This integral made me mad... we have learned some special techniques for integraing integrals involving multiple-valued functions, for instance z^a, where a is not an integer... The textbook is: Fundamentals on Complex Analysis with Application to Ehgineering and Science (third eddition) by E. B. Saff and A. D. Snider --------------------------------------------- Now comes the integral: I = Integrate[x^(a-1) / (x^2+x+1), x from 0 to inf], where 0 Now comes the integral: I = Integrate[x^(a-1) / (x^2+x+1), x from 0 to inf], where 0 =/= 1... Integrate, as usual, over a keyhole contour with radius R with an inner hole of radius e taken out to avoid the origin. Standard arguments show that this tends to I - J where I is the given integral and J is the integral where x^(a-1) is replaced by exp((a-1)(log x + 2 pi i)) = x^(a-1)exp(2 pi i(a-1)). Thus the contour integrals tend to (1 - exp(2 pi i (a-1))I = -2i sin(pi(a-1))exp(pi i(a-1)) I. The poles are simple at z = exp(2pi i/3) and z = exp(4 pi i/3). The residues of 1/(z^2 + z + 1) at these points are 1/[i sqrt(3)] and -1/[i sqrt(3)] and so the residues of the integrand are exp(2pi i(a-1)/3)/[i sqrt(3)] and -exp(4pi i(a-1)/3)/[i sqrt(3)]. Thus the countour integral is (2pi/sqrt(3))[exp(2pi i(a-1)/3) - exp(4pi i(a-1)/3)] =(2pi/sqrt(3))exp(pi i(a-1))(-2i sin(pi(a-1)/3)). We get I = (2pi/sqrt(3)) (sin(b)/sin(3b)) where b = pi(a-1)/3. > The book asked to show that this integral equals > 2*pi/sqrt(3)*cos((2*a*pi+pi)/6) / sin(a*pi)... Yuck! sin(b)/sin(3b) = sin(pi (a-1)/3)/sin(pi a - pi) = -cos(pi(a-1)/3 - pi/2)/sin(pi a) = -cos(pi a/3 - 5pi/6)/sin(pi a) = cos(pi a/3 + pi/6)/sin(pi a). OK? -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== Dear all, I spent almost 2 hours in finding its residue: f(z)=z^6 / (z^4+1)^2 at z0=exp(i*pi/4) and z1=exp(i*pi*3/4)... The reason for finding these residues is that I need to compute: I=Integrate[x^6 / (x^4+1)^2, x from -inf to inf] So I formed a close contour including the real axis and the counterclockwise arc on the upper half plane... Then I=2*pi*i*(Res[f(z), at z0] + Res[f(z), at z1]) But this is really hard to correctly compute manually(except using some software such as Mathematica or Matlab... :=) Anybody has a good idea how to do this efficiently? By the way, z0 and z1 are like mirrors to each other, what is the relationship between Res[f(z), at z0] and Res[f(z), at z1]... Even if there is some relationship between these two residues, computing a single Res[f(z), at z0] is horrible... Can anybody give me a hand? -Walala p.s. After many try, I finally used Matlab and admitted my failure... :=( //sigh ==== > Dear all, I spent almost 2 hours in finding its residue: f(z)=z^6 / (z^4+1)^2 at z0=exp(i*pi/4) and z1=exp(i*pi*3/4)... Blimey! Note that z^4 + 1 = ((z-z0) + z0)^4 + 1 = 4z_0^3(z - z0) + 6z0^2(z - z0)^2 + O((z - z0)^3) and z^6 = ((z-z0) + z0)^6 = z0^6 + 6 z_0^5(z - z0) + O((z - z0)^2). Let's write w for z - z0. Then f(z) = w^{-2} (z0^6 + 6 z0^5 w + O(w^2))/(4z0^3 + 6z0^2 w + O(w^2))^2 = w^{-2}(16 z0^6)^{-1} (z0^6 + 6 z0^5 w+O(w^2))(1 +(3/(2z0) w + O(w^2))^{-2} = w^{-2}(16 z0^6)^{-1} (z0^6 + 6 z0^5 w + O(w^2))(1 - (3/z0) w + O(w^2)) = something w^{-2} + 3/(16 z_0) w^{-1} + O(1) so residue is 3/(16 z0). Replace z0 by z1 to get residue 3/(16 z1) at z1. (If I'm not mistaken!) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== We are trying to explain to the company who creates this language we are using that the following is WRONG. As you can see, the following code produces 2 different results. They claim that infact .5 and .50 are different... any comments? 0010 LET A=119.35*8.5/100 0020 LET B=119.35*8.50/100 0030 PRINT a=,A 0040 PRINT b=,B -:run a= 10.15 b= 10.14 ==== > We are trying to explain to the company who creates this language we are > using that the following is WRONG. As you can see, the following code > produces 2 different results. They claim that infact .5 and .50 are > different... any comments? 0010 LET A=119.35*8.5/100 > 0020 LET B=119.35*8.50/100 > 0030 PRINT a=,A > 0040 PRINT b=,B -:run a= 10.15 > b= 10.14 a and b are 10.14475 Try this: 0050 let c=0.5 0060 let d=0.50 0070 if c-d=0 0080 then print Sorry guys, but .5 and .50 are NOT different ;-P 0090 else print Indeed, .5 and .50 are different 0100 endif Microsoft VB 6.0 sp5 says they are not different. Dirk Vdm ==== >We are trying to explain to the company who creates this language we are >using that the following is WRONG. As you can see, the following code >produces 2 different results. They claim that infact .5 and .50 are >different... any comments? I suppose that if it's their computer language they can define it how they like, but it's obviously wrong if 8.5 and 8.50 are supposed to represent real numbers in the usual way. You can imagine explanations for it: for example, that 8.50 has more digits than 8.5 so it is represented using a data type with more bits, so the whole expression is evaluated more precisely, but real computer languages generally have an explicit notation to let do this (e.g. in C you can use 3.5 or 3.5F to distinguish between double and float). The best solution is to avoid proprietary languages and stick to ones with a public definition. -- Richard -- FreeBSD rules! ==== > We are trying to explain to the company who creates this language we are > using that the following is WRONG. As you can see, the following code > produces 2 different results. They claim that infact .5 and .50 are > different... any comments? 0010 LET A=119.35*8.5/100 0020 LET B=119.35*8.50/100 0030 PRINT a=,A 0040 PRINT b=,B -:run a= 10.15 b= 10.14 On my calculator (not high tech, but oh well) 119.35*8.5/100 = 10.14475 119.35*8.50/100 = 10.14475 In both cases, there is something happening, which I suspect is rounding. 119.35*8.5=1014.475 I suspect they are doing the following roundings: 119.35*8.5=1014.5 119.35*8.50=1014.48 Then they are divinding by 100 and rounding again to get: 1014.5/100 = 10.15 1014.48/100 = 10.14 Based on this, it appears that their program is losing accuracy at an alarming rate by doing some optimizations for size/speed. They are clearly treating 8.5 and 8.50 differently, which is producing the erroneous result. I would go back to them and insist that they NOT round their answers internally, especially at such a low level of accuracy. -- Will Twentyman ==== > We are trying to explain to the company who creates this language we are > using that the following is WRONG. As you can see, the following code > produces 2 different results. They claim that infact .5 and .50 are > different... any comments? 0010 LET A=119.35*8.5/100 0020 LET B=119.35*8.50/100 0030 PRINT a=,A 0040 PRINT b=,B -:run a= 10.15 b= 10.14 My guess is that they use intermediate rounding to a number of decimals that depends on the input data. Looks pretty wacky to me, but if you compute the products first: 119.35*8.5 = 1014.475, round to one decimal before continuing (because of 8.5) 109.,35*8.50 = 1014.475, round to two decimals before continuing, you get the results they report. Obviously there is some rounding, otherwise they would (should) have reported a and b to five decimals. I don't believe that the internal data representation makes a difference, since the binary expansion of 8.5 truncates. ==== In the shop we have a tolerance block on each properly made drawing. If it's missing, and there isn't a tolerance on every dimensioned feature in the drawing, there's trouble. Like did they mean 0.375 or 3/8? And in physics, we consider this more thoroughly. It's simply the definition of the total differential dF of a function F of some numbers n1, n2, n3, etc. Let's say F = 3 * 5.1 / 8.1 To a shop machinist, the implied tolerance is 0.05. But to a physicist dF = 1 * 5.1 / 8/1 + 3 * 0.1 / 8.1 + 3 * 5.1 / 0.1 At least I think that's how they do it. It might be half that amount. In any case, it's not going to be 0.05, at least I don't think it will.... A lot of this can be taken care of by letting all numbers be rational floats with a two-value ration mantissa and an integer exponent. I don't know if any languages do that. Now in your example 119.35 seems to imply a round off or tolerance of 0.005, and 8.50 seems to imply a tolerance of 0.01. At least that's the way many engineers would read it. I think at this point I am going to go put some bananna peels in a plastic bag, wait a week, and then prophesy with the generated ethylene. You, on the other hand, need to get a machinist, an engineer, and a physicist together to discuss this problem. Oh, and a programmer and a computer scientist, too. Yours, Doug Goncz, Replikon Research, Seven Corners, VA Fair use and Usenet distribution without restriction or fee Civil and criminal penalties for circumvention of any embedded encryption ==== > In the shop we have a tolerance block on each properly made drawing. If it's > missing, and there isn't a tolerance on every dimensioned feature in the > drawing, there's trouble. Like did they mean 0.375 or 3/8? And in physics, we consider this more thoroughly. It's simply the definition of > the total differential dF of a function F of some numbers n1, n2, n3, etc. Let's say F = 3 * 5.1 / 8.1 To a shop machinist, the implied tolerance is 0.05. But to a physicist dF = 1 * 5.1 / 8/1 + 3 * 0.1 / 8.1 + 3 * 5.1 / 0.1 Close, but it doesn't look quite right. To first order, you'd want the partial derivitive of: x*y/z with respect to x multiplied by the tolerance in x with respect to y multiplied by the tolerance in y with respect to z multiplied by the tolerance in z. I think you screwed up the derivitive of the last term. And you're not being very careful about your tolerance convention. d/dx = 5.1/8.1 dx = .5 => .5 * 5.1/8.1 d/dy = 3/8.1 dy = .05 => .05 * 3 / 8.1 d/dz = 3*5.1/8.1^2 dz = .05 => .05 * 3*5.1/8.1^2 Then too, that 3 looks like a constant with an implied tolerance of 0. > Now in your example 119.35 seems to imply a round off or tolerance of 0.005, > and 8.50 seems to imply a tolerance of 0.01. At least that's the way many > engineers would read it. What? 119.35 = tolerance of .005 8.50 = tolerance of .010 Do you have some rule about rounding aggressively toward a last digit of zero so that tolerances get doubled for rounded values that end in an explicit zero? Perhaps we need another rule for multiplying by 2.5 for rounded values that end in 5 on the same basis. Advanced only halfway in jest: 119.35 has a tolerance of +/- .025 (could be 119.325 through 119.375) 119.34 has a tolerance of +/- .005 (could be 119.335 through 119.345) 119.30 has a tolerance of +/- .01 (could be 119.290 through 119.310) or 119.30 has a tolerance of +/- .025 (could be 119.275 through 119.325) The tolerance you can safely infer from a value is (in part) intimately related to the way you round off to arrive at that value. John Briggs ==== Sometimes it is understood that: 8.5 means a real number between 8.45 and 8.55 8.50 means a real number between 8.495 and 8.505 so we would not write them interchangeably. This possible interpretation is one reason I prefer to write 1/2 when I mean an exact number, and not 0.5 . -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== In sci.math, Mike Curry : > We are trying to explain to the company who creates this language we are > using that the following is WRONG. As you can see, the following code > produces 2 different results. They claim that infact .5 and .50 are > different... any comments? 0010 LET A=119.35*8.5/100 0020 LET B=119.35*8.50/100 0030 PRINT a=,A 0040 PRINT b=,B -:run a= 10.15 b= 10.14 > It probably depends on the method of conversion. If one does it correctly 8.5 = 8.50 = 0x4021000000000000. (IEEE representation. 1 = 0x3ff0000000000000. Subtract '400' from the first three hex digits to get the actual power of 2. The binary point follows, then a hidden 1, then the rest of the number. The first bit is the sign bit of the entire number; the modified exponent is always positive.) However, if one does something slightly stupid like the following: 8.00 = 0x4020000000000000 0.10 = 0x3fb999999999999a 0.50 = 0x3ffe7fffffffffff (oops, rounding goof) 8.50 = 0x4020ffffffffffff one now has a very slightly different number, and misconversion of that number may produce the anomalous result 8.4 or 8.49. (It turns out 0.1 * 5 = 0.5 exactly anyway, at least in Intel. So the bug probably lies elsewhere. I'd have to see what 0.01 * 50 is.) Such a thing actually happened to Microsoft. 0.01 = 0x3f847ae147ae147b which turns out to be very slightly more than 0.01. Note the repeating hexadecimal. However, 3.11 = 0x4008e147ae147ae1 3.10 = 0x4008cccccccccccd 3.11 - 3.10 = 0x3f847ae147ae1400 which is very slightly less than 0.01 because the 7b got chopped from precision loss. A naive calculator (which Windows had for awhile; the bug finally did get fixed) would represent the result as 0.00. Oops! :-) -- #191, ewill3@earthlink.net It's still legal to go .sigless. ==== I would appreciate some help in the following. One can define the Kronecker Delta as delta_(ij) = 0 if i =/= j delta_(ij) = 1 if i = j I think that one can generalize this definition to an arbitrary field. Is F is a field, 0_F is is identity for addition, and 1_F is is identity for multiplication, then one would define the Kronecker Delta as delta_(ij) = 0_F if i =/= j delta_(ij) = 1_F if i = j My question is: is this last definition (for a field) a standard one? Sorry my english. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com ==== > I would appreciate some help in the following. One can define the Kronecker Delta as delta_(ij) = 0 if i =/= j > delta_(ij) = 1 if i = j I think that one can generalize this definition to an arbitrary field. Is F > is a field, 0_F is is identity for addition, and 1_F is is identity for > multiplication, then one would define the Kronecker Delta as delta_(ij) = 0_F if i =/= j > delta_(ij) = 1_F if i = j My question is: is this last definition (for a field) a standard one? Yes. For instance, this is used implicitly at: http://www.wikipedia.org/wiki/Identity_matrix Jose Carlos Santos ==== > One can define the Kronecker Delta as > delta_(ij) = 0 if i =/= j > delta_(ij) = 1 if i = j > With i,j in N and 0,1 in R > I think that one can generalize this definition to an arbitrary field. Is F > is a field, 0_F is is identity for addition, and 1_F is is identity for > multiplication, then one would define the Kronecker Delta as > delta_(ij) = 0_F if i =/= j > delta_(ij) = 1_F if i = j My question is: is this last definition (for a field) a standard one? > The same definition, will do for any ring and can be generalized by loosing up on the domain N of i,j. It's not the definition, but what you do with it that makes it notable. Basically delta is the characteristic function of the diagonal of NxN. Thus you see, characteristic functions are another generalization of K's delta. Furthermore, in topology it may make no difference what 0 & 1, only that they adhere to the ring axiom 0 /= 1. Thus the notion of a characteristic function could be stretched to a map from a subset into a two element set. ==== Jamis Harres > Dear mathematicians, I want to inform all of you that James Harris completed his proof of > Riemann Hypothesis. I was struggling for hours, then God (my Father) whispered the > solution to me. (sic) If any newcomers to sci.math are wondering what the JSH commotion is about, some background can be found here: www.crank.net/harris.html Larry ==== Dear mathematicians, I want to inform all of you that James Harris completed his proof of Riemann Hypothesis. I was struggling for hours, then God (my Father) whispered the solution to me. (sic) I know we (humans) are not prepared for such an event, but the time has come... a genius has come! And our duty is to recognize him: for what he completed, for what he is completing, and for what he'll complete. Jamis Harres ==== > Dear mathematicians, I want to inform all of you that James Harris completed his proof of > Riemann Hypothesis. I was struggling for hours, then God (my Father) whispered the > solution to me. (sic) I know we (humans) are not prepared for such an event, but the time > has come... a genius has come! And our duty is to recognize him: for > what he completed, for what he is completing, and for what he'll > complete. Jamis Harres Dang, he has beat me again. First an elementary proof of FLT, and now just today I was putting the final touches on my elementary 2-page proof that the Riemann Hypothesis is independent of ZF+AC. Ramanujan thought that his results came from God. Shiva, I think, or maybe Krishna, but with all the avatars I get confused. The formulas for 1/pi just sort of popped into his head. I wonder who Jamis Harres really is... The alterego a regular sci.math poster no doubt... I wonder who... And then, when it comes down to a question of identity, I wonder who I am... Max Maximum ==== This is how I successfully refuted James' proof of FLT. 1. James constructs all his arguments on elements and operations within the ring of algebraic integers. (There is an implicit assumption that the algebraic integers form a ring.) 2. He reaches the conclusion that the ring of algebraic integers is incomplete. (This means, although cryptically stated, that the algebraic integers do *not* form a ring.) 3. Either his conclusion is wrong -- which it is -- and the algebraic integers *do* form a ring, a result which as been proven, (The sums and products of any elements in the ring are also elements of the ring.) 4. or the algebraic integers do *not* form a ring. (in which case his assumptions are false and the arguments are invalid.) QED -- 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 <3c65f87.0307100405.10f63d42@posting.google.com> <17a4a089.0307102258.65776332@posting.google.com> <3c65f87.0307110436.7e58d364@posting.google.com> <17a4a089.0307132213.3a6c171f@posting.google.com> <3c65f87.0307140624.15589c74@posting.google.com> <17a4a089.0307142310.7128d8f1@posting.google.com> <3c65f87.0307150943.2b082551@posting.google.com> <17a4a089.0307160041.6b6bbc40@posting.google.com> <3c65f87.0307160630.322f0a17@posting.google.com> ==== Your belief here is irrelevant as a proof begins with a truth then > proceeds by logical steps to a conclusion which then must be true. > Challenging a paper requires that you find that it did not begin with > a truth, or that you find a break in the logical chain. > Actually, almost the opposite is true. You can challenge the *proof* of a fact by finding a false premise or logical flaw, but the result may be true independent of these errors, if one can find another proof. To challenge a statement, you need to show that it is false. To do this, you need not _ever_ see the proof. If you can draw false conclusions from a result, it is false. The proof is absolutely irrelevant if it is incorrect. If I claim that the product of two algebraic integers is an algebraic integer, and give a proof (A. Magidin and I, if not others, have done so on sci.math recently, proofs can be found in, say Dummit & Foote's _Abstract Algebra_ or Atiyah and Macdonald's _Commutative Algebra_), if someone finds a flaw in the proof, this does _not_ invalidate the result. The result is invalidated when someone produces two explicit algebraic an airtight argument for the existence of two such algebraic integers, without explicitly writing them down. That said, a _courteous_ thing to do when one has found an erroneous statement is to help the author of that statement find the flaw in his premeses or logic... ====================================================================== Michael A. Van Opstall Padelford C-113 opstall@math.washington.edu http://www.math.washington.edu/~opstall/ ==== > [...] The behaviour for all the values of n < 67 mentioned in the previous post > follows the pattern of one of the preceding paragraphs, except that > I was unable to verify that there were no solutions when n=62 and n=64 > (I suppose the ranks are one but my software found no generator for the > curves in the little time I allotted, so I couldn't check to see whether > the generator lay on the bounded part of the curve or not.) dave The following message seemed to be only visible in The Math Forum: Author: Allan MacLeod The problem of the representation N = a/b + b/c + c/a with a,b,c integers ( not necessarily positive ) is discussed in the paper TWO MORE REPRESENTATION PROBLEMS by Andrew Bremner and Richard Guy published in Proc. Edinburgh Math. Soc. vol 40 1997 pp 1-17. N = 62 and N = 64 both have solutions but they are quite large. Allan MacLeod ==== You set up some very general definitions, derive a few easy theorems and --- voila! --- all of a sudden get a deep result about a very concrete problem. Some authors like to call this abstract nonsense, tongue in cheek of course because by solving the concrete problem the approach turns out not be nonsense after all. Where was the phrase abstract nonsense used for the first time, for a piece of math? Nemo ==== >Where was the phrase abstract nonsense used for the first time, for a >piece of math? See http://www.risc.uni-linz.ac.at/research/category/risc/catlist/gen-abs-nons Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >You set up some very general definitions, derive a few easy theorems and > --- voila! --- all of a sudden get a deep result about a very concrete >problem. Some authors like to call this abstract nonsense, tongue in cheek of >course because by solving the concrete problem the approach turns out >not be nonsense after all. Where was the phrase abstract nonsense used for the first time, for a >piece of math? Well, if I recall correctly, it was commonly used to describe Category Theory. Nemo Larry (this space unintentially left blank ..... ==== >>Where was the phrase abstract nonsense used for the first time, for a >>piece of math? >See >http://www.risc.uni-linz.ac.at/research/category/risc/catlist/gen-abs-nons I have not used the term before, but it is often the case that using unreasonable generalizations makes understanding almost trivial. In some case, even introducing something which seems irrelevant makes the problem understandable when it would not be otherwise. One rarely uses randomized decisions, but most basic theorems require them. -- 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, Deptartment of Statistics, Purdue University ==== > Let (A_n)_{n>=0} be defined by A_0=1, A_1=2 > and for n in {2,3,...