mm-19 some special techniques forintegraing integrals involving multiple-valued functions, for instance z^a,where a is not an integer...The textbook is: Fundamentals on Complex Analysis withApplication 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=/=1...The book asked to show that this integral equals2*pi/sqrt(3)*cos((2*a*pi+pi)/6) / sin(a*pi)...But I never can get this result... I doubt if the book has a mis-print...Because if I apply the special technique for multiple-valued functionintegrals, I get[1 - exp(i*2*pi(a-1))] * I = 2 * pi * i *{ Res[ z^(a-1) / (z^2+z+1), atz=exp(i*2/3*pi)] + Res[ z^(a-1) / (z^2+z+1), atz=exp(-i*2/3*pi)] }The reason for [1 - exp(i*2*pi(a-1))] is that the integration from 0 to infalong the real positive axis in the upper half plane has a different valuethan that from inf to 0 in the lower half plane along the real positiveaxis...But in my result, the [1 - exp(i*2*pi(a-1))] term factors out a termexp(i*pi*a), so my result isexp(-i*pi*a) * { 2*pi/sqrt(3)*cos((2*a*pi+pi)/6) / sin(a*pi) }Except this strange term, my result is the same as provided in the book...Am I right? Is the book wrong? Can anybody give me some solid hand on this?After many tries, I still get my current result... I am totally lost...-Walala 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 Rwith an inner hole of radius e taken out to avoid the origin.Standard arguments show that this tends to I - Jwhere I is the given integral and J is the integralwhere 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 integralstend 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 are1/[i sqrt(3)] and -1/[i sqrt(3)] and so the residues of the integrandare 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 getI = (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 residue:f(z)=z^6 / (z^4+1)^2at z0=exp(i*pi/4) and z1=exp(i*pi*3/4)...The reason for 'nding 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 >counterclockwisearc 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 somesoftware such as Mathematica or Matlab... :=)Anybody has a good idea how to do this ef'ciently? By the way, z0 and z1are 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 asingle Res[f(z), at z0] is horrible...Can anybody give me a hand?-Walalap.s. After many try, I 'nally used Matlab and admitted my failure... :=(//sigh 2 hours in 'nding its residue:f(z)=z^6 / (z^4+1)^2at 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 areusing that the following is WRONG. As you can see, the following codeproduces 2 different results. They claim that infact .5 and .50 aredifferent... any comments?0010 LET A=119.35*8.5/1000020 LET B=119.35*8.50/1000030 PRINT >a,A0040 PRINT >b,B-:runa= 10.15b= 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= 10.14a and b are 10.14475Try this:0050 let c=0.50060 let d=0.500070 if c-d=00080 then print >Sorry guys, but .5 and .50 are NOT different ;-P>0090 else print >Indeed, .5 and .50 are different>0100 endifMicrosoft 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 de'ne it howthey like, but it's obviously wrong if >8.5> and >8.50> are supposedto represent real numbers in the usual way.You can imagine explanations for it: for example, that 8.50 has moredigits than 8.5 so it is represented using a data type with more bits,so the whole expression is evaluated more precisely, but real computerlanguages 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 ?at).The best solution is to avoid proprietary languages and stick to oneswith a public de'nition.-- 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*8.5/1000020 LET B=119.35*8.50/1000030 PRINT >a,A0040 PRINT >b,B-:runa= 10.15b= 10.14On my calculator (not high tech, but oh well)119.35*8.5/100 = 10.14475119.35*8.50/100 = 10.14475In both cases, there is something happening, which I suspect is rounding.119.35*8.5=1014.475I suspect they are doing the following roundings:119.35*8.5=1014.5119.35*8.50=1014.48Then they are divinding by 100 and rounding again to get:1014.5/100 = 10.151014.48/100 = 10.14Based 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= 10.14My guess is that they use intermediate rounding to a number of decimalsthat depends on the input data. Looks pretty wacky to me, but if youcompute the products 'rst:119.35*8.5 = 1014.475, round to one decimal before continuing (because of8.5)109.,35*8.50 = 1014.475, round to two decimals before continuing,you get the results they report. Obviously there is some rounding, >otherwisethey would (should) have reported a and b to 've decimals.I don't believe that the internal data representation makes a difference,sincethe binary expansion of 8.5 truncates. =In the shop we have a tolerance block on each properly made drawing. If >it'smissing, and there isn't a tolerance on every dimensioned feature in thedrawing, there's trouble.Like did they mean 0.375 or 3/8?And in physics, we consider this more thoroughly. It's simply the de'nition >ofthe total differential dF of a function F of some numbers n1, n2, n3, etc.Let's say F = 3 * 5.1 / 8.1To a shop machinist, the implied tolerance is 0.05. But to a physicistdF = 1 * 5.1 / 8/1 + 3 * 0.1 / 8.1 + 3 * 5.1 / 0.1At least I think that's how they do it. It might be half that amount. In >anycase, 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 >?atswith a two-value ration mantissa and an integer exponent. I don't know if >anylanguages 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 manyengineers 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 >otherhand, need to get a machinist, an engineer, and a physicist together to >discussthis 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 feeCivil 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 >de'nition of> the total differential dF of a function F of some numbers n1, n2, n3, >etc.Let's say F = 3 * 5.1 / 8.1To a shop machinist, the implied tolerance is 0.05. But to a physicistdF = 1 * 5.1 / 8/1 + 3 * 0.1 / 8.1 + 3 * 5.1 / 0.1Close, but it doesn't look quite right. To 'rst order, you'd wantthe 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'renot being very careful about your tolerance convention.d/dx = 5.1/8.1 dx = .5 .5 * 5.1/8.1d/dy = 3/8.1 dy = .05 .05 * 3 / 8.1d/dz = 3*5.1/8.1^2 dz = .05 .05 * 3*5.1/8.1^2Then 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 .010Do you have some rule about rounding aggressively toward a last digitof zero so that tolerances get doubled for rounded values thatend in an explicit zero? Perhaps we need another rule for multiplyingby 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) intimatelyrelated 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 8.505so we would not write them interchangeably.This possible interpretation is one reason I prefer to write 1/2 when Imean 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*8.5/1000020 LET B=119.35*8.50/1000030 PRINT >a,A0040 PRINT >b,B-:runa= 10.15b= 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 'rst three hex digits to get the actualpower of 2. The binary point follows, then a hidden 1,then the rest of the number. The 'rst bit is the signbit of the entire number; the modi'ed exponent is alwayspositive.)However, if one does something slightly stupid like the following:8.00 = 0x40200000000000000.10 = 0x3fb999999999999a0.50 = 0x3ffe7fffffffffff (oops, rounding goof)8.50 = 0x4020ffffffffffffone now has a very slightly different number, and misconversionof 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 what0.01 * 50 is.)Such a thing actually happened to Microsoft.0.01 = 0x3f847ae147ae147bwhich turns out to be very slightly more than 0.01. Note therepeating hexadecimal.However,3.11 = 0x4008e147ae147ae13.10 = 