mm-7 === I forgot to mention the important fact that the Sfs (not super-S) areall equal-area shapes, but are not necessarily identical.(If they were not required to be equal area, we could just take onebig circle and add as many TINY circles as we want to to the edge ofthe big circle.)(sorry about starting a new thread)Original post below:------------De?e circleness (a concept probably already called something else)as being one of these three de?itions:For a (not necessarily?) convex closed-shape, S:alpha(S) = (area a maximum inscribed circle)/(area of S)beta(S) = (area of S)/(area a minimum circumscribed circle)gamma(S) = alpha(S)*beta(S) =(area a maximum inscribed circle)/(area a minimum circumscribedcircle)By maximum inscribed circle I mean the largest circle which can bedrawn so as to be completely contained within S.By minimum circumscribed circle I mean the smallest circle which canbe drawn so that S can be placed completely within it.So, this is really 3 puzzles -- which one puzzle being a result ofwhich de?ition of circleness we are using.Actually, this is really an in?ite number of puzzles, because I amwondering about the solutions for any positive integer n.Question (I am wondering and do not personally know the answer): For a ?ed n:What are the shapes, S(1), S(2),..., S(n), such that the product ofthe circlenesses of these Sfs, and this product multiplied in-turn bythe circleness of super-S (see below), all by one de?ition ofcircleness, is maximized?super-S is any simply-connected and topologically circular (ie. noholes) shape formed by arranging S(1), S(2),..., S(n) so that they areall touching and not overlapping.An example:3 squares with unit-area can be combined as: --- ---! ! ! --- --- ! ! ---(Hopefully, this comes out okay on your browser.)Each square has alpha(S) = pi/4.And alpha(super-S) = x (do not feel like ?uring it out now...)Then the product is:pi^3 *x /64, which I doubt is maximal. Leroy Quet === Portfolio of PAF as of 12AUG03:50 BCE 22.24 $1,112.0050 BLS 25.58 $1,279.001,050 BMY 26.19 $27,499.5050 DT 15.34 $767.0051,000 Q 4.05 $206,550.008,100 SBC 23.40 $189,540.002,100 SGP 16.00 $33,600.00380 VZ 35.85 $13,623.0080 WYE 44.29 $3,543.20realestate land 3APR03 of 3 lots $19,000realestate land 30JUL03 another lot $11,500art of lithographs & porcelain JAN-JUN03 for $12,000Well, today I gained 100 free shares of SBC plus some cash extras.Today, BMY had a 3 point Crossover compared to SBC of Maythis year where SBC was higher than BMY:I tried very diligently today to have that 3 point spread and I placedthe sell 1000 shares BMY at 26.15 and the buy 1100 SBC at 23.10and as the day progressed I sold BMY at 26.12 and bought the SBCat 23.40. I could not bear the thought of ending the day in cash.So, today was a Crossover day of about 3 points between SBC and BMYand I gained 100 free shares of SBC. And I expect that within the next3 months that SBC will be more than 3 points above BMY where I makethe switch over in the Crossover and gain at least 100 free new sharesofBMY.Life, really cannot be any easier than this.whole entire Universe is just one big atom where dotsof the electron-dot-cloud are galaxies === any compound rotation (initial torqueabout different axes) resolve to a single rotationabout a single axis, just as any two do;these problems can be solved by purely imaginary quaternions,such as pitch/yaw/roll of a craft. try throwing a bottle in the air with two axesofspin. > As to the main question, consider that any complex rotation can be > broken down into simple rotations about more than one axis. You can add > momentums together and you have one angular momentum vector for the entire > asteroid.--les ducs dfEnron!http://members.tripod.com/~american_almanac === > To envision this question, imagine I have a gyroscope with dozens> of different circles, or chambers, and I set them in motion, one by> one, and describe the complex rotation by a number of mathematical> function. Exactly how many times can I do this before I donft have> to describe it by any more numbers of mathematical functions? Or is> there a limit at all?Ifm sorry, but you faded there. What does it mean for a gyroscope> to have circles? What does it mean for a gyroscope to have chambers?> Are you talking about adding different gyrscopes with different> rotating wheels? I believe he means a single rotor, mounted inside a gimbal,which is in turn mounted inside another gimbal, and so forth,all turning in different directions simultaneously. He wantsto know whether there is some number of gimbals beyond whichadding yet another rotating gimbal has no effect on thecomplexity of the net motion of the central rotor. -- Jeff, in Minneapolis . === >any compound rotation (initial torque>about different axes) resolve to a single rotation>about a single axis, just as any two do;>these problems can be solved by purely imaginary quaternions,>such as pitch/yaw/roll of a craft.