} (n+1)(2n-1)A_n=-2(2n^4-5n^2+1)A_{n-1} - n^2(n-1)^3(2n+1)A_{n-2} . For instance A_2= - 8 , A_3= 44 ,... . Let W(0):=1 and for k in {1,2,...} we denote > W(k)= (0^2 +1)(1^2 +1)...((k-1)^2 +1). > I am interested in following questions : 1) Prove or disprove that there are integers C(k,n) > such that > A_n=Sum_{k=0 to k=n}(-1)^{n-k}C(k,n)W(k) , n=1,2,... . Are the A_n all integers? If not, the answer is NO, since the W(k) are all integers. If the A_n are all integers, then use W(0)=1 to solve this by taking C(0,n) = (-1)^n A_n, and all other C(k,n)=0. Of course, there are other solutions, too. Here is what I get for the first few equations: 2 = -C[0, 1] + C[1, 1] -8 = C[0, 2] - C[1, 2] + 2 C[2, 2] 44 = -C[0, 3] + C[1, 3] - 2 C[2, 3] + 10 C[3, 3] -200 = C[0, 4] - C[1, 4] + 2 C[2, 4] - 10 C[3, 4] + 100 C[4, 4] -6000 = -C[0, 5] + C[1, 5] - 2 C[2, 5] + 10 C[3, 5] - 100 C[4, 5] + 1700 C[5, 5] 528000 = C[0, 6] - C[1, 6] + 2 C[2, 6] - 10 C[3, 6] + 100 C[4, 6] - 1700 C[5, 6] + 44200 C[6, 6] > 2) It's true that for k=0,1,...,n the inequalities > (-1)^{n-k}C(k,n) > 0 are verifed ? In my solution, NO, since many of these are =0 3) To find an explicit form of A_n . ============== ==== > Let (A_n)_{n>=0} be defined by A_0=1, A_1=2 > and for n in {2,3,...} > (n+1)(2n-1)A_n=-2(2n^4-5n^2+1)A_{n-1} - n^2(n-1)^3(2n+1)A_{n-2} . > For instance A_2= - 8 , A_3= 44 ,... . > Let W(0):=1 and for k in {1,2,...} we denote > W(k)= (0^2 +1)(1^2 +1)...((k-1)^2 +1). > I am interested in following questions : > 1) Prove or disprove that there are integers C(k,n) > such that > A_n=Sum_{k=0 to k=n}(-1)^{n-k}C(k,n)W(k) , n=1,2,... . > Are the A_n all integers? If not, the answer is NO, since the W(k) are > all integers. If the A_n are all integers, then use W(0)=1 to solve > this by taking C(0,n) = (-1)^n A_n, and all other C(k,n)=0. > Of course, there are other solutions, too. > Here is what I get for the first few equations: > 2 = -C[0, 1] + C[1, 1] > -8 = C[0, 2] - C[1, 2] + 2 C[2, 2] > 44 = -C[0, 3] + C[1, 3] - 2 C[2, 3] + 10 C[3, 3] > -200 = C[0, 4] - C[1, 4] + 2 C[2, 4] - 10 C[3, 4] + 100 C[4, 4] > -6000 = -C[0, 5] + C[1, 5] - 2 C[2, 5] + 10 C[3, 5] - 100 C[4, 5] + 1700 C[5, 5] > 528000 = C[0, 6] - C[1, 6] + 2 C[2, 6] - 10 C[3, 6] + 100 C[4, 6] > - 1700 C[5, 6] + 44200 C[6, 6] > 2) It's true that for k=0,1,...,n the inequalities > (-1)^{n-k}C(k,n) > 0 are verifed ? > In my solution, NO, since many of these are =0 > 3) To find an explicit form of A_n . > ============== Yes, I assert that all A_n , n=0,1,2,..., are integers.Alex/Proposer ==== P(x) = (x+1)(x+2) > that if I stick in actual values for x, the factorization is gone? > > No, it's just the opposite. If you stick in actual values > for x, you may get factorizations which are not consistent > with the polynomial factorization. For example, you > let x = 2. P(x) = 12. The factorization that is consistent > with the polynomial factorization that you just gave is 12 = 3 * 4 = (2 + 1)*(2 + 2). There are, however, other factorizations: for example 2 * 6 > or 1 * 12. These are NOT consistent with the polynomial > factorization. That is true. Hopefully there's some progress being made Nora Baron!!! > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). > > See above. > The expression I use is > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > > Not at all. You don't lose any of the polynomial factorizations > when you evaluate. You gain some *new* ones. That is exactly what > happened in what I posted. Well let's consider what you actually *did* which was to put in values for a_1, a_2, and a_3, as if you could just pick them at will. However, above you admit that the factorization still remains even when I put in a value for x. And in fact, the x's and the value of the a's are independent anyway, as can be seen from (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) so your position that my picking a value for x, as I picked x=2, affected the a's is NOT algebra. That is, there is no rational way you could have supposed that my picking an actual value for x would affect the a's in such a way that you thought you could just pick values for the a's as you did in your post. So I have (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and to simplify in order to lessen the ability of posters to confuse because of the symbol load, I set f=5, u=1, and x=2. Of those choices *only* the selection of f=5, affected the value of the a's, not x, Nora Baron, so your focus on x is bogus. James Harris ==== Let me ask you a question, are you claiming that given P(x) = (x+1)(x+2) that if I stick in actual values for x, the factorization is gone? > No, it's just the opposite. If you stick in actual values >>for x, you may get factorizations which are not consistent >>with the polynomial factorization. For example, you >>let x = 2. P(x) = 12. The factorization that is consistent >>with the polynomial factorization that you just gave is >> 12 = 3 * 4 = (2 + 1)*(2 + 2). >>There are, however, other factorizations: for example 2 * 6 >>or 1 * 12. These are NOT consistent with the polynomial >>factorization. > That is true. Hopefully there's some progress being made Nora > Baron!!! > Ah good. You understand that simplifying can introduce extraneous solutions that do not help you analyze your original problem. > >Yes, if I have P(2)=12, it is true that you just see a number, but >notice that P(2) = 3(4). > See above. >The expression I use is (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) where v=-1+mf^2, and y=uf, and now you are arguing that the >factorization goes away simply because I set x=2? > Not at all. You don't lose any of the polynomial factorizations >>when you evaluate. You gain some *new* ones. That is exactly what >>happened in what I posted. > Well let's consider what you actually *did* which was to put in values > for a_1, a_2, and a_3, as if you could just pick them at will. Or perhaps you don't. However, above you admit that the factorization still remains even > when I put in a value for x. It remains, but not uniquely. You appear to have missed the entire point. Put in a value for x and you get *extra* *valid* factorizations. If you don't want them, don't plug a value in for x. > And in fact, the x's and the value of > the a's are independent anyway, as can be seen from (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) Nora pretty clearly discredited that statement in her post. so your position that my picking a value for x, as I picked x=2, > affected the a's is NOT algebra. That is, there is no rational way > you could have supposed that my picking an actual value for x would > affect the a's in such a way that you thought you could just pick > values for the a's as you did in your post. What part of her algebra is incorrect? At what point did she do something where the right side doesn't equal the left side? If she did something that is not algebra, there is a mistake. Where is it? So I have (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and to simplify in order to lessen the > ability of posters to confuse because of the symbol load, I set f=5, > u=1, and x=2. Of those choices *only* the selection of f=5, affected the value of > the a's, not x, Nora Baron, so your focus on x is bogus. Assigning x a value changed it to a completely different problem with MORE factorizations. One of them was inconvenient. I guess you didn't understand after all. If you don't like the results, point out the specific error in her work. Otherwise, you are wasting bandwidth. -- Will Twentyman ==== > [...] > | DISPROOF OF CLAIM > | > | Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are > | integers, and assume Q(x) is irreducible over the rationals. > | Assume that c = p * v, where p is a prime and v is an integer. > | Let a1, a2, and a3 be the roots of Q(x). Note that by > | definition, since Q(x) is monic, a1, a2, and a3 are algebraic > | integers. > | > | Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime > | to p. It seems to me that it should be possible to prove this without relying > upon the existence of the automorphism of the algebraic numbers you used. Keith Ramsay You are right about that. W. Dale Hall has produced an independent proof for a special case that is simpler to verify. However on balance I think the automorphism argument is the shortest and simplest to understand. Nora B. ==== > My latest instruction, thanks very much to the efforts of Keith and Nora, > has been in the area of mathematical proof - I always believed that proofs > belonged solely to the realm of geometry (showing one triangle to be the > same as the other) and now I have seen how proof pervades even > (especially ?) the most esoteric math (not that I believe, looking at other > threads in this NG, that this stuff is particularly esoteric). I'd say high-level math courses are almost entirely proofs. Same idea as in Euclidean geometry: Start with a small set of axioms, then build a powerful set of theorems on top of them. The stuff in JSH threads is algebraic number theory for the most part. I've always been fascinated with it, but never took a course, and for some reason it never quite sticks with me. If it did, I'd be able to produce those polynomials and factorizations at the drop of a hat like Nora et al do. > I wonder, in the spirit of JSH, whether that last sentence is syntactically > and gramatically correct ? I'm not sure if I have set the NG stuff correctly - if I've done this right > then only sci.math will see me ... I don't understand why JSH feels a need > to post to sci.*, alt.*, *.fr and so on ... if it's mathematics then it > belongs here doesn't it ? (unless it's specifically related to > undergraduate studies, the definition of which varies by country) Only James really knows the answer to that. It has something to do with seeking a fresh unbiased audience who will accept his proof, now that he knows all mathematicians are corrupt. That was the motivation for alt.math.undergrad anyway. I can't recall why alt.writing. One of the lovely intricacies of the mind of James Harris is that he never sees the irony between what he claims evil mathematicians do and what he actually does. For instance, he keeps insisting that mathematical proofs are accepted without inspection, that the world has blind faith in any math paper written. But what he also keeps insisting on is that we all accept his proof on blind faith, just because he says it's true. He hasn't yet figured out that the verification process, the careful scrutiny any published proof is exposed to, is exactly the thing that's stopped him and the thing that he says doesn't exist. You left alt.math.undergrad in the headers, so I trimmed it in your spirit of trying to limit the crossposts. - Randy ==== > >>[...] >>| DISPROOF OF CLAIM >>| >>| Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are >>| integers, and assume Q(x) is irreducible over the rationals. >>| Assume that c = p * v, where p is a prime and v is an integer. >>| Let a1, a2, and a3 be the roots of Q(x). Note that by >>| definition, since Q(x) is monic, a1, a2, and a3 are algebraic >>| integers. >>| >>| Now ASSUME, as you claim, that one of a1, a2, or a3 is coprime >>| to p. >>It seems to me that it should be possible to prove this without relying >>upon the existence of the automorphism of the algebraic numbers you used. >>Keith Ramsay > You are right about that. W. Dale Hall has produced an independent > proof for a special case that is simpler to verify. However on > balance I think the automorphism argument is the shortest and > simplest to understand. Nora B. I agree that the Galois theory argument is more informative, shortest, and [given even a glimmer of understanding of what Galois theory is all about], simplest to understand. Further, it has the advantage of admitting some amount of generalization, which my approach foregoes entirely. That said, I think there is a major hurdle in getting JSH to accept the fact that Galois theory is correct, and applicable in the context of the factorization of polynomials. He seemingly has a huge bug up in an unmentionable orifice, about the fact that Galois theory canonically deals with fields and field extensions, not realizing that the ring of integers of a number field enjoys some properties that arbitrary rings do not. I had vainly hoped that JSH would [irrespective of any acknowledgement of the hated source] take a look at the verification of those common factors that I gave, via direct polynomial multiplications, verify those multiplications for himself to see that I wasn't lying, and come to his senses about his argument. Well, I *did* strongly suspect that my errand was in vain, but I wanted to give him the benefit of the doubt, and failing that, give him enough opportunity to demonstrate his unwillingness to face reality. I have apparently been granted that second wish rather than the first. Dale ==== P(x) = (x+1)(x+2) > that if I stick in actual values for x, the factorization is gone? > No, it's just the opposite. If you stick in actual values > for x, you may get factorizations which are not consistent > with the polynomial factorization. For example, you > let x = 2. P(x) = 12. The factorization that is consistent > with the polynomial factorization that you just gave is > 12 = 3 * 4 = (2 + 1)*(2 + 2). > There are, however, other factorizations: for example 2 * 6 > or 1 * 12. These are NOT consistent with the polynomial > factorization. That is true. Hopefully there's some progress being made Nora > Baron!!! > This seems a little ironic. You were claiming previously that by evaluating the polynomial, one would lose the factor- ization associated with it. I pointed out that you had it exactly backwards - that by evaluating the polynomial, in general you introduce some factorizations that are different from the polynomial factorization. Of course you still have the polynomial factorization as well. Also of course I knew all this. So when you say there's some progress being made, it is evidently on your end, not on mine. Also as I pointed out previously, in showing the factorization of P(2), I was making the point that it does you no good whatsoever to try to simplify by substituting in an actual value for x, when you have no intention of considering any other factorization than the polynomial factorization. It seemed to me that you did not understand that little subtlety, but now perhaps you are getting it. It is not the central point anyway, so I don't really care on this one. > Yes, if I have P(2)=12, it is true that you just see a number, but > notice that P(2) = 3(4). > See above. > The expression I use is > (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > where v=-1+mf^2, and y=uf, and now you are arguing that the > factorization goes away simply because I set x=2? > Not at all. You don't lose any of the polynomial factorizations > when you evaluate. You gain some *new* ones. That is exactly what > happened in what I posted. Well let's consider what you actually *did* which was to put in values > for a_1, a_2, and a_3, as if you could just pick them at will. > No, I didn't pick them at will. I did retain exactly the *form* of the factorization, (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5), and of course I factored in such a way that a1, a2, and a3 were all divisible by 5. Since when you let x = 2, it was a perfectly valid factorization of an ordinary integer, not of a polynomial. You said it could not be factored in this form and I proved that it could. Again, this is a minor side-issue. If you still don't get, don't worry about it. The main things you should worry about are what you deleted. I will say them again in a different way that you may possibly be able to understand: 1. You are considering a degree 3 polynomial P(x) which is also a function of m. When m = 0, that polynomial becomes of first degree in x. You note that if the factorization is of the form (a1*x + 5)*(a2*x + 5)*(a3*x + 5), then when m = 0, to retain this form for the factorization, two of the a's, say, a1 and a2, must be zero. Of course zero is divisible by 5, so you can say that when m = 0, a1 and a2 are multiples of 5. Up to this point everything is OK. Then you make the great leap. You conclude that not only are a1 and a2 multiples of 5 when m = 0, they must be multiples of 5 for other values of m also. It goes without saying here that for different values of m, the values of a1 and a2 are different. You may consider them as functions of m, and it would make sense to denote them as a1(m) and a2(m). So you are saying: a1(m) and a2(m) are divisible by 5 when m = 0 therefore a1(m) and a2(m) are divisible by 5 for ALL OTHER values of m. Right? So where's the problem? Why is everyone being so obtuse about this? Because you do not have the slightest hint, not the slightest shred, not the faintest wisp of justfication for the word therefore. It is pure hunch, pure intuition, pure guess. And pure error. It is a false conclusion. You do not cite any general theorem or mathematical principle that justifies this. You just say it. You think it is obvious and everyone else should just endorse it on the dotted line. Saying it is enough, right? You say it, and as with so many other things you have said, everyone else should just shut up and believe it. No further proof needed, really. It's not enough. It's false. There is another slightly interesting issue here, related to the degeneracy and singularity that occurs when m = 0. I will post something on this later. 2. How do I know it's false? Because W. Dale Hall and I have given separate, independent proofs that your main claim is false. Since you have YET AGAIN deleted out the section of my post that contained my proof, and since you have not previously found any valid objection to it, I am going to give you yet another chance and reproduce it here. I am sure your many fans out there are beginning to be embarrassed by your failure to cope with this, and they will appreciate the fact that you are being given another shot at it. Right, fans ? Not to mention the many future math historians, when they starting writing your biography: ============================================================== JSH CLAIM: It is possible to find a 3rd degree polynomial with integer coefficients, monic and irreducible over the rationals, such that, if a1, a2, and a3 are the three roots, then at least one of a1, a2 or a3 is coprime in the algebraic integers to a prime integer which divides the constant term of the polynomial. DISPROOF OF CLAIM: Let Q(x) = x^3 + a*x^2 + b*x + c, where a, b, and c are integers and c = p*v, where p is a prime and v is another integer. Q(x) is clearly monic. Assume Q(x) is irreducible. Let a1, a2, and a3 be roots of Q(x). Note that by definition, a1, a2, and a3 are algebraic integers. You are claiming that at least one of a1, a2, or a3 is coprime to p. Assume a1 is coprime to p. By standard theory, there exists an automorphism F12 of the field of algebraic numbers such that: 1. F12 leaves the subfield of rational numbers fixed, i.e., if q is rational, F12(q) = q. 2. F12(a1) = a2. 