0x4008cccccccccccd3.11 - 3.10 = 0x3f847ae147ae1400which is very slightly less than 0.01 because the >7b> got choppedfrom precision loss. A naive calculator (which Windows hadfor awhile; the bug 'nally did get 'xed) would represent theresult as 0.00.Oops! :-)-- #191, ewill3@earthlink.netIt's still legal to go .sigless. =I would appreciate some help in the following.One can de'ne the Kronecker Delta as delta_(ij) = 0 if i =/= j delta_(ij) = 1 if i = jI think that one can generalize this de'nition to an arbitrary 'eld. Is Fis a 'eld, 0_F is is identity for addition, and 1_F is is identity formultiplication, then one would de'ne the Kronecker Delta as delta_(ij) = 0_F if i =/= j delta_(ij) = 1_F if i = jMy question is: is this last de'nition (for a 'eld) a standard one?Sorry my english. Jaime Gaspar ______________________________ Homepage: www.jaimegaspar.com I would appreciate some help in the following.One can de'ne the Kronecker Delta as delta_(ij) = 0 if i =/= j> delta_(ij) = 1 if i = jI think that one can generalize this de'nition to an arbitrary 'eld. Is >F> is a 'eld, 0_F is is identity for addition, and 1_F is is identity for> multiplication, then one would de'ne the Kronecker Delta as delta_(ij) = 0_F if i =/= j> delta_(ij) = 1_F if i = jMy question is: is this last de'nition (for a 'eld) a standard one?Yes. For instance, this is used implicitly at:http://www.wikipedia.org/wiki/Identity_matrixJose Carlos Santos One can de'ne 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 de'nition to an arbitrary 'eld. Is >F> is a 'eld, 0_F is is identity for addition, and 1_F is is identity for> multiplication, then one would de'ne the Kronecker Delta as> delta_(ij) = 0_F if i =/= j> delta_(ij) = 1_F if i = j>> My question is: is this last de'nition (for a 'eld) a standard one?>The same de'nition, will do for any ring and canbe generalized by loosing up on the domain N of i,j.It's not the de'nition, 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'sdelta. Furthermore, in topology it may make no difference what 0 & 1, onlythat they adhere to the ring axiom 0 /= 1. Thus the notion of acharacteristic function could be stretched to a map from a subset into atwo element set. Jamis 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.htmlLarry want to inform all of you that James Harris completed his proof ofRiemann Hypothesis.>I was struggling for hours, then God (my Father) whispered thesolution to me.> (sic)I know we (humans) are not prepared for such an event, but the timehas come... a genius has come! And our duty is to recognize him: forwhat he completed, for what he is completing, and for what he'llcomplete.Jamis Harres 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 HarresDang, he has beat me again. First an elementary proof of FLT, and nowjust today I was putting the 'nal touches on my elementary 2-pageproof that the Riemann Hypothesis is independent of ZF+AC.Ramanujan thought that his results came from God. Shiva, I think, ormaybe Krishna, but with all the avatars I get confused. The formulasfor 1/pi just sort of popped into his head.I wonder who Jamis Harres really is... The alterego a regularsci.math poster no doubt... I wonder who... And then, when it comesdown 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 aring.)2. He reaches the conclusion that the >ring of algebraic integers> isincomplete. (This means, although cryptically stated, that the algebraic integersdo *not* form a ring.)3. Either his conclusion is wrong -- which it is -- and the algebraicintegers *do* form a ring, a result which as been proven, (The sums and products of any elements in the ring are also elementsof the ring.)4. or the algebraic integers do *not* form a ring. (in which case his assumptions are false and the arguments areinvalid.)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 =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 'nd that it did not begin with> a truth, or that you 'nd a break in the logical chain.> Actually, almost the opposite is true. You can challenge the *proof* of a fact by 'nding a false premise or logical ?w, but the result may be true independent of these errors, if one can 'nd 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 'nds a ?w 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 'nd the ?w in his premeses or logic... ==Michael A. Van OpstallPadelford C-113opstall@math.washington.eduhttp://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.)daveThe following message seemed to be only visible in The Math Forum:Author: Allan MacLeod The problem of the representationN = a/b + b/c + c/awith a,b,c integers ( not necessarily positive )is discussed in the paperTWO MORE REPRESENTATION PROBLEMSby Andrew Bremner and Richard Guypublished 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 de'nitions, derive a few easy theorems and --- voila! --- all of a sudden get a deep result about a very concreteproblem.Some authors like to call this >abstract nonsense>, tongue in cheek ofcourse because by solving the concrete problem the approach turns outnot be nonsense after all.Where was the phrase >abstract nonsense> used for the 'rst time, for apiece of math?Nemo Where was the phrase >abstract nonsense> used for the 'rst time, for a>piece of math?See http://www.risc.uni-linz.ac.at/research/category/risc/catlist/ gen-abs-nonsRobert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 You set up some very general de'nitions, 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 'rst time, for a>piece of math?Well, if I recall correctly, it was commonly used to describe CategoryTheory.>>NemoLarry(this space unintentially left blank ..... >Where was the phrase >abstract nonsense> used for the 'rst time, for a>>piece of math?>See >http://www.risc.uni-linz.ac.at/research/category/risc/catlist /gen-abs-nonsI have not used the term before, but it is often thecase that using >unreasonable> generalizations makesunderstanding almost trivial.In some case, even introducing something which seemsirrelevant makes the problem understandable when itwould 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 viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Deptartment of Statistics, Purdue University Let (A_n)_{n>=0} be de'ned 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) areall integers. If the A_n are all integers, then use W(0)=1 to solvethis 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 'rst 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 =03) To 'nd an explicit form of A_n . > Let (A_n)_{n>=0} be de'ned 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 'rst 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 'nd an explicit form of A_n . > =Yes, I assert that all A_n , n=0,1,2,..., are integers.Alex/Proposer => 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 NoraBaron!!! > 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 valuesfor 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 evenwhen I put in a value for x. And in fact, the x's and the value ofthe 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 wayyou could have supposed that my picking an actual value for x wouldaffect the a's in such a way that you thought you could just pickvalues 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 theability 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 ofthe 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 speci'c 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> | de'nition, 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 independentproof 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. 