprehension! Shame on you people! The real crime here is not that FLT is being ignored -- sometimes it is only generations later that genius is understood -- but that sci.math is wasting so much of James's time with our nitpicking. James has proved FLT. It's over and done with. And sure, that would be enough for one lifetime, but think about how much more he has to offer! James has already begun work on the theory of prime numbers. I for one would like to see him bring this work to fruition in a proof of the Riemann Hypothesis. Then maybe we can talk about this whole mathematical messiah thing without the disruption of rude snickerings. ==== Can anyone help me identify the sequence of polynomials defined by the recurrence relation: p_n (x) = x p_{n-1} (x) + p_{n-2} (x) This leads to the following few terms: p_0 = 0 p_1 = 1 p_2 = x p_3 = 1 + x^2 p_4 = 2x + x^3 p_5 = 1 + 3x^2 + x^4 p_6 = 3x + 4x^3 + x^5 p_7 = 1 + 6x^2 + 5x^4 + x^6 -Lotofun ==== > Can anyone help me identify the sequence of polynomials defined by the > recurrence relation: p_n (x) = x p_{n-1} (x) + p_{n-2} (x) This leads to the following few terms: p_0 = 0 > p_1 = 1 > p_2 = x > p_3 = 1 + x^2 > p_4 = 2x + x^3 > p_5 = 1 + 3x^2 + x^4 > p_6 = 3x + 4x^3 + x^5 > p_7 = 1 + 6x^2 + 5x^4 + x^6 If you arrange the coefficients carefully, you get Pascal's triangle. -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== Can anyone help me identify the sequence of polynomials defined by the > recurrence relation: p_n (x) = x p_{n-1} (x) + p_{n-2} (x) This leads to the following few terms: p_0 = 0 > p_1 = 1 > p_2 = x > p_3 = 1 + x^2 > p_4 = 2x + x^3 > p_5 = 1 + 3x^2 + x^4 > p_6 = 3x + 4x^3 + x^5 > p_7 = 1 + 6x^2 + 5x^4 + x^6 If you arrange the coefficients carefully, you get > Pascal's triangle. p_{2m} (x) = sum_{k=0}^m C(m+k,2k) x^{2k} and p_{2m+1} (x) = sum_{k=0}^m C(m+k+1,2k+1) x^{2k+1} where C(n,j) is the binomial coefficient. ==== > Can anyone help me identify the sequence of polynomials defined by the > recurrence relation: p_n (x) = x p_{n-1} (x) + p_{n-2} (x) This leads to the following few terms: p_0 = 0 > p_1 = 1 > p_2 = x Let C = sqrt(x^2/4 + 1), A = x/2 + C, B = x/2 - C then p_n = (A^n - B^n)/(2C) ==== >> |There have been NO demonstrations to the contrary. If you have one, >> |give it. >> I've lost count of how many times they've presented a non-unit common >> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3 >> as usual are algebraic integers satisfying >> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). >> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5, >> and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy >> to post the _calculation_ again if we thought you'd read it. Huh? Are you sure you're claiming that the non-monic primitive >irreducible over Q supports your case? You are just saying words. 8a_1^2 + 4a_1 - 45 is not a non-monic primitive irreducible over Q. It's not even a polynomial. It's an algebraic integer, just like 8*(4)^2 + 4(4) - 45 is an integer. >Oh, I know. You're probably talking about that argument that Arturo >Magidin tossed at me in a post that I just refuted. Basically any such claims depend on the assertion that an algebraic >integer 'a' that is not a unit, must have a non-unit algebraic integer >factor. If a is not a unit, then a itself is a non-unit algebraic integer factor of itself. That's because this ring has a 1. If that's hard for you, think about integers: any integer other than 1, -1, is a factor of itself. >That's a sneaky little argument as it's getting the reader to assume >something not proven, which is that 'a' actually has a non-unit >algebraic integer factor because it's not a unit. > Your argument, on the other hand, always relies at the crucial point >> on the reader sharing your feeling that the kind of weird behavior >> the common divisors have as f and m vary just couldn't possibly >> be so. >> Keith Ramsay Well if you're telling the truth, fill in the gap I've pointed out in >YOUR argument. The gap is the assertion that given an algebraic integer 'a' which is >not a unit, that 'a' must have some non-unit factor *in the ring of >algebraic integers*. This is nonsense. >To protect readers from thinking it's simple, I remind them of the >ring of evens, that is the set of even numbers, where you have 2 not a >factor of 6, and not a unit. But 2 and 6 don't have factors in the >ring. That's because the ring doesn't have 1. Are you claiming that the ring of algebraic integers does not have a 1? ====================================================================== 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 ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== [.snip.] >If that's hard for you, think about integers: any integer other than >1, -1, is a factor of itself. That, of course, should be non-unit factor or nontrivial factor. ==== >|There have been NO demonstrations to the contrary. If you have one, >>|give it. >>I've lost count of how many times they've presented a non-unit common >>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3 >>as usual are algebraic integers satisfying >> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). >>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5, >>and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy >>to post the _calculation_ again if we thought you'd read it. > Huh? Are you sure you're claiming that the non-monic primitive > irreducible over Q supports your case? Oh, I know. You're probably talking about that argument that Arturo > Magidin tossed at me in a post that I just refuted. Basically any such claims depend on the assertion that an algebraic > integer 'a' that is not a unit, must have a non-unit algebraic integer > factor. > The factor that has been provided, r(a) = 8 a^2 - 4 a - 45 is > (1) an algebraic integer for any algebraic integer a, > (2) a divisor of both a and 5 for > a = -(any of the ai's in the above > factorization of 65x^3 - 12x + 1) The factorizations have already been given, but since you're so adept > at ignoring the evidence, I will provide them once again: q(a) = 8 a^2 - 76 a - 185 > r(a) = 8 a^2 - 4 a - 45 > s(a) = 4 a^2 - 37 a - 104 Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative* > of any of the ai's of the above factorization of 65x^3 - 12x + 1), the > following factorizations hold: q(a) r(a) = 5 > r(a) s(a) = a. Hmmm...the poster has tried to dupe *you* the reader, and it all has to do with the word unit. See below... > In fact, the minimal polynomial of this number (r(a) for -a = any of > the above ai's) is given as: MinPoly(r) = x^3 - 969 x^2 + 315 x + 5 The above facts prove that this a has a non-unit algebraic integer > as a factor. No it doesn't. What you can prove is that if 'a' is coprime to 5 it can't be a unit in the ring of algebraic integers, but as I've said the ring of algebraic integers is screwed up. Your algebraic manipulations will just keep running into that over and over again. > That's a sneaky little argument as it's getting the reader to assume > something not proven, which is that 'a' actually has a non-unit > algebraic integer factor because it's not a unit. > This is actually *much* easier to prove: the ring of algebraic > integers is closed under the extraction of roots of monic > polynomials with algebraic integer coefficients. In particular, > every equation: x^n - a = 0 That's not the equation you have. What you have is x^3 - 969 x^2 + 315 x + 5 and what you want to do is convince readers that none of its roots can be coprime to 5. But you cannot do that, and instead have maintained that proving that none of the roots can be a unit proves that one of them must have a non-unit factor in common with 5 in the ring of algebraic integers. I've just shot down that little trick, but you're squirming. > has algebraic integer roots, whenever a is an algebraic integer, > and n is a natural number. Thus, if a is an algebraic integer, and n is any natural number, > the number a^(1/n) is an algebraic integer, irrespective of which of the n roots you take. Duh, but you're not talking about such a simple expression as x^n - a = 0. You're talking about x^3 - 969 x^2 + 315 x + 5 = 0 and *repeatedly* posters like you try to fool readers on the newsgroups as you work to *convince* rather than get to the mathematical truth. The truth is that you're relying on the negative, which is that NONE of the roots of that expression can be units, to try and prove the positive, which is that then they all have a non-unit factor in common with 5 in the ring of algebraic integers. But that's beyond bogus as my point is that the ring of algebraic integers is screwed up to the extent that a root of that expression can be coprime to 5, and yet not be a unit. >Your argument, on the other hand, always relies at the crucial point >>on the reader sharing your feeling that the kind of weird behavior >>the common divisors have as f and m vary just couldn't possibly >>be so. >>Keith Ramsay > Well if you're telling the truth, fill in the gap I've pointed out in > YOUR argument. > No gap. That's what you want to convince readers. However, simply asserting something that is not true does not fill the gap. Again, what you have is the negative--certain numbers aren't units--and you want the positive--that they all have to have non-unit factors in common with 5. But you haven't proven the positive. It's a gap. Denying it won't fill it, so quit being lazy and try to fill it. > The gap is the assertion that given an algebraic integer 'a' which is > not a unit, that 'a' must have some non-unit factor *in the ring of > algebraic integers*. > As I mentioned above, the factor r that I provided *is* an algebraic > integer, and in the case that a is (-1) times any of the ai's in the > factorization cited, I have provided its minimal polynomial. It is > simple to verify from that minimal polynomial that it's an algebraic > integer and that it's not a unit in the ring of algebraic integers. See readers? You can catch the trick here as notice the poster said not a unit, and what I'm telling you is that these posters are working hard to convince YOU the reader. > To protect readers from thinking it's simple, I remind them of the > ring of evens, that is the set of even numbers, where you have 2 not a > factor of 6, and not a unit. But 2 and 6 don't have factors in the > ring. That's because the ring doesn't have 1. You have to go to a higher ring, in this case the ring of integers, to > escape that interesting little situation. Now the problem is different in the ring of algebraic integers, but > the ring is screwed up, and the fix is again a higher ring. What I want you to consider is the *possibility* that weird things can > happen with rings, when you're not careful enough, and that > mathematicians weren't careful enough with the ring of algebraic > integers. > The problem is with your ignorance. The ring of algebraic integers has > none of the shortcomings you seem to want it to have. It is your method > which is at fault, since it forces you to believe a falsehood, namely > that the ai's in that above factorization are coprime to 5. Now this poster is the one who tried to fool all of you, and now that I've caught him, he decides to hurl insults. These posters are intellectually lazy, and they're rude. But they're not lazy when it comes to posting as they *keep* posting now don't they? I'm curious about what you, the reader, thinks is the reason for their persistence. Why do YOU think they keep posting? James Harris ==== >> |There have been NO demonstrations to the contrary. If you have one, >> |give it. >> I've lost count of how many times they've presented a non-unit common >> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3 >> as usual are algebraic integers satisfying >> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). >> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5, >> and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy >> to post the _calculation_ again if we thought you'd read it. Huh? Are you sure you're claiming that the non-monic primitive >irreducible over Q supports your case? Oh, I know. You're probably talking about that argument that Arturo >Magidin tossed at me in a post that I just refuted. Basically any such claims depend on the assertion that an algebraic >integer 'a' that is not a unit, must have a non-unit algebraic integer >factor. That's a sneaky little argument as it's getting the reader to assume >something not proven, which is that 'a' actually has a non-unit >algebraic integer factor because it's not a unit. If that's really what is behind it, then it's easy to prove: if a is not a unit in a ring with 1, then a is a nonunit factor of a. Again, remember: DEFINITION. Let R be a commutative ring, and let a and b be elements of R. Then a is a factor of b (in R) if and only if there exists an element c in R such that a*c = b. Since a=1*a, then a is a factor of a. Since a is not a unit, a is a non-unit factor of a. Surely you agree with ->that<-. As to the refutation, you sure talked a lot, but you did not refute anything. Here's why you did not refute it: You are demanding a proof that if a1, a2, a3 are algebraic integers chosen so that 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1), then each of a1, a2, a3 have a non unit factor in common with 5. That is, we need to find algebraic integers r1,s1,q1,r2,s2,q2,r3,s3,q3 such that (1) r1*s1 = a1; r1*q1 = 5 (so r1 is a factor in common of a1 and 5); (2) r2*s2 = a2; r2*q2 = 5 (so r2 is a factor in common of a2 and 5); (3) r3*s3 = a3; r3*q3 = 5 (so r3 is a factor in common of a3 and 5); (4) none of r1, r2, r3 are units in the ring of algebraic integers. Surely, you agree, that would necessarily show that a1 has a non unit factor in common with 5 (namely r1); that a2 has a nonunit factor in common with 5 (namely r2); and that a3 has a nonunit factori n common with 5 (namely r3). The values were given explicitly: Note that -a1^3 - 12*a1^2 + 65 = 0 -a2^3 - 12*a2^2 + 65 = 0 -a3^3 - 12*a3^2 + 65 = 0. Let r1 = 8*a1^2 + 4*a1 - 54 q1 = 8*a1^2 + 76a1 - 185 s1 = -(4a1^2 + 37a1 - 104) r2 = 8*a2^2 + 4*a2 - 54 q1 = 8*a2^2 + 76a2 - 185 s1 = -(4a2^2 + 37a2 - 104) r3 = 8*a3^2 + 4*a2 - 54 q1 = 8*a3^2 + 76a3 - 185 s1 = -(4a3^2 + 37a3 - 104) Then a simple calculation using the fact that -a1^3 - 12*a1^2 + 65 = 0; -a2^3 - 12*a2^2 + 65 = 0; -a3^3 - 12*a3^2 + 65 = 0; shows that r1*q1 = r2*q2 = r3*q3 = 5. Another simple calculation shows that r1*s1 = a1 r2*s2 = a2 r3*s3 = a3. (The explicit calculations may be found either in Dale's original post, http://groups.google.com/groups?selm=3F148963.6000800%40farir.com or my more recent almost-verbatim-quote in http://groups.google.com/groups?selm=bh6666%24a7e%241%40agate.berkeley.edu ) Note also that all of r1, r2, r3, s1, s2, s3, q1, q2, a3 are algebraic integers. This simply because they are obtained from algebraic integers by addition and multiplication by integers and/or algebraic integers. So it is CERTAINLY the case that r1 is an algebraic integer which is a common factor, IN THE RING OF ALGEBRAIC INTEGERS, of a1 and 5; that r2 is an algebraic integer which is a common factor IN THE RING OF ALGEBRAIC INTEGERS, of a2 and 5; and that r3 is an algebraic integer which is a common factor IN THE RING OF ALGEBRAIC INTEGERS, of a3 and 5. We have not, in any way, left or been pushed out of the ring of algebraic integers. Finally, we just need to verify that none of r1, r2, or r3 are units in the ring of algebraic integers. If that is indeed the case, WE ARE DONE. And we are done, because we already know they are common factors, we only need to know they are not units in this ring. The word unit is shorthand. It means nothing more and nothing less than: DEF. Let R be a ring with 1. An element a in R is a unit (in R) if and only if there exists an element b IN R such that a*b=b*a = 1. In a commutative ring, of course, it is enough to check a*b=1. Then we have the following easy theorem, consequence of the theorem you accepted last December after 10 months of denying it: THEOREM. Let a<>0 be an algebraic integer, and let f(x) be a monic polynomial with integer coefficients, irreducible over Q, which has a as a root. Then a is a unit in the ring of all algebraic integers if and only if the constant term of f(x) is either 1 or -1. Proof. Let b be the unique complex number such that a*b=1. That is, b=1/a. By definition, a is a unit in the ring of all algebraic integers if, and only if, b is an algebraic integer. Let f(x) = x^n + ... + a_1*x + a_0. Then f(a) = 0. Multiplying through by b^n, we have 0 = b^nf(a) = b^n*a^n + ... + a_1*a*b^{n} + a_0*b^n = 1 + ... = a_1*b^{n-1} + a_0*b^n. Let g(x) = a_0*x^n + ... + a_{n-1}*x + 1. That means that g(b) = 0. Since any factorization of g(x) will produce a factorization of f(x) by substituting x^{-1} for x and clearing denominators, we see that g(x) is an irreducible polynomial with integer coefficients. Since the last term equals 1, it is therefore primitive. Therefore, b is an algebraic integer if and only if the leading term of g(x) is either 1 or -1. Since the leading term of g(x) is a_0, that says that b is an algebraic integer if and only if the constant term of f(x) is 1 or -1. IN summary: a is an algebraic integer unit if and only if b is an algebraic integer; b is an algebraic integer if and only if the constant term of f(x) is 1 or -1. Therefore, a is an algebraic integer unit if and only if the constant term of f(x) is 1 or -1. QED So one way to prove that none of r1, r2, r3 are units in the ring of all algebraic integers is to exhibit a monic polynomial with integer coefficients, irreducible over Q, with constant term different from 1 and -1, which has r1, r2, r3 as roots. (Not the only way: we could try finding a different one for each ri, but it so happens they are all roots of the same polynomial). And indeed, as Dale noted, if we let f(x) = x^3 - 969x^2 + 315x + 5, it so happens that f(r1)=f(r2)=f(r3)=0. The irreducibility over Q of f(x) may be checked by the rational root theorem: the only possible roots are 1, -1, 5, and -5, none of them are roots, and so we are done. Now, you can dance around the issue all you want, James. You can produce paragraph upon paragraph, post upon post, saying all sorts of things about problems and sneakiness and how it is impossible to show something or other. But you cannot get around one very simple issue: we have EXPLICITLY shown the factors, we have EXPLICITLY shown they are factors, we have EXPLICITLY shown they are not units, we have EXPLICITLY shown they are common factors. We have EXPLICITLY produced the factors you claim cannot exist. Anything else is sophistry. (Though I say we, the credit goes of course to Dale who actually took the time to make the calculations and initial verifications). ====================================================================== 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 ===================================================================== >|There have been NO demonstrations to the contrary. If you have one, >|give it. I've lost count of how many times they've presented a non-unit common >divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3 >as usual are algebraic integers satisfying 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5, >and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy >to post the _calculation_ again if we thought you'd read it. >> Huh? Are you sure you're claiming that the non-monic primitive >> irreducible over Q supports your case? >> Oh, I know. You're probably talking about that argument that Arturo >> Magidin tossed at me in a post that I just refuted. >> Basically any such claims depend on the assertion that an algebraic >> integer 'a' that is not a unit, must have a non-unit algebraic integer >> factor. >> The factor that has been provided, >> r(a) = 8 a^2 - 4 a - 45 >> is >> (1) an algebraic integer for any algebraic integer a, >> (2) a divisor of both a and 5 for >> a = -(any of the ai's in the above >> factorization of 65x^3 - 12x + 1) >> The factorizations have already been given, but since you're so adept >> at ignoring the evidence, I will provide them once again: >> q(a) = 8 a^2 - 76 a - 185 >> r(a) = 8 a^2 - 4 a - 45 >> s(a) = 4 a^2 - 37 a - 104 >> Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative* >> of