> try throwing a bottle in the air with two axes>ofspin.This is a common mistake. What is true is that for any twotimes t1 and t2, the change in orientation of a rigid body fromt1 to t2 can be described by a single rotation about a singleaxis. But itfs not true that the evolution of the orientation in time is described by rotations about a ?ed axis. Indeed,this wonft happen for a free motion when the body is non-symmetric (all three moments of inertia distinct) and the motion is not rotation about one of the principal axes.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === >This brings up a question: An irregular asteroid (Toutatis?) that was>visited by the Galileo or another spacecraft was described as having>a chaotic spin, or different spins on different axes. If angularmomentum>is a simple vector quantity, how is this possible?> Yes, Toutatis. See> The same angular momentum can be obtained by different motions.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2Robert,thanks for the clari?ations.I have one more question - and apologies, my math is extremely rusty here -Ifm fairly sure I knew this stuff at one point:Angular velocity about one axis can be described as a vector, which whenreferenced to the center of rotation is in the direction of the axis, withmagnitude equal to the angular velocity about that axis in a clockwisedirection (correct so far?).In the case of an object such as Toutatis, which I understand to bedescribed as rotating at different angular velocities about each of twodifferent axes, I donft think it can be described by a single vector. Ifmnot sure though.1. Is this correct? Are two vectors required here?2. If so, what is the maximum number of vectors required to describe thearbitrary rotations of a three dimensional object?Ifm thinking the answer is either one or three, but I donft know how toprove it.Many thanks in advance for any enlightenmentKrill === >This brings up a question: An irregular asteroid (Toutatis?) that wasvisited by the Galileo or another spacecraft was described as having>a chaotic spin, or different spins on different axes. If angular> momentum>is a simple vector quantity, how is this possible?> Yes, Toutatis. See> The same angular momentum can be obtained by different motions.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2Robert,> thanks for the clari?ations.I have one more question - and apologies, my math is extremely rusty here -> Ifm fairly sure I knew this stuff at one point:> Angular velocity about one axis can be described as a vector, which when> referenced to the center of rotation is in the direction of the axis, with> magnitude equal to the angular velocity about that axis in a clockwise> direction (correct so far?).Yes. > In the case of an object such as Toutatis, which I understand to be> described as rotating at different angular velocities about each of two> different axes, I donft think it can be described by a single vector. Ifm> not sure though.1. Is this correct? Are two vectors required here?Exactly what are you hoping to describe?The angular velocity at any particular time is a single vector. However, this changes as time goes on. The angular momentum staysconstant (in the absenceof torque), but if the object is not symmetric the moments of inertiaabout different axes are different, so the angular velocity andangular momentumdonft have to be in the same direction. Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > To envision this question, imagine I have a gyroscope with dozens> of different circles, or chambers, and I set them in motion, one by> one, and describe the complex rotation by a number of mathematical> function. Exactly how many times can I do this before I donft have> to describe it by any more numbers of mathematical functions? Or is> there a limit at all?Ifm sorry, but you faded there. What does it mean for a gyroscope> to have circles? What does it mean for a gyroscope to have chambers?> Are you talking about adding different gyrscopes with different> rotating wheels?> I believe he means a single rotor, mounted inside a gimbal,> which is in turn mounted inside another gimbal, and so forth,> all turning in different directions simultaneously. He wants> to know whether there is some number of gimbals beyond which> adding yet another rotating gimbal has no effect on the> complexity of the net motion of the central rotor.> -- Jeff, in Minneapolis> .Yes, that is exactly what I mean. THANK YOU!So, how far does this go? Does this go only up to 3, which I wouldassume is the limit in 3 dimensions, or does it go past that?(...Starblade Riven Darksquall...) === This brings up a question: An irregular asteroid (Toutatis?) that was>visited by the Galileo or another spacecraft was described as having>a chaotic spin, or different spins on different axes. If angular> momentum>is a simple vector quantity, how is this possible?> Yes, Toutatis. See> The same angular momentum can be obtained by different motions.> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel> University of British Columbia> Vancouver, BC, Canada V6T 1Z2Robert,> thanks for the clari?ations.I have one more question - and apologies, my math is extremely rusty here -> Ifm fairly sure I knew this stuff at one point:Angular velocity about one axis can be described as a vector, which when> referenced to the center of rotation is in the direction of the axis, with> magnitude equal to the angular velocity about that axis in a clockwise> direction (correct so far?).