3. If t is an algebraic integer, F12(t) is also an algebraic integer. Now since a1 is relatively prime to p, there exist algebraic integers r and s such that [1] r*a1 + s*p = 1. Now apply the automorphism F12 to both sides of [1]: F12(r)*F12(a1) + F12(s)*F12(p) = F12(1). By the properties above, F12(p) = p and F12(1) = 1. Moreover, r' = F12(r) is an algebraic integer, and s' = F12(s) is an algebraic integer. Finally, F12(a1) = a2. Thus one obtains: r'*a2 + s'*p = 1, which says: a2 and p are coprime in the algebraic integers. Similarly one shows that a3 and p are coprime. Therefore if one of a1, a2, or a3 is coprime to p, then they all are. But a1 * a2 * a3 = p * v. That is, p divides the product of a1, a2, and a3. Therefore p cannot be coprime to each of a1, a2, and a3. Putting all this together, one concludes that NONE of a1, a2, or a3 can be coprime to p. This directly contradicts your claim noted above. Please feel free to point out any errors or gaps in the proof I just gave. ============================================================== > However, above you admit that the factorization still remains even > when I put in a value for x. Yes, of course the original polynomial factorization is still there. I never said it wasn't. I just pointed out that when you evaluate, you get other factorizations also - factorizations which clearly, obviously violate your claims. > And in fact, the x's and the value of > the a's are independent anyway, as can be seen from (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) so your position that my picking a value for x, as I picked x=2, > affected the a's is NOT algebra. That is, there is no rational way > you could have supposed that my picking an actual value for x would > affect the a's in such a way that you thought you could just pick > values for the a's as you did in your post. > Wrong. The original a's were derived from the roots of a polynomial in x. When you chose x = 2, you were abandoning the polynomial. What you had was no longer a polynomial in x. When you talked about factoring it, you were factoring an ordinary number. You didn't say, OK, now I have an ordinary number, but I still want to factor it as if it were a polynomial in x. That would have been a silly statement to make, and it would have implied that your simplification was really quite pointless. The only sensible interpretation is that after you evaluated the polynomial, you were thinking about factoring an ordinary number. Now you are trying weasel out of that because you see it was obviously false. AGAIN: this is a minor side issue. If you STILL really don't get it, stop worrying about it. Your real big-time problems are the two I listed above. To summarize: A. I and Dale Hall have found separate proofs that your central claims are incorrect. Unless you can find an error in both our proofs, it really doesn't matter much what you say. There is an error in your proofs and an error in your thinking. B. I have found an explicit place in your argument where your thinking is incorrect. I have described it in excruciating detail: see above. So far you have either been unable to understand it or unwilling to understand it. It is conceivable that I am wrong and that you could convince me that your argument actually makes sense. Of course you would also need to show that my proofs and Dale's are incorrect, because they say your CONCLUSION is wrong, regardless of how you got there. However so far you have not made even a feeble attempt on either front. Basically you just say, If it's true when m = 0, it must be true for all m. Period, end of argument. I got news for you. That ain't a proof. Nora B. > So I have (v^3+1)x^3 -3vxy^2 + y^3 = (a_1 x + y)(a_2 x + y)(a_3 x + y) > > where v=-1+mf^2, and y=uf, and to simplify in order to lessen the > ability of posters to confuse because of the symbol load, I set f=5, > u=1, and x=2. Of those choices *only* the selection of f=5, affected the value of > the a's, not x, Nora Baron, so your focus on x is bogus. > James Harris ==== Why not just take a simpler approach with James? Just ignore him. Nobody really believes him anyway (with the *possible* exception of himself). Why bother entertaining him? (If nothing else, it gives him the misleading appearance of credibility.) ==== > Well I'm going to try and break it down even more to try and see if > y'all will accept the mathematics: > Ok I have > P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) > where f is a prime integer other than 3, and u is coprime to f, and > looking at that x, I see the possibility for the factorization > P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > That's what I've been putting up a lot where you see a LOT of symbols, > which seem to confuse people. The ring is algebraic integers, and let > me get rid of as many of those symbols as I can: > Let x=2, f=5, u=1, so that I have > P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) > which is what I put up earlier, but I'm wary about some of you still > finding that confusing, so I'll work it out more to get > P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is > P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), > and now you can see what the polynomial P(m) looks like without so > many symbols. > Now from before where I had x, I *still* have that > P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > Interesting. When the x's were left unspecified, > the form of the factorization was P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5). Now that you have substituted in x = 2, it is no > longer a factorization of a polynomial in x. It is > just a factorization as a product of three numbers. However the a's in general are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so x has *nothing* to do with the value of the a's. > So 2*a_1 + 5, for example, is just an algebraic > integer. The factorization is [1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5). > So then what you are asserting below is that if you factor > the number P(m) as in [1], then one of a1, a2, or > a3 must be coprime to 5. Right? Yup, one of them is clearly coprime to 5. The coprimeness result leads to the conclusion that they all must be coprime to 5 in the ring of algebraic integers, which is what's wacky, and shows a problem with the ring. > Let's take m = 1. Then P(m) = 25 * 4285. This can be factored as 5 * 5 * 4285, which yields a1 = 0, a2 = 0, and a3 = 2140. None of these is coprime to 5. End of story. That is not correct as you can't just pick values for the a's in that way. > Don't like a1 = a2 = 0 ? Other things work > too - e.g., a1 = -5, a2 = -5, a3 = 2140. None > coprime to 5, as before. Read on, however - Ok. > So > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 - 600 m^2 - 126m + 11). > Now setting m=0 gives me > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > Notice that > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > (Note: Some posters have gotten a lot of mileage out of calling that a > degenerate case, but they were just fooling you into forgetting your > basic algebra and what you know about polynomials. I think they did > so deliberately as the math isn't complicated.) > Now comes the question of what happens when m is NOT 0, and answering > that question requires looking again at > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 - 600 m^2 - 126m + 11) > and noticing that the constant term is 25(11), but you can divide off > that 25 to get P(m)/25 which gives you a constant term that's 11. And > 11 and 5 are coprime. That's very important. In fact, that's the > *key* fact which should stick in your mind. > So now looking at > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = > (5000m^3 - 600 m^2 - 126m + 11) > I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor > is a factor of the constant term, and in fact, it'd have a factor of > the constant term that is 5. So why would you think that the factor > of the constant term would move or change when m changes? > Well it can't. > Given that with 2 a_1 + 5, that 5 in there is a factor of the constant > term of > 25(5000m^3 - 600 m^2 - 126m + 11) > you *still* have a factor of the constant term when 25 is separated > off. > But now your constant term is 11. > That forces all the factors of 5 to go away from 2 a_1 + 5. > Now you may wish for there to be someway for some factors to remain, > but what actuall happens is you get > (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = > (5000m^3 - 600 m^2 - 126m + 11) > while posters have argued that *all* the a's would have some factor of > 5. > > Yep, sure. See above. With actual numbers! Wrong as I pointed out above as Nora Baron apparently never realized that the a's are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so she can't just pick, though you can see she tried. And in fact for m=1, where she actually picked values for the a's the cubic is a^3 + 72a - 13825 = 0 and none of her picks work. James Harris ==== > Fair bit of snippage. I've got some work to do now to follow this up > properly. > And again. > 25(5000m^3 - 600 m^2 - 126m + 11) That some of the a's are coprime to 5? It turns out that in the ring of algebraic integers they all are, > which is why the ring has problems. > a_1 a_2 a_3 = 25 (625 m^3 - 75 m^2 + 3m) For any integer m, a_1 a_2 a_3 is a multiple of 25. i.e. not coprime to 5. Perhaps you mean that none of the a's are coprime to 5? Which is trivially obvious and uninteresting after all. Sorry to have bothered you. > James Harris Phil Nicholson. ==== > >Well I'm going to try and break it down even more to try and see if >y'all will accept the mathematics: Ok I have P(m) = f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) where f is a prime integer other than 3, and u is coprime to f, and >looking at that x, I see the possibility for the factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). That's what I've been putting up a lot where you see a LOT of symbols, >which seem to confuse people. The ring is algebraic integers, and let >me get rid of as many of those symbols as I can: Let x=2, f=5, u=1, so that I have P(m) = 25 (8(625 m^3 - 75 m^2 + 3m)- 3(2)(-1 + 25m ) + 5) which is what I put up earlier, but I'm wary about some of you still >finding that confusing, so I'll work it out more to get P(m) = 25(5000m^3 - 600 m^2 + 24m + 6 - 150m + 5), which is P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), and now you can see what the polynomial P(m) looks like without so >many symbols. Now from before where I had x, I *still* have that P(m) = (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5). > Interesting. When the x's were left unspecified, >>the form of the factorization was >> P(m) = (a_1*x + 5)*(a_2*x + 5)*(a_3*x + 5). >>Now that you have substituted in x = 2, it is no >>longer a factorization of a polynomial in x. It is >>just a factorization as a product of three numbers. > However the a's in general are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so x has *nothing* to do with the value of the a's. You keep saying this as if it's obvious... Let's look at it. P(m)= f^2 ((m^3 f^4 - 3m^2 f^2 + 3m)x^3- 3(-1 + mf^2 )xu^2 + u^3 f) distribute in f^2 P(m) = (m^3 f^6 - 3m^2 f^4 + 3mf^2) x^3 - 3(-1+mf^2)xu^2 f^2 + u^3 f^3 using v=-1+mf^2, v^3+1 is the coefficient on x^3 P(m) = (v^3+1) x^3 - 3vxu^2 f^2 + u^3 f^3 This can be viewed as a polynomial in terms of x, or v, or x and v. Viewed as a polynomial in terms of x and reversing it's coefficients gives the expression you have listed. Here's the catch, keeping x as an unknown RESTRICTS the possible values of the a's. Watch what happens when you plug in f=5, u=1 P(m) = (v^3+1) x^3 - 75 vx + 125 where now v=-1+25m. Now plug in x=2 P(m) = 8(v^3+1) - 600 v + 125 P(m) = 8v^3 - 600 v + 133 where v=-1+25m This can *not* be viewed as a polynomial in terms of x. It can only be the a's as you have suggested. When you set x=2, you lose information about the a's, and introduce additional values. In this example you can clearly see the difference between having x as a variable, and x=2. You fundamentally change the nature of the problem. Worse, when x=2, P(m) is no longer a monic polynomial under any available interpretation. The roots of this polynomial need not be algebraic integers. > > >>So 2*a_1 + 5, for example, is just an algebraic >>integer. The factorization is >>[1] P(m) = (2*a1 + 5)*(2*a2 + 5)*(2*a3 + 5). >>So then what you are asserting below is that if you factor >>the number P(m) as in [1], then one of a1, a2, or >>a3 must be coprime to 5. Right? > Yup, one of them is clearly coprime to 5. The coprimeness result > leads to the conclusion that they all must be coprime to 5 in the ring > of algebraic integers, which is what's wacky, and shows a problem with > the ring. >>Let's take m = 1. Then >> P(m) = 25 * 4285. >>This can be factored as >> 5 * 5 * 4285, >>which yields a1 = 0, a2 = 0, and a3 = 2140. >>None of these is coprime to 5. End of story. > That is not correct as you can't just pick values for the a's in that > way. You lost information when you chose x, and the available assignments of the a's grew. Inconvenient, but true. >>Don't like a1 = a2 = 0 ? Other things work >>too - e.g., a1 = -5, a2 = -5, a3 = 2140. None >>coprime to 5, as before. >>Read on, however - > Ok. >So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be >written like r+c, where r=0, or r varies as the polynomial variable >varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), which equals 0, when m=0, so at least one of the a's must equal 0, >when m=0. And to get that factor that is 25, you must have two a's that go to 0, >when m=0. (Note: Some posters have gotten a lot of mileage out of calling that a >degenerate case, but they were just fooling you into forgetting your >basic algebra and what you know about polynomials. I think they did >so deliberately as the math isn't complicated.) Now comes the question of what happens when m is NOT 0, and answering >that question requires looking again at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11) and noticing that the constant term is 25(11), but you can divide off >that 25 to get P(m)/25 which gives you a constant term that's 11. And >11 and 5 are coprime. That's very important. In fact, that's the >*key* fact which should stick in your mind. So now looking at (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5)/25 = (5000m^3 - 600 m^2 - 126m + 11) >I know that if 2 a_1 + 5 has a factor of 5, when m=0, then that factor >is a factor of the constant term, and in fact, it'd have a factor of >the constant term that is 5. So why would you think that the factor >of the constant term would move or change when m changes? Well it can't. Given that with 2 a_1 + 5, that 5 in there is a factor of the constant >term of 25(5000m^3 - 600 m^2 - 126m + 11) you *still* have a factor of the constant term when 25 is separated >off. But now your constant term is 11. That forces all the factors of 5 to go away from 2 a_1 + 5. Now you may wish for there to be someway for some factors to remain, >but what actuall happens is you get (2 a_1/5 + 1)(2 a_2/5 + 1)(2 a_3 + 5) = (5000m^3 - 600 m^2 - 126m + 11) while posters have argued that *all* the a's would have some factor of >5. > Yep, sure. See above. With actual numbers! > Wrong as I pointed out above as Nora Baron apparently never realized > that the a's are given by a^3 + 3v a - (v^3+1)=0, where v=-1+mf^2, so she can't just pick, though you can see she tried. And in fact for > m=1, where she actually picked values for the a's the cubic is a^3 + 72a - 13825 = 0 and none of her picks work. > James Harris Your assertion about the a's only holds true when x is kept as a variable. See my argument above. Perhaps if you clearly defined which variables the a's are dependent on we could clear up a lot of this mess. The only thing you've made clear is that they should depend on m. Do they also depend on f? u? x? If so, you must be careful which letters you substitute values in for. It appears that you do NOT wish m to be a function of x. This means that you cannot include a simplification that includes choosing values for x. If you do, you have *over*simplified and changed the problem. -- Will Twentyman ==== [snip] > I had vainly hoped that JSH would [irrespective of any acknowledgement > of the hated source] take a look at the verification of those common > factors that I gave, via direct polynomial multiplications, verify > those multiplications for himself to see that I wasn't lying, and > come to his senses about his argument. I'm afraid that your hope was truly in vain. In the few instances where James tried to follow up a simple multiplication of a few binomials, he constistently got the exponents and the signs wrong. He is hopelessly sloppy in his algebra. I think if you want to gain some ground (no promises) you will have to post the exact values of the numbers a1, a2 and a3, and then derive the expressions which prove they are not coprime to 5 in the ring of algebraic integers so that they are public record. I don't think James is disposed to do this or has the ability. It is simpler to for him just to assert that you must be wrong, since his proof is irrefutable. In a previous post he even said not to bother citing errors in his proof because if any existed *he* would let everyone know about it. > Well, I *did* strongly suspect that my errand was in vain, but I > wanted to give him the benefit of the doubt, and failing that, give > him enough opportunity to demonstrate his unwillingness to face reality. I have apparently been granted that second wish rather than the first. Dale -- 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 ==== Your (and Will's) patience is admirable, but Quixotic. JSH has shown time and time again that he either fails to understand the nature of mathematical proof, or simply does not feel constrained by it. Observing your attempts to enlighten him reminds me of a Simpsons episode where Homer is talking to the dog, but from the dog's end it is just meaningless blah blah blah interspersed with occurrences of the only word the dog understands, his own name. Gib ==== Your (and Will's) patience is admirable, but Quixotic. JSH has shown > time and time again that he either fails to understand the nature of > mathematical proof, or simply does not feel constrained by it. Observing > your attempts to enlighten him reminds me of a Simpsons episode where > Homer is talking to the dog, but from the dog's end it is just > meaningless blah blah blah interspersed with occurrences of the only > word the dog understands, his own name. Gib > Being an educator, I like to maintain the belief that all people are capable of learning. James has shown a capacity for thinking about math, so with my (perhaps false but still comforting) belief and his professed willingness to learn, I am willing to explain as long as he listens. I also find it fascinating to observe the various ways in which he avoids listening and the various mental gymnastics he displays in defending his version of truth. Finally, it's good for me. I've knocked a lot of rust off the mental gears by following the arguments and presenting my own. -- Will Twentyman ==== | Why not just take a simpler approach with James? Just ignore him. | Nobody really believes him anyway (with the *possible* exception of | himself). Why bother entertaining him? (If nothing else, it gives | him the misleading appearance of credibility.) By now this suggestion has been made many times, and there hasn't been anything like the kind of general cooperation that it would take to end the discussion. If the only point were to prevent people from being taken in, I agree just letting him attempt to persuade people would be as effective as arguing with him. But there are other reasons why people keep at it. For one thing, sometimes it's interesting to try to convince someone who's exceptionally resistant to being convinced, and see how it it that they manage to hang on to their opinions anyway. I think a number of us just find it irritating to see someone making claims we know are wrong and not being corrected. I won't claim that's a rational irritation. Some people find it entertaining in other ways. The main thing that would convince me to stop would be if I thought it would be better for his own well-being not to have me as a distraction from his other issues. It seems possible to me that he'd prefer I stopped discussing his proofs with him too, but I don't know. Keith Ramsay ==== > | Why not just take a simpler approach with James? Just ignore him. > | Nobody really believes him anyway (with the *possible* exception of > | himself). Why bother entertaining him? (If nothing else, it gives > | him the misleading appearance of credibility.) By now this suggestion has been made many times, and there hasn't > been anything like the kind of general cooperation that it would > take to end the discussion. If the only point were to prevent > people from being taken in, I agree just letting him attempt to > persuade people would be as effective as