any of the ai's of the above factorization of 65x^3 - 12x + 1), the >> following factorizations hold: >> q(a) r(a) = 5 >> r(a) s(a) = a. Hmmm...the poster has tried to dupe *you* the reader, and it all has >to do with the word unit. See below... > In fact, the minimal polynomial of this number (r(a) for -a = any of >> the above ai's) is given as: >> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5 >> The above facts prove that this a has a non-unit algebraic integer >> as a factor. No it doesn't. Why not? There is an algebraic integer, called here r(a), with the following three properties: (1) It is not a unit in the ring of all algebraic integers; you apparently agree with that conclusion, but then engage in sophistry about a supposed problem with it. (2) There is an algebraic integer, called s(a), such that r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic integers. (3) There is an algebraic integer, called q(a), such that r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic integers. So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5. Why is it not the case that r(a) is a non-unit algebraic integer which is a common factor of a and 5? [.snip.] >What you have is x^3 - 969 x^2 + 315 x + 5 and what you want to do is convince readers that none of its roots can >be coprime to 5. There is no convincing that needs to be done. NONE Of the roots are units, and ALL the roots are factors of 5. You're done. >But you cannot do that, Correction: James Harris cannot do that because he does not know what coprime means. > and instead have maintained that proving that >none of the roots can be a unit proves that one of them must have a >non-unit factor in common with 5 in the ring of algebraic integers. They are each FACTORS of 5, and EACH non-units. Do you agree or disagree with (all happening in a commutative ring WITH 1): (a) If a is a factor of b, then a and b have a common factor (namely, a). (b) If a is a factor of b and is not a unit, then a is a non-unit common factor of a and b. (c) If is a factor of b and is not a unit, then a and b have a non-unit common factor (namely, a). >I've just shot down that little trick, but you're squirming. I do see a lot of squirming, but it's not coming from Dale. [.snip.] >The truth is that you're relying on the negative, which is that NONE >of the roots of that expression can be units, to try and prove the >positive, which is that then they all have a non-unit factor in common >with 5 in the ring of algebraic integers. You are completely, totally, utterly lost and confused. The roots themselves are factors of 5. Each root is a factor of itself, and a factor of 5. Therefore, each root is a common factor of itself and 5. And since each of them is not a unit, then we have EXHIBITED a non-unit common factor of each with 5. The roots are not the original numbers we were interested in (the a's), they are the common factors that have been produced. [.snip.] >> As I mentioned above, the factor r that I provided *is* an algebraic >> integer, and in the case that a is (-1) times any of the ai's in the >> factorization cited, I have provided its minimal polynomial. It is >> simple to verify from that minimal polynomial that it's an algebraic >> integer and that it's not a unit in the ring of algebraic integers. See readers? You can catch the trick here as notice the poster said >not a unit, and what I'm telling you is that these posters are >working hard to convince YOU the reader. Are you claiming that non-unit and not a unit are not the same thing? [.snip.] ====================================================================== 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 ====================================================================== Arturo Magidin magidin@math.berkeley.edu ==== >|There have been NO demonstrations to the contrary. If you have one, >>|give it. >>I've lost count of how many times they've presented a non-unit common >>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3 >>as usual are algebraic integers satisfying >> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). >>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5, >>and likewise for a_2 and a_3 by substituting them for a_1. We'd be happy >>to post the _calculation_ again if we thought you'd read it. >Huh? Are you sure you're claiming that the non-monic primitive >irreducible over Q supports your case? Oh, I know. You're probably talking about that argument that Arturo >Magidin tossed at me in a post that I just refuted. Basically any such claims depend on the assertion that an algebraic >integer 'a' that is not a unit, must have a non-unit algebraic integer >factor. >The factor that has been provided, >> r(a) = 8 a^2 - 4 a - 45 >>is >> (1) an algebraic integer for any algebraic integer a, >> (2) a divisor of both a and 5 for >> a = -(any of the ai's in the above >> factorization of 65x^3 - 12x + 1) >>The factorizations have already been given, but since you're so adept >>at ignoring the evidence, I will provide them once again: >> q(a) = 8 a^2 - 76 a - 185 >> r(a) = 8 a^2 - 4 a - 45 >> s(a) = 4 a^2 - 37 a - 104 >>Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative* >>of any of the ai's of the above factorization of 65x^3 - 12x + 1), the >>following factorizations hold: >> q(a) r(a) = 5 >> r(a) s(a) = a. > Hmmm...the poster has tried to dupe *you* the reader, and it all has > to do with the word unit. See below... > >In fact, the minimal polynomial of this number (r(a) for -a = any of >>the above ai's) is given as: >> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5 >>The above facts prove that this a has a non-unit algebraic integer >>as a factor. > No it doesn't. What you can prove is that if 'a' is coprime to 5 it > can't be a unit in the ring of algebraic integers, but as I've said > the ring of algebraic integers is screwed up. > No, no one can prove that. After all, a unit *is* coprime to every number! Let u be a unit algebraic integer, and let p be any algebraic integer. Then there is an algebraic integer v for which uv = 1, and we get: u v + 0 p = 1, where 0 is zero. This establishes the fact that a UNIT algebraic integer is coprime to every algebraic integer, in the ring of algebraic integers. A very slight rewording proves the result for a unit of any commutative ring. No wonder you think the ring of algebraic integers is screwed up! Let's get the language straight. I have been clear in this that 'a' refers to (-1) times any one of the coefficients ai in the factorization 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). I claim that *each* of those numbers a has a non-unit algebraic integer factor in common with 5. PROOF: The above expressions q(a),r(a),s(a) all yield algebraic integers for any algebraic integer a. When, in addition, a is a root of the polynomial x^3 - 12 x^2 + 65 then the products are as I mentioned above: q(a) r(a) = 5 r(a) s(a) = a. Your claim: > > No it doesn't. What you can prove is that if 'a' is > coprime to 5 it can't be a unit in the ring of algebraic > integers, but as I've said the ring of algebraic integers > is screwed up. amounts to this: PARAPHRASE: It doesn't matter whether anyone provides algebraic integers q,r, and s for which 5 = q*r a = r*s it doesn't prove that a and 5 are not coprime. However, it *does* prove that. Assume otherwise, that a is coprime to 5. Then there exist algebraic integers u and v for which a u + 5 v = 1 and the above factorizations show this: rs u + qr v = 1 so r( su + qv ) = 1, and we deduce that r is a unit. However, it's not one of your fanciful Unit with no inverse hallucinations. In fact, the above equation provides us with an inverse for r. Based on its construction, it *must be* an algebraic integer: u and v are specified to be algebraic integers, and q,r,s were already specified as algebraic integers. Thus, r must be an algebraic integer unit, in the ordinary sense: an algebraic integer whose inverse is *also* an algebraic integer. However, I show below that the minimal polynomial for r precludes it being a unit. > Your algebraic manipulations will just keep running into that over and > over again. > I've given the proof before, and it's only cut 'n' paste, so here it is, once again: Given: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 A quick bit of arithmetic will show the following: q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325 r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680 and: (64 x + 128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320 (32 x + 72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680 and so, if you care to compare: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x, where p(x) = x^3 - 12 x^2 + 65. Note that this immediately shows that for z = any root of p(x), q(z)*r(z) = 5, r(z)*s(z) = z. To derive the minimal polynomial: >> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5 let mpr(x) = x^3 - 969 x^2 + 315 x + 5. First, expand the polynomial mpr(r(x)): mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5 = (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2 + 315 (8 x^2 - 4 x - 45) + 5 = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 Next, multiply these two polynomials: p(x) = x^3 - 12 x^2 + 65, and w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808 to get this: p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 Notice the equality mpr(r(x)) = p(x)*w(x). That means for every value of x, the polynomial you get by computing r(x), then evaluating mpr(x) at that value, is equal to the product of p(x) and w(x). In short, if x is a root of p(x) (in particular, this is true for x = -ai for any of your ai's), mpr(r(x)) = 0. Thus, the r's are roots of the polynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5 This cubic polynomial has integer coefficients, is monic, and has no integer roots, thus it's irreducible. Its roots cannot be units in the ordinary sense (i.e., with inverses within the ring of algebraic integers), because the constant term is a non-unit. So, what can I conclude at this point? 1. each of the a's has an algebraic integer factor in common with 5: 5 = q(a)*r(a) a = r(a)*s(a). where q(a) = 8 a^2 - 76 a - 185 r(a) = 8 a^2 - 4 a - 45 s(a) = 4 a^2 - 37 a - 104 These values q(a),r(a),s(a) are all algebraic integers. 2. If a and 5 were coprime, then there would be an inverse *in the ring of algebraic integers* for r(a). 3. The minimal polynomial mpr(x) for r(a) is: mpr(x) = x^3 - 969 x^2 + 315 x + 5 That's a sneaky little argument as it's getting the reader to assume >something not proven, which is that 'a' actually has a non-unit >algebraic integer factor because it's not a unit. >This is actually *much* easier to prove: the ring of algebraic >>integers is closed under the extraction of roots of monic >>polynomials with algebraic integer coefficients. In particular, >>every equation: >> x^n - a = 0 > That's not the equation you have. > Idiot! I'm showing that your alleged something not proven, as I quote: ... it's getting the reader to assume something not proven, which is that 'a' actually has a non-unit algebraic integer factor because it's not a unit. is hogwash! EVERY NON-UNIT ALGEBRAIC INTEGER HAS NON-UNIT FACTORS AMONG THE ALGEBRAIC INTEGERS! I'll repeat that: Take any algebraic integer K that is NOT a unit. Then take the equation x^n - K = 0. and solve it. The roots are algebraic integers, and they are ALSO non-unit factors of K. What was it you claimed was unproven: that 'a' actually has a non-unit algebraic integer factor because it's not a unit. What did I just find: a non-unit algebraic integer factor K^(1/n), for an ARBITRARY non-unit algebraic integer K. Does that address this tiny little point? Can we get off that one then? You're wrong, I'm right. > What you have is x^3 - 969 x^2 + 315 x + 5 and what you want to do is convince readers that none of its roots can > be coprime to 5. > This is ridiculous. What was to be shown was this: NONE OF THE COEFFICIENTS ai IN THE FACTORIZATION 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). IS COPRIME TO 5. I showed, for each a, two factorizations: 5 = q(a) r(a) a = r(a) s(a). The common factor r has minimal polynomial x^3 - 969 x^2 + 315 x + 5 I don't have to show *ANYTHING* about that polynomial: its roots are by definition algebraic integers, and the inverses of those roots have minimal polynomial: 5 x^3 + 315 x^2 - 969 x + 1 and so none of them is a unit. > But you cannot do that, and instead have maintained that proving that > none of the roots can be a unit proves that one of them must have a > non-unit factor in common with 5 in the ring of algebraic integers. > Idiot. I have displayed the factor r(a), that you have claimed does not exist. That later polynomial is the minimal polynomial for the common factor r(a), which shows that r(a) is not a unit. Geez, it's just high school algebra, what's the confusion? I have just exhibited the common factor. I have shown such a factor for *each* of those roots. I have also shown the minimal polynomial for those common factors, and used it to show that the common factor is not a unit algebraic integer. > I've just shot down that little trick, but you're squirming. > Please show me an error. You've had probably three months of looking at the derivation, in one form or another, and *THAT* is all you can come up with? I see that you could be confused, since all this is just simple numbers, and anyone could be confused with numbers. >has algebraic integer roots, whenever a is an algebraic integer, >>and n is a natural number. >>Thus, if a is an algebraic integer, and n is any natural number, >>the number >> a^(1/n) >>is an algebraic integer, irrespective of which of the n roots you take. > Duh, but you're not talking about such a simple expression as x^n - a = 0. > No, I was showing you that ANY non-unit algebraic integer has non-unit algebraic integer factors, in fact will have infinitely many of them. > You're talking about x^3 - 969 x^2 + 315 x + 5 = 0 and *repeatedly* posters like you try to fool readers on the > newsgroups as you work to *convince* rather than get to the > mathematical truth. > I have the truth up there. You are now acting to deny it. That's OK, since we have been through YEARS of seeing you behave like a little boy who has pooped his britches. No, I don't have poop in my britches, um, I grew a tail! > The truth is that you're relying on the negative, which is that NONE > of the roots of that expression can be units, to try and prove the > positive, which is that then they all have a non-unit factor in common > with 5 in the ring of algebraic integers. > Hey, WHAT negative: 5 = q(a)*r(a) a = r(a)*s(a). Where's the negative? This is what I've shown, and you still can't see through your own hatred of mathematicians to recognize that it plainly shows your error. > But that's beyond bogus as my point is that the ring of algebraic > integers is screwed up to the extent that a root of that expression > can be coprime to 5, and yet not be a unit. > Where did I claim something's not a unit? Not in the above pair of expressions, in which the root is shown NOT to be coprime to 5. No, it was in ANOTHER expression altogether: x^3 - 969 x^2 + 315 x + 5 where I argue that none of the roots of THIS polynomial can be a unit. ... stuff deleted ... >>No gap. > That's what you want to convince readers. However, simply asserting something that is not true does not fill the > gap. > I made the following assertions. The proofs, up to verification of a few polynomial expansions, can be seen above, but the interested reader can produce these products explicitly, without any form of influence on my part. ASSERTION 1: Given: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104 we have q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x, where p(x) = x^3 - 12 x^2 + 65. Corollary: q(a)r(a) = 5, r(a)s(a) = a Therefore a and 5 have the common factor r(a) in the algebraic integers. ASSERTION 2: Given the polynomials mpr(x) = x^3 - 969 x^2 + 315 x + 5 r(x) = 8 x^2 - 4 x - 45 p(x) = x^3 - 12 x^2 + 65. w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808 we have: mpr(r(x)) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 and p(x)w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520 Therefore, for a = any root of p(x), mpr(r(a)) = 0, so mpr(x) is a polynomial for which r(a) is a root. mpr(x) is easily seen to be irreducible, and so is the minimal polynomial of r(a). Corollary: NONE OF THE FACTORS r(a) is a unit in the ring of algebraic integers. > Again, what you have is the negative--certain numbers aren't > units--and you want the positive--that they all have to have non-unit > factors in common with 5. > I have shown as much, but you refuse to look at it: a and 5 have common factors in the ring of algebraic integers at least one of those common factors (r(a)) is not a unit So, you're confusing what has to be shown non coprime to 5, with what has to be shown not to be a unit: a is not coprime to 5 r(a) is not a unit. > But you haven't proven the positive. > Of course I have, but you got your ears all plugged up and refuse to listen. > It's a gap. Denying it won't fill it, so quit being lazy and try to fill it. > You should say that to yourself once in a while. The gap is the assertion that given an algebraic integer 'a' which is >not a unit, that 'a' must have some non-unit factor *in the ring of >algebraic integers*. >As I mentioned above, the factor r that I provided *is* an algebraic >>integer, and in the case that a is (-1) times any of the ai's in the >>factorization cited, I have provided its minimal polynomial. It is >>simple to verify from that minimal polynomial that it's an algebraic >>integer and that it's not a unit in the ring of algebraic integers. > See readers? You can catch the trick here as notice the poster said > not a unit, and what I'm telling you is that these posters are > working hard to convince YOU the reader. > What? Now you're claiming that I have attempted to fool some unnamed class of readers by asserting something's not a unit? Anyone can scroll back to verify *exactly* what I am claiming. I've labelled it clearly. To protect readers from thinking it's simple, I remind them of the >ring of evens, that is the set of even numbers, where you have 2 not a >factor of 6, and not a unit. But 2 and 6 don't have factors in the >ring. That's because the ring doesn't have 1. You have to go to a higher ring, in this case the ring of integers, to >escape that interesting little situation. Now the problem is different in the ring of algebraic integers, but >the ring is screwed up, and the fix is again a higher ring. What I want you to consider is the *possibility* that weird things can >happen with rings, when you're not careful enough, and that >mathematicians weren't careful enough with the ring of algebraic >integers. >The problem is with your ignorance. The ring of algebraic integers has >>none of the shortcomings you seem to want it to have. It is your method >>which is at fault, since it forces you to believe a falsehood, namely >>that the ai's in that above factorization are coprime to 5. > Now this poster is the one who tried to fool all of you, and now that > I've caught him, he decides to hurl insults. > Please stop mischaracterizing a proof as an attempt at fraud. Just because YOU couldn't follow the fairly simple arithmetic I've provided, and just because it showed that everything you've been doing for, what is it, 8 years now(?), has been for no good reason at all? You have perpetually misunderstood the mathematics you're attempting to work with, and have, to date, NEVER accepted even the smallest bit of responsibility for knowing what it is you're talking about. It is not an insult to say that the problem with all this is that you are ignorant of the subject. That is the simple truth. Have I insulted you? How? By saying you're ignorant of algebra? Can you demonstrate that you know some nontrivial algebraic topic, say Galois theory? What about any of the [other] standard theorems about fields and their extensions? To claim that you have caught me, you should at least provide some statement that you take issue with. All I saw above was your claim that I was making the argument: If 'a' isn't