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 entirelyproofs. Same idea as in Euclidean geometry: Start with asmall set of axioms, then build a powerful set of theoremson top of them.The stuff in JSH threads is algebraic number theory for themost part. I've always been fascinated with it, but never tooka course, and for some reason it never quite sticks with me.If it did, I'd be able to produce those polynomials andfactorizations 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 speci'cally related to > undergraduate studies, the de'nition of which varies by country)Only James really knows the answer to that. It has somethingto do with seeking a fresh unbiased audience who willaccept his proof, now that he knows all mathematiciansare corrupt. That was the motivation for alt.math.undergradanyway. I can't recall why alt.writing.One of the lovely intricacies of the mind of James Harris isthat he never sees the irony between what he claims evilmathematicians do and what he actually does. For instance,he keeps insisting that mathematical proofs are acceptedwithout inspection, that the world has blind faith inany math paper written. But what he also keeps insistingon is that we all accept his proof on blind faith, justbecause he says it's true. He hasn't yet 'gured out thatthe veri'cation process, the careful scrutiny any publishedproof is exposed to, is exactly the thing that's stoppedhim and the thing that he says doesn't exist.You left alt.math.undergrad in the headers, so I trimmed itin 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>>| de'nition, 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 isall about], simplest to understand. Further, it has the advantage ofadmitting some amount of generalization, which my approach foregoesentirely.That said, I think there is a major hurdle in getting JSH to acceptthe fact that Galois theory is correct, and applicable in the contextof the factorization of polynomials. He seemingly has a huge bug upin an unmentionable ori'ce, about the fact that Galois theorycanonically deals with 'elds and 'eld extensions, not realizingthat the ring of integers of a number 'eld enjoys some propertiesthat arbitrary rings do not.I had vainly hoped that JSH would [irrespective of any acknowledgementof the hated source] take a look at the veri'cation of those commonfactors that I gave, via direct polynomial multiplications, verifythose multiplications for himself to see that I wasn't lying, andcome to his senses about his argument.Well, I *did* strongly suspect that my errand was in vain, but Iwanted to give him the bene't of the doubt, and failing that, givehim enough opportunity to demonstrate his unwillingness to face reality.I have apparently been granted that second wish rather than the 'rst.Dale => 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!!!> 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 differentfrom the polynomial factorization. Of course you still havethe polynomial factorization as well. Also of course I knewall 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 factorizationof P(2), I was making the point that it does you no good whatsoeverto try to >simplify> by substituting in an actual value for x,when you have no intention of considering any other factorizationthan the polynomial factorization. It seemed to me that you didnot understand that little subtlety, but now perhaps you aregetting 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 wereall divisible by 5. Since when you let x = 2, it was a perfectlyvalid factorization of an ordinary integer, not of a polynomial.You said it could not be factored in this form and I proved thatit could. Again, this is a minor side-issue. If you still don't get, don'tworry 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 'rst 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 just'cation 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 justi'es 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 'nd a 3rd degree polynomial with integer coef'cients, 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 de'nition, 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 'eld of algebraic numbers such that: 1. F12 leaves the sub'eld of rational numbers 'xed, 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 apolynomial in x. When you chose x = 2, you were abandoningthe polynomial. What you had was no longer a polynomial inx. When you talked about factoring it, you were factoringan ordinary number. You didn't say, >OK, now I have an ordinary number, but I still want to factor it as if itwere a polynomial in x.> That would have been a silly statement to make, and it would have implied that your>simpli'cation> was really quite pointless. The only sensibleinterpretation is that after you evaluated the polynomial, youwere thinking about factoring an ordinary number. Now youare trying weasel out of that because you see it was obviously false. AGAIN: this is a minor side issue. If you STILL reallydon't get it, stop worrying about it. Your real big-timeproblems 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 'nd 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 ofhimself). Why bother entertaining him? (If nothing else, it giveshim 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> 'nding 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 unspeci'ed,> 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 resultleads to the conclusion that they all must be coprime to 5 in the ringof algebraic integers, which is what's wacky, and shows a problem withthe 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 thatway. > 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 realizedthat 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 form=1, where she actually picked values for the a's the cubic is a^3 + 72a - 13825 = 0and 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.> > (2 a_1 + 5)(2 a_2 + 5)(2 b_3 + 11) 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 HarrisPhil 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>'nding 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 unspeci'ed,>>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^2P(m) = (m^3 f^6 - 3m^2 f^4 + 3mf^2) x^3 - 3(-1+mf^2)xu^2 f^2 + u^3 f^3using v=-1+mf^2, v^3+1 is the coef'cient on x^3P(m) = (v^3+1) x^3 - 3vxu^2 f^2 + u^3 f^3This 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 coef'cients 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=1P(m) = (v^3+1) x^3 - 75 vx + 125 where now v=-1+25m.Now plug in x=2P(m) = 8(v^3+1) - 600 v + 125P(m) = 8v^3 - 600 v + 133 where v=-1+25mThis 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 = 0and none of her picks work.> James HarrisYour assertion about the a's only holds true when x is kept as a variable. See my argument above. Perhaps if you clearly de'ned 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 simpli'cation that includes choosing values for x. If you do, you have *over*simpli'ed and changed the problem.-- Will Twentyman JSH would [irrespective of any acknowledgement> of the hated source] take a look at the veri'cation 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 simplemultiplication of a few binomials, he constistently got the exponents and >the signs wrong. He is hopelessly sloppyin his algebra. I think if you want to gain some ground (no promises) you >will have to post the exact values of thenumbers a1, a2 and a3, and then derive the expressions which prove they are >not coprime to 5 in the ring ofalgebraic 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 posthe even said not to bother citing errors in his >proof> because if any >existed *he* would let everyone know aboutit.