arguing with him. But > there are other reasons why people keep at it. For one thing, sometimes it's interesting to try to convince someone > who's exceptionally resistant to being convinced, and see how it it > that they manage to hang on to their opinions anyway. I think a number > of us just find it irritating to see someone making claims we know > are wrong and not being corrected. I won't claim that's a rational > irritation. Some people find it entertaining in other ways. The main thing that would convince me to stop would be if I thought > it would be better for his own well-being not to have me as a > distraction from his other issues. It seems possible to me that he'd > prefer I stopped discussing his proofs with him too, but I don't know. Keith Ramsay Perhaps I am a little behind on this discussion, but why can't we just give James a polynomial and see if he can make his technique work? Working from a polynomial you made yourself around your method (which if I checked correctly cannot be factored with integers anyway) is quite different than trying to apply it in practice. It also seems that - though a bit of work - the traditional p/q approach and sign examination are pretty simple. They are also fairly straightforward to implement programmatically. (I wonder if James has seen this?) ==== Your (and Will's) patience is admirable, but Quixotic. JSH has shown > time and time again that he either fails to understand the nature of > mathematical proof, or simply does not feel constrained by it. > Observing your attempts to enlighten him reminds me of a Simpsons > episode where Homer is talking to the dog, but from the dog's end it is > just meaningless blah blah blah interspersed with occurrences of the > only word the dog understands, his own name. Gib JSH is intelligent enough to do basic algebra pretty much through the quadratic equation and has learned random facts past that - e.g., he now knows what algebraic integers are - but he has not absorbed what it means to construct a complete rigorous proof. In the present controversy the problem is that he has an overpowering intuition that there must be a formula connecting the roots of a polynomial equation with the constant term - in fact he is right, in the sense that for polynomials through 4th degree, there are formulas involving radicals that specify the roots in terms of the coefficients. He believes that such formulas extend to degenerate cases. He then finds a formula-type relationship for a degenerate case, and then assumes that this formula must hold in nondegenerate cases as well. He cannot conceive that in nondegenerate cases, the formula generalizes in any but the obvious way. Ideally what we would do to prove this is incorrect is actually write down the roots of his polynomial and show how each root shares algebraic integer factors with prime divisors of the constant term. Incredibly enough, this is not completely simple even in the case of a quadratic. In the case he is interested in, the cubic, the formulas for the roots are quite complicated and showing directly that they include algebraic integer factors of the constant term is a horrible mess. It is much easier to prove it indirectly. That is what I have done using automorphisms and what W. Dale Hall has done in a quite different The automorphism argument is, I think, a nice example of how sometimes having some theoretical superstructure can create a shorter, more easily understood path to the answer than a direct frontal assault. Sometimes the long way round the mountain is actually the shorter than trying to drill your way through it. JSH essentially refuses to look at either of our arguments. That is the other problem. Although not stupid, he has truly enormous ego investment in his argument. He does not see that it has a flaw or gap. He thinks that bit about generalizing from the degenerate case is the natural and obvious thing to do, because he has that simple formula (a_1/5). He would rather conclude that there is something wrong with the definition of algebraic integers or there is a flaw in Galois theory than look really critically and rigorously at his own proof. However I actually don't think this can continue forever. I think we will eventually find a way through his armor. It has happened before. Simpson's dog: my recollection of that is that it came from a Far Side cartoon. In one panel the dog's owner is saying something like No, Ginger! You must not bark in the house, Ginger! Do you hear me, Ginger?, and the dog hears Blah, Ginger! Blah blah blah, Ginger! Blah blah blah, Ginger! I wonder which came first - Far Side, or the Simpson's version? In any case, yes, JSH has reinvented this also. Nora B. ==== certainly, but can he Complete the Square? it's not really a property of tetragona per se, but the diagram will help, either way. (I prefer the lunes proof of the pythagoreean th.; and, I finally realized what the spatial analog is, a couple o'weeks, ago .-) > JSH is intelligent enough to do basic algebra pretty much through > the quadratic equation and has learned random facts past that - e.g., --A church-school McCrusade (Blair's ideals?): Harry-the-Mad-Potter want's US to kill Iraqis?... http://www.tarpley.net/bush25.htm (Thyroid Storm ch.) http://www.rwgrayprojects.com/synergetics/plates/plates.html http://quincy4board.homestead.com/files/curriculum/Cosmo.PCX ==== > (A lot of the usual stuff snipped.) > So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 = 0 and conclude that at least one of the a_i's, say a_1, is zero. Putting this into your equation following the line Now setting m=0 gives me will then give you 5(2 a_2 +5)(2 a_3 + 5) = 25(11), and I can divide both sides by 5. Now, why couldn't it be that a_2 and a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by 5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you arrive at your statement, And to get that factor that is 25, you must -- Mark Thornquist ==== > (A lot of the usual stuff snipped.) > So > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = > 25(5000m^3 - 600 m^2 - 126m + 11). > Now setting m=0 gives me > (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). > Now let's say you accept that any factor of a polynomial can be > written like r+c, where r=0, or r varies as the polynomial variable > varies, while c remains constant and is a factor of the constant term. > Notice that > a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), > which equals 0, when m=0, so at least one of the a's must equal 0, > when m=0. > And to get that factor that is 25, you must have two a's that go to 0, > when m=0. > Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 > = 0 and conclude that at least one of the a_i's, say a_1, is zero. > Putting this into your equation following the line Now setting m=0 > gives me will then give you 5(2 a_2 +5)(2 a_3 + 5) = 25(11), and I can divide both sides by 5. Now, why couldn't it be that a_2 and > a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by > 5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you > arrive at your statement, And to get that factor that is 25, you must Sure. I'm considering non-polynomial factors of P(m), and I have from my lemma that any such factor can be written as r+c, where r=0, or varies with m, while c remains constant and is a factor of the constant term. So with g_1 = (2 a_1 + 5), and since one of the a's MUST equal 0, when m=0, selecting a_1 as the one gives me g_1 = 5, so c=5, and of course r = g_1 - c, and as m varies, r varies, while, of course, at m=0, it also equals 0. Now then I notice that P(m)/25 has a constant term that is coprime to 5, as P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11. So when that 25 goes, then a factor has to come out of g_1 as well. That is, looking at P(0)/25 = 11, I have that the constant term is coprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out. It's like P(m) = g_1 g_2 g_3, and I checked at P(0), to find that at that point g_1=5. But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5 goes. But when m doesn't equal 0, that only handles one of the 5's as 25 has two factors where each is 5. Therefore, ONE other of the a's must go to 0, when m=0, and it also has a factor that is 5 which separates out. That's why the lemma is KEY, and is the linchpin of the argument. I've also used P(m) = 25 Q(m) to explain, which can help you by letting you consider factors of Q(m). For instance, if you have h_1 = 2a_1/5 + 1 as a factor of Q(m), you can check at m=0, to find that h_1=1. But let's say that h_1 = 2a_1/s + 5/s where s is some non unit factor of 5, but 5/s is not coprime to 5. Then at m=0, you'd have h_1 = 5/s, which contradicts with Q(0)=11. Now you MAY wish to forget that a_1 = 0 from before with P(m) but doing so is not logical. If someone wishes any portion of what I just showed you expanded on, then please point out which section. I'm quite willing to explain in detail, but I can't read your mind. You need to tell me where you're not sure, so that I can give you more details at that point. James Harris ==== > >>(A lot of the usual stuff snipped.) >So (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(5000m^3 - 600 m^2 - 126m + 11). Now setting m=0 gives me (2 a_1 + 5)(2 a_2 + 5)(2 a_3 + 5) = 25(11). Now let's say you accept that any factor of a polynomial can be >written like r+c, where r=0, or r varies as the polynomial variable >varies, while c remains constant and is a factor of the constant term. Notice that a_1 a_2 a_3 = 25 (8(625 m^3 - 75 m^2 + 3m)), How do you figure? a_1, a_2, a_3 are non-polynomial functions of m. They don't have to be anything so simple as this. which equals 0, when m=0, so at least one of the a's must equal 0, >when m=0. And to get that factor that is 25, you must have two a's that go to 0, >when m=0. >Let's see if I've got this straight. You note that at m=0, a_1 a_2 a_3 >>= 0 and conclude that at least one of the a_i's, say a_1, is zero. >>Putting this into your equation following the line Now setting m=0 >>gives me will then give you >>5(2 a_2 +5)(2 a_3 + 5) = 25(11), >>and I can divide both sides by 5. Now, why couldn't it be that a_2 and >>a_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by >>5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how you >>arrive at your statement, And to get that factor that is 25, you must > Sure. I'm considering non-polynomial factors of P(m), and I have from > my lemma that any such factor can be written as r+c, where r=0, or > varies with m, while c remains constant and is a factor of the > constant term. So with g_1 = (2 a_1 + 5), and since one of the a's MUST equal 0, > when m=0, selecting a_1 as the one gives me See objection above regarding this. Note: you have never addressed the objections to your assumption that non-polynomial factors behave like polynomial factors. g_1 = 5, so c=5, and of course r = g_1 - c, and as m varies, r varies, while, > of course, at m=0, it also equals 0. Now then I notice that P(m)/25 has a constant term that is coprime to > 5, as P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11. So when that 25 goes, then a factor has to come out of g_1 as well. That is, looking at P(0)/25 = 11, I have that the constant term is > coprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out. It's like P(m) = g_1 g_2 g_3, and I checked at P(0), to find that at that point > g_1=5. But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5 > goes. But when m doesn't equal 0, that only handles one of the 5's as 25 has > two factors where each is 5. Therefore, ONE other of the a's must go > to 0, when m=0, and it also has a factor that is 5 which separates > out. Why? I'll suggest an alternative: a_2 = -2, a_3 = 25. Along with a_1=0 you end up with: g_1=5, g_2=1, g_3=55 That's why the lemma is KEY, and is the linchpin of the argument. I've also used P(m) = 25 Q(m) to explain, which can help you by > letting you consider factors of Q(m). For instance, if you have h_1 = 2a_1/5 + 1 as a factor of Q(m), you can check at m=0, to find that h_1=1. But let's say that h_1 = 2a_1/s + 5/s where s is some non unit factor of 5, but 5/s is not coprime to 5. Then at m=0, you'd have h_1 = 5/s, which contradicts with Q(0)=11. Not if h_2 = s/5 and h_3 = 11 Now you MAY wish to forget that a_1 = 0 from before with P(m) but > doing so is not logical. If someone wishes any portion of what I just showed you expanded on, > then please point out which section. I'm quite willing to explain in > detail, but I can't read your mind. You need to tell me where you're > not sure, so that I can give you more details at that point. Ok, I'd love to see why non-polynomial a_i must have the product you claimed. I've provided counter-examples elsewhere and you have yet to address them. > James Harris -- Will Twentyman ==== [...] |Now then I notice that P(m)/25 has a constant term that is coprime to |5, as | | P(m) = 25(5000m^3 - 600 m^2 - 126m + 11), so | | P(m)/25 = 5000m^3 - 600 m^2 - 126m + 11. | |So when that 25 goes, then a factor has to come out of g_1 as well. You could say what you mean more precisely if you quit using metaphors like factors coming out of terms. You appear again to be indicating that there must exist f1, f2, and f3 which are divisors of 5, which have the property that f1 divides g_1, f2 divides g_2, and f3 divides g_3, and f1*f2*f3=25, so that P(m)/25 = (g_1/f1)(g_2/f2)(g_3/f3). If this is what you mean, your way of putting it is not so good. |That is, looking at P(0)/25 = 11, I have that the constant term is |coprime to 5, and as g_1 at 0 IS 5 that means a 5 separates out. Why do you think that the factor coming out of each g is constant as m varies? Note that yet again, your explanation pulls uniformity in m out of a hat without stating any general principle to justify it. In the original cubic under consideration, it turns out (for reasons you aren't describing here) that there are divisors of f which one can divide out of g_1, g_2, and g_3, so that when the quotients are multiplied together, one gets P(m)/f^2. But these divisors of f vary with m! The way you've described your g's, not saying anything particular about how they vary from one value of m to another, there's no reason why they have to be in any particular order (except at m=0), or why the order can't vary from one value of m to another. If g_1, g_2, and g_3 satisfy your equation, then so do g'_1 = {g_1 if m <> 1 {g_2 if m = 1 g'_2 = {g_2 if m <> 1 {g_3 if m = 1 g'_3 = {g_3 if m <> 1 {g_1 if m = 1. Whatever the common factors of g_1(1), g_2(1) and g_3(1) with f were, they're a permutation from the common factors of g'_1(1), g'_2(1), and g'_3(1). The only way it could possibly make sense to conclude that they share factors with f in a consistent way going from one value of m to another, is if you imposed some additional condition on them that would guarantee the ordering of the three. [...] |If someone wishes any portion of what I just showed you expanded on, |then please point out which section. I'm quite willing to explain in |detail, but I can't read your mind. You need to tell me where you're |not sure, so that I can give you more details at that point. It's not a matter of not being sure. Your error is consistent. This expansion of this step repeats essentially the same error as the step being questioned had. Aren't you getting tired of doing that? Let's try an example. Let g1(m) and g2(m) satisfy for each m the identity x^2 - x + 3m = (x-g1(m))(x-g2(m)) for each value of x. Thus the product of g1 and g2 is always divisible by 3. But (using the fact that the algebraic integers are a Bezout domain, so that two algebraic integers have a GCD in the algebraic integers) what are the common factors of g1 and g2 with 3? Well, g1 and g2 are (1+-sqrt(1-12m))/2. When m=0, one of them is 1 and the other of them is 0. There, 0 has a GCD of 3 with 3, and 1 has a GCD of 1 with 3. But generally, when m is an integer the GCD in the algebraic integers of (1+sqrt(1-12m))/2 with 3 is a number of the form [a + b*(1+sqrt(1-12m))/2] ^{1/k} where a, b, and k are integers that depend on m. For example, GCD(3, (1+sqrt(-11))/2) = (1+sqrt(-11))/2 GCD(3, (1+sqrt(-23))/2) = [2-sqrt(-23)]^{1/3} and so on with no simple, obvious pattern. It took more calculation than I expected to compute that last one, and I'd be interested to know whether there's a pattern in the answers that I haven't recognized. If m is an irrational algebraic integer, for a and b one might need instead algebraic integers which are expressible as polynomials with rational coefficients in m. Keith Ramsay ==== > There are a lot of interesting documents here and I may have missed > something but I can't find any ideas about how to model affine > transformations. > Have you come across a document which covers this? > Where I am stuck is, do I have to use 3,4 or 5 D vectors? Which part of the > multvector holds the translation? And what goes in the other parts of the > multivector? > Martin Unfortunately, I don't have much time to answer your technical > questions, but I think you couldn't do any better than to start with > the paper http://modelingnts.la.asu.edu/pdf/CompGeom-ch1.pdf and take particular notice of the section Linearizing the Euclidean > group on page 20. As I recall, the vectors are in 4 dimensions. Patrick don't know if this helps, but if one constructs the clifford algebra on a quadratic space V(Q), then the unit ball of the resultant even subalgebra rotates V(Q) in the form you gave. so unit complex numbers rotate R^2, and unit quaternions rotate R^3. M.T. ==== Could someone please help me solve an algebra problem I found in a book: A survey of 1500 individuals found that 43% listen to radio news reports, 45% listen to TV news reports, and 36% read a daily newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. ==== > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > x+u+v+t=43 y+u+w+t=45 z+v+w+t=36 max(x+y+z+v+u+w+t) Question, is constraint x+y+z+v+u+w+t <=100 also required? ==== > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. 