> I really donft know... Ifve looked up the mathematical de?ition ofangular momentm, and I ?d that sometimes it doesnft point in adirection parallel to its axis. Like if itfs a spinning sphere, and Iits origin and the regular momentum vector, but I chose a point on thetop or bottom, it will be pointing diagonally. At least Ifm assumingorigin means the actual center rather than simply something having todo with the curvature, which would make a lot more sense if it was.> In the case of an object such as Toutatis, which I understand to be> described as rotating at different angular velocities about each of two> different axes, I donft think it can be described by a single vector. Ifm> not sure though.> I think if you de?e it one way, it can... but if you de?e itanother way it canft. Ifm not so sure about this though.> 1. Is this correct? Are two vectors required here?> 2. If so, what is the maximum number of vectors required to describe the> arbitrary rotations of a three dimensional object?> Thatfs what Ifm asking!And I think angular momentum is just one vector, but that thecomplexity of rotations is much more than one vector... and may not bea vector at all!> Ifm thinking the answer is either one or three, but I donft know how to> prove it.Many thanks in advance for any enlightenmentKrill(...Starblade Riven Darksquall...) === must they be the three orthogonal axes? > try throwing a bottle in the air with two axes>ofspin.This is a common mistake. What is true is that for any two> times t1 and t2, the change in orientation of a rigid body from> t1 to t2 can be described by a single rotation about a single> axis. But itfs not true that the evolution of the orientation > in time is described by rotations about a ?ed axis. Indeed,> this wonft happen for a free motion when the body is > non-symmetric (all three moments of inertia distinct) and the motion is > not rotation about one of the principal axes.--Dec.2000 ?WANDf Chairman Paul OfNeill, reelected to Board. Newsish?http://www.rand.org/publications/randreview/issues/rr .12.00/http://members.tripod.com/~american_almanac === >A better way of putting it is this: What is the most complex rotation>function or class of rotation functions you can make?>A (pseudo) vector.> ...>No limit. Its angular momentum is a single vector quantity.> This brings up a question: An irregular asteroid (Toutatis?) that was> visited by the Galileo or another spacecraft was described as having> a chaotic spin, or different spins on different axes. If angular momentum> is a simple vector quantity, how is this possible?> --> -MikeSuppose that a rigid body possesses three orthogonal angularmoments of inertia such that I1 < I2 < I3. Any component ofrotation about I2 is unstable. [Old Man] === >must they be the three orthogonal axes?What do you mean by they? Any massive body has a (not necessarilyunique) set of three orthogonal principal axes, de?ed by eigenvectors of the moment of inertia tensor. When there is adouble eigenvalue, you can choose any two orthogonal vectors inthe two-dimensional eigenspace for that eigenvalue as principal axes.In the case where the eigenvalues are all distinct, a rotation about a non-principal axis can not be a free motion, i.e. it will require torque to maintain it. This follows from the Euler equations.Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === > Let {a(k)} be any sequence of integers where the sum below converges.Let {b(j)} be a sequence such that:> sum{j=0 to oo} b(j) x^j /j! =exp(A_n(x)),> where A_n(x) = sum{j>=2, GCD(j-1,n)=1} a(j) x^j /j!,where this sum is over all integers j, j >= 2, where> n is coprime to (j-1), n = any ?ed positive integer.> Then:b(j) is an integer sequence, and (the main result...):> n divides b(n+1).> A generalization:m also divides b(m+1),for ALL positive integers m where every prime which divides m also divides n.(obvious, but noteworthy)Leroy Quet === |> Let {a(k)} be any sequence of integers where the sum below converges.|> Let {b(j)} be a sequence such that:|> |> sum{j=0 to oo} b(j) x^j /j! =|> |> exp(A_n(x)),|> |> where A_n(x) = |> |> sum{j>=2, GCD(j-1,n)=1} a(j) x^j /j!,|> |> where this sum is over all integers j, j >= 2, where|> n is coprime to (j-1), n = any ?ed positive integer.|> |> Then:|> |> b(j) is an integer sequence, and (the main result...):|> |> n divides b(n+1).||A generalization:||m also divides b(m+1),||for ALL positive integers m where every prime which divides m also divides n.