> Well, I *did* strongly suspect that my errand was in vain, but I> wanted to give him the bene't 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 'rst.>> 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 'nd 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'tbeen anything like the kind of general cooperation that it wouldtake to end the discussion. If the only point were to preventpeople from being taken in, I agree just letting him attempt topersuade people would be as effective as arguing with him. Butthere are other reasons why people keep at it.For one thing, sometimes it's interesting to try to convince someonewho's exceptionally resistant to being convinced, and see how it itthat they manage to hang on to their opinions anyway. I think a numberof us just 'nd it irritating to see someone making claims we knoware wrong and not being corrected. I won't claim that's a >rational>irritation. Some people 'nd it entertaining in other ways.The main thing that would convince me to stop would be if I thoughtit would be better for his own well-being not to have me as adistraction from his other issues. It seems possible to me that he'dprefer 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 'nd 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 'nd 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 RamsayPerhaps I am a little behind on this discussion, but why can't we justgive James a polynomial and see if he can make his technique work? Working from a polynomial you made yourself around your method (whichif I checked correctly cannot be factored with integers anyway) isquite different than trying to apply it in practice.It also seems that - though a bit of work - the traditional p/qapproach and sign examination are pretty simple. They are also fairlystraightforward to implement programmatically. (I wonder if James hasseen 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 presentcontroversy the problem is that he has an overpowering intuition thatthere must be a formula connecting the roots of a polynomial equationwith the constant term - in fact he is right, in the sense that forpolynomials through 4th degree, there are formulas involving radicalsthat specify the roots in terms of the coef'cients. He believes that such formulas extend to degenerate cases. He then 'nds a formula-typerelationship for a degenerate case, and then assumes that this formula must hold in nondegenerate cases as well. He cannot conceive that innondegenerate cases, the formula generalizes in any but the obvious way. Ideally what we would do to prove this is incorrect is actuallywrite down the roots of his polynomial and show how each root sharesalgebraic integer factors with prime divisors of the constant term. Incredibly enough, this is not completely simple even in the case of aquadratic. In the case he is interested in, the cubic, the formulas for the roots are quite complicated and showing directly that they includealgebraic 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 sometimeshaving some theoretical superstructure can create a shorter, more easilyunderstood path to the answer than a direct frontal assault. Sometimesthe long way round the mountain is actually the shorter than tryingto 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 ?w or gap. He thinks that bit about generalizing from the degenerate case is thenatural 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 de'nition of algebraic integers or there is a ?w 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 'nd a way through his armor. It has happened before. Simpson's dog: my recollection of that is that it came from aFar 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! Blahblah blah, Ginger! Blah blah blah, Ginger!> I wonder whichcame 'rst - 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, butthe diagram will help, either way.(I prefer the >lunes> proof of the pythagoreean th.; and,I 'nally 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/'les/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=0gives me> will then give you5(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 anda_3 are both divisible by sqrt(5)? Why couldn't a_2 be divisible by5^(1/3) and a_3 divisible by 5^(2/3)? I guess I'm not seeing how youarrive 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 you5(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 mustSure. I'm considering non-polynomial factors of P(m), and I have frommy lemma that any such factor can be written as r+c, where r=0, orvaries with m, while c remains constant and is a factor of theconstant 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 to5, 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 iscoprime 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 'nd that at that pointg_1=5.But, P(m)/25 = g_1 g_2 g_3/25, and as P(0)/25=11, at m=0, so the 5goes.But when m doesn't equal 0, that only handles one of the 5's as 25 hastwo factors where each is 5. Therefore, ONE other of the a's must goto 0, when m=0, and it also has a factor that is 5 which separatesout.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 byletting you consider factors of Q(m). For instance, if you have h_1 = 2a_1/5 + 1as a factor of Q(m), you can check at m=0, to 'nd that h_1=1.But let's say that h_1 = 2a_1/s + 5/swhere 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) butdoing 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 indetail, but I can't read your mind. You need to tell me where you'renot 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 'gure? 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 meSee 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 'nd 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=55That'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 + 1as a factor of Q(m), you can check at m=0, to 'nd that h_1=1.But let's say that h_1 = 2a_1/s + 5/swhere 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 = 11Now 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 metaphorslike factors >coming out> of terms. You appear again to be indicating >thatthere must exist f1, f2, and f3 which are divisors of 5, which have theproperty that f1 divides g_1, f2 divides g_2, and f3 divides g_3, andf1*f2*f3=25, so that P(m)/25 = (g_1/f1)(g_2/f2)(g_3/f3). If this is whatyou 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 asm varies? Note that yet again, your explanation pulls uniformity in mout of a hat without stating any general principle to justify it.In the original cubic under consideration, it turns out (for reasons youaren't describing here) that there are divisors of f which one can divideout of g_1, g_2, and g_3, so that when the quotients are multipliedtogether, 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 howthey vary from one value of m to another, there's no reason why they haveto be in any particular order (except at m=0), or why the order can't varyfrom one value of m to another. If g_1, g_2, and g_3 satisfy yourequation, 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), andg'_3(1). The only way it could possibly make sense to conclude that theyshare factors with f in a consistent way going from one value of m toanother, is if you imposed some additional condition on them that wouldguarantee 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 stepbeing 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 theidentity 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 >thattwo algebraic integers have a GCD in the algebraic integers) what are thecommon 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 theother of them is 0. There, 0 has a GCD of 3 with 3, and 1 has a GCD of 1with 3. But generally, when m is an integer the GCD in the algebraicintegers 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 thatdepend 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 calculationthan I expected to compute that last one, and I'd be interested to knowwhether 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 insteadalgebraic integers which are expressible as polynomials with rationalcoef'cients in m.Keith Ramsay There are a lot of interesting documents here and I may have missed> something but I can't 'nd any ideas about how to model af'ne> 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?