36%. There's nothing in the statement of the problem that forbids newspaper-readers from also listening to radio and watching TV. But there can't be any more than the smallest group size. Norm ==== > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > What have you done so far? [I'm assuming it is homework...] ==== Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > x+u+v+t=43 > y+u+w+t=45 > z+v+w+t=36 max(x+y+z+v+u+w+t) Terminology misunderstanding. What is the maximum possible number of people such that each is involved into all three activities: max(t) What is the maximum possible number of people such that each is involved into any of those three activities max(x+y+z+v+u+w+t) > Question, is constraint x+y+z+v+u+w+t <=100 also required? I think that constraint x+y+z+v+u+w+t <=100 is redundant, but can't formally prove it:-( ==== > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > This is a simple linear programming problem. Define seven variables: r => The number who listen to radio but don't watch TV or read a paper. rt => The number who listen to radio and watch TV but don't read a paper. rp => The number who listen to radio and read a paper but don't watch TV. rtp => The number who do all three. Similarly for t, p, and tp. Then we have the following program: max: rtp; radio: r + rt + rp + rtp = 0.43 * 1500; tv: t + rt + tp + rtp = 0.45 * 1500; paper: p + rp + tp + rtp = 0.36 * 1500; everybody: r + t + p + rt + rp + tp + rtp = 1500; I don't see a particularly nicer formulation. For what it's worth, this formulation is readily solved by lp_solve. ==== Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > This is a simple linear programming problem. Define seven variables: r => The number who listen to radio but don't watch TV or read a paper. > rt => The number who listen to radio and watch TV but don't read a paper. > rp => The number who listen to radio and read a paper but don't watch TV. > rtp => The number who do all three. Similarly for t, p, and tp. Then we have the following program: max: rtp; radio: r + rt + rp + rtp = 0.43 * 1500; > tv: t + rt + tp + rtp = 0.45 * 1500; > paper: p + rp + tp + rtp = 0.36 * 1500; > everybody: r + t + p + rt + rp + tp + rtp = 1500; I don't see a particularly nicer formulation. For what it's worth, this > formulation is readily solved by lp_solve. How about (.36 * 1500) = 540 Y'all are making this too hard. Read the question and think about it. ==== >> Could someone please help me solve an algebra problem I found in a >> book: >> A survey of 1500 individuals found that 43% listen to radio news >> reports, 45% listen to TV news reports, and 36% read a daily >> newspaper. What is the maximum possible number that do all three? >> Assume that each individual does at least one of these activities. >> x+u+v+t=43 >> y+u+w+t=45 >> z+v+w+t=36 >> max(x+y+z+v+u+w+t) Terminology misunderstanding. What is the maximum possible number of people such that each is involved >into all three activities: max(t) What is the maximum possible number of people such that each is involved >into any of those three activities max(x+y+z+v+u+w+t) > Question, is constraint >> x+y+z+v+u+w+t <=100 >> also required? I think that constraint x+y+z+v+u+w+t <=100 is redundant, but can't formally >prove it:-( > But from Assume that each individual does at least one of these activities you get x+y+z+u+v+w+t = 100, wich is definitely not redundant. -- Wim Benthem ==== Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > This is a simple linear programming problem. Define seven variables: r => The number who listen to radio but don't watch TV or read a paper. > rt => The number who listen to radio and watch TV but don't read a paper. > rp => The number who listen to radio and read a paper but don't watch TV. > rtp => The number who do all three. Similarly for t, p, and tp. Then we have the following program: max: rtp; radio: r + rt + rp + rtp = 0.43 * 1500; > tv: t + rt + tp + rtp = 0.45 * 1500; > paper: p + rp + tp + rtp = 0.36 * 1500; > everybody: r + t + p + rt + rp + tp + rtp = 1500; I don't see a particularly nicer formulation. For what it's worth, this > formulation is readily solved by lp_solve. How about (.36 * 1500) = 540 Y'all are making this too hard. Read the question and think about it. Assume r=30 % t=30 p=30 rp=0 tp=0 tr=0 rpt=10 and reverse engineer the problem to radio = 40% tv = 40% paper = 40% Apply your method. Got 10%? You are making it too easy. ==== > Could someone please help me solve an algebra problem I found in a >> book: >> A survey of 1500 individuals found that 43% listen to radio news >> reports, 45% listen to TV news reports, and 36% read a daily >> newspaper. What is the maximum possible number that do all >three? >> Assume that each individual does at least one of these activities. >> This is a simple linear programming problem. >> Define seven variables: >> r => The number who listen to radio but don't watch TV or read a >paper. >> rt => The number who listen to radio and watch TV but don't read a >paper. >> rp => The number who listen to radio and read a paper but don't >watch TV. >> rtp => The number who do all three. >> Similarly for t, p, and tp. >> Then we have the following program: >> max: rtp; >> radio: r + rt + rp + rtp = 0.43 * 1500; >> tv: t + rt + tp + rtp = 0.45 * 1500; >> paper: p + rp + tp + rtp = 0.36 * 1500; >> everybody: r + t + p + rt + rp + tp + rtp = 1500; >> I don't see a particularly nicer formulation. For what it's worth, >this >> formulation is readily solved by lp_solve. >How about (.36 * 1500) = 540 Y'all are making this too hard. Read the question >and think about it. > you're forgetting that everyone does at least one of the three activities, if you let 36% do all three than you have only a further 7% that can listen to the radio, and 9% that can watch TV. The four equations above are really necessary, and are easy to solve by substiting the values of r, t and p from the first three in the last equation. -- Wim Benthem ==== > >> Could someone please help me solve an algebra problem I found in a >> book: >> A survey of 1500 individuals found that 43% listen to radio news >> reports, 45% listen to TV news reports, and 36% read a daily >> newspaper. What is the maximum possible number that do all > three? >> Assume that each individual does at least one of these activities. >> This is a simple linear programming problem. >> Define seven variables: >> r => The number who listen to radio but don't watch TV or read a > paper. >> rt => The number who listen to radio and watch TV but don't read a > paper. >> rp => The number who listen to radio and read a paper but don't > watch TV. >> rtp => The number who do all three. >> Similarly for t, p, and tp. >> Then we have the following program: >> max: rtp; >> radio: r + rt + rp + rtp = 0.43 * 1500; >> tv: t + rt + tp + rtp = 0.45 * 1500; >> paper: p + rp + tp + rtp = 0.36 * 1500; >> everybody: r + t + p + rt + rp + tp + rtp = 1500; >> I don't see a particularly nicer formulation. For what it's worth, > this >> formulation is readily solved by lp_solve. > How about (.36 * 1500) = 540 Gee, thanks. I prefer my version, which shows clearly where the numbers come from. You're welcome to work up your own version. > Y'all are making this too hard. Read the question > and think about it. There is nothing hard about my version. Also, it's correct, unlike a number of the comments I see being tossed about through this thread. To make life a bit easier on everybody, the answer is 180. I hope that's not giving away too much. The rest of the answer looks like this: The number of people who listen to radio exclusively is 465. The number of people who watch TV exclusively is 495. The number of people who read the paper exclusively is 360. Nobody does exactly two of the three. The number of people do all three is 180. It's easy to verify that these numbers satisfy the constraints of the problem. The trick is to prove that 180 is the maximum who could do all three. ==== Suppose we have 3 sets, A,B,C with n(A)=45, n(B)=43 and n(C)=36 Let * mean intersection Let n(A*B*C)=36. Then clearly A*C=B*C= empty set. Since n(A)-n(B)=2 we put 2 into A*B'*C' Now since 43-36=7 we put 7 into A*B*C' Clearly 36 is the largest size of A*B*C so .36(1500)=540 is the answer. > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > ==== We can generalize this one. Let there be 3 sets, A,B and C with n(A)=a, n(B)=b and n(C)=c where a<=b<=c. Let n(A*B*C)=a (where * means intersection) So we immediately get n(A*B)=n(A*C)= empty set. let n(B*C)=b-a, n(B*A'*C')=0 and n(C*B'*A')=c-b Then the smaller of a,b,c is the most that can go into the triple intersection!! > Suppose we have 3 sets, A,B,C with n(A)=45, n(B)=43 and n(C)=36 > Let * mean intersection > Let n(A*B*C)=36. > Then clearly A*C=B*C= empty set. > Since n(A)-n(B)=2 we put 2 into A*B'*C' > Now since 43-36=7 we put 7 into A*B*C' > Clearly 36 is the largest size of A*B*C > so .36(1500)=540 is the answer. > Could someone please help me solve an algebra problem I found in a > book: A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. > ==== To make the rtp maximum, shouldn't we simply set rp, rt, and tp to 0? Minimalizing them would logically maximalize rtp. radio: r + rtp = 0.43 * 1500; tv: t + rtp = 0.45 * 1500; paper: p + rtp = 0.36 * 1500; I still don't know how to solve this by hand though. > Could someone please help me solve an algebra problem I found in a > book: > A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? > Assume that each individual does at least one of these activities. > > This is a simple linear programming problem. Define seven variables: r => The number who listen to radio but don't watch TV or read a paper. > rt => The number who listen to radio and watch TV but don't read a paper. > rp => The number who listen to radio and read a paper but don't watch TV. > rtp => The number who do all three. Similarly for t, p, and tp. Then we have the following program: max: rtp; radio: r + rt + rp + rtp = 0.43 * 1500; > tv: t + rt + tp + rtp = 0.45 * 1500; > paper: p + rp + tp + rtp = 0.36 * 1500; > everybody: r + t + p + rt + rp + tp + rtp = 1500; I don't see a particularly nicer formulation. For what it's worth, this > formulation is readily solved by lp_solve. ==== >To make the rtp maximum, shouldn't we simply set rp, rt, and tp to 0? >Minimalizing them would logically maximalize rtp. radio: r + rtp = 0.43 * 1500; >tv: t + rtp = 0.45 * 1500; >paper: p + rtp = 0.36 * 1500; I still don't know how to solve this by hand though. *I'll leave out the factor of 1500 in the rest of this) You forgot r + t + p + rt + rp + tp + rtp = 1 if you then write the equations in this form (1) r = 0.43 - rp - rt - rtp (2) t = 0.45 - rt - tp - rtp (3) p = 0.36 - rp - tp - rtp and substitute these in the last equation: (0.43 - rp - rt - rtp) + (0.45 -rt - tp - rtp) + (0.36 - rp - tp - rtp) + rt + rp + tp + rtp = 1 wich gives (4) 2 * rtp = 0.24 - rp - rt - rp Furthermore, all of r,p,t,rp,rt,rp and rtp are >= 0 > A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? Assume that each individual does at least one of these activities. The maximum possible number that do all three is 1500. This of course requires the assumption that quite a few of them lied to the survey. But then again any solution requires us to make some assumption about the reaction of the people to the survey, the competence and honesty of the people taking the survey, etc., etc. -- ==== For Pete's sakes, it's a math problem. When you were in elementary school and were given word problems, did you grill the teacher with questions inquiring about every possible assumption used in the word problem? Nevertheless, let's clarify: assume all the individuals responded correctly and honestly. > A survey of 1500 individuals found that 43% listen to radio news > reports, 45% listen to TV news reports, and 36% read a daily > newspaper. What is the maximum possible number that do all three? > Assume that each individual does at least one of these activities. The maximum possible number that do all three is 1500. This of course > requires the assumption that quite a few of them lied to the survey. > But then again any solution requires us to make some assumption > about the reaction of the people to the survey, the competence > and honesty of the people taking the survey, etc., etc. ==== > For Pete's sakes, it's a math problem. When you were in elementary > school and were given word problems, did you grill the teacher with > questions inquiring about every possible assumption used in the word > problem? No, I didn't - but, you know, I've learned a few things since then. One thing I've learned is that in a math problem the assumptions are of paramount importance. Another thing I've learned is not to do other people's homework for them, but give them something to think about, instead. They are under no obligation to think, but, then again, I was under no obligation to do their homework, was I? -- ==== Prompt please where it is possible to find algorithm of the numerical decision of stochastic Shrodinger equation with casual potential having zero average and delta ö correlated in space and time? The equation: i*a*dF/dt b*nabla*F-U*F=0 where i - imaginary unit, d/dt - partial differential on time, F=F (x, t) - required complex function, nabla - Laplas operator, U=U (x, t)- stochastic potential. Delta-correlated potential =A*delta(x-x`) *delta(t-t`) . where delta - delta-function of Dirack, A ö const, <> - simbol of average, Zero average: =0 Gaussian distributed P(U)=C*exp(U^2/delU^2) Where C, delU - constants. ==== Prompt please where it is possible to find algorithm of the numerical decision of stochastic Shrodinger equation with casual potential having zero average and delta ö correlated in space and time? The equation: i*a*dF/dt b*nabla*F-U*F=0 where i - imaginary unit, d/dt - partial differential on time, F=F (x, t) - required complex function, nabla - Laplas operator, U=U (x, t)- stochastic potential. Delta-correlated potential =A*delta(x-x`) *delta(t-t`) . where delta - delta-function of Dirack, A ö const, <> - simbol of average, Zero average: =0 Gaussian distributed P(U)=C*exp(U^2/delU^2) Where C, delU - constants. ==== Prompt please where it is possible to find algorithm of the numerical > decision of stochastic Shrodinger equation with casual potential > having zero average and delta ö correlated in space and time? The equation: > i*a*dF/dt b*nabla*F-U*F=0 where > i - imaginary unit, > d/dt - partial differential on time, > F=F (x, t) - required complex function, > nabla - Laplas operator, > U=U (x, t)- stochastic potential. > Delta-correlated potential =A*delta(x-x`) > *delta(t-t`) . > where delta - delta-function of Dirack, A ö const, <> - > simbol of average, > Zero average: =0 > Gaussian distributed P(U)=C*exp(U^2/delU^2) > Where C, delU - constants. Alexey, Since nobody answers i will try to help (even if i can not give you a definit answer). Perhaps you may repeat your question either in sci.math.num-analysis or a physics group. Further there are books of Peter Kloeden on numerical methods for stochastic differential equations. You can find them at www.amazon.com for example and look at the content to see whether they would be be helpfull. One is http://www.amazon.com/exec/obidos/tg/detail/-/3540540628/ref= pm dp ln b 2/104-4516719-6687142?v=glance&s=books&vi=contents (bring it in 1 line to have the URL) Mainly he works with Maple, his homepage is http://www.math.uni-frankfurt.de/~numerik/kloeden/maplesde/ You also can look at http://www.google.com/advanced search?hl=en or http://www.google.com/advanced search?hl=ru If you do so please not that the spelling is Schr.9adinger or Schroedinger. ==== > Leroy's ingenious analytic methods are all very well and good. However, I recommend adopting a more empirical, probabilistic approach. n! is the mean time it takes to get all numbers in ascending order when > tickets numbered 1 to n are randomly jumbled up and picked sequentially. So if you do this experiment often enough, you will have a perfect statistical > estimate of n! And note - as you are trying to estimate an INTEGER, you can > be suitably assured when you have the EXACT answer - most unusual in stats. > For the really advanced, you might like to consider drawing n from m, > where m > n. This makes use of the fact that n! = (P^m_n)/(C^m_n) . Impressive! > PERFECT!! Factorials are indeed the number of ways of picking things one at a time. The number of ways you can pick three coloured balls one at a time is 6: Red Green Blue Red Blue Green Green Red Blue Green Blue Red Blue Red Green Blue Green Red Similarly, the number of ways of picking 2 one at a time is 2! The number of ways of picking 1 one at a time is 1! The number of ways of picking 0 one at a time is 1 (or 0!) because the Way is NOT TO PICK. The number of ways of picking HALF a ball ONE at a time is ROOT-PI over TWO. Explain.......? > ----------------------------------------------------------------------------- - > Bill Taylor W.Taylor@math.canterbury.ac.nz > ----------------------------------------------------------------------------- - > He's the sort of fellow that uses statistics like a drunk uses a lamp-post, > for support rather than illumination. > ----------------------------------------------------------------------------- - ARE statistics DAMNED LIES? Charles Douglas Wehner ==== > The theme of this message thread may be summarized by: All 4-partite graphs are 4-colorable. > A graph is 4-C if and only if it is 4-partite Since there is an obvious lack of interest in the alternative > approach; it is unnecessary to discuss it further! Perhaps so, but I would like to add that I have found an elementary proof of the four color theorem that is so short that (as an exercise in pandering to the curiosity of the masses) I found an experienced etcher who etched the entire proof on the head of a pin. Maxissimo ==== > The theme of this message thread may be summarized by: All 4-partite graphs are 4-colorable. > A graph is 4-C if and only if it is 4-partite Since there is an obvious lack of interest in the alternative > approach; it is unnecessary to discuss it further! Perhaps so, but I would like to add that I have found an elementary > proof of the four color theorem that is so short that (as an exercise > in pandering to the curiosity of the masses) I found an experienced > etcher who etched the entire proof on the head of a pin. Maxissimo Ha! Just yesterday, I found one which is so small it could be inscribed with a very high energy laser on an up or down quark! I am attempting to shrink it further, down to the size of the component strings, but so far I keep losing parts of the proof in the quantum foam. ...tonyC ==== > The theme of this message thread may be summarized by: All 4-partite graphs are 4-colorable. > A graph is 4-C if and only if it is 4-partite Since there is an obvious lack of interest in the > alternative > approach; it is unnecessary to discuss it further! Perhaps so, but I would like to add that I have found an > elementary > proof of the four color theorem that is so short that (as > an exercise > in pandering to the curiosity of the masses) I found an > experienced > etcher who etched the entire proof on the head of a pin. Maxissimo Ha! Just yesterday, I found one which is so small it could > be inscribed with a very high energy laser on an up or down > quark! I am attempting to shrink it further, down to the > size of the component strings, but so far I keep losing > parts of the proof in the quantum foam. ...tonyC To tonyC. I cannot find the point to this posting! Is there one? ==== As I've received no analytical objections to the following post I'm appending several historical observations. Planck's Constant Previously in the thread Angular Momentum in Rotating Bodies, I >presented an analytical framework for the interpretation of dr/dt in >circular rotation of a point mass m at velocity v and radius r. No one >I know of agrees with my interpretation of dr/dt. However, in the >interests of further establishing this general framework, I would like >to pursue general developement of the idea which culminates in the >analytical definition of Planck's constant. We begin by noting that in cases of circular rotation at constant >angular velocity we have a centripetally directed dr/dt acting on >point mass m of a magnitude equal to tangential velocity v. This is >what causes the rotation of v and produces r as a consequence of >rotation. We then integrate dr/dt along r which produces 1/2 mvr/2pi with units >of measure equal to rr/t. Now, I have been cautioned on several >occasions not to suggest that this quantity represents angular >momentum in conventional terms and I agree. Perhaps we should simply >call it rotational momentum to prevent confusion. What we notice immediately however is that it bears the same form as >the quantity mvr corresponding to Planck's constant. However, we have >to straighten certain things out in this connection. In conventional macro angular rotation such as flywheels we have a >centripetal dr/dt and tangential v which are equal to each other. They >are effectively bound up through tensile forces internal to the body >undergoing rotation. In celestial angular mechanics on the other hand >we have a wide variety of potential dr/dt's and tangential orbital >velocities operating in various combinations. different situation. The tangential velocity of rotation v is constant >under all circumstances. In other words, v = c. Thus dr/dt operates >mass. second) times an analytical masslet, m0 (kg-sec) and interpret the >quantity mvr as a multiple of nm0vr. Further we can interpret r as a >function of c/n such that Planck's constant = m0cc. In other words, m0 >is roughly on the order of 10^-50 kg-sec in magnitude and Planck's >constant corresponds to the multiple of m0 and the square of the >velocity of light. We notice several things about rotational momentum. In linear motion >at constant velocity rotational momentum is zero because dr/dt and mvr >are both zero. And in circular rotation at a constant angular velocity >rotational momentum is constant because mvr is constant. This >represents the analytical distinction between circular and linear >motion. Further we notice that dr/dt can be of any magnitude. It is not bound >by the constancy of the velocity of light as an upper limit because it >doesn't go anywhere. It only produces rotation in relation to actual >tangential motion v = c. mass and radius of rotation are inversely proportional, that is that > Linear versus Analytical Mechanics One of the really unfortunate aspects of Newton's choice of a linear frame of reference for the analysis of mechanics is that r is poorly defined and t is not defined at all. In other words, r is only defined in direction and t is not defined by any consideration pertinent to the analytical frame of reference. And this had a pernicious impact on the subsequent development of angular mechanics as well as relativistic considerations and quantum mechanics in the twentieth century. The problem is that r and t and their combinations are all we have to work with. Taken to the second level of compounding we have six combinations: r, 1/t, r/t, r/tt, rr/t, and rr/tt. However, in the linear analytical frame of reference the next to last combination rr/t was overlooked because there is no apparent application for it in linear mechanical contexts. On the other hand, in angular frames of reference we have applications for all combinations and all the elements are well defined. The radius of rotation is well defined in terms of direction and magnitude and time is well