||(obvious, but noteworthy)I like this one. I donft know how you prove it, though. It doesnft seem allthat obvious to me. Perhaps Ifm missing something simple.By computing a few of them, it seems that b_k is the sum over all of theways of partitioning k distinct elements of the product over the elements*of the partition of a_i where i is the number of elements in it. [*Therefsgot to be a more descriptive term than element. I was about to writepartitions in the partition but thatfs worse. The partition correspondsto an equivalence relation, and I mean the individual equivalence classes.There should be a good term for them.]It looks like thatfs not so hard to prove by induction. The value of thek-th derivative of e^A at x=0 gives b_k. The k-th derivative itself is thesum over the equivalence relations on a k element set of e^A * (d/dx)^{i1}(A) * ... * (d/dx)^{it}(A)where i1,i2,...,it are the numbers of elements in the equivalence classes.Taking the derivative of that term gives us t+1 terms which correspond tothe t+1 partitions of a k+1 element set that restrict to the givenpartition on the ?st k elements. The term we get by taking the derivativeof (d/dx)^{ij}(A) corresponds to adding the k+1-st element to the j-thequivalence class. The term we get by taking the derivative of e^A (whichgives Afe^A) corresponds to creating an additional partition to put thek+1-st element into. And of course the value of the ij-th derivative ofA at x=0 is a_ij, while A(0)=0 and so e^A(0)=1.It seems that the result is true because the coef?ient of each termis separately divisible by m. The partitions of m+1 elements can betransformed by cyclic permutation of the ?st m elements. This preservesthe number of elements in the equivalence classes. So the number ofequivalence relations having i1,i2,...,it elements in their equivalenceclasses is a multiple of m unless at least one of them is ?ed by acyclic permutation of the ?st m elements.The only ones which are ?ed are ones where the equivalence class thatcontains the m+1-st element is ?ed, since the k+1-st element is ?ed.So if there are i_j elements in that equivalence class, for it to be ?edthere must be a common factor between i_j-1 and m. The orbits of anontrivial cyclic permutation of m elements have equal order, all dividingm. But since A_n(x) is a sum only over the j where (j-1,n)=1, that amountsto setting a_ij=0 when (j-1,n)>1. So for the nonzero terms, the number ofassociated equivalence relations is actually divisible by m.Ifm curious to know in what terms youfve been thinking about it.Keith Ramsay === in the korea high school,(high school = as the front step of university, 3 years period)All students are requested to be at the school before 7:30 every morning.and go to the home after 9:00 or 10:00 night.All students have spend great part of time studying in school.i am concerned about attending school time and ending time in your country. === To what country are you refering? USA?3:00pm.Lurch> in the korea high school,> (high school = as the front step of university, 3 years period)> All students are requested to be at the school before 7:30 every morning.> and go to the home after 9:00 or 10:00 night.> All students have spend great part of time studying in school.> i am concerned about attending school time and ending time in yourcountry.> === oh...envious USAitfs time is equal to korea elementary school time. === > in the korea high school,(high school = as the front step of university, 3 years period)All students are requested to be at the school before 7:30 every morning.and go to the home after 9:00 or 10:00 night.All students have spend great part of time studying in school.i am concerned about attending school time and ending time in your country. And end the regular schools hours at 2:30 to 3:30 PM. Most schools have some sort of optional after-school hour events: like sports, tutoring, accelerated classes, or work placement for a few hours after that. But, I donft think many go later than 5-6 pm. But, many American students look for alternatives to University Studies also. Since the cost or American Universoties is incredible to begin with. And many American University curriculums donft even begin to satisify the job requirements that industry needs. === European high schools (a.k.a. secondary schools) start at 8am and endsomewhere around 3pm (in Ireland and England itfs more like 9am to 4pm. Itfstotally beyond me why someone from Korea (north or south) would have anyidea that their school system is superior to anywhere else. Especially whenone considers that the rest of the world, the US in particular, has toprevent the Koreans from killing each other.If Bush wins another term as president my Korean friend, you wonft need toworry about high school opening hours any longer; therefll be shag all leftin your country once he gets to test his new weaponry on you.davidoff404 === in the korea high school,(high school = as the front step of university, 3 years period)All students are requested to be at the school before 7:30 every morning.and go to the home after 9:00 or 10:00 night.All students have spend great part of time studying in school.i am concerned about attending school time and ending time in your country.In Britain 9 am to 4 pm (it varies) weekdays only. But there will behomework for the evenings and weekends. Some pupils will choose toattend extra-curricula clubs etc. Are you at school in North Korea orSouth Korea? Whichever, I hope that you ?d time to enjoy yourself aswell.