a unit, then 'a' can't be coprime to 5. where my argument was this: A. Here are two products: a,q,r,s are all algebraic integers: 5 = q*r a = r*s Then, unless 5 is a unit, then a and 5 have a non-unit common factor in the ring of algebraic integers B. Here is the minimal polynomial for r(a). mpr(x) = x^3 - 969 x^2 + 315 x + 5 Thus, r(a) is not a unit C. Thus, a and 5 share a non-unit algebraic integer factor. > These posters are intellectually lazy, and they're rude. > Um, intellectually lazy? I think you just have a little file of all the epithets that have come your way, and choose one that seems to fit. Unfortunately, I have no grounds for claiming not to have been intellectually lazy. Oh, I did find those factors you said were impossible, didn't I? And that other cubic a couple of weeks ago, where you said that no one could provide the monic polynomial for which something was a root. I gave you that polynomial. So, I guess that I'm intellectually lazy, yet can do what mankind cannot possibly do, by your own estimation! I'm SUPERMAN!!!! WOO HOOOO!!! Mere mortals, step aside! I can solve your problems, even those that mankind cannot possibly do, and not break a mental sweat. SOOOO PPERRRRRR...MAAAAAAANNNNNNN!!!!!! Yesssssssssssssssssss!!!!!!!!!!! I gotta get one of those cape thingies, some blue tights, with the red codpiece, er, maybe I can be Clark Kent? I never did like those superhero gymnasts' costumes, capes, tiaras, big S on the shirt. It always seemed a bit, er, sissy for my tastes. Well, I have time to reflect on that one for awhile. I have crushed your argument. It has nowhere to go, now that we can all see how it predicts stuff that is plain wrong. It's fine for symbols, you can push little a's and b's and u's and v's around on your paper all day long, and once you come to plugging real values in for the variables it says that the a's are coprime to 5. That's got to be embarassing. Yeah, there's no insult on your side. At least I've been factually correct: you ARE ignorant, and parade that ignorance around day after day. In addition, I haven't been running around proclaiming to have shown something to be true, all the while ignoring a direct calculation that shows it to be false. Who has been doing that? Now that you're pretending to address the issue (since it won't go away), who is it who is continuing to mis-label every thing, out of some vain belief that if it fools the old JSHter, it'll fool someone else on earth. I think for these purposes, you will just have to settle on being THE MOST GULLIBLE MAN ON EARTH. No one else will ever believe you, and you will end up at the age of 50 a broken little man who couldn't let it go when it was clear he was done. > But they're not lazy when it comes to posting as they *keep* posting > now don't they? > Let's do a post count for me vs. you WDH: 693 JSH: 4180 if I count JSH-related posts, I have 322. About half of what I contribute to sci.math goes to JSH nonsense. Already I'm ashamed of myself. Boy, I'm just a-burnin' up that internet, don't you think? Next baseless accusation? > I'm curious about what you, the reader, thinks is the reason for their > persistence. > Why do YOU think I keep posting? I've already given my reasons: you are a punk you leech of the help of others and treat your benefactors shamefully. you have an inflated sense of your own worth, originality, talent, intelligence you behave as though yours is the only thought that's worth having you bully others you clog the newsgroup with ill will, and lead other discussions into the ground you despise what it is that I have spent the better part of my life doing. you cast aspersions at honorable people, and make idle threats you are a cynical abuser of the very concept of intellectual debate, you are a crybaby, your useage of the English language is substandard, even for a southerner. you pretend to lecture those who are your betters (in terms of comprehension, intuition, and experience). > Why do YOU think they keep posting? > James Harris Dale ==== [snip hysterical rant] Is your nose starting to grow? If you had any credibility left, which you don't, your own posts over the last few days would have gone a long way toward demolishing it. You continue to ignore those posts from others which disprove your claim that the ring of algebraic integers is flawed. The flaw is entirely between your ears. Just to recap (for the record): Every root of a monic polynomial with integer coefficients divides the constant term within the ring of algebraic integers. That is, the quotient obtained by dividing the constant term by any root is an algebraic integer. The above is easy to prove, and has been shown repeatedly by more than one poster, but so far seems beyond your comprehension. Furthermore, you have not shown a single example of a number which *should be* in the ring of algebraic integers, but which is *left out*. Tsk, tsk. That doesn't look good for you, James. You have claimed repeatedly that such numbers exist, but you have never produced a single one! At this point, what you have claimed is a 'linchpin of my FLT proof' has become a 'nail in your FLT proof's coffin'. -- 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 ==== The factor that has been provided, >> r(a) = 8 a^2 - 4 a - 45 >> is >> (1) an algebraic integer for any algebraic integer a, >> (2) a divisor of both a and 5 for >> a = -(any of the ai's in the above >> factorization of 65x^3 - 12x + 1) >> The factorizations have already been given, but since you're so adept >> at ignoring the evidence, I will provide them once again: >> q(a) = 8 a^2 - 76 a - 185 >> r(a) = 8 a^2 - 4 a - 45 >> s(a) = 4 a^2 - 37 a - 104 >> Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative* >> of any of the ai's of the above factorization of 65x^3 - 12x + 1), the >> following factorizations hold: >> q(a) r(a) = 5 >> r(a) s(a) = a. Hmmm...the poster has tried to dupe *you* the reader, and it all has >to do with the word unit. See below... > In fact, the minimal polynomial of this number (r(a) for -a = any of >> the above ai's) is given as: >> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5 >> The above facts prove that this a has a non-unit algebraic integer >> as a factor. No it doesn't. Why not? There is an algebraic integer, called here r(a), with the > following three properties: (1) It is not a unit in the ring of all algebraic integers; you > apparently agree with that conclusion, but then engage in > sophistry about a supposed problem with it. It is true that r(a) is not a unit in the ring of algebraic integers. Now what are the actual expressions? They are q(a) r(a) = 5 r(a) s(a) = a. where I simply note that r(a) may not have non-unit factors in common with 5, which *should* make q(a) have a factor of 5 ***in the ring of algebraic integers***, but the ring of algebraic integers is screwed up. In trying to prove that r(a) must have a non-unit factor in common with 5, you rely on the fact that it can't be a unit ***in the ring of algebraic integers***!!! The trick is to act as if proving the negative, proves that it must have some non-unit factor in common with 5, but there's a fascinating error with the ring of algebraic integers. (2) There is an algebraic integer, called s(a), such that > r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic > integers. Well, what if 'a' doesn't share non-unit factors in common with 5? Then both r(a) and s(a) don't either. That *should* leave q(a) with a factor of 5, but the ring is screwed up. In order to try and prove that it does have a factor of 5, you have to try and claim that it must because otherwise r(a) would be a unit. But the ring of algebraic integers is screwed up, which is my point. > (3) There is an algebraic integer, called q(a), such that > r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic > integers. However, how do you know that r(a) must share non-unit factors in common with 5? Ultimately your claim must be that it's because it's not a unit in the ring of algebraic integers!!! > So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5. Why is it not the case that r(a) is a non-unit algebraic integer which > is a common factor of a and 5? For readers, these posters are working to *convince* not get to the bottom of the problem. If they cared about the truth, then they only need consider my proof of the problem with algebraic integers. Proofs don't duel. If they're correct, then they'd be able to find an error with my argument, but it has no error as the ring of algebraic integers IS screwed up, so instead they work to convince YOU the reader. > [.snip.] >What you have is x^3 - 969 x^2 + 315 x + 5 and what you want to do is convince readers that none of its roots can >be coprime to 5. There is no convincing that needs to be done. NONE Of the roots are > units, and ALL the roots are factors of 5. You're done. No, you have a gap. The gap is that now you need to prove that the roots share non-unit factors in common with 5 in the ring of algebraic integers. >But you cannot do that, Correction: James Harris cannot do that because he does not know what > coprime means. Which is the semantic argument. > and instead have maintained that proving that >none of the roots can be a unit proves that one of them must have a >non-unit factor in common with 5 in the ring of algebraic integers. They are each FACTORS of 5, and EACH non-units. Which does NOT prove that they each have a non-unit factor in common with 5, because the ring of algebraic integers is flawed. > Do you agree or disagree with (all happening in a commutative ring > WITH 1): (a) If a is a factor of b, then a and b have a common factor > (namely, a). (b) If a is a factor of b and is not a unit, then a is a non-unit > common factor of a and b. (c) If is a factor of b and is not a unit, then a and b have a > non-unit common factor (namely, a). Hmmm...I can see that you're dedicated at working to convince the audience, so I'll give a demonstration to show them *how* you're working. Consider abc = 5, where a = sqrt(5), b=sqrt(5)(-1+sqrt(-3))/2, c=(-1-sqrt(-3))/2, but imagine that you're talking to someone who doesn't know about radicals, as all they know about are integers. Now neither 'a' nor 'b' is an integer, but this person is using d=ab, and they now see cd = 5 and tell you that 'c' and 'd' must be factors of 5 ***in the ring of integers***. You say nope. Well they come back, and argue, and point out that neither is a unit ***in the ring of integers***. You say, yup, that is correct. Then they come back and say, well, cd = 5, neither is a unit in the ring, so they must be factors of 5. And you say, nope, they are not. But the ring of algebraic integers is VERY screwed up, so that only gives you an idea. >I've just shot down that little trick, but you're squirming. I do see a lot of squirming, but it's not coming from Dale. What mathematicians can do, because it's such an odd error, is keep casting doubt, and running away from the proof of the error. Probably for most of you the idea that you could have abc=5 where neither 'a', 'b', nor 'c' is a factor of 5, in the ring of algebraic integers, seems nonsensical. However, it's an esoteric problem in an esoteric branch of mathematics, which has been there for over a *hundred* years. If mathematicians want to confuse most of you about it, they can. > [.snip.] The truth is that you're relying on the negative, which is that NONE >of the roots of that expression can be units, to try and prove the >positive, which is that then they all have a non-unit factor in common >with 5 in the ring of algebraic integers. You are completely, totally, utterly lost and confused. The roots themselves are factors of 5. Each root is a factor of > itself, and a factor of 5. Therefore, each root is a common factor of > itself and 5. And since each of them is not a unit, then we have > EXHIBITED a non-unit common factor of each with 5. The roots are not > the original numbers we were interested in (the a's), they are the > common factors that have been produced. > > [.snip.] And again the trick is that use of the word unit as in fact the factors are NOT units ***in the ring of algebraic integers***. In a higher, more complete ring, one of them IS a unit, while two of them have a factor that is sqrt(5). >> As I mentioned above, the factor r that I provided *is* an algebraic >> integer, and in the case that a is (-1) times any of the ai's in the >> factorization cited, I have provided its minimal polynomial. It is >> simple to verify from that minimal polynomial that it's an algebraic >> integer and that it's not a unit in the ring of algebraic integers. See readers? You can catch the trick here as notice the poster said >not a unit, and what I'm telling you is that these posters are >working hard to convince YOU the reader. Are you claiming that non-unit and not a unit are not the same > thing? [.snip.] Nope. For readers who don't realize how this ring business can be confusing, consider my example with the ring of evens. That ring doesn't have 1 in it, as 1 is odd, so 2 and 6 do not share factors with each other. However, you go to the higher ring--integers--and they do. Mathematicians have this problem where they didn't think there was one, until I pushed it into the open. Rather than be fascinated by it, and accept the mathematics, you can see posters like Arturo Magidin instead working to convince YOU. Why does he have to keep replying to me in posts? Because I'm right. If he goes away, and isn't around to confuse readers, he probably realizes that eventually I might convince some of you to follow the mathematical logic. That explains why posters like Arturo Magidin are so dedicated. They need to hang around to keep YOU confused. James Harris ==== |Well if you're telling the truth, fill in the gap I've pointed out in |YOUR argument. | |The gap is the assertion that given an algebraic integer 'a' which is |not a unit, that 'a' must have some non-unit factor *in the ring of |algebraic integers*. One obvious answer is that 'a' is a factor of itself. If you mean nontrivial factor, then the square roots of 'a' are non-unit factors of 'a'. If 'a' is an algebraic integer, then it's a root of a monic polynomial with integer coefficients x^n + c_1*x^{n-1} + ... + c_n = 0. If b^2=a, then b is a root of the monic polynomial with integer coefficients x^{2n} + c_1*x^{2n-2} + ... + c_n = 0. We know b is not a unit in the algebraic integers, because if b were a unit then by definition there would be some algebraic integer c such that bc=1. But then we would have 1=(bc)^2 =b^2c^2 = ac^2. Because c is an algebraic integer, then so is c^2. That would imply a is a unit, but we assumed to start with that a was not a unit. If you want an explicit proof that if c is an algebraic integer then so is c^2: suppose c is the root of a polynomial x^m + d_1*x^{m-1} + ... + d_m = 0 with integer coefficients. Then we can group the even terms on one side and the odd terms on the other: d_m + d_{m-2}*x^2 + ... = -(d_{m-1} + d_{m-3}*x^2 + ...)*x. Square both sides: (d_m+d_{m-2}*x^2+...)^2 = (d_{m-1}+d_{m-3}*x^2+...)*x^2. This is now a polynomial in x^2. Substituting y for x^2 we get a monic polynomial with integer with integer coefficients which has c^2 as a root. |To protect readers from thinking it's simple, I remind them of the |ring of evens, that is the set of even numbers, where you have 2 not a |factor of 6, and not a unit. But 2 and 6 don't have factors in the |ring. | |That's because the ring doesn't have 1. Before reading this past year's discussion, I hadn't realized what the definition of coprime was in such cases. So you sometimes learn something new. (Not that this definition was ever important to anything I had to do.) |You have to go to a higher ring, in this case the ring of integers, to |escape that interesting little situation. | |Now the problem is different in the ring of algebraic integers, but |the ring is screwed up, and the fix is again a higher ring. | |What I want you to consider is the *possibility* that weird things can |happen with rings, when you're not careful enough, and that |mathematicians weren't careful enough with the ring of algebraic |integers. I'll do that if you agree to do something much easier. Consider the possibility that weird things are causing the common factors which you believe to be independent of m to depend upon m, but that you weren't careful enough. Keith Ramsay <3c65f87.0308110729.3f68620@posting.google.com> <3F37D13C.40008@farir.com> <3c65f87.0308111350.19f0910e@posting.google.com> <3F38305E.4050008@farir.com> ==== > your useage of the English language is > substandard, even for a southerner. Don't be an ass. At least, not in the same sentence in which you misspell usage and fail to capitalize Southerner. (Now, where's my requisite typo in this spelling flame?) -- Jesse Hughes [I]t's the damndest thing. There's something wrong with every last one of you, and I *never* thought that was a possibility. But now I feel it's the only reasonable conclusion. --JSH sees some sorta light ==== >The integration problem >int p Beta(p|a,b) dp >where a and b are the parameters for the beta distribution, has the >exceedingly nice solution >a / (a + b). >Is there a somewhat nice (tho certainly not AS nice!) solution >available to the related integration problem: >int log(p) Beta(p|a,b)dp >If so, it would be very useful for something I'm working on. I'd be >very grateful for any help! It is Psi(a) - Psi(a+b). Psi is the derivative of the logarithm of Gamma. Just differentiate the equation int p^{a-1}(1-p)^{b-1}dp = Gamma(a)*Gamma(b)/Gamma(a+b) with respect to a. -- 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 ==== >The integration problem int p Beta(p|a,b) dp >where a and b are the parameters for the beta distribution, has the > exceedingly nice solution [....] >[Herman Rubin=.. It is Psi(a) - Psi(a+b). Psi is the derivative of > the logarithm of Gamma. Just differentiate the > equation > int p^{a-1}(1-p)^{b-1}dp = Gamma(a)*Gamma(b)/Gamma(a+b) > with respect to a. (a)_k=a(a+1)...(a+k-1) . k=1,2,... , (a)_0:=1 -(Pochhammer Symbol) F_{2,1}(a,b;c; z)=F(a,b;c; z)= =SUM_{k=0 to k=infty}(a)_k*(b)_k*z^k/((c)_k*k!) -(Gauss Hypergeometric Series) F_{3,2}(a,b,c;p,q; z)= =SUM_{k=0 to k=infty}(a)_k*(b)_k*(c)_k*z^k/((p)_k*(q)_k*k!) -(a generalization of Gauss Hypergeometric Series) G(z)=Gamma Function , B(p,q)=Beta Function , Psi(z)=d(G(z))/dz =G'(z)/G(z)= the Logarithmic Derivative of Gamma Function, and for a>0 , b>0 (1) T(a,b;f;z)= =(1/B(a,b))*Integral_{t=0 to t=1}t^{b-1}*(1-t)^{a-1}*f(zt) dt (Beta Transform of f) GAry need t a ,,closed form for (#) Z:= T(a,b; ln(.); 1) = ?? For instance, let us note that (2) ln(z) =-(1-z)*F(1;1;2;1-z) , |z-1|< 1 , and (3) F_{3,2}(a,a_1,a_2;a+b,b_1; z)= T(a,b;H;z) with H(x)=F(a_1,a_2;b_1; x) . STEP 1; select in (3) a_1=a_2=1 , b_1=2 and z=1 . According to (1)-(2) one finds for a>0 , b>0 (3') F_{3,2}(a,1,1;a+b,2; 1) = = -(1/B(a,b))*Integral_{t=0 to t=1}t^{b-1}*(1-t)^{a-1}*(1-t)*ln(1-t) dt = = -(1/B(a,b))*Integral_{t=0 to t=1}t^a*(1-t)^{b-1}*ln(t) dt = = - (a/(a+b))* T(a+1,b;ln(.) ; 1) . Therefore (see (#)) Z=T(a,b;ln(.);1)= - ((a+b-1)/(a-1))* F_{3,2}(a-1,1,1;a+b-1,2; 1) , a>1,b>0 . STEP 2; Use the fact that for A=/=1 , C-A > 0 , we have (5) F_{3,2}(A,1,1; C,2; 1)=(C-1)/(A-1))*( Psi(C-1) - Psi(C-A) ) . If we choose in (5) A=a-1 , C=a+b-1 and then using (4), we give ============================================================================ = Z =((a+b-1)*(a+b-2)/(a-1)(a-2))*( Psi(b)-Psi(a+b-1) ) , a>1 ,a=/=2, b>0 . ============================================================================ == Note that Psi(1)=-gamma=-(Euler constant) and Psi(n+1)= 1+1/2+...+1/n - gamma , Psi(1/2)=-gamma -2*ln(2) , Psi (z+n)-Psi(z)= 1/z + 1/(z+1) + ... +1/(z+n-1) , n being a positive integer. I hope that there are no many misprints. However, Herman solution is nice ============ ==== Many thanks for your responses. My setup (based on the python language running on OS X) doesn't seem to have a readily-available psi function. There is one in SciPy, but I haven't been able to get SciPy to compile on OS X. I wonder if there's another way to calculate it that uses less-exotic functions. I may have to run a simulation to get it... Many thanks again -- hopefully I'll be able to get access to a psi function if there's no other way. Gary http://www.garyrobinson.net ==== wow, the internet is really an awesome tool. thanks to both you fellows very much. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBKFmp204548; ==== Do you still have this book? - J. J. Barton, L. R. Nackman, Scientific and Engineering C++: An Introduction with Advanced Techniques and Examples, Addison Wesley, 1994, $25 {Book Condition: New}. If so, what is you lowest price? Antonio Ferrao Neto X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBKLX4328498; ==== idoit ==== quatium bomb @ multidemtional website @ http://www.speedypc.20m.com ==== I am trying to solve poisson equation using buneman's algorithm on a single machine (no need of any parallelization). I have read buzbee, golub and nielson; and stoer and bulirsch regarding this. While solving poisson equation I have to use the nine-point formula. For the usual 5-point formula the offdiagonal block matrix is just identity matrix. Buzbee's paper and stoer's book discuss only this case. I could not find any reference when the off-diagonal terms are non-identity matrices. Could someone point me to the right direction? will the same stability conditions hold for 9-point formula also? what are the necessary changes to be done...? I have derived the equations by myself, but would like to check with some one/paper before I start coding this stuff. thanks raju ==== There are 'ready to use' formulas in Abramowitz & Stegun, limited in the numbers of points used for the Lagrange method. Are there any readings [on the web?] using polynomials instead of the concrete solutions? Moreover i would like to find something for non-equal spaced points assuming my functions are very smooth [like 2nd derivative ~ exp(polynomial)]. Any good hints in the sense above? --- ==== There are 'ready to use' formulas in Abramowitz & Stegun, > limited in the numbers of points used for the Lagrange > method. Are there any readings [on the web?] using polynomials > instead of the concrete solutions? Moreover i would like to find something for non-equal > spaced points assuming my functions are very smooth > [like 2nd derivative ~ exp(polynomial)]. Any good hints > in the sense above? --- Use Gram polynomials to fit either equally-spaced or unequally space data in the least-squares sense. This IS a filtering method, of course, if you use polynomials of (much) lower degree than the number of data. Then use a standard algorithm to compute the derivative of the fitting polynomial. (See, e.g. F.S. Acton, Numerical Methods that (Usually) Work, AMS pub.) -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. ==== > There are 'ready to use' formulas in Abramowitz & Stegun, > limited in the numbers of points used for the Lagrange > method. Are there any readings [on the web?] using polynomials > instead of the concrete solutions? Moreover i would like to find something for non-equal > spaced points assuming my functions are very smooth > [like 2nd derivative ~ exp(polynomial)]. Any good hints > in the sense above? You might want to investigate Savitzky-Golay filters which not only support taking derivatives but smoothing (least squares) as well. For unequally spaced points you may have to be content with fitting splines (or polynomials) and taking derivatives from them. -- 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 ==== Dear all, At the moment I use NAG routines to solve sets of non-linear equations. Untill now I was able to solve sets of at maximum 5000 equations. However, I have to go to at least 30000-50000. Does anybody has a good tip for a solver able to handle such systems? Henk ==== > Dear all, At the moment I use NAG routines to solve sets of non-linear equations. > Untill now I was able to solve sets of at maximum 5000 equations. However, I > have to go to at least 30000-50000. Does anybody has a good tip for a solver > able to handle such systems? > Henk http://www-unix.mcs.anl.gov/petsc/petsc-2/ Hans Mittelmann ==== > I have prepared > http://informatik.uibk.ac.at/users/c703211/solok.jpg > as an example for the GSN-SOR, residual ~ 1e-5 after 18 iterations. Compare this to > http://informatik.uibk.ac.at/users/c703211/solnok.jpg > You can recognize the small oscillations as the moire pattern. > This was made with Jacobi, residual ~ 25 after 17 iterations, changing > hardly thereafter. Which damping factor do you use? The standard 5-point stencil needs damped Jacobi for smoothing (a factor 0.5 works well). Nicolas. ==== > Strange. Jacobi is very local, so if anything it should be good at > precisely losing the oscillatory components of the error, while really > slow at finding the low frequencies. Indeed. (If it is correctly damped.) Nicolas. ==== hy all, i'm searching a good (=VERY FAST) algorithm to compute eigenvalue of a matrix in skyline format in order to study it and reimplement it (don't ask me why, please... :) Any help is appreciated. Very many thanks, Attilio Gelosa. ==== There always seem to be interesting discussions here but very little mention of application (presumably to avoid divulging what your company is scheming). So, to satisfy my curiosity, I would like to ask what kind of work you all do that requires knowledge of num-analysis? I.E. why are you reading this newsgroup? Personally, I'm an engineering student with an interest in this kind of stuff. While I have had a reasonable amount of exposure to engineering work, I have never come across anyone requiring num-analysis knowledge outside of academia. So, let's hear it. CM ==== You have had very little experience In Engineering we use numerical methods for almost anything. You may notice the large engineering companies like say Boeing have humongous libraries of proprietrary routines. But there is a lot of info out on the net and at various site that hold fortran and other codes. So there is really tonnes of stuff, Paul > There always seem to be interesting discussions here but very little mention > of application (presumably to avoid divulging what your company is > scheming). So, to satisfy my curiosity, I would like to ask what kind of work you all > do that requires knowledge of num-analysis? I.E. why are you reading this > newsgroup? Personally, I'm an engineering student with an interest in this kind of > stuff. While I have had a reasonable amount of exposure to engineering work, > I have never come across anyone requiring num-analysis knowledge outside of > academia. So, let's hear it. CM > ==== Chris, > There always seem to be interesting discussions here but very > little mention of application...So, to satisfy my curiosity, > I would like to ask what kind of work you all do that requires > knowledge of num-analysis? I.E. why are you reading this > newsgroup? I write a circuit analysis program used for analog circuit design. It's used both in-house(we're an IC manufacturer) and it is given away to customers as a general purpose SPICE program to virtually demo the company's IC products. I recently asked here for pointers on symbolic dimensional analysis of expressions (i.e., to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K* time)*time as Joule-seconds). Julian N. gave some useful pointers and the code is now release as http://LTspice.linear- tech.com/software/swcadiii.exe. (BTW, Julian, I see your point about Forth being ideal for recursively reducing the symbolic expressions, but C++ returning String objects on the stack was fortunately just as effective and just as easy to handle the memory management of all those strings.) --Mike ==== Chris, There always seem to be interesting discussions here but very > little mention of application...So, to satisfy my curiosity, > I would like to ask what kind of work you all do that requires > knowledge of num-analysis? I.E. why are you reading this > newsgroup? I write a circuit analysis program used for analog circuit design. > It's used both in-house(we're an IC manufacturer) and it is given > away to customers as a general purpose SPICE program to virtually > demo the company's IC products. I recently asked here for > pointers on symbolic dimensional analysis of expressions (i.e., > to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K* > time)*time as Joule-seconds). Julian N. gave some useful pointers > and the code is now release as http://LTspice.linear- > tech.com/software/swcadiii.exe. (BTW, Julian, I see your point > about Forth being ideal for recursively reducing the symbolic > expressions, but C++ returning String objects on the stack was > fortunately just as effective and just as easy to handle the > memory management of all those strings.) --Mike First, thanks for the acknowledgement of my advice. I should send a bill if it was that useful ;-) Seriously, my only objection to using C++ is the HUMUNGUOUS size of the resulting code. I guarantee that, had you used a Windoze- based Forth such as Win32Forth or SwiftForth, the code size would have been way smaller than 4+ Mb, probably would have allowed user modification (although companies rarely want to do this even for stuff they give away for free), and just as much GUI functionality or more. Maybe using OpenGL. Win32Forth was written by a guy who got tired of sweating the M$ PDK for C++ apps. And I am sure it would have run about as fast. Certainly SwiftForth would. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. ==== Julian, > There always seem to be interesting discussions here but very > little mention of application...So, to satisfy my curiosity, > I would like to ask what kind of work you all do that requires > knowledge of num-analysis? I.E. why are you reading this > newsgroup? I write a circuit analysis program used for analog circuit design. > It's used both in-house(we're an IC manufacturer) and it is given > away to customers as a general purpose SPICE program to virtually > demo the company's IC products. I recently asked here for > pointers on symbolic dimensional analysis of expressions (i.e., > to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K* > time)*time as Joule-seconds). Julian N. gave some useful pointers > and the code is now release as http://LTspice.linear- > tech.com/software/swcadiii.exe. (BTW, Julian, I see your point > about Forth being ideal for recursively reducing the symbolic > expressions, but C++ returning String objects on the stack was > fortunately just as effective and just as easy to handle the > memory management of all those strings.) First, thanks for the acknowledgement of my advice. I should send a > bill if it was that useful ;-) Seriously, my only objection to using C++ is the HUMUNGUOUS size > of the resulting code. I guarantee that, had you used a Windoze- > based Forth such as Win32Forth or SwiftForth, the code size would > have been way smaller than 4+ Mb, probably would have allowed user > modification (although companies rarely want to do this even for > stuff they give away for free), and just as much GUI functionality > or more. Maybe using OpenGL. Win32Forth was written by a guy who > got tired of sweating the M$ PDK for C++ apps. And I am sure it > would have run about as fast. Certainly SwiftForth would. C++ can also be used to make efficient use of object code size, but it isn't a trend after which many people aspire. I certainly do in the interest of esthetics and code execution speed. BTW, I found that the MS compiler can make the smallest complete executable of all PC-based compilers I tried. While LTspice is now 4.65MB(slightly larger if compiled to use P4's 128 bit width data types to simultaneously do two FLOPS on two 64bit double precision numbers), years ago LTspice was just over 1.4MB, as I recall. In that 1.4MB there was a complete SPICE engine, a waveform viewer(that had it's own 64bit virtual address space for waveform data) a schematic capture program, a SMPS synthesizer, about 30 small compilers(circuit simulation is ultimately a complier problem these days as I see it) and MFC statically linked in for a multi-threaded GUI. Yep, people were surprised to see the small execution size. Today's 4.65MB version bloated in that it has now contains two complete SPICE solvers(one uses a different SPARSE matrix package that was incompatible with the normal solver) and very large MOSFET device evaluation code for various of experimental MOSFET's devices. However, the only penalty for the size is the download time and by industry standards for this field, it's still a lean, thin, and mean executable. It's distributed as a 5MB self-extracting gzip'ed executable that contains 1500+ files in addition to the executable. My guess is that it wound not be practical to implement such a project in Forth. --Mike ==== > I have never come across anyone requiring num-analysis knowledge outside of > academia. So, let's hear it. CM > Two words. Cleve Moler. ==== CenterSpace Software, a leading provider of enterprise class numerical component libraries for the .NET platform, today announced the release of NMath Stats. NMath Stats is part of CenterSpace Software's NMath product suite, which provides object-oriented components for mathematical, engineering, scientific, and financial applications on the .NET platform. NMath Stats provides functions for statistical computation, including descriptive statistics, probability distributions, combinatorial functions, multiple linear regression, hypothesis testing, and analysis of variance (ANOVA). Fully compliant with the Microsoft Common Language Specification, all NMath Stats routines are callable from any .NET language, including C#, Visual Basic.NET, and Managed C++. A free 14-day evaluation version is available for download from the CenterSpace website at http://www.centerspace.net, along with complete user documentation, performance benchmarks, a whitepaper, and code examples. ==== hi, ALL, I am wondering whether AMG likes banded matrix than general sparse matrix. Two similar linear systems. (generated from poisson equations) A * x = b A_* x = b They share the same RHS, and exact solution, but different matrix (A, and A_). In fact, x = { 1, 1, 1,......,1 }(Trans) A is a banded matrix. A_ is very similar to A, except several entries out of the band. If we combine those entries to diagonal elements, we will get A. It seems that AMG working on A_ gives better result than working on A. Could anyone tell me which kind of matrix AMG like to see? lin ==== > [...] > Right now, I am interested in doing the following: setup the > model equations using Maple and convert them into C or MATLAB routines > and solve them. Final aim is to be able to develop detailed models (a > highly nonlinear system of PDEs) quickly using Maple, but solve them > in MATLAB. Can somebody please let me know whether this is practical > and if possible can you provide me some Maple code as an example to > start with. See the ?CodeGeneration,Matlab help page. Some more examples are in http://www.mapleapps.com/categories/maple9/html/CodeGeneration.html. -- Thomas Richard Maple Support Scientific Computers GmbH http://www.scientific.de ==== I have recently discovered generalizations of the complex-conjugate for a number of algebras. http://library.wolfram.com/infocenter/MathSource/4894 contains a Mathematica notebook, Hoopalgebras.ma (designed to be comprehensible to non-Mathematica users) that investigates many algebras and includes 47 conjugate formulations. Briefly, Hoops are all the division algebras (including C, H, O, Clifford algebras, etc.) that conserve one or more size on multiplication and division. The sizes are the real factors of the inverse of the multiplication table determinant, so Det[A] * Det[B] = Det[A*B] where the Dets are calculated from this multiplication table after mapping with the vectors. Multiplying a vector by its conjugate is usually employed to find the squared length of the vector. It also gives a vector with elements that define these sizes. Complex algebra provides the archetype. Multiply the complex number {a,b} (==a + b i) by its complex-conjugate (a,-b} (==a - b i) using complex arithmetic and the result is {a^2+b^2, 0}; the real component of the product is both the squared length and the conserved size; the complex component is zero. The sizes are always divisors of the inverse, e.g. the complex inverse of {a,b} is {a/(a^2+b^2, -b/(a^2+b^2)}. The inverse goes to infinity (division by zero) when a size goes to zero. Quaternion & octonion conjugate products also have a single non-zero element (the sum of the squared elements), which is the (scalar) conserved property and the divisor in the inverse. R, C, H, & O are the only real algebras without divisors of zero because their sizes are the sum of the squared elements, which (with real coefficients) can only be zero in the trivial case. (The only other operations with sums of squared elements as divisors are the higher Cayley - Dickson algebras such Sedenions, but they have lost any form of associativity.) I know of no other algebras that conserve the sum of the squares of their elements, but a few have a signed sum of their elements as a size (and so have real divisors of zero - non-trivial sets of elements can give a zeroed size). Their conjugate products also have a single non-zero element. E.g.:- (1) The Pauli-sigma-matrix algebra product of {a,b,c,d} & {a,-b,-c,-d} is {a^2 -b^2 - c^2 - d^2, 0,0,0}, with size a^2 - b^2 - c^2 - d^2. (2) The Clifford(2) product of {a,b,c,d} & {a,-b,-c,-d} is {a^2 - b^2 + c^2 - d^2, 0,0,0}, with size a^2 - b^2 + c^2 - d^2. I use {a,b,...} as the initial vector in the following examples. Some algebras have quartic sizes that split into compact expressions as signed sums of squared quadratics. I have found several such compact expressions by searching for a conjugate and confirming that the sums of squares of the product elements define conserved sizes. E.g. the Clifford(3) conjugate {a, -b, -c, -d, -e, -f, -g, h} gives a product with scalar element a^2 -b^2 -c^2 +d^2 -e^2 +f^2 +g^2 -h^2 & trivector element -2d e +2c f -2b g +2a h, the other elements being zero. The (single, quartic) conserved factor is the sum of the squares of these two terms. I have found three other Clifford(3) conjugates, each with four non-zero terms; signed sums of their squares provides different formulations of the same conserved factor. Most algebras have several sizes, so their inverses split into partial fractions. All the above algebras have a single size; they are degenerate because their inverses do not split into partial fractions. I have found plex-conjugates for 40 other algebras; these usually give a product containing many zero coefficients, with the few non-zero elements being related to quadratic or quartic conserved sizes, though the relationship is not always obvious. In algebras with ternary symmetry the conjugate has reversed pairs of elements (instead of negated elements), and the product has repeated (instead of zero) elements. The C3 group defines a division algebra in which the product of {a,b,c} and {a,c,b} is {a^2+b^2+c^2, a b +a c +b c, a b +a c +b c} whilst the sizes are {a+b+c, a^2 +b^2 +c^2 -(a b +a c +b c)}. The quadratic size is the sum of the unrepeated and one of the repeated terms. Linear sizes (a+b+c in this case) obviously cannot be given by conjugate multiplication. Plex-conjugates appear to be a new concept that raises some interesting research problems. Do all hoops have them? I have failed to find them for C5, D3, C7, D4, D3C2 & A4 and a few small signed-table algebras; I have only looked at larger algebras where I expected an obvious conjugate. Is there an extension to triple products to generate cubic factors? A4 is the smallest group with a cubic size. Complex conjugation is a special case of a more general property of some algebras, which gives a squared length for a vector because of the simple form of the sizes of C, H, & O. Roger Beresford. Facts do not cease to exist because they are ignored. T. H. Huxley. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBMBIP319625; ==== Can anyone suggest a book (or internet source) that describes in detail how to construct (and read in) a Maple library? I have dozens of Maple procedures in a *.txt file that I read into Maple every time I begin a session. I want to somehow construct a library for these procedures and also make documentation files that could be accessed through the Maple help menus. I believe I am really asking two separate questions here: (1) How to build a library (given the procedures are already constructed and currently sit in a large *.txt file). (2) How to build your own documentation Help menu -- just as Maple has Help menus for its various functions. Diane Evans Mathematics Assistant Professor at Rose-Hulman Institute of Technology ==== Can anyone suggest a book (or internet source) that describes in > detail how to construct (and read in) a Maple library? I have dozens > of Maple procedures in a *.txt file that I read into Maple every time > I begin a session. I want to somehow construct a library for these > procedures and also make documentation files that could be accessed > through the Maple help menus. I believe I am really asking two > separate questions here: (1) How to build a library (given the procedures are already > constructed and currently sit in a large *.txt file). (2) How to build your own documentation Help menu -- just as Maple > has Help menus for its various functions. > Diane Evans > Mathematics Assistant Professor at > Rose-Hulman Institute of Technology I forwarded this to comp.soft-sys.math.maple in hopes a maven there could answer. -- Julian V. Noble Professor Emeritus of Physics jvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^ http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince. X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBMGtSf10747; ==== I have a proof for Goldbach's Conjecture that is prominently posted in directories throughout the web. It is most notably found at Florida State's virtual library under: www.math.fsu.edu/Science/Specialized as In Defense of Mr. Fermat. ==== On Wed, 4 Feb 1998 10:01:12 -0500 (EST) in Geometry-Research, But Bucky Fuller, though a wonderful architect, was close to crazy, as his books clearly show. He regularly used phrases (such as the fundamental structure of the plane is hexagonal) that sound wonderful but have no meaning. In fact it's a mere matter of convenience whether we use orthogonal or hexagonal coordinates - neither is intrinsically better than the other; it's just that some coordinate-systems are better suited to some problems than others. I've used dozens of different coordinate-systems in my life, as have most other professional mathematicians, and I prefer not to waste time by muttering meaningless mumbo-jumbo to show how one is somehow more moral than the others. --- I have read some things that say the rect in rectilinear means correct and it implies rectilinear Cartesian coordinates are somehow righteous. Bucky Fuller insisted that academia, all over the globe, does not agree with John Conway about the merits of non-orthogonal coordinate systems, and that is what I've seen in everything I've read in the last forty years except Bucky Fuller's writing. Could it be that the top people in academia don't know what is being taught to the public and the students? I was asked to show examples of how the Synergetics coordinate system can make things clearer or easier than Cartesian coordinates. That's a job for mathematicians (teachers) who agree that some coordinate-systems are better suited to some problems than others. If you pull string through equal length soda straws, the triangle holds its shape and the square does not. The tetrahedron with four triangular faces holds its shape and the cube with six square faces does not. Why? The only author I've read who has written about the importance of the stability of the triangle is Buckminster Fuller. He is very persuasive that the stability of the triangle is very important; it is a primary fact of existence, it comes before other facts of geometry. He is very persuasive that if there is no triangulation there is no structure in anything. Definition mathematics: Math`e*matics, n. [F. math['e]matiques, pl., L. mathematica, sing., Gr. ? (sc. ?) science. See Mathematic, and -ics.] That science, or class of sciences, which treats of the exact relations existing between quantities or magnitudes, and of the methods by which, in accordance with these relations, quantities sought are deducible from other quantities known or supposed; the science of spatial and quantitative relations. Note: Mathematics embraces three departments, namely: 1. Arithmetic. 2.Geometry, including Trigonometry and Conic Sections. 3. Analysis, in which letters are used, including Algebra, Analytical Geometry, and Calculus. Each of these divisions is divided into pure or abstract, which considers magnitude or quantity abstractly, without relation to matter; and mixed or applied, which treats of magnitude as subsisting in material bodies, and is consequently interwoven with physical considerations. Source: Webster's Revised Unabridged Dictionary, © 1996, 1998 MICRA, In --- The Cartesian coordinate system should be considered part of pure or abstract mathematics, and the Synergetics coordinate system part of mixed or applied mathematics, even though the Synergetics coordinate system was developed in pure principle. R. Buckminster Fuller thought that the mistake most people make is to leave out relevant parameters because they want to be brief; they want leave out relevant parameters but was not redundant, and, he is the only writer I've read who has even tried to do that. A small summary of Synergetics Coordinates would be misleading because it would leave out relevant parameters. Here are a few URLs to prepare you to read RBF's books Synergetics 1 and 2. A brief description of Synergetics coordinates at: http://mathworld.wolfram.com/SynergeticsCoordinates.html The Mathematica notebook SynergeticsApplication7 at: http://library.wolfram.com/infocenter/MathSource/600/ or the SynergeticsApplication7 notebook as html at: http://users.adelphia.net/~cnelson9/ R. Buckminster Fuller's Synergetics 1 and 2 at: http://www.rwgrayprojects.com/synergetics/synergetics.html Cliff Nelson ==== > I just looked in a second algebra book on my shelf, > Birkhoff and MacLane, A Brief Survey of Modern Algebra. > Page 17. In fact it say > of the two possible g.c.d.'s +-d for a and b, the > positive one is often denoted by the symbol . > (a , b) ... And here is another place.. > http://mathworld.wolfram.com/GreatestCommonDivisor.html > I'm a mathematician, and again I can insure you I never > saw this notation before. Which means for me a couple > of (integers or polynomials). Maybe it's a local US notation. Its not US specific. I am a mathematician too, and (a,b) is the standard notion in mathematics for the gcd of two integers. Pick up any book on elementary number theory and it will contain this notation. eg. I just picked Niven, Zuckerman and Montgomery of my bookshelf, and there on page 7 is the notation (a,b) for gcd. graham ==== > So to throw some wood in this fire, I would say that Java (for me) > is a very good language to develop CAS software. This depends on what you mean by good language. Java has some strengths -- garbage collection, runs everywhere, JIT compilers, etc. But to me it also has some weaknesses (esp for symbolic code): no algebraic data types, very weak type system, OO is a stupid idea (this one is a personal peeve). YMMV graham ==== > The most popular CAS seems to have a C kernel and > a proprietary extension language that has an algebraic syntax. > If the designers of these systems would design them today, it > would be interesting to know what they would do... They could answer that question here if they are reading this! > My guess is that they would write in C because their programmers know > C. Ok I will bite. I was one of the designers of Magma. It has a C kernel and a proprietary extension language.If I could do it again what would I do? Before answering let me be clear that these are my personal opinions and not those of the Magma group (I no longer work for them). I would do one of two things :- a) start with Lisp, Caml, or (S)ML, and build my CA system as a set of libraries on top of such languages. b) (more ambitiously) start totally from scratch, and rethink everything in the design of CA systems. I suspect the end result of this would be a language similar to Lisp or SML, but with some area specific add ons. I would not repeat the C kernel plus proprietary language design. graham ==== > The most popular CAS seems to have a C kernel and > a proprietary extension language that has an algebraic syntax. > If the designers of these systems would design them today, it > would be interesting to know what they would do... They could answer that question here if they are reading this! > My guess is that they would write in C because their programmers know > C. Ok I will bite. I was one of the designers of Magma. It has a C kernel > and a proprietary extension language.If I could do it again what would > I do? Before answering let me be clear that these are my personal > opinions and not those of the Magma group (I no longer work for them). I would do one of two things :- a) start with Lisp, Caml, or (S)ML, and build my CA system as a set of > libraries on top of such languages. I wouldn't do that. > b) (more ambitiously) start totally from scratch, and rethink everything > in the design of CA systems. I suspect the end result of this would be > a language similar to Lisp or SML, but with some area specific add > ons. I would not repeat the C kernel plus proprietary language design. I would program the kernel in assembly meaning 64-bit assembly because of it becoming a near future for all of us. Lisp, C and SML seem to be somewhat outdated at this time. Interface for Windows can be also designed in they just can't provide a normal performance now and certainly won't be able to do that in the future. Dr. Alec Mihailovs http://webpages.shepherd.edu/amihailo/ ==== mixed up the columns in my notes, and I was differentiating when I was supposed to be integrating. I'm very sorry. B. ==== I'm trying to prove mathematically that: For all primes p, 2^p - 1 is prime. True or False. I know that the proposition is false, because 2^11 - 1 = 2047 which is divisible by 23 & 89. What I cant figure out is how to go about proving it. So far I have tried by Induction and now Disproof, but I have to admit that my algebra isn't that good after 25 years away from high school. I've been trying to prove that if p = x*y (i.e. p is composite) that 2^p-1 is composite. (provided p>2) It follows that if p is not composite (i.e. prime) then 2^p-1 is also not composite (i.e. prime). Problem is, that that logic proves the original proposition as True ... which I know to wrong. Can someone give me some guidance or URLs? -- Luke ------ Q: What does FAQ stand for? A: We are Frequently Asked this Question, and we have no idea. ------ ==== I'm trying to prove mathematically that: For all primes p, 2^p - 1 is prime. True or False. I know that the proposition is false, because 2^11 - 1 = 2047 which is > divisible by 23 & 89. > What I cant figure out is how to go about proving it. You just did! One counterexample, such as p = 11, is enough to falsify the general statement. ==== I'm trying to prove mathematically that: For all primes p, 2^p - 1 is prime. True or False. I know that the proposition is false, because 2^11 - 1 = 2047 which is > divisible by 23 & 89. > What I cant figure out is how to go about proving it. You just proved it's false. The proposition said for ALL primes p...; you only have to find a single counterexample to prove that the proposition is false. End of story. You don't need to go any further. So far I have tried by Induction and now Disproof, but I have to admit > that my algebra isn't that good after 25 years away from high school. I've been trying to prove that if p = x*y (i.e. p is composite) that > 2^p-1 is composite. (provided p>2) It follows that if p is not composite (i.e. prime) then 2^p-1 is also > not composite (i.e. prime). Problem is, that that logic proves the > original proposition as True ... which I know to wrong. Can someone give me some guidance or URLs? ==== This has been asked as an assignment question for Discrete Maths that I am doing over the x-mas break (I use the word break loosely!) I know that p=11 is a counter example, but is simply stating that adequate proof from an academic point of view? My experience to date says they want a more professional/academic explanation. Perhaps I'm barking up the wrong tree insofar as there isn't a way of proving it mathematically? It's been many years (over 2 decades) since I've done any serious algebra,and I'm a little surprised at how rusty I am. Any further guidance ... ideas? Luke. I'm trying to prove mathematically that: For all primes p, 2^p - 1 is prime. True or False. I know that the proposition is false, because 2^11 - 1 = 2047 which is > divisible by 23 & 89. > What I cant figure out is how to go about proving it. You just proved it's false. The proposition said for ALL primes p...; > you only have to find a single counterexample to prove that the > proposition is false. End of story. You don't need to go any further. ==== > I know that p=11 is a counter example, but is simply stating that > adequate proof from an academic point of view? Yes. > My experience to date > says they want a more professional/academic explanation. Perhaps I'm barking up the wrong tree insofar as there isn't a way of > proving it mathematically? You (and conceivably they) may be confusing different sorts of propositions. Can you please reproduce EXACTLY what the question says? The precise wording may be crucial. ==== Everett, >> While LTspice is now 4.65MB(slightly larger if >> compiled to use P4's 128 bit width data types to >> simultaneously do two FLOPS on two 64bit double >> precision numbers), years ago LTspice was just >> over 1.4MB, as I recall. In that 1.4MB there was a >> complete SPICE engine, a waveform viewer(that had >> it's own 64bit virtual address space for waveform >> data) a schematic capture program, a SMPS >> synthesizer, about 30 small compilers(circuit >> simulation is ultimately a complier problem these >> days as I see it) and MFC statically linked in for >> a multi-threaded GUI. Yep, people were surprised >> to see the small execution size. Today's 4.65MB >> version bloated in that it has now contains two >> complete SPICE solvers(one uses a different SPARSE >> matrix package that was incompatible with the normal >> solver) and very large MOSFET device evaluation >> code for various of experimental MOSFET's devices. >> However, the only penalty for the size is the >> download time and by industry standards for this >> field, it's still a lean, thin, and mean executable. >> It's distributed as a 5MB self-extracting gzip'ed >> executable that contains 1500+ files in addition to >> the executable. My guess is that it wound not be >> practical to implement such a project in Forth. We put men on the moon with probably less code than > that in all the computers involved, ground and mobile. >> I sure hope so, it's about half a million lines of densely >> written C++ code. The simulator is far beyond 60's >> technology. It does full chip, transistor level >> simulation of IC's much larger than were available during >> the moon shots. It's routinely used for IC's with 10,000 >> transistors solved in a single matrix 100,000 elements on >> a side. Technology moves on, even if it feels like all >> the good engineering has already been done. I'm just trying to fathom how it takes so much code to > simulate transistors. One line of code per electron? Eh? I already described what else besides transitor evaluation is in that 4.6MB two posts back. The size of the transitor evalution code is probably less than a 1MB. It handles 14 different types of transitors, each with object code sizes ranging from less than 20K for a unified bipolar device to 80K for lastest Berkeley release of BSIM4. --Mike ==== In Von Neumann's stability analysis, each element of the solution, u_j^n is written as: u_j^n = A^n(k) e^{i k j dx} eq(1) I really don't understand this. What I *DO* understand is that we can expand the solution u(x,t) in a Fourier series: u(x,t) = sum_k A^n(k) e^{i k x} eq(2) where, at least numerically, x = j dr and t = n dt. What I *THINK* I understand is that we can play some orthogonality games (borrowing Dirac's notation and surpressing the time index for clarity): |u> = sum_k A(k) |e_k> = sum_k A(k) = A(j) eq(3) Although eq(3) looks like eq(1), is not really u_j. At least, I don't think it is. Can someone explain how and why you get eq(1) from eq(2)? I'm looking for something intuitive if possible. Anything to help understand why we write down eq(1) to begin Von Neumann's stability analysis. Pete ==== Merry Christmas! I have some questions to the statistic application amos. I have some specific questions about modelling latent variables. Which one would the right News Group be? ==== Data encryption 360 degrees rotation document 90 degrees and encryption on every angel then change it two binary code and fold it over like a piece of paper then having the one's and zero cancel each other out. if you written a very long letter and then change it two binary code it would look like this 01010101010101010101010 10010101010101010101010 01010101001010101010010 00010101000101010101010 10010101010100101010101 would equal = 01 01010101010100001100101 01001010101010101010111 11110111001101010101010 01010101010101010101010 10101010101010101010101 if you took the piece of paper and folded it and folded it and folded it the 0 and 1 would cancel each out and if you keep folding the piece of paper too the smallest you would have 4 numbers left if 1+1 = nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 01 now if the key new the folding times you could send 2 bytes over the internet and unzip a 100 zetabyte program you computer could store all the programs ever written but just need the key to unzip then you could us this for SETI for signals or can you imagine a computer processor that would be 1.8 Hz but run like 100 million zeta hz you could use the new 64 bit process second side to unzip while the front side processes. or use this for the matrix or quantum computing or supercomputer. 