defined in analytical terms as whatever time is needed for 2pi radians of rotation. The rr/t combination is also well defined in angular terms. However, in extrapolating the idea of rr/t from linear to angular contexts in classical mechanics, whoever devised the analytical approach made the mistake of trying to emulate linear mechanics in the sense of explaining rotation as a linear progression of r instead of a simple radial v in combination with tangential v. This is more akin to an anachronistic pre Newtonian view of mechanics. Kepler thought that some force of angels was needed to keep planets in orbit around the sun and regarded that force as tangential in direction. Newton on the other hand recognized that the only force needed was centripetal in nature and not tangential. But whoever devised the analytical considerations underlying angular mechanics apparently never considered the Newtonian perspective and presumably relied on the pre Newtonian rationale. Thus we wind up with a conceptual schism among the various realms of angular mechanics. On the one hand we have orbital angular mechanics, the macro realm of ordinary angular mechanics, and the micro realm of quantum effects. And unfortunately there is no conceptual integration among them. We are convinced that all represent mechanical realms but we have no basis for comprehending each in terms of the others. Orbital angular mechanics represents the realm of remote interactions dealt with in terms of inverse square centripetal forces and tangential orbital velocities. Whereas the macro realm of ordinary angular mechanics deals with linear analogs such as moments of inertia instead of mass, torque instead of force, and angular acceleration and velocity instead of their linear analogs. The micro realm of angular mechanics on the other hand is dealt with on the merely descriptive basis of formalisms. This is the realm of quantum mechanics - QM - or as I prefer to call it quantum magic where things don't seem to happen for any definite mechanical reason at all. However with the redefinition of macro angular momentum and Planck's constant in circular rotation we are at last in a position to understand the mechanical differences among the realms in conceptual terms. The micro realm of quantum effects is one of constant tangential velocity of rotation v = c and a variable radial dr/dt. The macro realm of ordinary angular mechanics on the other hand is one in which the tangential velocity of rotation is variable but tangential v = radial dr/dt and both are kept in strict synchronization by internal tensile forces. And finally orbital angular mechanics is defined by various combinations of tangential v and radial dr/dt. This is normally thought of in celestial terms but in point of fact applies equally to the atomic realm as well. ==== > Planck's Constant Previously in the thread Angular Momentum in Rotating Bodies, I > presented an analytical framework for the interpretation of dr/dt in > circular rotation of a point mass m at velocity v and radius r. No one > I know of agrees with my interpretation of dr/dt. However, in the > interests of further establishing this general framework, I would like > to pursue general developement of the idea which culminates in the > analytical definition of Planck's constant. The form of dr/dt follows directly from the vector definition of r. >There is no freedom for further definitions. Lester has delusions >of competency. He needs to solve his conceptual problems with >vector calculus before assuming the pulpit on quantum mechanics. >[Old Man] Well, actually the form of dr/dt follows directly from the definition > of v not r. dr/dt doesn't follow from the definition of v, it IS the definition of v. dr/dt is the derivative of r, therefore its form follows from the definition of r, and the definition of derivative. > I have been roundly chastized repeatedly for considering > dr in isolation. I hope that even you can see that r(t) is not dr in isolation. Given r(t), given what it means to take a derivative, the derivative of r(t) is a vector which does not necessarily point in the direction of r. Circular motion about x=3 (as opposed to z=0, just for variety): r = (3, cos(w*t), sin(w*t)) The derivative at all times t is v = dr/dt = (0, -w*sin(w*t), w*cos(w*t)) You say differently? What do you get for the derivative? At time t = 0, the vectors are: r(0) = (3, 1, 0) v(0) = (0, 0, w) I think even you can see those two vectors are not in the same direction. In fact, they are perpendicular. r(0) lies in the (x,y) plane. v(0) points up, in the +z direction. You disagree with this? What numerical value do you get for v(0)? > So, I have a suggestion: get your story straight. > before trying to admonish others. You have garbled Old Man's story as well as everybody else's. Using dr in isolation has nothing to do with the process of taking r(t), taking its derivative, and writing down v(t). As he said, the form of v(t) is completely specified by the definition of r(t). There is no room for interpreting (0,0,w) as being in other directions than +z. - Randy ==== > Planck's Constant Previously in the thread Angular Momentum in Rotating Bodies, I > presented an analytical framework for the interpretation of dr/dt in > circular rotation of a point mass m at velocity v and radius r. OK, a point mass rotating around a central point a fixed distance away (r). > No one > I know of agrees with my interpretation of dr/dt. Since you fixed r in the first paragraph, we know dr/dt = 0. What's the problem? ==== >> Planck's Constant >> >> Previously in the thread Angular Momentum in Rotating Bodies, I >> presented an analytical framework for the interpretation of dr/dt in >> circular rotation of a point mass m at velocity v and radius r. OK, a point mass rotating around a central point a fixed distance away (r). > No one >> I know of agrees with my interpretation of dr/dt. Since you fixed r in the first paragraph, we know dr/dt = 0. What's the problem? Not exactly. I fixed r in the first paragraph not infinitesimal r. Radial dr/dt = tangential v. ==== There's two big problems with your interpretation. First off, the a lepton) is virtual, meaning it's nonreal. Simple calculus (which you may not have even done correctly) cannot give you the correct interpretation of a qfp. Second, you took v=c, but you seem to use 'c' interchangeably between c representing constant and c representing the speed of light. This is a very big no-no. Please be more clear when you write this stuff out. The very foundations of your argument are false. (...Starblade Riven Darksquall...) <3f1d4c78.15468236@netnews.att.net> ==== In message <3f1d4c78.15468236@netnews.att.net>, Lester Zick [...] One of the really unfortunate aspects of Newton's choice of a linear >frame of reference for the analysis of mechanics is that r is poorly >defined and t is not defined at all. In other words, r is only defined >in direction and t is not defined by any consideration pertinent to >the analytical frame of reference. You mean, Newtonian mechanics is invariant under translation in space and time, and rotation? The consequences of this symmetrical deficiency are (to put it conservatively) interesting. Don't you just love irony? -- Richard Herring <3f0cc224.42180505@netnews.att.net> ==== In message <3f0cc224.42180505@netnews.att.net>, Lester Zick > Planck's Constant Previously in the thread Angular Momentum in Rotating Bodies, I > presented an analytical framework for the interpretation of dr/dt in > circular rotation of a point mass m at velocity v and radius r. No one > I know of agrees with my interpretation of dr/dt. However, in the > interests of further establishing this general framework, I would like > to pursue general developement of the idea which culminates in the > analytical definition of Planck's constant. >>The form of dr/dt follows directly from the vector definition of r. >>There is no freedom for further definitions. Lester has delusions >>of competency. He needs to solve his conceptual problems with >>vector calculus before assuming the pulpit on quantum mechanics. >>[Old Man] Well, actually the form of dr/dt follows directly from the definition >of v not r. Well, actually no. The form of dr/dt follows from the definition of derivative, d()/dt, and the definition of whatever () happens to be, the vector r in this case. dr/dt is tautologously the definition of v. > I have been roundly chastized repeatedly for considering >dr in isolation. So, I have a suggestion: get your story straight. >before trying to admonish others. P. K. B. -- Richard Herring ==== >The form of dr/dt follows directly from the vector definition of r. >There is no freedom for further definitions. Lester has delusions >of competency. He needs to solve his conceptual problems with >vector calculus before assuming the pulpit on quantum mechanics. In fact you could drop the word vector in that last sentence. -- Richard Herring ==== >There's two big problems with your interpretation. First off, the >a lepton) is virtual, meaning it's nonreal. Simple calculus (which you >may not have even done correctly) cannot give you the correct >interpretation of a qfp. It's certainly real enough to measure. I guess we'll just have to take your word for it. Second, you took v=c, but you seem to use 'c' interchangeably between >c representing constant and c representing the speed of light. This is >a very big no-no. Please be more clear when you write this stuff out. Of course an angular mechanic has no difficulty using v in the same way. The very foundations of your argument are false. This is certainly good to know. > ==== >In message <3f1d4c78.15468236@netnews.att.net>, Lester Zick [...] >>One of the really unfortunate aspects of Newton's choice of a linear >>frame of reference for the analysis of mechanics is that r is poorly >>defined and t is not defined at all. In other words, r is only defined >>in direction and t is not defined by any consideration pertinent to >>the analytical frame of reference. You mean, Newtonian mechanics is invariant under translation in space >and time, and rotation? The consequences of this symmetrical >deficiency are (to put it conservatively) interesting. Don't you just love irony? > I never irony while the strike is hot. <3f1d4c78.15468236@netnews.att.net> <3f1ea618.24033274@netnews.att.net> ==== In message <3f1ea618.24033274@netnews.att.net>, Lester Zick >In message <3f1d4c78.15468236@netnews.att.net>, Lester Zick >>[...] One of the really unfortunate aspects of Newton's choice of a linear >frame of reference for the analysis of mechanics is that r is poorly >defined and t is not defined at all. In other words, r is only defined >in direction and t is not defined by any consideration pertinent to >the analytical frame of reference. >>You mean, Newtonian mechanics is invariant under translation in space >>and time, and rotation? The consequences of this symmetrical >>deficiency are (to put it conservatively) interesting. >>Don't you just love irony? >I never irony while the strike is hot. Neither knowest Noether. -- Richard Herring ==== >>There's two big problems with your interpretation. First off, the >>a lepton) is virtual, meaning it's nonreal. Simple calculus (which you >>may not have even done correctly) cannot give you the correct >>interpretation of a qfp. > It's certainly real enough to measure. I guess we'll just have to take > your word for it. > >>Second, you took v=c, but you seem to use 'c' interchangeably between >>c representing constant and c representing the speed of light. This is >>a very big no-no. Please be more clear when you write this stuff out. > Of course an angular mechanic has no difficulty using v in the same > way. > >>The very foundations of your argument are false. > This is certainly good to know. > Yes, the first step to recovery is accepting that you are ill. Dabbling with quantum mechanics in your state is dangerous - even classical mechanics are risky. Fortunately, your condition is not contagious, but you must avoid abstractions, especially attempts at mathematics. Take some time off, rest in bed and repeat twice a day with feeling - Physics is for nerds. Stay off usenet until your symptoms subside and then restrict yourself to lurking on sci.agriculture. -- Joe Legris ==== > Circular motion about x=3 (as opposed to z=0, > just for variety): r = (3, cos(w*t), sin(w*t)) > Small point of clarification, though everybody but Lester probably already knew what I meant: This is circular motion in the y-z plane around the point (3,0,0). My standard example has been something like r = (cos(w*t), sin(w*t), 0) which is circular motion in the xy plane around the origin. - Randy ==== >> Planck's Constant > Previously in the thread Angular Momentum in Rotating Bodies, I > presented an analytical framework for the interpretation of dr/dt in > circular rotation of a point mass m at velocity v and radius r. OK, a point mass rotating around a central point a fixed distance away (r). No one > I know of agrees with my interpretation of dr/dt. Since you fixed r in the first paragraph, we know dr/dt = 0. What's the problem? Not exactly. I fixed r in the first paragraph not infinitesimal r. > Radial dr/dt = tangential v. Close :). You mean to say (I think) that you fixed magnitude(r) in the first paragraph not the position vector r. ==== ... stuff deleted ... > Linear versus Analytical Mechanics One of the really unfortunate aspects of Newton's choice of a linear > frame of reference for the analysis of mechanics is that r is poorly > defined and t is not defined at all. In other words, r is only defined > in direction and t is not defined by any consideration pertinent to > the analytical frame of reference. And this had a pernicious impact on the subsequent development of > angular mechanics as well as relativistic considerations and quantum > mechanics in the twentieth century. > Golly, that must be why physics doesn't work at all! And to think that I used to believe that we had computers and airplanes and rockets and satellites and all that stuff, due in large part to models based on the theories of physics. What's worse is that all that angular motion stuff (you know, rotating objects, orbital mechanics, the quantization of angular momentum, the whole schmear). Yet, my bike always worked just fine, and they've even managed to send satellites out to visit planets and all. I wonder how they figured out how to throw the things up there just right, so that they would go to the right place, and how they knew from the start, just how long it would take? Probably just a lucky guess, right? At any rate, it's good to be rid of those pesky wrong-brained ideas. ... stuff deleted ... > Dale ==== >> >> Planck's Constant >> >> Previously in the thread Angular Momentum in Rotating Bodies, I >> presented an analytical framework for the interpretation of dr/dt in >> circular rotation of a point mass m at velocity v and radius r. >>OK, a point mass rotating around a central point a fixed distance away (r). >> No one >> I know of agrees with my interpretation of dr/dt. >>Since you fixed r in the first paragraph, we know dr/dt = 0. What's the problem? >> >> Not exactly. I fixed r in the first paragraph not infinitesimal r. >> Radial dr/dt = tangential v. >> >> Close :). You mean to say (I think) that you fixed magnitude(r) in >the first paragraph not the position vector r. ==== >> >> >There's two big problems with your interpretation. First off, the >a lepton) is virtual, meaning it's nonreal. Simple calculus (which you >may not have even done correctly) cannot give you the correct >interpretation of a qfp. >> >> >> It's certainly real enough to measure. I guess we'll just have to take >> your word for it. >> >Second, you took v=c, but you seem to use 'c' interchangeably between >c representing constant and c representing the speed of light. This is >a very big no-no. Please be more clear when you write this stuff out. >> >> >> Of course an angular mechanic has no difficulty using v in the same >> way. >> >The very foundations of your argument are false. >> >> >> This is certainly good to know. >> Yes, the first step to recovery is accepting that you are ill. Dabbling >with quantum mechanics in your state is dangerous - even classical >mechanics are risky. Fortunately, your condition is not contagious, but >you must avoid abstractions, especially attempts at mathematics. Take >some time off, rest in bed and repeat twice a day with feeling - >Physics is for nerds. Stay off usenet until your symptoms subside and >then restrict yourself to lurking on sci.agriculture. > This puts me in mind of that old line from Ben Casey - if anyone remembers the show. He said of psychologists that he never met one who didn't need one. The fascinating thing is that everyone seems to to keep coming back for more. Apparently they need the eggs. And I've managed to collect quite a number of ovarians and quite a few returnees. Prussing just came back for more as well as Hall. I'm expecting Hogg any day now. And of course Green and Herring and yourself. Needless to say more should be arriving any time. Perhaps we can all undergo shrinkology together. I'll bring my bow tie and mustache. ==== ... stuff deleted ... > Linear versus Analytical Mechanics >> >> One of the really unfortunate aspects of Newton's choice of a linear >> frame of reference for the analysis of mechanics is that r is poorly >> defined and t is not defined at all. In other words, r is only defined >> in direction and t is not defined by any consideration pertinent to >> the analytical frame of reference. >> >> And this had a pernicious impact on the subsequent development of >> angular mechanics as well as relativistic considerations and quantum >> mechanics in the twentieth century. >> Golly, that must be why physics doesn't work at all! And to think that I >used to believe that we had computers and airplanes and rockets and >satellites and all that stuff, due in large part to models based on the >theories of physics. What's worse is that all that angular motion stuff (you know, rotating >objects, orbital mechanics, the quantization of angular momentum, the >whole schmear). Yet, my bike always worked just fine, and they've even >managed to send satellites out to visit planets and all. I wonder how >they figured out how to throw the things up there just right, so that >they would go to the right place, and how they knew from the start, just >how long it would take? Probably just a lucky guess, right? At any rate, it's good to be rid of those pesky wrong-brained ideas. I know how you feel. It's so good to have you back online critiquing ideas. I'm sure there are more to come. ==== > >> Wrong, sort of. >> The most general definition of an analytic function (on an open subset of >> C) requires only that it is differentiable (in the complex sense). On the >> other hand, for pedagogical purposes it's common to start with the >> assumption it's continuously differentiable, and only later (or perhaps >> never) prove Goursat's Theorem which says that differentiability alone is >> enough. I don't know of a text that uses Cauchy-Riemann, rather >> than complex differentiability, as the definition of analytic function. I took a course in complex analysis last semester. We used the book >Basic Complex Analysis by J.E. Marsden. Marsden defined analyticity >by the complex differentiability requirement. My professor went on >to make a terminological distinction: he called this class of >functions holomorphic. We were not entitled to call holomorphic >functions analytic until we proved equivalence (He said analytic >functions were functions which are equal to power series which >converge on the region in question) . We went on to prove that >holomorphic functions were C^oo with the Cauchy Integral formula. >Then we proved Taylor's theorem and thus the equivalence of the >holomorphicity and analyticity definitions. Good for him! What the word analytic really means is indeed given by a power series, and it's an amazing fact that a function that's differentiable in the complex sense is in fact given by a power series. Making the distinction is a good idea, because for example people often speak of a function f : R -> R as being analytic if it's given by a power series (locally); in _that_ context differentiability, even infinite differentiability, does not imply analyticity. (I should mention that this last notion is often called real-analytic, just because analytic is so often taken as a synonym for holomorphic.) In the same vein one might say that entire doesn't really mean analytic in the whole plane, it really means given by a single power series in the entire plane, and that turns out to be equivalent. But it's not equivalent in other contexts: Ex: Give an example of an analytic f : R -> R such that the Taylor