-- G.C. === Hot-girl, please ignore this nastiness!GeorgeEuropean high schools (a.k.a. secondary schools) start at 8am and end> somewhere around 3pm (in Ireland and England itfs more like 9am to 4pm. Itfs> totally beyond me why someone from Korea (north or south) would have any> idea that their school system is superior to anywhere else. Especially when> one considers that the rest of the world, the US in particular, has to> prevent the Koreans from killing each other.If Bush wins another term as president my Korean friend, you wonft need to> worry about high school opening hours any longer; therefll be shag all left> in your country once he gets to test his new weaponry on you.davidoff404-- G.C. === > So how many of you have designed your own numerals? Sixteen should> cover the most used bases (including the cancerous one based on the> ?st and third prime). Surely you donft use the terribly designed> Arabic ones? I have my own set brewing myself, just wanted to check> in on all of yours.Sixteen? Given up on the idea of mixed base number systems? Maybe youshould re-?l your bong. === In sci.math, Thinkit<2db4f40b.0308110544.1fa88aaa@posting.google.com>:> So how many of you have designed your own numerals? Sixteen should> cover the most used bases (including the cancerous one based on the> ?st and third prime). Surely you donft use the terribly designed> Arabic ones? I have my own set brewing myself, just wanted to check> in on all of yours.Color me slightly puzzled but, apart from esoteric usessuch as making up an alien script suitable for moviessuch as Independence Day or shows such as Star Trek andDoctor Who I fail to see the point in replacing 0-9A-F,the traditional digits for computer hexadecimal work.On the other hand, the digits are decidedly ad hoc, exceptperhaps for ?1f. The Chinese are a little more logical,at least for the ?st three; 1 horizontal stroke, 2horizontal strokes, 3 horizontal strokes. The Romansused 1-4 vertical strokes; the V might have been atally, the X a doubletally. The ancient Greek systemwas largely alphabetic.I did create for my amusement a 10-digit set where the numberof strokes equals the digit, except for 0; a blank justdidnft make sense as a digit. :-) However, I now have noidea where Ifve put it, but 4 was a square, 3 was a triangle,for example. Or perhaps 3 was 3 strokes; I donft remember now.A 15-stroke last digit would probably be a little complicatedbut not that complicated. However, Ifm not familiar withkanjii; the only other glyph I know is woman.Ifm not sure how the Arabic numeral set can be construed asbadly designed, although I canft say itfs well-designedeither. For various uses (e.g., Braille) of course itfshighly inappropriate, mostly because one canft tell thedifference between say 8 and 9, or 8 and 0. But thatfshandled by using a different font: raised dot patterns.Nor is it clear how 10 is cancerous. At best, 10 ispernicious, because of a biological accident; Naturecould have given us the far more logical (numerologicallyspeaking) base of 12, for example. But how do 10fsmultiply and mutate? :-)Still, itfs an intriguing question.-- #191, ewill3@earthlink.netItfs still legal to go .sigless. === > So how many of you have designed your own numerals? Sixteen should> cover the most used bases (including the cancerous one based on the> ?st and third prime). Surely you donft use the terribly designed> Arabic ones? I have my own set brewing myself, just wanted to check> in on all of yours.> Color me slightly puzzled but, apart from esoteric uses> such as making up an alien script suitable for movies> such as Independence Day or shows such as Star Trek and> Doctor Who I fail to see the point in replacing 0-9A-F,> the traditional digits for computer hexadecimal work.Well, there is one possible reason: using the symbols A-F as digits, inprogramming languages or in algebra, will interfere with the use of theletters A-F as variables; this often necessitates the awkward use ofpre?es 0x (etc.) or the suf? h.Now, I think the poster has gone out on a limb by claiming the Arabicnumerals are terribly designed; in fact, the Arabic numerals are veryconvenient for writing (requiring very few strokes). But, admittedly therefsdoesnft really seem to be anything logical about their design; for example,6 and 9 are rotations of each other, yet they donft seem to possess anyspecial numerical relationship to warrant this. Also, the digits 1, 4, and 7are peculiar in that they are drawn using only lines (the other digitsinvolve curves), yet there doesnft seem to be any logical reason for this.There is also a nice property of base 10 that most bases do not have. Now,this may not seem signi?ant, but let me explain it anyhow ... Most anyonewho has done some computational number theory has no doubt heard of theright shift algorithm of ?ding the GCD of two numbers. This algorithmusually applies when the numbers are stored in binary format, but it canalso be performed in decimal, or any other base, with an appropriatemodi?ation. Itfs best illustrated with an example: consider ?ding theGCD of 71 and 1891. Recall the basic fact of GCDfs that you can add orsubract multiples of one number to the other and this will not change theirGCD (i.e., GCD(a,b) = GCD(a,b+na) ). For example, we can subtract a 71 awayfrom 1891 to get GCD(71,1891) = GCD(71,1820)Now, notice that 1820 ends in a zero; we can chop off this zero (i.e.,right shifting) and this will not change the GCD (because