> MartinUnfortunately, 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 paperhttp://modelingnts.la.asu.edu/pdf/CompGeom-ch1.pdfand take particular notice of the section >Linearizing the Euclidean> group> on page 20. As I recall, the vectors are in 4 dimensions.Patrickdon't know if this helps, but if one constructs the clifford algebraon a quadratic space V(Q), then the unit ball of the resultant evensubalgebra rotates V(Q) in the form you gave. so unit complex numbersrotate R^2, and unit quaternions rotate R^3. M.T. =Could someone please help me solve an algebra problem I found in abook:>A survey of 1500 individuals found that 43% listen to radio newsreports, 45% listen to TV news reports, and 36% read a dailynewspaper. 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=43y+u+w+t=45z+v+w+t=36max(x+y+z+v+u+w+t) Question, is constraintx+y+z+v+u+w+t <=100also 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 forbidsnewspaper-readers from also listening to radio and watching TV. But therecan'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 involvedinto all three activities>:max(t)>What is the maximum possible number of people such that each is involvedinto 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 >formallyprove 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.De'ne 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, thisformulation 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 allthree?>>> Assume that each individual does at least one of these activities.> This is a simple linear programming problem.>> De'ne seven variables:>> r The number who listen to radio but don't watch TV or read apaper.> rt The number who listen to radio and watch TV but don't read apaper.> rp The number who listen to radio and read a paper but don'twatch 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) = 540Y'all are making this too hard. Read the questionand 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 de'nitely 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 allthree?>>> Assume that each individual does at least one of these activities.> This is a simple linear programming problem.>> De'ne seven variables:>> r The number who listen to radio but don't watch TV or read apaper.> rt The number who listen to radio and watch TV but don't read apaper.> rp The number who listen to radio and read a paper but don'twatch 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) = 540Y'all are making this too hard. Read the questionand think about it.Assumer=30 %t=30p=30rp=0tp=0tr=0rpt=10and reverse engineer the problem toradio = 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.>> De'ne 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 canlisten to the radio, and 9% that can watch TV. The four equations above are really necessary, and are easy to solveby substiting the values of r, t and p from the 'rst 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.>> De'ne 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 my version, which shows clearly where the numbers >comefrom. 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 numberof 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 >notgiving 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 theproblem. The trick is to prove that 180 is the maximum who could do allthree. =Suppose we have 3 sets, A,B,C with n(A)=45, n(B)=43 and n(C)=36Let * mean intersectionLet 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*Cso .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-bThen the smaller of a,b,c is the most that can go into the tripleintersection!!> 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.De'ne 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 forgotr + t + p + rt + rp + tp + rtp = 1if 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 - rtpand 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 = 1wich gives(4) 2 * rtp = 0.24 - rp - rt - rpFurthermore, 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 elementaryschool and were given word problems, did you grill the teacher withquestions inquiring about every possible assumption used in the wordproblem?Nevertheless, let's clarify: assume all the individuals respondedcorrectly 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 'nd algorithm of the numericaldecision of stochastic Shrodinger equation with casual potentialhaving zero average and delta 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 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 'nd algorithm of the numericaldecision of stochastic Shrodinger equation with casual potentialhaving zero average and delta 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 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 'nd algorithm of the numerical> decision of stochastic Shrodinger equation with casual potential> having zero average and delta correlated in space and >time?The equation:> i*a*dF/dt b*nabla*F-U*F=0where> 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 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 notgive you a de'nit answer). Perhaps you may repeat yourquestion either in sci.math.num-analysis or a physics group.Further there are books of Peter Kloeden on numerical methodsfor stochastic differential equations. You can 'nd them atwww.amazon.com for example and look at the content to seewhether 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 ishttp://www.math.uni-frankfurt.de/~numerik/kloeden/maplesde/ you do so please not that the spelling is Schr>.9adinger orSchroedinger. 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 atime.The number of ways you can pick three coloured balls one at a time is6:Red Green BlueRed Blue GreenGreen Red BlueGreen Blue RedBlue Red GreenBlue Green RedSimilarly, 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-PIover 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.> >- 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-partiteSince 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 elementaryproof of the four color theorem that is so short that (as an exercisein pandering to the curiosity of the masses) I found an experiencedetcher 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 anelementary> proof of the four color theorem that is so short that (asan exercise> in pandering to the curiosity of the masses) I found anexperienced> etcher who etched the entire proof on the head of a pin.>> MaxissimoHa! Just yesterday, I found one which is so small it couldbe inscribed with a very high energy laser on an up or downquark! I am attempting to shrink it further, down to thesize of the component strings, but so far I keep losingparts 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.>> MaxissimoHa! 