64 bit. 1+1 = nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 dont forrget to use < and > signes 0>1+1<0 HTTP://WWW.SPEEDYPC.20M.COM ==== > please send me any source code for pid controller Companies that want to sell embedded controller chips have such things: http://www.microchip.com/1010/suppdoc/appnote/all/an532/index.htm ==== please send me any source code for pid controller Would its transfer fcn, y/u = Kp + Ki/s + Kd*s help you roll your own? ==== Has anyone been able to compile the BLAS and LAPACK libraries with Intel Fortan 8.0?. What were the compilation flags that you used? If this has been discussed before, could you direct me to a link/discussion thread ? -Nyarlathotep ==== > Has anyone been able to compile the BLAS and LAPACK libraries with Intel > Fortan 8.0?. What were the compilation flags that you used? Instead of blas you should really use www.netlib.org/atlas, which gives a much higher performance. Regardless your compiler. V. -- homepage: cs utk edu tilde lastname ==== > given a 2-d pdf with P = 0 for x<0 OR y<0 OR x>1 OR y>1. > i need to find an algorithm which divides this domain [(0, 0) - (1, > 1)] into n areas (e.g. polygons) of equal probabilities, i.e. each > area's probability will be 1/n. I wouldn't be surprise if this is a known problem; maybe this can be considered a variation on the Voronoi tesselation. Try searching for weighted tessellation or something. Considering this abstractly, there are n equations (area_1 = 1/n, ..., area_n = 1/n) and the number of variables is equal to the twice the number of vertices. In general the number of variables will be greater than the number of equations, so you have an underdetermined system. That should help you find an appropriate solution. It seems like there should be a suitable modification of Newton's method for undetermined systems; I've cross-posted to sci.math.num-analysis -- any ideas there? If I'm not mistaken, derivatives of the mass of the pdf over a polygon with respect to its vertices do exist, assuming the pdf is at least continuous. I believe the derivative of the mass with respect to the x coordinate of one vertex, a, is something like 1/2 dx int_0^1 pdf(xy_1(t)) dt + 1/2 dx int_0^1 pdf(xy_2(t)) dt with xy_1(t) = (1-t) a + t b, and xy_2(t) = (1-t) a + t c, where b and c are the other two vertices which share an edge with a. Hmm, I might have goofed that up, you'll want to double check. In any event, the point is that the derivatives can be computed without too much trouble -- even if the integration in the derivative can't be computed symbolically, it is (hopefully) not difficult to compute it numerically; it's only one dimensional. > actually, i've done this with a genetic algorithm but it's not > efficient at all. i want to know if there exists a better (perhaps > more elegant) solution to this problem. The advantage of randomized algorithms such as genetic algorithms or simulated annealing is that they're applicable to problems, such as combinatorial problems, which have no notion of which way to go. If you can tell which way to go, you are usually better off using a method that can exploit that information. For what it's worth, Robert Dodier -- If I have not seen as far as others, it is because giants were standing on my shoulders. -- Hal Abelson ==== > I can't prove that this equation : y^2=x^3 + 7 > has no integer solutions. > Can you help me? It's fairly easy to show that x must be odd. By considering the equivalent form y^2 + 1 = (x + 2)*(x^2 - 2*x + 4) you can show that y^2 + 1 must be divisible by a prime congruent to 3 mod 4, and from that derive a contradiction. Geoff. ---------------------------------------------------------------------------- - Geoff Bailey (Fred the Wonder Worm) | Programmer by trade -- ftww@maths.usyd.edu.au | Gameplayer by vocation. ---------------------------------------------------------------------------- - ==== Prime numbers are actually very important in todays world. Those and the fact that you cant easily factor big numbers that is! Thats how encryption works. If I get get two really big prime numbers, nd multiply them together then I have a whopping great number that I can use in a formula that noone will be able to break (hopefully) since they wont be able to figure out what the initial two numbers were. For a really good explanation of this goto the sci.crypt newsgroup and read the FAQ there. For practical applications, the ATM you get your money from uses encryption. Online transactions use encrypted data, your communication sattelites, your ISP. Anywhere where there is a password your prolly gonna find encryption. Hope this helps - its pretty basic. Mathew B As a person of *extremely* limited mathematical education could someone > please explain to me the significance and importance of prime numbers? > How are they employed in real-world applications? > big H ==== Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y. But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I can assume that the expression is a difference of squares. Dennis ==== >Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y. But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I can >assume that the expression is a difference of squares. ??? I'm sorry, but I don't understand how you get what you said in your second paragraph. What operations did you perform? a = 2; b = (x+y); a^2 - b^2 = 4 - (x+y)^2. Factors are a+b = 2+x+y and b = 2-x-y. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Modern cyberspace is a deadly festering swamp, teeming with dangerous programs such as viruses, worms, Trojan horses, and licensed Microsoft software that can take over your computer and render it useless. --Dave Barry ==== You almost answer your own question... 4-(x+y)^2 = (2 + x + y)(2 - x - y) Which can perhaps better be seen as: (2 + (x + y))(2 - (x + y)) I hope that this helps - if it does, then please visit: www.brainbashers.com?postcard Kev ==== > Can anyone help alleviate some difficulty of mine in understanding the > details of the real projective plane and a few of its homeomorphic > spaces? What is the significance of the addition of the infinite line > to the projection? One way to think about it is that the addition of the infinite line and the infinite point removes some special cases from the geometry. For example, if you don't have the infinite point, then two lines intersect in one of zero points, depending on whether they are parallel. If you add the infinite point, then every pair of lines intersects in exactly one point -- the infinite point is the intersection of all pairs of parallel lines. Likewise, once you add the infinite point, if you want to maintain the property that for every line and every point there is a line that passes through that point that's parallel to the line, you need to add the infinite line. > I would assume that a better understanding of the > idea of the thing would assist me in understanding certain > homeomorphisms. For instance, its being homeomorphic to a sphere with > antipodal points identified. If you call each great circle of the sphere a line then each pair of lines intersects in two antipodal points of the sphere. Now you call that pair a point and suddenly you have a geometry which satisfies the same constraints: every pair of lines gives you a point, one pair of points gives you a line, etc. > Or a disc with boundary points identified. Or, in what seems the most confusing analogy, the set of > all lines in R^3. How can that be the case? I'm just at a total loss > here. For lines in R^3, you call each line a point and each plane a line and then you again have the same property (you do have to choose a suitable way to define the line that passes through two points that represent skew lines). might be useful. hth meeroh -- If this message helped you, consider buying an item from my wish list: ==== confuse everyone with this message: > ... So what is the status of automatic theorem >> proving? Have any great breakthroughs been made using this approach? Here's a place at least to begin looking for the answers to your >questions: http://gtps.math.cmu.edu/tps.html > Maybe someone knows how to compile this thing with GNU CLisp for Windows? -- |E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15,===================; |E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--| |E1M3_:43__E1M7_:14__E3M1_:43__E4M2_:52__END__:37;================[4*72] ==== ... So what is the status of automatic theorem > proving? Have any great breakthroughs been made using this approach? Here's a place at least to begin looking for the answers to your > questions: http://gtps.math.cmu.edu/tps.html thanks ==== All messages from thread Message 1 in thread (a) Show that I_n = (4n -2)*q^2*I_n-1 - p^2*q^2*I_n-2 for n>= 2 where pi = p/q (b) Deduce that I_n is an integer (c) show that 0 < I_n < (p/q)*(p/2)^(2n) /n! n = 0, 1, 2, .... (d) If (p/q)*(p/2)^(2n) /n! < 1, deduce that pi is irrational. I am stuck at (c) at the moment, (d) seems to be quite easy once you got I try to present some hints. Following notation is used : I[f]= Integral_{t=a to t=x}f(t) dt E(t)|_{a,x}=E(b)-E(x) W^{(p)}= p-th derivative of W (if exists) C(z,k)= z(z-1)(z-2)...(z-k+1)/k! M(n,k)=C(n,k)*(2n-k)!/(2n)! ================================================================ J_n= Integral {from -pi/2 to pi/2}(((pi^2)/4 - t^2)^n)*cos(t) dt. ================================================================= Suppose that p is a positive integer and W,F are in C^{p}[a,b] and x is in [a,b]. The repeated integration by parts formula (so called ,,Green-Lagrange rule ) is I[W*F^{(p)}]=(-1)^p*I[W^{(p)}*F]+ (1) +SUM_{k=0 to k=p-1}(-1)^{p-1-k}W^{(p-1-k)}(t)F^{(k)}(t)|_{a,x} . STEP I. Select in (1) the following input data : W(t)=W_1(t)= (t-a)^n*(x-t)^n , F= f in C^{2n+1}[a,b] . You find (2) f(x)= f(a)+ + SUM_{k=1 to k=n}M(n,k)(x-a)^k*(f^{(k)}(a)-(-1)^k*f^{(k)}(x))+ r_n(f;x) where the ,,remainder-term r_n(f;x) is r_n(f;a,x)=((-1)^n/(2n)!)* I(W*f^{(2n+1)})= =(see Mean-value theorem for Riemann integrals)= (2') =(-1)^n*(x-a)^{2n+1}f^{(2n+1)}(z)/(C(2n,n)*(2n+1)!) for some z=z(a,x,f) in [a,x] . Identity (2') may be called ,,HERMITE FORMULA (1878), see [3]. This formula has a long history. After me, he was rediscovered in 1915 by K. Petr [11] 1940 by Nikola Obresckhoff [10] (in a more general form) 1947 by I. Niven [9] 1948 by P.M. Hummel and C.L.Seebeck (as in [10]) 1957 by D.V.Ionescu (as in [10]). For application see [8]. STEP II. In (2) let x= pi/2 , a=-pi/2, i.e. [a,x]=[-pi/2,pi/2] and W(t)=W_2(t)=((pi^2)/4 - x^2)^n , f(t)= sin(t) . r_n(sin;-pi/2,pi/2)=(1/(2n)!)*J_n = =pi^{2n+1}*cos(z)/(C(2n,n)*(2n+1)!), that integral J_n verifies (3) J_n=pi^{2n+1}*(n!)^2*cos(z_1)/(2n+1)! for some z_1 in [-pi/2,pi/2]. In fact it may be justified that z_1 is in (-pi/2,pi/2). Suppose that pi=p/q . According to (3) (4) J_n= p^{2n+1}*(n!)^2*cos(z_1)/((2n+1)!*q^{2n+1}). =================== Further use ....(2). Perhaps you need also the remark that =================== (5) sqrt(n*pi)/4^n < (n!)^2/(2n)! = 1/C(2n,n) < sqrt(n*pi+pi/2)/4^n for n=1,2,... . Inequalities (5) may be called ,,WALLIS INEQUALITIES. Possible literature about this subject : REFERENCES [1] AIGNER M., ZIEGLER G.M., Proofs from THE BOOK. Springer-Verlag , Berlin,1998. [2] GROSSWALD E., Bessel Polynomials. Lecture Notes in Mathematics- 698,Springer Verlag, Berlin, 1978. [3] HERMITE Ch., Sur la formule d'interpolation de Lagrange. Journ.f.Reine Angew. Mathematik 84(1878), 70-79. [4] HUMMEL P.M., SEEBECK C.L., A generalization of Taylor's expansion. Amer.Math.Monthly , 56(1949) 243-247. [5] IONESCU D.V., Numerical Quadratures , (Romanian) Editura tehnica, 1957. [6] IWAMOTO Y., A proof that pi^2 is irrational. J.Osaka Institute of Science and Technology , 1, (1949) 147-1148. [7] KOKSMA J.F., On Niven's proof that pi is irrational. Nieuw Archiv Wiskunde (2) , 23, (1949) 39 . [8] ...., Computation of some elementary functions (I), (Romanian), Gazeta Matematica (seria A), VII, 1, (1986) 15-26. [9] NIVEN I., A simple proof that pi is irrational. Bulletin Amer.Math.Soc., 53, (1947), 509. [10] OBRESCHKOFF N., Neue Quadraturformeln. Abhandlugen der Preussischen Akademie der Wissenschaften, 4, 1940, 1-20. [11] PETR K., O jedne formuli pro numerichy vypost urcitych integralu, Casopis pro pertovani Matematiks 44 (1915) 454-455. ==== Let f:[0,1] X [0,1] - > R be defined by f(x,y) = 0 if x is irrational = 0 if x is rational, y is irrational = 1/q if x is rational, y = p/q (in lowest terms) Prove that the function f is Riemann integrable, and that int [f,[0,1] X [0,1]] = 0 (Spivak, Calculus on Manifolds, problem 3-7.) According to the text (as well as standard defintions/theorems), the function is integrable iff for any epsilon > 0 there exists a partition P of [0,1] X [0,1] such that U(f,P) - L(f,P) < epsilon. It is easy to see that for any partition of [0,1] X [0,1] all Lower Sums L(f,P) = 0 . The problem I am having is with the proper setup of the partition so that the upper sums U(f,P) can be shown to be arbitrarily small. We know for example that for say 1, 1/2, 1/3, ..., 1/k the function will take on all of these values (and beyond) so we will need to construct the rectangles of the partition P small enough so that U(f,P) will be arbitrarily small. I have a preliminary attempt which constructs two sums: U(f,P) = 1*v_1 + (1/2)*v_2 + (1/3)*v_3 + ... + (1/k)*v_k and v_1 + v_2 + ... + v_k = 1. The v_i each represent the area of each rectangle in the partition so that there sum must be equal to 1. Each term in the first equation represents the supremum of the function on the rectangle whose area is v_i times that area (i.e. (1/r)*v_r). The goal is to make the value of U(f,P) arbitrarily close to 0 while maintaining the second relation (that the sum of all the v_i is 1). Does anyone have a specific construction that would solve this? ==== Let epsilon>0 be given. Look at this set: Y={y in [0,1] : y is rational and y=p/q, where q < 2/epsilon}. Notice that this is a finite set!!! Write Y={y_1,...,y_n}. Choose a number delta > 0 such that: delta < epsilon/(4n) and the sets [y_1-delta,y_1+delta],...,[y_n-delta,y_n+delta] do not overlab. (This can be done since Y is finite.) For k=1,...,n let P_k=[0,1] X [y_k-delta,y_k+delta]. P_0=[0,1] X [0,1] union(P_1,...,P_k). ( means setminus.) Now P_0,....,P_n is a partion of [0,1] X [0,1]. If (x,y) is in P_0 and y=a/b then b>=2/epsilon, otherwise y would be en Y. So 1/b < epsilon/2. Therefore U(f,P_0)Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y. But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I can >assume that the expression is a difference of squares. ??? I'm sorry, but I don't understand how you get what you said in your second paragraph. What operations did you perform? a = 2; b = (x+y); a^2 - b^2 = 4 - (x+y)^2. Factors are a+b = 2+x+y and b = 2-x-y. -- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.com Modern cyberspace is a deadly festering swamp, teeming with dangerous programs such as viruses, worms, Trojan horses, and licensed Microsoft software that can take over your computer and render it useless. --Dave Barry ==== You almost answer your own question... 4-(x+y)^2 = (2 + x + y)(2 - x - y) Which can perhaps better be seen as: (2 + (x + y))(2 - (x + y)) I hope that this helps - if it does, then please visit: www.brainbashers.com?postcard Kev ==== > Can anyone help alleviate some difficulty of mine in understanding the > details of the real projective plane and a few of its homeomorphic > spaces? What is the significance of the addition of the infinite line > to the projection? One way to think about it is that the addition of the infinite line and the infinite point removes some special cases from the geometry. For example, if you don't have the infinite point, then two lines intersect in one of zero points, depending on whether they are parallel. If you add the infinite point, then every pair of lines intersects in exactly one point -- the infinite point is the intersection of all pairs of parallel lines. Likewise, once you add the infinite point, if you want to maintain the property that for every line and every point there is a line that passes through that point that's parallel to the line, you need to add the infinite line. > I would assume that a better understanding of the > idea of the thing would assist me in understanding certain > homeomorphisms. For instance, its being homeomorphic to a sphere with > antipodal points identified. If you call each great circle of the sphere a line then each pair of lines intersects in two antipodal points of the sphere. Now you call that pair a point and suddenly you have a geometry which satisfies the same constraints: every pair of lines gives you a point, one pair of points gives you a line, etc. > Or a disc with boundary points identified. Or, in what seems the most confusing analogy, the set of > all lines in R^3. How can that be the case? I'm just at a total loss > here. For lines in R^3, you call each line a point and each plane a line and then you again have the same property (you do have to choose a suitable way to define the line that passes through two points that represent skew lines). might be useful. hth meeroh -- If this message helped you, consider buying an item from my wish list: ==== confuse everyone with this message: > ... So what is the status of automatic theorem >> proving? Have any great breakthroughs been made using this approach? Here's a place at least to begin looking for the answers to your >questions: http://gtps.math.cmu.edu/tps.html > Maybe someone knows how to compile this thing with GNU CLisp for Windows? -- |E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15,===================; |E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--| |E1M3_:43__E1M7_:14__E3M1_:43__E4M2_:52__END__:37;================[4*72] ==== ... So what is the status of automatic theorem > proving? Have any great breakthroughs been made using this approach? Here's a place at least to begin looking for the answers to your > questions: http://gtps.math.cmu.edu/tps.html thanks ==== All messages from thread Message 1 in thread (a) Show that I_n = (4n -2)*q^2*I_n-1 - p^2*q^2*I_n-2 for n>= 2 where pi = p/q (b) Deduce that I_n is an integer (c) show that 0 < I_n < (p/q)*(p/2)^(2n) /n! n = 0, 1, 2, .... (d) If (p/q)*(p/2)^(2n) /n! < 1, deduce that pi is irrational. I am stuck at (c) at the moment, (d) seems to be quite easy once you got I try to present some hints. Following notation is used : I[f]= Integral_{t=a to t=x}f(t) dt E(t)|_{a,x}=E(b)-E(x) W^{(p)}= p-th derivative of W (if exists) C(z,k)= z(z-1)(z-2)...(z-k+1)/k! M(n,k)=C(n,k)*(2n-k)!/(2n)! ================================================================ J_n= Integral {from -pi/2 to pi/2}(((pi^2)/4 - t^2)^n)*cos(t) dt. ================================================================= Suppose that p is a positive integer and W,F are in C^{p}[a,b] and x is in [a,b]. The repeated integration by parts formula (so called ,,Green-Lagrange rule ) is I[W*F^{(p)}]=(-1)^p*I[W^{(p)}*F]+ (1) +SUM_{k=0 to k=p-1}(-1)^{p-1-k}W^{(p-1-k)}(t)F^{(k)}(t)|_{a,x} . STEP I. Select in (1) the following input data : W(t)=W_1(t)= (t-a)^n*(x-t)^n , F= f in C^{2n+1}[a,b] . You find (2) f(x)= f(a)+ + SUM_{k=1 to k=n}M(n,k)(x-a)^k*(f^{(k)}(a)-(-1)^k*f^{(k)}(x))+ r_n(f;x) where the ,,remainder-term r_n(f;x) is r_n(f;a,x)=((-1)^n/(2n)!)* I(W*f^{(2n+1)})= =(see Mean-value theorem for Riemann integrals)= (2') =(-1)^n*(x-a)^{2n+1}f^{(2n+1)}(z)/(C(2n,n)*(2n+1)!) for some z=z(a,x,f) in [a,x] . Identity (2') may be called ,,HERMITE FORMULA (1878), see [3]. This formula has a long history. After me, he was rediscovered in 1915 by K. Petr [11] 1940 by Nikola Obresckhoff [10] (in a more general form) 1947 by I. Niven [9] 1948 by P.M. Hummel and C.L.Seebeck (as in [10]) 1957 by D.V.Ionescu (as in [10]). For application see [8]. STEP II. In (2) let x= pi/2 , a=-pi/2, i.e. [a,x]=[-pi/2,pi/2] and W(t)=W_2(t)=((pi^2)/4 - x^2)^n , f(t)= sin(t) . r_n(sin;-pi/2,pi/2)=(1/(2n)!)*J_n = =pi^{2n+1}*cos(z)/(C(2n,n)*(2n+1)!), that integral J_n verifies (3) J_n=pi^{2n+1}*(n!)^2*cos(z_1)/(2n+1)! for some z_1 in [-pi/2,pi/2]. In fact it may be justified that z_1 is in (-pi/2,pi/2). Suppose that pi=p/q . According to (3) (4) J_n= p^{2n+1}*(n!)^2*cos(z_1)/((2n+1)!*q^{2n+1}). =================== Further use ....(2). Perhaps you need also the remark that =================== (5) sqrt(n*pi)/4^n < (n!)