all wefre doingis removing a factor of 10, which is valid because 71 has no common factorswith 10), so: GCD(71,1820) = GCD(71,182)At this point, if we recognize that 71 is prime, we can stop here andannounce the GCD is 1. Now, suppose the example had been to ?d GCD(31,1787). Then we proceed by adding 3 times 31 to 1787 GCD(31,1787) = GCD(31,1880) = GCD(31,188) = GCD(31,47) = 1As you can see, this method can be used to test if a prime divides aparticular number, and it is much quicker to do it this way than to performa division. The great calculator George Bidder used a variation of thismethod when factoring 5-digit integers (see p. 273 inhttp://users.lk.net/~stepanov/mnemo/biddere.html).The special thing about decimal is that in carrying out the right shiftmethod, assuming we have divided out all powers of 2 and 5 from bothnumbers, at any point we can proceed by adding or subtracting the smallernumber or 3 times the smaller number to the larger number and make thelarger number end in a 0. In other bases, it is not quite so convenient; inhexadecimal, for instance, we may have to add/subtract the smaller number,or 3, 5, or 7 times that number. In dozenal, we have to add/subtract thesmaller number or 5 times that number. So, we see that in this respect,decimal has a slight advantage over the other bases. Now, granted, theadvantage is very slim (particularly against dozenal), but it seemsworthwhile to point out a good property of base ten since it is so rare tosee (I think itfs fairly clear that overall dozenal is somewhat superior todecimal).- Brent === Let b[1] = c[1] = 1;Let, {b[k]} and {c[k]} be such that,for every integer m >= 2,b[m] = (1/m) sum{k=1 to m} c[GCD(k,m)]andc[m] = sum{k=1 to m} b[GCD(k,m)]. (?GCDf is the Greatest Common Divisor function.)What is the closed-form (ie. well-known non-recursive de?ition) forthe two sequences?This puzzle is too close to an old math-puzzle that I posted once(http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe= off&threadm=b4be2fdf.0212211642.bf65366%40posting.google.com& rnum=3&prev=), and the solution much too simple,for me to wait to post the solution. So it is below:||V||V||V||V||V||Vb[m] = the number of positive divisors of m.c[m] = the sum of positive divisors of m.Leroy Quet === No I disagree. if f(n) = O(g(n)) i.e.0 <= f(n) <= C g(n), where C is some constantthen you can clearly see that g(n) = Omega(f(n)) and so this holds for the lower bounds of g(n) = Theta(f(n))=> C f(n) <= g(n)Although I agree this wonft hold for g(n) <= C2 f(n)but donft you think itfs correct saying that g(n) = Theta(f(n)) is true?Anderson> A quickie:Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n))cheers,Anderson> f(n), g(n), and h(n) can be any positive functions.. by that > de?ition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture > true :)> cheers,> -Andre> what I actually meant was that if what you said was the case, then > the following would also hold:> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants> => f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e.> f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n)> and this would then mean that:> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n))> and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + > g(n) + h(n))> There is no contradiction between these. If itfs Theta, then itfs also> Omega.> Letfs back up a bit. You said> For three positive functions, I believe it can be graphically shown > that the minimum of either of these functions will not (and cannot) > have an asymptotic lower bound of these functions altogether.> To which I replied> No. It _can be_ true that min{f(n), g(n), h(n)} = > Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)).> So my statement was that in this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n))> Now, as you noticed, more than that is true in this case:> min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n))> But thatfs ?e: saying itfs Theta(...) means that itfs both O(...) > and Omega(...).> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada V6T 1Z2> === Hmm.. I guess youfre right. But yes, it *may* hold true for:g(n) <= C f(n) as g(n) may be equal to f(n) at some value of C and n.-Andre> No I disagree. if f(n) = O(g(n)) i.e.0 <= f(n) <= C g(n), where C is some constantthen you can clearly see that g(n) = Omega(f(n)) and so this holds for > the lower bounds of> g(n) = Theta(f(n))=> C f(n) <= g(n)Although I agree this wonft hold for g(n) <= C2 f(n)but donft you think itfs correct saying that g(n) = Theta(f(n)) is true?> Anderson> A quickie:> Prove or disprove f(n) = O(g(n)) implies g(n)=Theta(f(n))> cheers,> Anderson> f(n), g(n), and h(n) can be any positive functions.. by that > de?ition f(n) = Theta(g(n)) = Theta (h(n)) proves the conjecture > true :)> cheers,> -Andre> what I actually meant was that if what you said was the case, then > the following would also hold:> f(n) <= C2 g(n), f(n) <= D2 h(n) where C2 and D2 are constants> => f(n) + g(n) + h(n) >= (1 + 1/C2 + 1/D2) f(n), i.e.