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. ...tonyCTo tonyC.I cannot 'nd the point to this posting! Is there one? =As I've received no analytical objections to the following post I'mappending 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 de'nition 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 ?wheels 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 MechanicsOne of the really unfortunate aspects of Newton's choice of a linearframe of reference for the analysis of mechanics is that r is poorlyde'ned and t is not de'ned at all. In other words, r is only de'nedin direction and t is not de'ned by any consideration pertinent tothe analytical frame of reference.And this had a pernicious impact on the subsequent development ofangular mechanics as well as relativistic considerations and quantummechanics in the twentieth century.The problem is that r and t and their combinations are all we have towork with. Taken to the second level of compounding we have sixcombinations: r, 1/t, r/t, r/tt, rr/t, and rr/tt. However, in thelinear analytical frame of reference the next to last combination rr/twas overlooked because there is no apparent application for it inlinear mechanical contexts.On the other hand, in angular frames of reference we have applicationsfor all combinations and all the elements are well de'ned. The radiusof rotation is well de'ned in terms of direction and magnitude andtime is well de'ned in analytical terms as whatever time is neededfor 2pi radians of rotation.The rr/t combination is also well de'ned in angular terms. However,in extrapolating the idea of rr/t from linear to angular contexts inclassical mechanics, whoever devised the analytical approach made themistake of trying to emulate linear mechanics in the sense ofexplaining rotation as a linear progression of r instead of a simpleradial 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 inorbit around the sun and regarded that force as tangential indirection. Newton on the other hand recognized that the only forceneeded was centripetal in nature and not tangential. But whoeverdevised the analytical considerations underlying angular mechanicsapparently never considered the Newtonian perspective and presumablyrelied on the pre Newtonian rationale.Thus we wind up with a conceptual schism among the various realms ofangular mechanics. On the one hand we have orbital angular mechanics,the macro realm of ordinary angular mechanics, and the micro realm ofquantum effects. And unfortunately there is no conceptual integrationamong them. We are convinced that all represent mechanical realms butwe have no basis for comprehending each in terms of the others.Orbital angular mechanics represents the realm of remote interactionsdealt with in terms of inverse square centripetal forces andtangential orbital velocities. Whereas the macro realm of ordinaryangular mechanics deals with linear analogs such as moments of inertiainstead of mass, torque instead of force, and angular acceleration andvelocity instead of their linear analogs.The micro realm of angular mechanics on the other hand is dealt withon the merely descriptive basis of formalisms. This is the realm ofquantum mechanics - QM - or as I prefer to call it quantum magic wherethings don't seem to happen for any de'nite mechanical reason at all.However with the rede'nition of macro angular momentum and Planck'sconstant in circular rotation we are at last in a position tounderstand the mechanical differences among the realms in conceptualterms.The micro realm of quantum effects is one of constant tangentialvelocity of rotation v = c and a variable radial dr/dt.The macro realm of ordinary angular mechanics on the other hand is onein which the tangential velocity of rotation is variable buttangential v = radial dr/dt and both are kept in strictsynchronization by internal tensile forces.And 'nally orbital angular mechanics is de'ned by variouscombinations of tangential v and radial dr/dt. This is normallythought of in celestial terms but in point of fact applies equally tothe 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 de'nition of Planck's constant.>>The form of dr/dt follows directly from the vector de'nition of r.>There is no freedom for further de'nitions. 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 de'nition> of v not r.dr/dt doesn't >follow from the de'nition of v>, itIS the de'nition of v. dr/dt is the derivative of r,therefore its form follows from the de'nition of r,and the de'nition 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 notnecessarily 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 notin 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 youget 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 everybodyelse's. Using dr in isolation has nothing to do withthe process of taking r(t), taking its derivative,and writing down v(t). As he said, the form of v(t)is completely speci'ed by the de'nition of r(t).There is no room for interpreting (0,0,w) as beingin other directions than +z. - Randy Planck's ConstantPreviously 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 'xed distance away >(>r>).> No one> I know of agrees with my interpretation of dr/dt. Since you 'xed >r> in the 'rst 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 'xed distance away >(>r>).> No one>> I know of agrees with my interpretation of dr/dt. >>Since you 'xed >r> in the 'rst paragraph, we know dr/dt = 0. What's >the problem?Not exactly. I 'xed r in the 'rst paragraph not in'nitesimal r.Radial dr/dt = tangential v. =There's two big problems with your interpretation. First off, thea lepton) is virtual, meaning it's nonreal. Simple calculus (which youmay not have even done correctly) cannot give you the correctinterpretation of a qfp.Second, you took v=c, but you seem to use ?' interchangeably betweenc representing constant and c representing the speed of light. This isa 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>de'ned and t is not de'ned at all. In other words, r is only de'ned>in direction and t is not de'ned 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 >de'ciency> 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 de'nition of Planck's constant.>>The form of dr/dt follows directly from the vector de'nition of r.>>There is no freedom for further de'nitions. 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 de'nition>of v not r.Well, actually no.The form of dr/dt follows from the de'nition of >derivative>, d()/dt, and the de'nition of whatever () happens to be, the vector >r> in this case.>dr/dt> is tautologously the de'nition 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 de'nition of r.>There is no freedom for further de'nitions. 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 takeyour word for it.>>Second, you took v=c, but you seem to use ?' 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 dif'culty using v in the sameway.>>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>>de'ned and t is not de'ned at all. In other words, r is only de'ned>>in direction and t is not de'ned 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 >>de'ciency> 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>de'ned and t is not de'ned at all. In other words, r is only de'ned>in direction and t is not de'ned 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>>>de'ciency> 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 ?' 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 dif'culty using v in the same> way.> >>The very foundations of your argument are false.> This is certainly good to know.> Yes, the 'rst 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 clari'cation, though everybody butLester probably already knew what I meant: This iscircular 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 theorigin. - 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 'xed distance away >(>r>).