series for f centered at the origin has finite radius of convergence. >Alex Solla >Junior >Reed College ************************ David C. Ullrich ==== In the same vein one might say that entire doesn't > really mean analytic in the whole plane, it really > means given by a single power series in the entire > plane, and that turns out to be equivalent. But > it's not equivalent in other contexts: Ex: Give an example of an analytic f : R -> R such > that the Taylor series for f centered at the origin > has finite radius of convergence. > Uhm, Sum (from zero to infinity) of x^n? Yeah, the geometric series ought to work. It converges on the interval (-1, 1). So the radius of convergence is 1. Easy. :) I can't really think of any non-trivial ones, though. Alex Solla Junior Reed College ==== > In the same vein one might say that entire doesn't > really mean analytic in the whole plane, it really > means given by a single power series in the entire > plane, and that turns out to be equivalent. But > it's not equivalent in other contexts: > Ex: Give an example of an analytic f : R -> R such > that the Taylor series for f centered at the origin > has finite radius of convergence. > > Uhm, Sum (from zero to infinity) of x^n? Yeah, the geometric series ought to work. It converges on the > interval (-1, 1). So the radius of convergence is 1. Easy. :) I > can't really think of any non-trivial ones, though. But it is not analytic R -> R. Has a pole at x=1. Alex Solla > Junior > Reed College ==== >> >> In the same vein one might say that entire doesn't >> really mean analytic in the whole plane, it really >> means given by a single power series in the entire >> plane, and that turns out to be equivalent. But >> it's not equivalent in other contexts: >> >> Ex: Give an example of an analytic f : R -> R such >> that the Taylor series for f centered at the origin >> has finite radius of convergence. >> Uhm, Sum (from zero to infinity) of x^n? That doesn't define a function that's analytic on all of R. of R if and only if there exists a function F, holomorphic in an open set in the plane containing R, such that f is the restriction of F to R.) >Yeah, the geometric series ought to work. It converges on the >interval (-1, 1). So the radius of convergence is 1. Easy. :) I >can't really think of any non-trivial ones, though. Alex Solla >Junior >Reed College ************************ David C. Ullrich ==== > In the same vein one might say that entire doesn't > really mean analytic in the whole plane, it really > means given by a single power series in the entire > plane, and that turns out to be equivalent. But > it's not equivalent in other contexts: > Ex: Give an example of an analytic f : R -> R such > that the Taylor series for f centered at the origin > has finite radius of convergence. Uhm, Sum (from zero to infinity) of x^n? That doesn't define a function that's analytic on all of R. of R if and only if there exists a function F, holomorphic > in an open set in the plane containing R, such that > f is the restriction of F to R.) >Yeah, the geometric series ought to work. It converges on the >interval (-1, 1). So the radius of convergence is 1. Easy. :) I >can't really think of any non-trivial ones, though. Alex Solla >Junior >Reed College ************************ David C. Ullrich SPOILER ... | | V | | V | | V | | V | | V | | V | | V | | V f(x) = 1 / (1+x^2) should do it. It is the restriction to R of F(z) = 1 / (1+z^2), a function that is analytic except for 2 poles at i and -i. Therefore, f can be expanded in a Taylor series about any point x0, and the radius of convergence will be sqrt(x0^2+1). In particular, the Taylor series about the origin has radius of convergence 1. - EM ==== I am in posession of the function: f(z)=c^z, c<>0, c in C and its iterates: f^(n)(z)={f(f^(n-1)(z)), iff n > 1, f(z), iff n = 1}, and I am interested in solving the complex equation: f^(n)(z)=z. (1) The above cannot be solved analytically, but all the iterates are analytic for c <> 0. Therefore the following Maple code works: >f:=z->c^z; >s:=series((f@@5)(z),z=c,5); >p:=convert(s,polynom); >c:=1.2; >solve(p=z,z); I am getting 5 solutions, of which one is indeed a solution of (1). My questions: 1) Are the rest of the solutions (polynomial roots) completely useless for the original equation? 2) Can I perhaps optimize the code to filter out the polynomial solutions which are not solutions of (1)? 3) Can I improve on the above to get other solutions as well? 4) Can I improve on the above to increase the accuracy of the found solutions to (1), without increasing the degree of the resultant polynomial too much? -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. ==== > I am in posession of the function: > f(z)=c^z, c<>0, c in C and its iterates: > f^(n)(z)={f(f^(n-1)(z)), iff n > 1, f(z), iff n = 1}, > and I am interested in solving the complex equation: > f^(n)(z)=z. (1) The above cannot be solved analytically, but all the iterates are > analytic for c <> 0. Therefore the following Maple code works: >f:=z->c^z; >s:=series((f@@5)(z),z=c,5); >p:=convert(s,polynom); >c:=1.2; >solve(p=z,z); 3) Can I improve on the above to get other solutions as well? > 4) Can I improve on the above to increase the accuracy of the found > solutions to (1), without increasing the degree of the resultant > polynomial too much? compute the series of the inverse function directly: s:= series(RootOf((f@@5)(z) = y, z), y = c, 5); y=c is not the only possible choice for the center of the series. Different centers will lead to different solutions. You can comfortably accomodate far more than 5 terms in this series. ==== |>I am in posession of the function: |>f(z)=c^z, c<>0, c in C and its iterates: |>f^(n)(z)={f(f^(n-1)(z)), iff n > 1, f(z), iff n = 1}, |>and I am interested in solving the complex equation: |>f^(n)(z)=z. (1) |>The above cannot be solved analytically, but all the iterates are |>analytic for c <> 0. Therefore the following Maple code works: |>>f:=z->c^z; |>>s:=series((f@@5)(z),z=c,5); |>>p:=convert(s,polynom); |>>c:=1.2; |>>solve(p=z,z); |>I am getting 5 solutions, of which one is indeed a solution of (1). |>My questions: |>1) Are the rest of the solutions (polynomial roots) completely useless |>for the original equation? Yes, because the series approximation is only good for z near c. |>2) Can I perhaps optimize the code to filter out the polynomial |>solutions which are not solutions of (1)? You can use fsolve with an interval around c. But you may as well do this with the original equation: > Digits:= 15; fsolve((f@@5)(z)=z, z=c-0.1 .. c + 0.1); 1.25773454137653 Note BTW that this is actually a solution of f(z)=z, not just of (f@@5)(z)=z. |>3) Can I improve on the above to get other solutions as well? > fsolve((f@@5)(z)=z, z=c + 0.1 .. 100); There's another solution of f(z)=z at approximately 14.7674583809828. I haven't found any complex solutions of (f@@5)(z)=z. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >|>I am in posession of the function: >|>f(z)=c^z, c<>0, c in C and its iterates: >|>>c:=1.2; >There's another solution of f(z)=z at approximately 14.7674583809828. >I haven't found any complex solutions of (f@@5)(z)=z. Well, again it suffices to solve f(z)=z. Write z = u+vi, match magnitudes and angles of f(z) and z, eliminate u, plot the function of v and look for a match. I get f(z) = z when z is about 20.94645536 + 40.45731830 I This method finds all the solutions of f(z) = z, and they form a nice sequence in the plane. Fixed points of iterates of f look like they won't permit such an easy reduction to functions of one variable. They also look to be a numerical-analysis nightmare... dave ==== > f(z)=c^z, c<>0, c in C and its iterates: > f^(n)(z)={f(f^(n-1)(z)), iff n > 1, f(z), iff n = 1}, > and I am interested in solving the complex equation: > f^(n)(z)=z. (1) ^^^ My earlier answer ignored the fact that you were looking for fixed points. The series of the inverse function will probably be of no more use than the series of the original for finding fixed points. ==== |>> f(z)=c^z, c<>0, c in C and its iterates: |>> f^(n)(z)={f(f^(n-1)(z)), iff n > 1, f(z), iff n = 1}, |>> and I am interested in solving the complex equation: |>> f^(n)(z)=z. (1) |> ^^^ |>My earlier answer ignored the fact that you were looking for fixed points. |>The series of the inverse function will probably be of no more use than |>the series of the original for finding fixed points. On the other hand, you can get a series for the fixed point as a function of c: > series(RootOf(c^x-x,x,1),c=1,5); 2 3 4 5 1 + c - 1 + (c - 1) + 3/2 (c - 1) + 7/3 (c - 1) + O((c - 1) ) Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== [snip] >I haven't found any complex solutions of (f@@5)(z)=z. Well, again it suffices to solve f(z)=z. No, it doesn't suffice to solve f(z)=z. The fixed points of f(z) are essentially all known, and can be expressed analytically as: e^[-LW(k,-Log(c))], k in Z, depending on the LambertW branch. All these fixed points of f are automatically fixed points of f^(n)(z), but the converse is not true. A fixed point of f^(n)(z) is not necessarily a fixed point of f(z). In particular, the (perhaps complex) points of a p-cycle may be fixed points of f^(kp)(z), k in Z, but not of f(z). And such p-cycles _do_ exist. Certain real points on (0, e^(-e)) for example, are known to be a 2-cycle, so they are fixed points of f^(2)(x), but they are not fixed points of f(x): [snip] > dave -- Ioannis http://users.forthnet.gr/ath/jgal/ ___________________________________________ Eventually, _everything_ is understandable. ==== >>|>I am in posession of the function: >>|>f(z)=c^z, c<>0, c in C and its iterates: >>|>>c:=1.2; >There's another solution of f(z)=z at approximately 14.7674583809828. >>I haven't found any complex solutions of (f@@5)(z)=z. >Well, again it suffices to solve f(z)=z. Write z = u+vi, match magnitudes >and angles of f(z) and z, eliminate u, plot the function of v and >look for a match. I get f(z) = z when z is about > 20.94645536 + 40.45731830 I >This method finds all the solutions of f(z) = z, and they form a >nice sequence in the plane. >Fixed points of iterates of f look like they won't permit such an >easy reduction to functions of one variable. They also look to be a >numerical-analysis nightmare... z = 15.384972737405246873 + 0.42943156081631657205 I (approximately) is a fixed point of f@@5 but not of f. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >There's another solution of f(z)=z at approximately 14.7674583809828. >I haven't found any complex solutions of (f@@5)(z)=z. >Well, again it suffices to solve f(z)=z. z = 15.384972737405246873 + 0.42943156081631657205 I (approximately) >is a fixed point of f@@5 but not of f. I was responding to the question of whether the fixed points had to be real (they don't, as we have both now noted). I was not claiming that all fixed points of f^(5) would also be fixed points of f, merely that the fixed points of f are among the fixed points of f^(5). dave ==== I need help for this problem: (i) Find the number of four letter words that can be formed from the letters of the word HISTORY. (ii) How many of them contain only consonants? (iii) How many of them begin and end in a consonant? (iv) How many of them begin with a vowel? (v) How many of them contain the letter Y? (vi) How many of them begin with T and end in a vowel? (vii) How many of them begin with T and also contain S? (viii) How many of them contain both vowels? ==== > I need help for this problem: (i) Find the number of four letter words that can be formed > from the letters of the word HISTORY. > (ii) How many of them contain only consonants? > (iii) How many of them begin and end in a consonant? > (iv) How many of them begin with a vowel? > (v) How many of them contain the letter Y? > (vi) How many of them begin with T and end in a vowel? > (vii) How many of them begin with T and also contain S? > (viii) How many of them contain both vowels? > Find definitions of combination, permutation and multiplication principle. Those will answer most of these. To start with, (i) How many letters do you have? How many will you use? Does order matter? (ii) How many consanants do you have? How many will you use? Does order matter? (viii) How many vowels are there? How many consanants are there? How many will you use? Do these without order. How many ways can you order your collection of vowels/consonants? Note: All of these problems boil down to looking at them the right way. Once you can do that, the answer is trivial. -- Will Twentyman ==== In math most equations treat time as just >another spatial coordinate, which suggests >time can be reversed in time Time can be reversed in time ..What the hell is your problem? >This contradicts the second law which suggests >that ....nothing in the universe is reversible. > The second law of what? Your Mom? Maybe you should be a little more specific about what the hell you are talking about... and not just assume that we will infer The 2nd law of Thermodynamics. There's other second laws, you know... Like Newton's... >A model of reality must have as it's axioms >those qualities that are observed. Nature >and the universe operate as iterated maps >into itself, yet math treats it as if the >output is proportional to the input. >The universe is non-linear where the final >state is independent of the initial conditions >yet all our sciences are built upon defining >initial conditions to understand the future. You have absolutly no idea what you are talking about, do you? Biaaaatch. >Is mathematics, as a model of reality, fundamentally >flawed? Sure. Yes. No. Who cares. > Shouldn't any model of reality have the >same traits as what is being modeled? >It appears to me math in inconsistent with >reality in every primary aspect. Maybe because you just can't do math... because you're stupid. >The frame of reference of classical science is >wrong. Does that mean something? >Reducing to understand the whole is >a fundamental error, science should be >rebuilt from scratch using the inverse >perspective. Great Idea!!! Why don't you get started on that, buddy. >Our primary axioms You mean, YOUR primary axioms... don't try to drag the rest of us down with you. >need >to be constructed from the largest scale >sciences, not the smallest. Cosmology >and astrophysics for example, as opposed >to quantum and other reductionist disciplines. >Global system properties should be studied >first in order to later understand the components. >Science was established within a frame of >reducing to understand out of necessity, that >no longer is the case. >Imho. Imho your mom is easy >Jonathan P.S. You suck ==== >>In math most equations treat time as just >>another spatial coordinate, which suggests >>time can be reversed in time >Time can be reversed in time >..What the hell is your problem? >>This contradicts the second law which suggests >>that ....nothing in the universe is reversible. >The second law of what? Your Mom? >Maybe you should be a little more specific >about what the hell you are talking about... and not >just assume that we will infer The 2nd law of >Thermodynamics. There's other second laws, you know... >Like Newton's... And in Statistical Mechanics, the Second Law of Thermodynamics is no longer a hard and fast rule. In statistical mechanics, the entropy will tend to increase as time goes on as a consequence of the sheer weight of numbers. There are so many more options for which the entropy is higher rather than lower that the entropy has a very high chance of increasing and a very low chance of decreasing. If the system is in a very low entropy state, then the states in its past will tend to have a higher entropy than the present state. And so we see a symmetry with respect to inversion in time for that specific case. David McAnally -------------- ==== In math most equations treat time as just another spatial coordinate, which suggests time can be reversed in time and the equation will work as well. This contradicts the second law which suggests that ....nothing in the universe is reversible. A model of reality must have as it's axioms those qualities that are observed. Nature and the universe operate as iterated maps into itself, yet math treats it as if the output is proportional to the input. The universe is non-linear where the final state is independent of the initial conditions yet all our sciences are built upon defining initial conditions to understand the future. Is mathematics, as a model of reality, fundamentally flawed? Shouldn't any model of reality have the same traits as what is being modeled? It appears to me math in inconsistent with reality in every primary aspect. The frame of reference of classical science is wrong. Reducing to understand the whole is a fundamental error, science should be rebuilt from scratch using the inverse perspective. Our primary axioms need to be constructed from the largest scale sciences, not the smallest. Cosmology and astrophysics for example, as opposed to quantum and other reductionist disciplines. Global system properties should be studied first in order to later understand the components. Science was established within a frame of reducing to understand out of necessity, that no longer is the case. Imho. Jonathan s ==== > In math most equations treat time as just > another spatial coordinate, which suggests > time can be reversed in time and the equation will > work as well. > Indeed, and equations of physics and quantum mechanics also. > This contradicts the second law which suggests > that ....nothing in the universe is reversible. > My my, an antique paradox of a past century. > Is mathematics, as a model of reality, fundamentally flawed? No, math makes no such claim to model reality nor does physics even tho their favorite pass time is to make models of reality. Your discussion would likely receive less boredom at sci.physics than here. ==== In math most equations treat time as just > another spatial coordinate, which suggests > time can be reversed in time and the equation will > work as well. Indeed, and equations of physics and quantum mechanics also. This contradicts the second law which suggests > that ....nothing in the universe is reversible. My my, an antique paradox of a past century. Is mathematics, as a model of reality, fundamentally flawed? No, math makes no such claim to model reality nor does physics even tho > their favorite pass time is to make models of reality. Your discussion > would likely receive less boredom at sci.physics than here. To paraphrase: We don't need no steenking reality. ==== Is mathematics, as a model of reality, fundamentally > flawed? Mathematics is reality. Mathematics is not a model of non-mathematics (but some writers used reality as a synonym for non-mathematics). -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen ==== > In math most equations treat time as just >another spatial coordinate, which suggests >time can be reversed in time and the equation will >work as well. >Indeed, and equations of physics and quantum mechanics also. >This contradicts the second law which suggests >that ....nothing in the universe is reversible. >My my, an antique paradox of a past century. >Is mathematics, as a model of reality, fundamentally flawed? > >No, math makes no such claim to model reality nor does physics even tho >>their favorite pass time is to make models of reality. Your discussion >>would likely receive less boredom at sci.physics than here. >> >To paraphrase: We don't need no steenking reality. > Yeah, but what do you do with Wigner? How do you explain the unreasonable effectiveness of mathematics? Jon Miller ==== > Yeah, but what do you do with Wigner? How do you explain the > unreasonable effectiveness of mathematics? Wigner was too close to see mathmatics clearly. See an essay in: Jerry P. King, _The Art of Mathematics_ -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ ==== the Butterfly Effect is a crock of initial conditions, since we have to worry about the hurricane-producing potential of *groups* of *lepidopterae* and their life-cycles. what you refer to is known in engineering as phase space co-ordinates, whose fad-meaning was established by the reputation of Einstein by his teacher's mere suggestion (Minkowski coords.). the same problems of uncertainty apply to the observation of cosmology, in that a telescope is essentially a microscope, with the (conceptual) objective reversed. > In math most equations treat time as just > another spatial coordinate, which suggests > The universe is non-linear where the final > state is independent of the initial conditions > perspective. Our primary axioms need > to be constructed from the largest scale > sciences, not the smallest. Cosmology > and astrophysics for example, as opposed http://www.channel1.com/users/bobwb/synergetics/photos/sk3prsm.html ---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 ==== In math most equations treat time as just > another spatial coordinate, which suggests > time can be reversed and the equation will > work as well. > Indeed, and equations of physics and quantum mechanics also. > This contradicts the second law which suggests > that ....nothing in the universe is reversible. > My my, an antique paradox of a past century. > Is mathematics, as a model of reality, fundamentally flawed? > No, math makes no such claim to model reality nor does physics even tho > their favorite pass time is to make models of reality. So you seem to agree that math is only self-consistent, but not necessarily consistent with reality. Then that begs the question, what should we do differently in order to have a science that ...does properly model the real world? Towards the end of my post I suggest that this 'ancient paradox' has a solution and it lies with our chosen frame of reference. A proper model of the real world is being created by simply inversing the frame of reference classical science has been constructed within. Which is a task any mathematician should find trivial. The implications of doing so, however, are as powerful as the task is trivial. > Your discussion > would likely receive less boredom at sci.physics than here. This topic applies to ...all subjects with equal validity whether in science, religion, arts or the humanities. I am trying to describe a new universal theory of organization called complexity science. Jonathan An Introduction to Complex Systems Torsten Reil, Department of Zoology, University of Oxford http://users.ox.ac.uk/~quee0818/complexity/complexity.html The study of complex systems has gained increasing attention in recent years, in such diverse disciplines as economics, life science, sociology, physics and chemistry. The multidisciplinary approach taken by its students has revealed a surprisingly high degree of applicability of the concepts to the different fields. Behaviour of biological systems seems to be mirrored in that of economic ones... Center for the Study of Complex Systems at the University of Michigan http://www.pscs.umich.edu/ The Center is based on the recognition that many different kinds of systems which include self-regulation, feedback or adaptation in their dynamics, may have a common underlying structure despite their apparent differences. Moreover, these deep structural similarities can be exploited to transfer methods of analysis and understanding from one field to another. Chaos at Maryland http://www-chaos.umd.edu/ Chaos is a multidisciplinary science, and this is reflected in the fact that the members of the group are affiliated with diverse departments and institutes The Complexity & Artificial Life Research Concept for Self-Organizing Systems http://www.calresco.org/index.htm Here we will introduce the integrating sciences of Complex Systems and of ALife together with related systems areas. We'll also pursue the wider social implications of these transdisciplinary theories of self-organization on mind, art, spirit and life as it could be... PERCEPTION of an Object costs Precise the Object's loss. Perception in itself a gain The Object Absolute is nought, Perception sets it fair, And then upbraids a Perfectness That situates so far By E Dickinson s ==== > I am trying to find the area of the intersection of 2 sphere caps, more > precisely: > Consider a sphere of radius one. Consider two circles on the surface > of the sphere drawn using a compass with radius set at r. What is the > area of the overlapping region of the two circles in terms of r and > the angle between the centers of the two circles ? > Best, > Aslan > I used the Gauss-Bonnet theorem and got the following formula, where R > is the intrinsic radius of the disks, d is the intrinsic distance > between their centers, and A is the area of the intersection of the > disks. I assume that R <= PI/2 and d <= PI. The case when R > PI/2 can > easily be handled by computing the complementary area. Note that r as > you defined it is equal to 2*sin(R/2). > If 0 <= d <= 2*R then > A = > 2*PI > - 2 * cos(R) * arccos(2*(tan(d/2)/tan(R))^2 - 1) > - 2 * arccos(1 - 2*(s(d/2)/s(R))^2) > If 2*R < d <= PI then A = 0 (since the disks do not overlap in that > case). > John Mitchell The function s in the third line of the formula is my shorthad for > sin. I forgot to expand it when typing the formula into the > newsreader. Sorry about that. Written out in full: A = > 2*PI > - 2 * cos(R) * arccos(2*(tan(d/2)/tan(R))^2 - 1) > - 2 * arccos(1 - 2*(sin(d/2)/sin(R))^2) John Mitchell [Note: I attempted to submit this to sci.math last night, with a cross-posting to sci.math.research (where the same question appeared), but it somehow ended up going only to sci.math.research, so I'm making a separate post to sci.math.] You can simplify the formula for the area somewhat by using the formulas arccos(2*x^2 - 1) = 2*arccos(x) and arccos(1 - 2*x^2) = PI - 2*arccos(x), for 0 <= x <= 1 (these follow from the half-angle formula for cosine), to get: A = 4*arccos(sin(d/2)/sin(R)) - 4*cos(R)*arccos(tan(d/2)/tan(R)) It's also not hard to show, by applying this formula to the complementary disks, that the same formula is valid for PI/2 < R <= PI - d/2. In other words, this formula is valid for R <= PI and d <= PI provided that d/2 <= R <= PI - d/2. In fact, the real part of the right side of the formula (with a naturally chosen branch) gives the correct result for R <= PI and d <= PI (i.e., even when the disks or their complements are disjoint). John Mitchell ==== > I am trying to find the area of the intersection of 2 sphere caps, more > precisely: Consider a sphere of radius one. Consider two circles on the surface > of the sphere drawn using a compass with radius set at r. What is the > area of the overlapping region of the two circles in terms of r and > the angle between the centers of the two circles ? > Best, > Aslan I used the Gauss-Bonnet theorem and got the following formula, where R > is the intrinsic radius of the disks, d is the intrinsic distance > between their centers, and A is the area of the intersection of the > disks. I assume that R <= PI/2 and d <= PI. The case when R > PI/2 can > easily be handled by computing the complementary area. Note that r as > you defined it is equal to 2*sin(R/2). If 0 <= d <= 2*R then > A = > 2*PI > - 2 * cos(R) * arccos(2*(tan(d/2)/tan(R))^2 - 1) > - 2 * arccos(1 - 2*(s(d/2)/s(R))^2) If 2*R < d <= PI then A = 0 (since the disks do not overlap in that > case). John Mitchell The function s in the third line of the formula is my shorthad for > sin. I forgot to expand it when typing the formula into the > newsreader. Sorry about that. Written out in full: A = > 2*PI > - 2 * cos(R) * arccos(2*(tan(d/2)/tan(R))^2 - 1) > - 2 * arccos(1 - 2*(sin(d/2)/sin(R))^2) John Mitchell [Note: I attempted to submit this to sci.math last night, with a > cross-posting to sci.math.research (where the same question appeared), > but it somehow ended up going only to sci.math.research, so I'm making > a separate post to sci.math.] You can simplify the formula for the area somewhat by using the > formulas arccos(2*x^2 - 1) = 2*arccos(x) and arccos(1 - 2*x^2) = PI - > 2*arccos(x), for 0 <= x <= 1 (these follow from the half-angle formula > for cosine), to get: A = 4*arccos(sin(d/2)/sin(R)) - 4*cos(R)*arccos(tan(d/2)/tan(R)) It's also not hard to show, by applying this formula to the > complementary disks, that the same formula is valid for PI/2 < R <= PI > - d/2. In other words, this formula is valid for R <= PI and d <= PI > provided that d/2 <= R <= PI - d/2. In fact, the real part of the > right side of the formula (with a naturally chosen branch) gives the > correct result for R <= PI and d <= PI (i.e., even when the disks or > their complements are disjoint). John Mitchell Nicely done. I checked that your result gives approximately the planar result for small circles. ==== I can't reproduce your c=3965 result. c^2=15721225, which is expressible in > 13 different ways as the sum of 2 non-zero squares. The best combination of > these 13 solutions is (715*3900)+(715*3900)=2*2788500=5577000 compared with > (1525*3660)=5581500, so a*b+x*y-u*v=-4500. What you probably computed was c^2=3965 which is not a square. c^2=3965 > would indeed give your solution (11*62+22*59-43*46)/2=1 minded brute force search has no reasonable chance of finding a solution. Hugo Pfoertner You have reason. I made a mistake taking the wrong numbers as legs of triangles. My best results was with hipotenuse = 15025. ab + xy = 111012312 ; uv=11018648 difference= 6336 , relative error = 3x10^(-5). I do not understand why you speak of 13 solutions of TWO triangles. I think we must look for solutions of THREE or more triangles with the same hypotenuse. In this case the maximum I found were 6 solutions for hypotenuses < 30000. Luis Rodriguez ==== ? ? ? ? I can't reproduce your c=3965 result. c^2=15721225, which is expressible in ? ? 13 different ways as the sum of 2 non-zero squares. The best combination of ? ? these 13 solutions is (715*3900)+(715*3900)=2*2788500=5577000 compared with ? ? (1525*3660)=5581500, so a*b+x*y-u*v=-4500. ? ? ? ? What you probably computed was c^2=3965 which is not a square. c^2=3965 ? ? would indeed give your solution (11*62+22*59-43*46)/2=1 ? ? ? ? minded brute force search has no reasonable chance of finding a solution. ? ? ? ? Hugo Pfoertner ? ? You have reason. I made a mistake taking the wrong numbers as legs of ? triangles. ? My best results was with hipotenuse = 15025. ab + xy = 111012312 ; ? uv=11018648 ? difference= 6336 , relative error = 3x10^(-5). ? I do not understand why you speak of 13 solutions of TWO triangles. I ? think we must look for solutions of THREE or more triangles with the ? same hypotenuse. In this case the maximum I found were 6 solutions for ? hypotenuses ? 30000. ? Luis Rodriguez What I had in mind with 13 solutions were the 13 different ways to represent c^2=15721225 as sums of two squares. We have then to check n^2*(n-1)/2 different ways to combine the n sums of square solutions into a*b+x*y-u*v. For 13 ways we have to check 1014 combinations admitting also {a,b}={x,y} For your best solution c=15025 c^2=225750625 can be written in 7 ways as the sum of 2 non-zero squares: a b c a*b 1903 14904 15025.000 28362312 3465 14620 15025.000 50658300 4207 14424 15025.000 60681768 6000 13775 15025.000 82650000 7420 13065 15025.000 96942300 9015 12020 15025.000 108360300 9617 11544 15025.000 111018648 You should probably check your program if it really searches for a minimum of relative error. Hugo Pfoertner ==== In talk.bizarre, fresh meat punching bag McD5575971 > Dear boy? What are you trying to prove now? > I swear this newsgroup is full of the most > cynical people on hell's creation. In standard usage, that would be in hell's creation. Not even close. How sad for you that Usenet is absolutely _full_ of smart people low on humility because they have little reason to be humble, so that you, being a stupid person easily humiliated, and in dire need of humility to fit your station in life, are going to be running into them wherever you go, and coming away each time resoundingly bloodied but some sad how forever unenlightened. Stay away from sci.math, for example, if you don't want to meet the mensanator: http://members.aol.com/rotanasnem/poster.htm xanthian, noting that while a word to the wise suffices, a whole novel typed at a fool has no detectable effect, as the present case is proving. -- ==== > In talk.bizarre, fresh meat punching bag McD5575971 > ........... xanthian, noting that while a word to the wise suffices, a whole novel > typed at a fool has no detectable effect, as the present case is > proving. However, a whole novel (say, War and Peace) thrown at a fool may have a noticable effect (if it makes contact). Martin Cohen ==== >> In talk.bizarre, fresh meat punching >> bag McD5575971 [drivel] >> xanthian, noting that while a word to >> the wise suffices, a whole novel >> typed at a fool has no detectable >> effect, as the present case is >> proving. > However, a whole novel (say, War and > Peace) thrown at a fool may have a > noticable effect (if it makes > contact). Hmmm, the start of a useful notion: how about if you see if you can get _War and Peace_ (a very good read) engraved on the head of an anvil, and my orbital trebuchet crew and I will see if we can get it engraved in mirror reversed type on the forehead of the above-ridiculed miscreant? xanthian, a.k.a. captain, my captain. Some impressions never go away, others bloom to plasma as soon as they are created: mayfly education mechanisms. -- ==== Alon Amit > After presenting the well-known theorem of Chevalley-Warning, Ireland > & Rosen (A Classical Introduction to Modern Number Theory) mention > that Warning was able to prove that N >= q^(n-d) Here N is the number of solutions of f(x1,...,xn)=0 over GF(q) and > d=deg(f). If we can prove the case n=d+1, the general case will follow by induction: For each k in GF(q), define a polynomial f_k(x_1, ..., x_{n-1}) as f(x_1, ..., x_{n-1}, k x_n}. By the inductive hypothesis, plus the fact deg(f_k)<=d, each f_k has at least q^(n-1-d) zeros, and there are q possibilities for k, et voila. I don't see any simple proof for the case n=d+1, but it does follow from Ax's result q^s | N for any positive integer s < n/d by just letting s=1. > The reference given is the original 1936 paper of Warning, > which is a bit hard to track down... This one might be particularly tough :)) The thing comes from a seminar at the University of Hamburg during the Third Reich. Larry ==== Larry Hammick ... > For each k in GF(q), define a polynomial > f_k(x_1, ..., x_{n-1}) > as > f(x_1, ..., x_{n-1}, k x_n}. Sorry, I bungled the notation. That last line should say f(x_1, ..., x_{n-1}, k x_{n-1}). LH ==== > Larry Hammick > ... > For each k in GF(q), define a polynomial > f_k(x_1, ..., x_{n-1}) > as > f(x_1, ..., x_{n-1}, k x_n}. > Sorry, I bungled the notation. That last line should say > f(x_1, ..., x_{n-1}, k x_{n-1}). > LH whenever x_{n-1} happens to be 0, so is k x_{n-1}, so for different values of k you may get the same solution (x_1, ..., x_{n-1}, k x_{n-1}). Therefore I don't think you can deduce that the q^{n-d-1} solutions to the reduced polynomial actually yield q^{n-d} (different) solutions. Or am I missing something? - A. A. ==== Alon Amit whenever x_{n-1} happens to be 0, so is k x_{n-1}, so for different > values of k you may get the same solution (x_1, ..., x_{n-1}, k > x_{n-1}). Therefore I don't think you can deduce that the q^{n-d-1} > solutions to the reduced polynomial actually yield q^{n-d} > (different) solutions. True :( I should have been more careful. Maybe one of the maestros over at sci.math.research is familiar with Warning's result. Still, somehow it doesn't look very deep or difficult; I'll have another go at proving it myself. Larry ==== >G is isomprhic to R under addition: this is a vector >space over Q of uncountable dimension. It has *lots* >of automorphisms .... And some of them are not divisible. For example, let {b_i} be a Hamel basis of R over Q, and consider the automorphism A of R that maps b_1 to 2 b_1 and leaves all other b_i fixed. Suppose B is any automorphism of R that commutes with A. Then A(B b_1) = B A b_1 = 2 B b_1. But the only members of R such that A x = 2 x are rational multiples of b_1, so B b_1 = r b_1 for some rational r. In particular, since there are no rationals r such that r^n = 2, A has no n'th root for any integer n > 1. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 ==== >G is isomprhic to R under addition: this is a vector >space over Q of uncountable dimension. It has *lots* >of automorphisms .... And some of them are not divisible. For example, > let {b_i} be a Hamel basis of R over Q, and consider the automorphism > A of R that maps b_1 to 2 b_1 and leaves all other b_i fixed. Suppose > B is any automorphism of R that commutes with A. Then > A(B b_1) = B A b_1 = 2 B b_1. But the only members of R such that > A x = 2 x are rational multiples of b_1, so B b_1 = r b_1 for > some rational r. In particular, since there are no rationals r such > that r^n = 2, A has no n'th root for any integer n > 1. > Is it true (regardless of whether G is abelian) that if G = H + K, and Aut(H) is not divisible, then Aut(G) is not divisible? --------------------------------------------------- C Brown Systems Designs Multimedia Environments for Museums and Theme Parks --------------------------------------------------- ==== >> Is there infinite group G such that the automorphism group Aut(G) is >> divisible ? I don't know what a divisible group is, > but I would have guessed that Aut(Q) = Q^* was divisible. A divisible (mutiplicative) abelian group H is one for > which for all a in H and positive integers n, there is b in > H with b^n = a. So Q^* is not divisible: b^2 = -1 is not soluble in Q^*. Also not divisible at x^2 = 2. > Let G be the multiplicative group of positive reals. > Is G divisible and does aut(G) = G ? Yes. No. G is isomprhic to R under addition: this is a vector > space over Q of uncountable dimension. It has *lots* > of automorphisms .... This is off topic but let me ask this : Is it true that a finitely generated group cannot be divisible ? ==== in message <36abe133.0307211545.58f4752d@posting.google.com>: > This is off topic but let me ask this : Is it true that a finitely generated group cannot be divisible ? It's certainly true for abelian groups, at least. Every divisible abelian group is a direct sum of quasicyclic groups Z(p^inf) and/or rational groups (Q,+), none of which are finitely generated. -- Jim Heckman ==== Background: A triple , where X is a set, g is a unary operation in X (that is, a function on X into X), and x_0 is an element of X is a unary system. Let be a unary system. The set of descendants of x_0 under g (in symbols D_gx_0) is the intersection of all subsets A of X, such that x_0 in A and xg in A whenever x in A (this later requirement will often be phrased as A is closed under g). [Set Theory & Logic -Stoll] My question: I can't imagine any more than one subset of X satisfying the above conditions (x_0 in A and A closed under g) so taking the intersection of all such sets (when I can only conceive of one) seems odd. Am I missing something? In the case that I am could you please show a simple example. ==== >Background: A triple , where X is a set, g is a unary >operation in X (that is, a function on X into X), and >x_0 is an element of X is a unary system. Let be a unary system. The set of descendants >of x_0 under g (in symbols D_gx_0) is the intersection of >all subsets A of X, such that x_0 in A and xg in A whenever >x in A (this later requirement will often be phrased as >A is closed under g). [Set Theory & Logic -Stoll] My question: I can't imagine any more than one subset of X satisfying >the above conditions (x_0 in A and A closed under g) so >taking the intersection of all such sets (when I can only >conceive of one) seems odd. Am I missing something? >In the case that I am could you please show a simple >example. Yes, you are missing something. Let's consider, for example, the set X = N x N where N is the set of nonnegative integers. Let g:X->X be the function defined as follows: given (a,b) in X, add one to every nonzero entry; call the result g(a,b). Now let x_0 = (1,1). Now let's consider sets which contain x_0 and are closed under g. One such set is X itself. Another such set is the diagonal: {(a,a) : a in N}. Another such set is the diagonal minus (0,0): {(a,a): a in N, a>0} Another such set is {(a,b): a,b in N, a>=b} Another such set is {(a,b): a,b in N, a<=b} and many more such sets. ====================================================================== 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 ==== >Background: A triple , where X is a set, g is a unary >operation in X (that is, a function on X into X), and >x_0 is an element of X is a unary system. Let be a unary system. The set of descendants >of x_0 under g (in symbols D_gx_0) is the intersection of >all subsets A of X, such that x_0 in A and xg in A whenever >x in A (this later requirement will often be phrased as >A is closed under g). Eech, I hate that notation - note that I'm going to write g(x) for what you call xg here. >[Set Theory & Logic -Stoll] My question: I can't imagine any more than one subset of X satisfying >the above conditions (x_0 in A and A closed under g) so >taking the intersection of all such sets (when I can only >conceive of one) seems odd. Am I missing something? >In the case that I am could you please show a simple >example. Let X = R, the real numbers. Define g by g(x) = x for all x. Let x_0 = 0. Then {x_0} is closed under g, isn't it? ************************ David C. Ullrich