> f(n) <= 1/(1 + 1/C2 + 1/D2)(f(n)+g(n)+h(n)> and this would then mean that:> min(f(n), g(n), h(n)) = Theta(f(n) + g(n) + h(n))> and in my understanding not: min(f(n), g(n), h(n)) = Omega(f(n) + > g(n) + h(n))> There is no contradiction between these. If itfs Theta, then itfs also> Omega.> Letfs back up a bit. You said> For three positive functions, I believe it can be graphically shown > that the minimum of either of these functions will not (and cannot) > have an asymptotic lower bound of these functions altogether.> To which I replied> No. It _can be_ true that min{f(n), g(n), h(n)} = > Omega(f(n)+g(n)+h(n)), namely if f(n) = Theta(g(n)) = Theta(h(n)).> So my statement was that in this case, where f(n) = Theta(g(n)) and > f(n) = Theta(h(n)), min(f(n),g(n),h(n)) = Omega(f(n)+g(n)+h(n))> Now, as you noticed, more than that is true in this case:> min(f(n),g(n),h(n)) = Theta(f(n)+g(n)+h(n))> But thatfs ?e: saying itfs Theta(...) means that itfs both O(...) > and Omega(...).> Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada V6T 1Z2> === > a^3 + 3va^2 - (v^3 + 1) = 0> is also true as an equation within the system.> You *have* to consider the entire system, rather than try to pull> pieces out of it, and claim a single piece is the entire thing.the entire P(m).A point to expressing it this way is to show that the m=0 case reducesit to.a^3 -3a^2 = 0You use two distinct arguments to conclude that a_3 is coprime to f.One is that to show that a_3 is coprime to f at m=0, and to concludethat therefore a_3 is coprime to f at m<>0. You donft use any mathematicalargument to do this. By expressing the afs directly, itfs clear that yourconclusion is invalid.The other way is by claiming that b_1*b_2*a_3, which is an algebraic integercoprime to f, implies that a_3 is coprime to f. This is only necessarily truewhen b_1 and b_2 are algebraic integers. Furthermore you do thiswithout any mathematical arguments.This is all notwithstanding the numerous claims you are being shown in otherthreads that none of the afs are coprime to 5.And those problems arenft dependent on on how f^2 divides off. There is noproblem with your algebra in dividing the f^2 off the way you do, and I have notany any point argued with it. I agree that your m and f are independent,just most interesting when f is coprime to 3 and m.Itfs just your conclusions that need proving, in the face of countless allegeddisproofs that are straightforward to verify, but so far not refuted.> Why wonft you at least admit the m dependency your position would> require on how f^2 divides off?> Repeatedly youfve claimed youfre not asserting such a dependency,> which goes against the math.No I wonft and no it doesnft.> James HarrisNote that your entire Advance Polynomial Factorization paper hingeson the unproven assertion:> Given that P(m) has a factor f^2 that separates off, two of the gfs> should have a factor of f which would force two of the afs to have a> factor that is f.> Therefore, that leaves one factor coprime to fThis is directly relevant to my arguments. Itfs in your interests to ? this upproperly and show the mathematics before you can expect to make headway. Youwonft be published with that statement in the paper.Phil Nicholson === > a^3 + 3va^2 - (v^3 + 1) = 0> is also true as an equation within the system.> You *have* to consider the entire system, rather than try to pull> pieces out of it, and claim a single piece is the entire thing.the entire P(m).Thatfs a false statement, which I can easily show to be false, byputting in some numbers. Using f=1, x=2, u=1, with P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)I have P(m) = 8m^3 - 24m^2 + 24m + 6 - 6m + 1 so P(m) = 8m^3 - 24m^2 + 18m + 7.But a^3 + 3va^2 - (v^3 + 1) = 0 is a^3 + 3(-1+m) -((-1+m)^3+1) = 0. Now then, would you mind explaining how you can use the secondexpression to get P(m)? > A point to expressing it this way is to show that the m=0 case reduces> it to.> a^3 -3a^2 = 0You use two distinct arguments to conclude that a_3 is coprime to f.> One is that to show that a_3 is coprime to f at m=0, and to conclude> that therefore a_3 is coprime to f at m<>0. You donft use any mathematical> argument to do this. By expressing the afs directly, itfs clear that your> conclusion is invalid.Why?You put in a lot into one paragraph and I think you just hoped to tossout a conclusion which *you* want readers to believe, possibly hopingthat youfd not have to elaborate. I want you to explain yourself indetail.What I will do is point out what youfre trying to deny, which is that f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)is an expression that represents a complex system that you canftreduce to a particular polynomial.Even a^3 + 3va^2 - (v^3 + 1)is still not a polynomial because you have variable coef?ients.What I do is take P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)and use m=0 to ?d the constant coef?ient.Now therefs nothing improper or invalid about that, and of course, theconstant term of the expression viewed as a polynomial with respect tom, will be given by m=0.That gives me P(0) = u^2 f^2(3x + uf).Now you can also multiply the expression out, group terms with m, andeventually come upon the same expression as your last coef?ient, butyou can also just use m=0.