>> No one> I know of agrees with my interpretation of dr/dt. >>Since you 'xed >r> in the 'rst paragraph, we know dr/dt = 0. What's >the problem?Not exactly. I 'xed r in the 'rst paragraph not in'nitesimal r.> Radial dr/dt = tangential v.Close :). You mean to say (I think) that you 'xed magnitude(r) inthe 'rst paragraph not the position vector r. = ... stuff deleted ...> Linear versus Analytical MechanicsOne 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> de'ned and t is not de'ned at all. In other words, r is only de'ned> in direction and t is not de'ned 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 Iused to believe that we had computers and airplanes and rockets andsatellites and all that stuff, due in large part to models based on thetheories of physics.What's worse is that all that angular motion stuff (you know, rotatingobjects, orbital mechanics, the quantization of angular momentum, thewhole schmear). Yet, my bike always worked just 'ne, and they've evenmanaged to send satellites out to visit planets and all. I wonder howthey 'gured out how to throw the things up there just right, so thatthey would go to the right place, and how they knew from the start, justhow 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 'xed distance away >(>r>).>> No one>> I know of agrees with my interpretation of dr/dt. >>Since you 'xed >r> in the 'rst paragraph, we know dr/dt = 0. What's >the problem?>> >> Not exactly. I 'xed r in the 'rst paragraph not in'nitesimal r.>> Radial dr/dt = tangential v.>> >> >>Close :). You mean to say (I think) that you 'xed magnitude(r) in>the 'rst 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 ?' 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 dif'culty using v in the same>> way.>> >The very foundations of your argument are false.>> >> >> This is certainly good to know.>> >>Yes, the 'rst 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 anyoneremembers the show. He said of psychologists that he never met one whodidn't need one.The fascinating thing is that everyone seems to to keep coming backfor more. Apparently they need the eggs. And I've managed to collectquite a number of ovarians and quite a few returnees. Prussing justcame 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 moreshould be arriving any time. Perhaps we can all undergo shrinkologytogether. 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>> de'ned and t is not de'ned at all. In other words, r is only de'ned>> in direction and t is not de'ned 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 'ne, and they've even>managed to send satellites out to visit planets and all. I wonder how>they 'gured 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 critiquingideas. I'm sure there are more to come. >> Wrong, sort of.>> The most general de'nition 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 de'nition of analytic function. >>I took a course in complex analysis last semester. We used the book>>Basic Complex Analysis> by J.E. Marsden. Marsden de'ned 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 de'nitions.Good for him! What the word analytic >really means> is indeed>given by a power series>, and it's an amazing fact that a functionthat's differentiable in the complex sense is in fact given bya power series.Making the distinction is a good idea, because for examplepeople often speak of a function f : R -> R as being analyticif it's given by a power series (locally); in _that_ contextdifferentiability, even in'nite differentiability, does notimply analyticity. (I should mention that this last notionis often called >real-analytic>, just because >analytic> isso often taken as a synonym for >holomorphic>.)In the same vein one might say that >entire> doesn'treally mean >analytic in the whole plane>, it reallymeans >given by a single power series in the entireplane>, and that turns out to be equivalent. Butit's not equivalent in other contexts:Ex: Give an example of an analytic f : R -> R suchthat the Taylor series for f centered at the originhas 'nite 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 'nite radius of convergence.> Uhm, Sum (from zero to in'nity) of x^n?Yeah, the geometric series ought to work. It converges on theinterval (-1, 1). So the radius of convergence is 1. Easy. :) Ican't really think of any non-trivial ones, though.Alex SollaJuniorReed 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 'nite radius of convergence.> > Uhm, Sum (from zero to in'nity) 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 'nite radius of convergence.>> >>Uhm, Sum (from zero to in'nity) of x^n?That doesn't de'ne a function that's analytic on all of R.of R if and only if there exists a function F, holomorphicin an open set in the plane containing R, such thatf 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 'nite radius of convergence.>>Uhm, Sum (from zero to in'nity) of x^n?That doesn't de'ne 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. UllrichSPOILER ...||V||V||V||V||V||V||V||Vf(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 >seriesabout 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 areanalytic 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 uselessfor the original equation?2) Can I perhaps optimize the code to >'lter out> the polynomialsolutions 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 foundsolutions to (1), without increasing the degree of the resultantpolynomial too much?-- Ioannishttp://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 cancomfortably 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 >'lter out> the polynomial|>solutions which are not solutions of (1)?You can use fsolve with an interval around c. But you may as well dothis with the original equation:> Digits:= 15; fsolve((f@@5)(z)=z, z=c-0.1 .. c + 0.1); 1.25773454137653Note 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.caDepartment 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 suf'ces to solve f(z)=z. Write z = u+vi, match magnitudesand angles of f(z) and z, eliminate u, plot the function of v andlook for a match. I get f(z) = z when z is about 20.94645536 + 40.45731830 IThis method 'nds all the solutions of f(z) = z, and they form anice sequence in the plane. Fixed points of iterates of f look like they won't permit such aneasy reduction to functions of one variable. They also look to be anumerical-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 'xed points.The series of the inverse function will probably be of no more use thanthe series of the original for 'nding 'xed 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 'xed points.|>The series of the inverse function will probably be of no more use than|>the series of the original for 'nding 'xed points.On the other hand, you can get a series for the 'xed 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.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 complex solutions of (f@@5)(z)=z.Well, again it suf'ces to solve f(z)=z.No, it doesn't suf'ce to solve f(z)=z. The 'xed points of f(z) areessentially all known, and can be expressed analytically as:e^[-LW(k,-Log(c))], k in Z, depending on the LambertW branch.All these 'xed points of f are automatically 'xed points of f^(n)(z),but the converse is not true. A 'xed point of f^(n)(z) is notnecessarily a 'xed point of f(z). In particular, the (perhaps complex)points of a p-cycle may be 'xed points of f^(kp)(z), k in Z, but not off(z). And such p-cycles _do_ exist.Certain real points on (0, e^(-e)) for example, are known to be a2-cycle, so they are 'xed points of f^(2)(x), but they are not 'xedpoints of f(x):|>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 suf'ces 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 'nds 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 'xed point of f@@5 but not of f.Robert Israel israel@math.ubc.caDepartment 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 suf'ces to solve f(z)=z.