^2/(2n)! = 1/C(2n,n) < sqrt(n*pi+pi/2)/4^n for n=1,2,... . Inequalities (5) may be called ,,WALLIS INEQUALITIES. Possible literature about this subject : REFERENCES [1] AIGNER M., ZIEGLER G.M., Proofs from THE BOOK. Springer-Verlag , Berlin,1998. [2] GROSSWALD E., Bessel Polynomials. Lecture Notes in Mathematics- 698,Springer Verlag, Berlin, 1978. [3] HERMITE Ch., Sur la formule d'interpolation de Lagrange. Journ.f.Reine Angew. Mathematik 84(1878), 70-79. [4] HUMMEL P.M., SEEBECK C.L., A generalization of Taylor's expansion. Amer.Math.Monthly , 56(1949) 243-247. [5] IONESCU D.V., Numerical Quadratures , (Romanian) Editura tehnica, 1957. [6] IWAMOTO Y., A proof that pi^2 is irrational. J.Osaka Institute of Science and Technology , 1, (1949) 147-1148. [7] KOKSMA J.F., On Niven's proof that pi is irrational. Nieuw Archiv Wiskunde (2) , 23, (1949) 39 . [8] ...., Computation of some elementary functions (I), (Romanian), Gazeta Matematica (seria A), VII, 1, (1986) 15-26. [9] NIVEN I., A simple proof that pi is irrational. Bulletin Amer.Math.Soc., 53, (1947), 509. [10] OBRESCHKOFF N., Neue Quadraturformeln. Abhandlugen der Preussischen Akademie der Wissenschaften, 4, 1940, 1-20. [11] PETR K., O jedne formuli pro numerichy vypost urcitych integralu, Casopis pro pertovani Matematiks 44 (1915) 454-455. ==== Let f:[0,1] X [0,1] - > R be defined by f(x,y) = 0 if x is irrational = 0 if x is rational, y is irrational = 1/q if x is rational, y = p/q (in lowest terms) Prove that the function f is Riemann integrable, and that int [f,[0,1] X [0,1]] = 0 (Spivak, Calculus on Manifolds, problem 3-7.) According to the text (as well as standard defintions/theorems), the function is integrable iff for any epsilon > 0 there exists a partition P of [0,1] X [0,1] such that U(f,P) - L(f,P) < epsilon. It is easy to see that for any partition of [0,1] X [0,1] all Lower Sums L(f,P) = 0 . The problem I am having is with the proper setup of the partition so that the upper sums U(f,P) can be shown to be arbitrarily small. We know for example that for say 1, 1/2, 1/3, ..., 1/k the function will take on all of these values (and beyond) so we will need to construct the rectangles of the partition P small enough so that U(f,P) will be arbitrarily small. I have a preliminary attempt which constructs two sums: U(f,P) = 1*v_1 + (1/2)*v_2 + (1/3)*v_3 + ... + (1/k)*v_k and v_1 + v_2 + ... + v_k = 1. The v_i each represent the area of each rectangle in the partition so that there sum must be equal to 1. Each term in the first equation represents the supremum of the function on the rectangle whose area is v_i times that area (i.e. (1/r)*v_r). The goal is to make the value of U(f,P) arbitrarily close to 0 while maintaining the second relation (that the sum of all the v_i is 1). Does anyone have a specific construction that would solve this? ==== Let epsilon>0 be given. Look at this set: Y={y in [0,1] : y is rational and y=p/q, where q < 2/epsilon}. Notice that this is a finite set!!! Write Y={y_1,...,y_n}. Choose a number delta > 0 such that: delta < epsilon/(4n) and the sets [y_1-delta,y_1+delta],...,[y_n-delta,y_n+delta] do not overlab. (This can be done since Y is finite.) For k=1,...,n let P_k=[0,1] X [y_k-delta,y_k+delta]. P_0=[0,1] X [0,1] union(P_1,...,P_k). ( means setminus.) Now P_0,....,P_n is a partion of [0,1] X [0,1]. If (x,y) is in P_0 and y=a/b then b>=2/epsilon, otherwise y would be en Y. So 1/b < epsilon/2. Therefore U(f,P_0) Will you help me in solving the following ODE k-th order non omog. > differential equation? let u denote a real function of one real variable , say x. > u^(k) (x) = 1/x Pheraps i need some iterative formula... > u^(k-1) = log x + c u^(k-2) = x log x - x + cx + d = x log x + cx + d, where new c = c-1 etc. You'll want integral x^n log x dx = [x^(n+1) log x]/(n+1) - x^(n+1) / (n+1)^2 and some elboe grease for the final result. ==== could u plz integrate this for me? 2/(4x^2+4x+5)^0.5 the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc) i dont get wher the ln come from? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > could u plz integrate this for me? 2/(4x^2+4x+5)^0.5 > the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc) i dont get wher the ln come from? > First, complete the square on the denominator 4x^2+4x+5 4(x^2+x)+5 1/4 -1 4(x+1/2)^2+4 Let u=x+1/2, du=dx 2/(4u^2+4) Integral(1/u^2+1) du Arctan(u)+C Arctan(x+1/2)+C or ArcSinh(x+1/2)+C (The latter was what I got when I ran it through Mathematica). David Moran -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== The ln came from the fact that: If f(x) = 1/x then the integral of f(x) = ln(x) > could u plz integrate this for me? 2/(4x^2+4x+5)^0.5 > the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc) i dont get wher the ln come from? > -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== HELP! I have been searching for a software program that would provide diagnostic information on multiple math strands. I am a remedial math teacher. I am looking for a program to help me quickly assess the knowledge and skill a student has in math, as well as give me some sort of narrative report. We are currently using the KeyMath Test -- but would like lomething a little more current. Any suggestions? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== hi- We are testing a small interactive learning lab so students can experiment with the quadratic equation. It is for a pre-algebra type curriculum. It runs on Mac OS X v10.3. If you would like to run it and give any feedback I'd be happy to provide a download URL and license key for it. TIA- -lance Lance Bland VVI 888-VVI-PLOT http://www.vvi.com -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== I need to determine a percentage for a SALES application. $168,000.00. They grew in sales by a total of $28,231.00. Do I take sales) that gives me .8319 and say that they grew by 17%???? When I take the lowest number and multply by 20% I get $27,953.00. How does the business world determine a true percentage for this type of scenario?? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Has anyone used Qwizdom? It looks like a great way to get the students involved with learning. Using the remotes!! Before I try to urge my school to buy this software I wanted to know if anyone has used it and how they feel about it?!! -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== I have been wondering on whether there exist methods to find the eigenvalues of a matrix other than solving the characteristic polynomial equation. Do there exist elementary row operation methods that will find the real eigenvalues of a matrix? I have been using a site that will find the eigenvalues of a matrix and I'm wondering how it does it because I'm pretty sure its not doing it by solving the characteristic equation. Jeremy ==== > I have been wondering on whether there exist methods to find the eigenvalues > of a matrix other than solving the characteristic polynomial equation. Do > there exist elementary row operation methods that will find the real > eigenvalues of a matrix? Sort of - see below. > I have been using a site that will find the eigenvalues of a matrix and I'm > wondering how it does it because I'm pretty sure its not doing it by solving > the characteristic equation. That is correct. There are ways to find eigenvalues (approximately) without first getting the characteristic polynomial. Google QR decomposition to see one way. is a start. It is perhaps counter-intuitive but Math packages, AFAIK, find the characteristic polynomial by first finding the eienvalues. -- Paul Sperry Columbia, SC (USA) ==== What I like about adding to Rick Decker's example is that it makes it easy. In case you missed it, Rick Decker is a professor at Hamilton College (do a web search to find out about his school) who posted a *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally found that I could modify it slightly to suit my purposes, and now confident in your understanding--even as undergrads--of independence between variables, I feel I can finally show you how I have been defending mathematics against people working to undermine it! Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't have x as a factor are revealed, and not surprisingly, 7y(2y) matches up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(14)y, and picking a_1(y,y) = sqrt(14)y, you have (5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that regardless of anyone's feelings on the matter, it's not possible for the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those factors would have to be a factor of 2y, as that term is both visible and independent of x. So you can see my conclusion about what happens when you divide both sides by 7, when x does not equal y, follows from some basic algebra. What I like here is that quadratics are simpler than cubics to show the result, the example I modified was presented first by Rick Decker, a professor at Hamilton College, and my conclusion follows from independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that if you supposed that you were in the ring of algebraic integers (notice I made no mention of a ring before now) then you run into a problem specific to that ring, where dividing both sides by 7 pushes you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of that problem is actually a fascinating story, and now you can see how mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== > A while back I found these algebraic tools for finding out about roots > of a polynomial using another polynomial, but the problem is that > apparently (from the arguing) mathematicians had never found my > technique before, and not having had it, they'd come up with some > false assumptions about roots of polynomials. So not surprisingly, worldwide, mathematicians when contacted about my > work have tried to run away from it. I've tried to explain it before, > but it is just complicated enough that it seems to shoot past people, > but recently a Rick Decker, a professor at Hamilton College came up > with a *quadratic* example to try and refute my position, but I found > that by modifying his example, I could use it to explain, knowing that > quadratic are simple enough that there's less room for confusion. Consider (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are independent variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > CORRECTION: Oh hey, now that I have another variable, I can set *it* to 0 to see what's left over as well, so letting y=0, gives a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, so you can actually separate out further, where I'll introduce c_1(x,y) and c_2(x,y) to get (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Want more advanced polynomial factorization? Then check out my blog archives: James Harris ==== > A while back I found these algebraic tools for finding out about roots > of a polynomial using another polynomial, but the problem is that > apparently (from the arguing) mathematicians had never found my > technique before, and not having had it, they'd come up with some > false assumptions about roots of polynomials. So not surprisingly, worldwide, mathematicians when contacted about my > work have tried to run away from it. I've tried to explain it before, > but it is just complicated enough that it seems to shoot past people, > but recently a Rick Decker, a professor at Hamilton College came up > with a *quadratic* example to try and refute my position, but I found > that by modifying his example, I could use it to explain, knowing that > quadratic are simple enough that there's less room for confusion. Consider (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are independent variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Oh hey, now that I have another variable, I can set *it* to 0 to see what's left over as well, so letting y=0, gives a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, so you can actually separate out further, where I'll introduce c_1(x,y) and c_2(x,y) to get (5(1+sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1-sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Want more advanced polynomial factorization? Then check out my blog archives: James Harris ==== [cut] > Want more advanced polynomial factorization? [cut] > James Harris Since you asked, I would like to see your factorization of x^3-x+8 ==== >[...] So why would mathematicians argue against such a simple result? Pride? Fear? What if you were simply _wrong_? I mean of course you can't be wrong, anyone can see from the record that while you make occaisional slips like anyone else, when you insist for months that something's true it< turns out to be true, once the dust has settled. Of course you're not wrong, but think about it for a second: if you _were_ wrong that would be another possible reason why mathematicians would say you were wrong. >Want more advanced polynomial factorization? Then check out my blog archives: James Harris ************************ David C. Ullrich ==== > A while back I found these algebraic tools for finding out about roots > of a polynomial using another polynomial, but the problem is that > apparently (from the arguing) mathematicians had never found my > technique before, and not having had it, they'd come up with some > false assumptions about roots of polynomials. So not surprisingly, worldwide, mathematicians when contacted about my > work have tried to run away from it. I've tried to explain it before, > but it is just complicated enough that it seems to shoot past people, > but recently a Rick Decker, a professor at Hamilton College came up > with a *quadratic* example to try and refute my position, but I found > that by modifying his example, I could use it to explain, knowing that > quadratic are simple enough that there's less room for confusion. Consider (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are independent variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Obviously, now the coefficients of the terms that have y but don't > have x as a factor are revealed, and not surprisingly, 7y(2y) matches > up with what you see on the right side with 7(2y^2). That's the simple result which mathematicians can't handle. So instead posters fighting the mathematical reality have made claims > that here would mean that that 7 divides as some kind of factor of x, > which is to say that depending on what value x has, 7 divides through > what's on the left hand side in different ways. Mathematicians, you see, believe that if the quadratic a^2 - (x - y)a + 7(x^2 + xy) doesn't have *integer* roots, then *both* roots would have to share > factors with 7. > Correct. That's what we believe. > If you ask them why you'll probably get a lot of complicated > explanations in return, and they're very defensive on the issue from > what I've gathered, as well as being wrong. > Let the roots be a1 and a2. Suppose, as Harris claims, that not both of them share factors with 7. That means essentially that one of them has no factors in common with 7, which implies that the other one must be *divisible* by 7. Say a1 is divisible by 7. That means that b1 = a1/7 is an algebraic integer. Of course a1 = 7*b1. Since a1 is a root of the polynomial, so is 7*b1. Therefore 7^2*b1^2 - (x - y)*7*b1 + 7*(x^2 + x*y) = 0. Divide through by 7: 7*b1^2 - (x - y)*b1 + (x^2 + x*y) = 0. Now, in general this polynomial in b1 is irreducible, non-monic, and primitive. These properties are all true for most choices of x and y. For example, x = 5 and y = 3; x = 1, y = 0; x = 1, y = -3; etc., etc.. This means (by a basic theorem in algebraic number theory) that b1 in general cannot be an algebraic integer. This contradicts the statement that a1 is divisible by 7, and therefore it also contradicts Harris's statement that one of the roots must have no factors in common with 7. Now: is that a complicated explanation? Is it defensive? Is it *wrong* ? > Here, for physicists it's easy to see that their wrong, as x and y are > INDEPENDENT of each other, so there's no way with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). that (5b_2(x,y) + 2y) can in general have any factor of 7. It's just not mathematically possible, as if it did that factor would > show up multiplied times y, along with 2. Notice you *do* see that > factor showing up with (5a_1(x,y) + 7y). > Just proving what we have said repeatedly: the only way you have ever learned to factor is by inspection. Good old high-school math. > Importantly, there is a special case here when x=y, as then the > quadratic becomes a^2 - (x - y)a + 7(x^2 + xy) = a^2 + 14y^2. At that value the equation belongs to a different family as the > solution then is identical to that from a^2 + 7(x^2 + xy) when x=y. > This is not a special case. This is just an example of what was described above, where the polynomial is irreducible. > So why would mathematicians argue against such a simple result? Pride? Fear? > We argue against your claim for one very simple reason: It is wrong. > Want more advanced polynomial factorization? > Good idea, though what I give below is not exactly advanced. Assume, as is usually the case, that x^2 + x*y is coprime to 7. Here is the right way to factor analyze divisibility of the roots of a^2 - (x - y)*a + 7 *(x^2 + x*y) with respect to 7: Let w1 = GCD(a1, 7) and w2 = GCD(a2, 7). Then both a1 and 7 are divisible by w1, and both a2 and 7 are divisible by w2. By the argument above, neither w1 nor w2 can be equal to 7 and neither can be a unit in the algebraic integers. Therefore, in direct contrast to what Harris claimed above, both of the roots a1 and a2 have nonunit factors in common with 7. Do you see any errors in this ? Nora B. > Then check out my blog archives: James Harris ==== [snip] > Mathematicians, you see, believe that if the quadratic a^2 - (x - y)a + 7(x^2 + xy) doesn't have *integer* roots, then *both* roots would have to share > factors with 7. If I wanted to know what mathematicians believe, I would ask a mathematician and not take your word for it. You have a prior history of misrepresenting the position that other posters, including mathematicians, have taken. Generally you engage in these misrepresentations maliciously and dishonestly, as is revealed by your posting record. -- 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 ==== What I like about adding to Rick Decker's example is that it makes it easy. In case you missed it, Rick Decker is a professor at Hamilton College (do a web search to find out about his school) who posted a *quadratic* in an attempt to refute some of my conclusions. Well, I found his example grabbed me, for various reasons, and finally found that I could modify it slightly to suit my purposes, and now confident in your understanding--even as undergrads--of independence between variables, I feel I can finally show you how I have been defending mathematics against people working to undermine it! Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't have x as a factor are revealed, and not surprisingly, 7y(2y) matches up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(14)y, and picking a_1(y,y) = sqrt(14)y, you have (5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that regardless of anyone's feelings on the matter, it's not possible for the terms *independent* of x, to actually have a dependency on x. So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those factors would have to be a factor of 2y, as that term is both visible and independent of x. So you can see my conclusion about what happens when you divide both sides by 7, when x does not equal y, follows from some basic algebra. What's fascinating is that you can also consider what happens when y=0 to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). What's fascinating here is that completely broken is *any* idea that there's a variable factor in common with 7 that divides differently dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. What I like here is that quadratics are simpler than cubics to show the result, the example I modified was presented first by Rick Decker, a professor at Hamilton College, and my conclusion follows from independence between independent variables. That is to doubt me here you need to doubt that concept in mathematics itself--the concept of independence between variables. Of course, the problem for the old view on algebraic integers, is that if you supposed that you were in the ring of algebraic integers (notice I made no mention of a ring before now) then you run into a problem specific to that ring, where dividing both sides by 7 pushes you out of the ring of algebraic integers. It might seem esoteric and distant as a problem, but my discovery of that problem is actually a fascinating story, and now you can see how mathematicians *worldwide* answer the question: What do you do when your paradigm is forced to shift by a discoverer who thought outside of the box? Want more? Then read my other threads on sci.math or check out my blog archives: James Harris ==== [propaganda clipped] Here's the quadratic that Decker gave: (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - 1)a + 7(x^2 + x). And my modification was just to add in the variable y, to get (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the a's are roots of > > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are *independent* variables, as I haven't related > them to each other in any way. That second polynomial a^2 - (x - y)a + 7(x^2 + xy) is the polynomial being analyzed by considering (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2). You find it from 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2 where what's on the right side is just a regrouping of terms from > what's on the left, as you can find out by simplifying the right side. Notice that with (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on the > left side, but that contradicts with 7(2y^2) on the right! So what gives? Well the problem is that x is in the way, and a simple technique in > analysis is to clear out one variable by setting it to 0. Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have a_2(x,y) = b_2(x,y) - y, so (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly. Obviously, now the coefficients of the terms that have y but don't > have x as a factor are revealed, and not surprisingly, 7y(2y) matches > up with what you see on the right side with 7(2y^2). So if x does not equal y then it follows that (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2 but if x-y=0, then you have a^2 + 7(x^2 + xy), defining the a's, and you can directly calculate that a = +/- sqrt(14)y, and picking > a_1(y,y) = sqrt(14)y, you have The zeros of a in a^2 + 7(x^2 + xy), with x = y, are NOT +/- sqrt(14)y, they are +/- sqrt(-14)y. (5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2). Now considering when x does not equal y it's easy enough to see that > regardless of anyone's feelings on the matter, it's not possible for > the terms *independent* of x, to actually have a dependency on x. And whom are you alleging has made such a claim? Or is this just another of your straw men? So, with (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) if (5b_2(x,y) + 2y) had any factors in common with 7, then those > factors would have to be a factor of 2y, as that term is both visible > and independent of x. It is quite possible in the ring of algebraic integers for 7 and 2y, for y an integer, always to have a common non-unit factor, at least until you or someone else proves it to be impossible. Note that the common factor may be different for different values of y. So you can see my conclusion about what happens when you divide both > sides by 7, when x does not equal y, follows from some basic algebra. So basic, that it is wrong in the more advanced context! What's fascinating is that you can also consider what happens when y=0 > to separate out further, so letting y=0, I have a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). What's fascinating here is that completely broken is *any* idea that > there's a variable factor in common with 7 that divides differently > dependent on the value of x, as you have the lead coefficients 1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant. Same fault as above. No one but JSH is insisting that the common factors of 7 and a_1 and of 7 and a_2 be independent of either x or y. > [more propaganda snipped] ==== Both of you gave truly wonderful posts - greatly appreciated. ==== hi all, I assume that a bezier patch has 16 control points and 4 corners control points are attached onto the surface. It implies that there are 4 bezier curves on the edges of the patch. My question is how can the middle 4 control points change the shape of the surface? newbie... ==== > I assume that a bezier patch has 16 control points and 4 corners control > points are attached onto the surface. It implies that there are 4 bezier > curves on the edges of the patch. My question is how can the middle 4 > control points change the shape of the surface? I don't understand your question. They change the shape because that's how Bezier patches are defined. Can you try to elaborate on what exactly is confusing you? meeroh -- If this message helped you, consider buying an item from my wish list: ==== It looks like there was more to adding in that y as a variable than was realized as I saw the usual suspects replying--yet again--trying to attack the mathematics, attacking algebra, as if they could still convince all of you. However, the other shoe fell as I set y=0 giving a^2 - xa + 7x^2 = 0, so a = (1+/-sqrt(-27))x/2, and introducing c_1(x,y) and c_2(x,y) I then have (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2) from my modification of Rick Decker's example. Now I guess that Decker is a good enough mathematician to recognize that now it's completely over, but I wonder about the others. Will they *dare* step out now despite the mathematics clearly showing that I've been right all along? Let's see, as I will admit that I'm now curious about what these people will do, and how many of you will still trust and believe them. That is, I'm curious about you, and what mathematics means to you, by how you *act* not what you say. James Harris ==== > When are we finally going to see the Tablet PC operating systems and > software give us math persons the holy grail? It's where we write out everything on a tablet, and then push a couple > of buttons and the software transforms the entire document into a form > that looks like we went through all the trouble of typesetting it when > we didn't! Better yet, the document would be compatible and > transformable into and from LaTeX! (I understand that MathType and > LaTeX are now convertible.) The software companies have already given the language persons this > capability (write it out, push some buttons and it looks like it was > typed). So what about us math persons? Math typesetting has historically been much more difficult than plain text. The math formulae have a two-dimensional structure that requires more careful placement, and spacing and positioning carry more of the meaning than in plain text. The market for math typesetting is also much smaller than the market for plain text, so there has been less incentive for industry to develop tools. (Consider that the software that does the best typesetting for math, TeX, has been in the public domain for over 20 years, but the software industry is still putting out third-rate math typesetting tools when they could freely use the algorithms and code in the TeX program.) -- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karplus life member (LAB, Adventure Cycling, American Youth Hostels) Effective Cycling Instructor #218-ck (lapsed) Professor of Computer Engineering, University of California, Santa Cruz Undergraduate and Graduate Director, Bioinformatics Affiliations for identification only. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > Math typesetting has historically been much more difficult than plain > text. The math formulae have a two-dimensional structure that > requires more careful placement, and spacing and positioning carry > more of the meaning than in plain text. The market for math typesetting is also much smaller than the market > for plain text, so there has been less incentive for industry to > develop tools. (Consider that the software that does the best > typesetting for math, TeX, has been in the public domain for over 20 > years, but the software industry is still putting out third-rate math > typesetting tools when they could freely use the algorithms and code > in the TeX program.) > I'd say that the industry underestimates how many math knowledge workers there are. Have they included the very many teachers and especially students world-wide who teach and learn any math at all, at any level at all? Maybe the market is much smaller percentage-wise, but what businessperson would knowingly want to walk away from all those potential customers? Paul -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > It's not about how often we state it. It's about how often we apply it > and even about how often we can apply it. For any expression involving > both operations, this axiom is lurking in the background as the main > tool for creating equivalent expressions. Haven't you ever wondered > why polynomials are so common and why sums of products are far more > useful throughout math than products of sums? I don't know about the relative usefulness of products of sums vs. > sums of products. After so much work in arithmetic of poly's (adding > them, multiplying them, combining like terms) we come to factoring, > where the goal is to write P(x) as (x-r1)(x-r2)...(x-rn) in place of > the usual x^n+...+a0. Those factorizations can be pretty useful. > I was just quoting and agreeing with Michael Hamm. Giving precedence to multiplication over addition in notation is > giving deference to this lurker. What about set union (U) and intersection (^)? We have two > distributive laws: > 1) (a^b)U c = (a U c)^(b U c) > 2) (a U b)^c = (a^c) U (b^c) Now there is no preference. So, a^b U c must be decided by convention? > In a field there is no distribution of addition over multiplication. > The analogy you create therefore doesn't hold. The fact that there is > distributivity in a field in only one way in my view opens the door > for a bias towards a certain way of doing things in terms of operation > precedence. The natural numbers don't constitute a field either. Nor do they form a ring. But their algebraic structure is very similar to that for the power set of a given set X, with the operations of union and intersection. The operations are closed, associative, and commutative. The main difference is that there are two distributive laws instead of one. But, I have seen in several books that intersection precedes union in the order of operations. It's arbitrary. If we were inclined to write all polynomials (over an appropriate ring) in factored form, we would perhaps choose to perform additions first. > Imagine if there were only one type of distributivity in logic and set > theory. What do you think would happen in terms of operation > precedence? > Paul -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > The natural numbers don't constitute a field either. Nor do they form > a ring. But their algebraic structure is very similar to that for the > power set of a given set X, with the operations of union and > intersection. The operations are closed, associative, and commutative. > The main difference is that there are two distributive laws instead of > one. But, I have seen in several books that intersection precedes > union in the order of operations. It's arbitrary. If we were inclined > to write all polynomials (over an appropriate ring) in factored form, > we would perhaps choose to perform additions first. Let's stick to the natural numbers if you wish, and so examine the argument given by Ron. (I wanted to start with the field structure, he with the natural numbers. Maybe he was right to do so. According to Kronecker, the natural numbers are the work of God, but everything else is the work of Humanity.) I do not see how a decision is arbitrary when one choice repeatedly leads to a contradiction and the other never does. Let multiplication in the natural numbers be understood as repeated addition of the same number: a + a + . . . + a (for n instances of a) = (a*1) + (a*1) + . . . + (a*1) (for n instances of a*1) = a*(1 + 1 + . . . + 1) (for n instances of 1) = a*n Then giving default precedence to addition repeatedly yields contradiction, but giving default precedence to multiplication never does. For example: 3 + 5 + 5 = 13 3 + 5 * 2 = 16 (addition gets default precedence) 3 + 5 * 2 = 13 (multiplication gets default precedence) To anticipate: Of course we can give precedence to addition if we want, such as in (3 + 5) * 2 = 16, but this is about default precedence. On expressions involving both addition and multiplication with no parentheses denoting precedence, this is about having a default precedence grammar that avoids contradiction. Cordially, Paul -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Depends on the question of course, but most ask what is the probability of something happening? So the answer would be: The probability of X happening is Y. Funny you should ask, I just had my final exam in prob and stat...I got an A in the class!! woohoo!! John > How do you explain your answer to a probability question > -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== I am an elementary PE teacher that has been asked to be on a committee to come up with ideas for a Math Week to be held in Feb 2004. We would like for each day to have a theme. I am basically at a loss right now as this is not really my area of expertise. I would love any input -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== >I am an elementary PE teacher that has been asked to be on a committee >to come up with ideas for a Math Week to be held in Feb 2004. We would >like for each day to have a theme. I am basically at a loss right now >as this is not really my area of expertise. I would love any input Here's a few topics that might be fun to expand: Prime days. Special badges for those who are 5, 7, 11, or 13. Tangram contests. Lots of interesting math facts come out of tangrams. Probability games. Guess the number of M&Ms in a Halloween-size bag, guess the color distribution, plot the distribution as bar graphs, simple equations where you add the total per color to get the overall total. Tetrahedral kites. I've taught that to 6th graders, and all 120 students in the sixth grade were able to build their kites in one hour, even with a short introduction on the history of the tetrahedron and tetrahedral kites. It took a ratio of one adult per 10 students to get enough help to keep the kids on track, and some of the special needs kids needed individual help. Jill Britton has her pre-service students work in groups of two or three to take it into classrooms in BC. For a much better selection of ideas, pick up one of Jill Britton's books or look at her website: http://camosun.bc.ca/~jbritton/Home.htm (The page is not responding as I type, but the pages for each book have tons of links to activities associated with each chapter of the book.) -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Does anyone know of a university offering a 5-8 Math Education course (preferrably on-line)? I live in south-central Minnesota--and need that one course in order to complete my 5-12 Math Teaching program. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== I need help on my hmwk. but it is to hard to understand -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== in part: > I need help on my hmwk. but it is to hard to understand The answer to number 3 is 54. After all, it doesn't matter what aorder you take the partials in. Hope this helps, AM, Math, Wash. U. St. Louis I've been erasing too much UBE. msh210@math.wustl.edu Of a reply, then, if you have been cheated, -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > I need help on my hmwk. but it is to hard to understand > Well, if it were easy you'd probably not need help. Could you be more specific about: 1. Which problem(s) you're having trouble with. NB: you usually get better responses if you confine it to no more than two or three at a time. I suspect not many readers are interested in doing your homework for you -- but quite a few are willing to explain mathematical concepts and problem solving techniques. 2. What you have done to solve the problems or at least your thoughts are on how you think they should be approached. - Sox -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== In a financial arrangement you are promised $1000 the first day and each day after that you will recieve 35% of the previous days amount. When one days amount drops below $1, you stop getting paid from that day on. What day is the day you would receive no payment and what is your total income? Use a formula for the nth partial sum of a geometric sequence. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== if i reviewed 30 charts and 13 were compliant what would be the percent. Please show me how to calculate. -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== > if i reviewed 30 charts and 13 were compliant what would be the > percent. Please show me how to calculate. Percent means out of 100. If you had 100 charts, how many of them would be compliant? -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== >if i reviewed 30 charts and 13 were compliant what would be the >percent. Please show me how to calculate. > 100 times the quantity of objects which qualify divided by total quantity of objects. If want percent compliant, use 100* (number compliant /number of total) If want percent non-compliant, use 100*(number noncompliant/number of total) G C -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Has anyone tried qualitytutors.com? I have heard that teachers make good money tutoring there. Sam -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== My son is currently in Algebra 2 and he can't figure out the following problem (nor can I): sqrt(14)/(sqrt(6) + sqrt(7) - 1) -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== Below I'll point out a link between number theory and music theory, specifically modular arithmetic modulo 12 and the 12-tone musical system. It also links to the standard 12-hour clock if you want a reference. On the piano, starting in any key you want, we'll move up or down the keyboard a number of notes at a time, including both black and white keys. Here are some options: 1) Move up one note, this interval called a semitone, then another, etc. 2) Move down one note, this interval called a semitone, then another, etc. 3) Move up a perfect fourth (5 notes), then another perfect fourth (5 more notes), etc. 4) Move down a perfect fourth, then another perfect fourth, etc. 5) Move up a perfect fifth (7 notes), then another perfect fifth (7 more notes), etc. 6) Move down a perfect fifth, then another perfect fifth, etc. 7) Move up a major seventh (11 notes), then another major seventh (11 more notes), etc. 8) Move down a major seventh, then another major seventh, etc. In each case, you'll cycle through all twelve keys exactly once. Does this remind you of a certain something? It's modular arithmetic. In modulo 12, it's sometimes called clock arithmetic. In almost all Western music, there are 12 keys as in tones. On the piano, if we start on any key, say a middle C, and play every note, one at a time upwards, the next C we hit is the 12th note up. This interval is an octave. Another 12 notes up, another C. Another octave. This occurs regardless of which of the 12 keys we start in. We thus label the keys so that we can have modular arithmetic, modulo 12. The 12 keys, appropriately labeled, function as the 12 congruence or residue classes mod 12. To apply case 1) above: Let's choose C, and label it 0 mod 12. A perfect 4th up from any note is 5 notes up. A perfect 4th up from C is F, which we label 5 mod 12: 12. The intervals mentioned above are generators of this additive group, because repeated addition cycles through all of the keys (congruence classes mod 12). Each of the intervals mentioned above has order 12. The interval of a minor sixth, 8 notes (or units) up (or down), has order 3, because the greatest common divisor of 8 and 12 is 4 and 12/4 = 3. This means that going up (or down) a minor sixth, an interval covering 8 notes (or units), at a time cycles through only 3 keys (congruence classes mod 12). Starting at C, these would be C (labeled 0 mod 12), A-flat (G# or G-sharp) (labeled 8 mod 12), and E (labeled 4 mod 12). (And then of course C again, but that is back to 0 mod 12). A minor third, which is a 3-note interval, is of order 4. All four notes obtained by repeatedly going up a minor third played together form a diminished chord. Wagner LOVED this chord as a so-called passing chord. Properly used, it creates rich, dark sonorities in the music. Now you get the idea. Show this to interested students. Number theory and music theory. Have fun! Cordially, Paul -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html ==== symmetry/half-turn symmetry and the affine equivalence of cubic polynomials that I've recently redone a little for re-publication. Both results use transformation geometry and algebraic function notation and can be nicely illustrated with Sketchpad as it allows students to see how any cubic polynomial can be translated, sheared and/or stretched. Though probably more suited as enrichment, the results should be accessible to most high school students and can be downloaded directly from: http://mzone.mweb.co.za/residents/profmd/cubic.pdf http://mzone.mweb.co.za/residents/profmd/cubeaffine.pdf A zipped Sketchpad sketch (4 KB) that could be used to demonstrate or allow students to investigate these results by themselves can be downloaded from: http://mzone.mweb.co.za/residents/profmd/cubic.zip I'd be very interested & happy to hear from anyone regarding any experiences you may have with students investigating these properties. Note that the above Sketchpad sketches certainly need some adaptation and improvement for direct use in a class. Michael de Villiers -- newsgroup website: http://www.thinkspot.net/k12math/ newsgroup charter: http://www.thinkspot.net/k12math/charter.html