>>z = 15.384972737405246873 + 0.42943156081631657205 I (approximately)>is a 'xed point of f@@5 but not of f.I was responding to the question of whether the 'xed pointshad to be real (they don't, as we have both now noted). I wasnot claiming that all 'xed points of f^(5) would also be'xed points of f, merely that the 'xed points of f areamong the 'xed 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 de'nitions 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 speci'cabout what the hell you are talking about... and notjust 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 'nal>state is independent of the initial conditions>yet all our sciences are built upon de'ning>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>?wed? 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'tdo 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>'rst 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>JonathanP.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 speci'c>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 Thermodynamicsis 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 thesystem is in a very low entropy state, then the states in itspast 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 speci'c case.David McAnally-- =In math most equations treat time as justanother spatial coordinate, which suggeststime can be reversed in time and the equation willwork as well.This contradicts the second law which suggeststhat ....nothing in the universe is reversible.A model of reality must have as it's axiomsthose qualities that are observed. Natureand the universe operate as iterated mapsinto itself, yet math treats it as if theoutput is proportional to the input.The universe is non-linear where the 'nalstate is independent of the initial conditionsyet all our sciences are built upon de'ninginitial conditions to understand the future.Is mathematics, as a model of reality, fundamentally?wed? Shouldn't any model of reality have thesame traits as what is being modeled?It appears to me math in inconsistent withreality in every primary aspect.The frame of reference of classical science iswrong. Reducing to understand the whole isa fundamental error, science should berebuilt from scratch using the inverseperspective. Our primary axioms needto be constructed from the largest scalesciences, not the smallest. Cosmologyand astrophysics for example, as opposedto quantum and other reductionist disciplines.Global system properties should be studied'rst in order to later understand the components.Science was established within a frame ofreducing to understand out of necessity, thatno longer is the case.Imho.Jonathans 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 ?wed?No, math makes no such claim to model reality nor does physics even thotheir favorite pass time is to make models of reality. Your discussionwould 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 ?wed?>> 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> ?wed? Mathematics is reality.Mathematics is not a model of non-mathematics(but some writers used >reality> as a synonym fornon-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 ?wed?> >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 Butter? Effect is a crock of >initial conditions, sincewe 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 spaceco-ordinates,> whose fad-meaning was establishedby the reputation of Einstein by his teacher's mere suggestion(>Minkowski coords.>).the same problems of >uncertainty> applyto the observation of cosmology, in thata 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 'nal> 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 opposedhttp://www.channel1.com/users/bobwb/synergetics/photos/ sk3prsm.html---Dec.2000 ?AND' Chairman Paul O'Neill, reelectedto 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 ?wed?> 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 notnecessarily consistent with reality.Then that begs the question, what should we do differently in orderto have a science that ...does properly model the real world?Towards the end of my post I suggest that this ?ncient paradox'has a solution and it lies with our chosen frame of reference.A proper model of the real world is being createdby simply inversing the frame of reference classicalscience has been constructed within. Whichis a task any mathematician should 'nd trivial.The implications of doing so, however, areas 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 whetherin science, religion, arts or the humanities.I am trying to describe a new universal theory oforganization called complexity science.JonathanAn Introduction to Complex SystemsTorsten Reil, Department of Zoology, University of Oxfordhttp://users.ox.ac.uk/~quee0818/complexity/ complexity.html>The study of complex systems has gained increasing attentionin recent years, in such diverse disciplines as economics, lifescience, sociology, physics and chemistry. The multidisciplinaryapproach taken by its students has revealed a surprisingly highdegree of applicability of the concepts to the different 'elds.Behaviour of biological systems seems to be mirrored in thatof economic ones...>Center for the Study of Complex Systemsat the University of Michiganhttp://www.pscs.umich.edu/>The Center is based on the recognition that many different kinds ofsystems which include self-regulation, feedback or adaptation in theirdynamics, may have a common underlying structure despite theirapparent differences. Moreover, these deep structural similaritiescan be exploited to transfer methods of analysis and understandingfrom one 'eld to another.>Chaos at Marylandhttp://www-chaos.umd.edu/>Chaos is a multidisciplinary science, and this is re?cted in the factthat the members of the group are af'liated with diverse departmentsand institutes>The Complexity & Arti'cial Life Research Conceptfor Self-Organizing Systemshttp://www.calresco.org/index.htm>Here we will introduce the integrating sciences of Complex Systemsand of ALife together with related systems areas. We'll also pursuethe wider social implications of these transdisciplinary theories ofself-organization on mind, art, spirit and life as it could be...>>PERCEPTION of anObject costsPrecise the Object's loss.Perception in itself a gainThe Object Absolute is nought,Perception sets it fair,And then upbraids a PerfectnessThat situates so far>By E Dickinsons I am trying to 'nd 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 de'ned 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 MitchellThe 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 across-posting to sci.math.research (where the same question appeared),but it somehow ended up going only to sci.math.research, so I'm makinga separate post to sci.math.]You can simplify the formula for the area somewhat by using theformulas 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 formulafor 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 thecomplementary 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 <= PIprovided that d/2 <= R <= PI - d/2. In fact, the real part of theright side of the formula (with a naturally chosen branch) gives thecorrect result for R <= PI and d <= PI (i.e., even when the disks ortheir complements are disjoint).John Mitchell I am trying to 'nd 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 isthe> area of the overlapping region of the two circles in terms of rand> 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