> The other way is by claiming that b_1*b_2*a_3, which is an algebraic integer> coprime to f, implies that a_3 is coprime to f. This is only necessarily true> when b_1 and b_2 are algebraic integers. Furthermore you do this> without any mathematical arguments.Herefs the mathematical argument.Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f).Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the bfs must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)proving that w_1 w_2 must be coprime to 3, if f is coprime to 3, whichleaves b_3 = 3, or a unit multiple of 3.Essentially objections to how f^2 divides off now come down toclaiming that the wfs are functions of m, but consider that w_1 w_2 iscoprime to f, when m=0, if f is coprime to 3.But that was an arbitrary choice, so let f=3.Now it can be shown that w_1 w_2 is coprime to 3 without regard to m.That is, the wfs can now be shown to all be constant with regard to m,so they have the same value no matter what the value of m is, so theyare also constant with f coprime to 3.Introducing a_1, a_2, and a_3, that is seen by considering P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)with f=3, as then you have P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 dividesoff as a constant so it is without a dependency on m.Now looking at P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)I can again check at m=0, to see P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),which forces w_3 to have a factor that is 3.Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf)where youfll notice that the bfs are algebraic integers with m=1,f=sqrt(2), but thatfs a special case as generally they are not, whichshows a problem with the ring of algebraic integers.> This is all notwithstanding the numerous claims you are being shown in other> threads that none of the afs are coprime to 5.Ifm not interested in claims. Ifm interested in proofs.> And those problems arenft dependent on on how f^2 divides off. There is no> problem with your algebra in dividing the f^2 off the way you do, and I have not> any any point argued with it. I agree that your m and f are independent,> just most interesting when f is coprime to 3 and m.> Itfs just your conclusions that need proving, in the face of countless alleged> disproofs that are straightforward to verify, but so far not refuted.I have refuted them.> Why wonft you at least admit the m dependency your position would> require on how f^2 divides off?> Repeatedly youfve claimed youfre not asserting such a dependency,> which goes against the math.No I wonft and no it doesnft.Well then, do you or do you not believe that m=0 is a special case? > Note that your entire Advance Polynomial Factorization paper hinges> on the unproven assertion:Given that P(m) has a factor f^2 that separates off, two of the gfs> should have a factor of f which would force two of the afs to have a> factor that is f.> Therefore, that leaves one factor coprime to fThis is directly relevant to my arguments. Itfs in your interests to ? this up> properly and show the mathematics before you can expect to make headway. You> wonft be published with that statement in the paper.Phil NicholsonWell, as usual readers, you see a poster working to convince, and inthis case itfs with something tossed at you out of the blue.In my paper Advanced Polynomial Factorizaton, you have P(m) = g_1 g_2 g_3 g_1 = (a_1 x + uf), g_2 = (a_2 x + uf), g_3 =(a_3 x + uf)and I prove that one of the gfs must be coprime to f, which *should*force two of the gfs to each have a factor that is f, for reasonsexplained in detail, but the ring of algebraic integers has thisweird, esoteric problem, which means that youfre forced out of thatring.James Harris === > let p(x)=x^2 + Ax + B in R[x]> if factor ring R[x]/(p(x)) is ?ld ,> ?d basis of ?ld R[x]/(p(x)) on R> ------------------------------------> i think that basis is {1,x}> but in solution paper,> deg p(x) =2> a exist such that p(a)=0> thus {1,a} is basis> -------------------------------> what do you think about it ?Well, x is an element of R[x], but not of R[x]/(p(x)). Let pi:R[x] - R[x]/(p(x))given by g(x) -> g(x) + (p(x)) be the canonical projection map. The a theyrefer to is pi(x) in R[x]/(p(x)). Notice that with a = pi(x), a^2 + Aa + Bis equal to pi(x^2 + Ax + B) which is 0in the ?ld R[x]/(p(x)). This is thesense in which they mean p(a) = 0. Itseems in the solution they are representingthe ?ld R[x]/(p(x)) with a more notationallyconvenient representation (such as {ca + d | c and d are in R and a is such thatp(a) = 0}). Note also that 1 is not the polynomial 1 in R[x] but rather pi(1).Anyway, thinking of a as pi(x), we see thatit is true that {1,a} is linearly independent.Hope that helps,Hugh === >Let me revise that a little bit:let p(x)=x^2 + ax + b in R[x]>First we should note that there are too many afs>in the notation - letfs change p to p(x)=x^2 + Ax + B.>if factor ring R[x]/(p(x)) is ?ld ,>?d basis of ?ld R[x]/(p(x)) on R>------------------------------------>i think that basis is {1,x}>but in solution paper,>deg p(x) =2>a exist such that p(a)=0>thus {1,a} is basis>------------------------------->what do you think about it ?>_If_ basis of ?ld R[x]/(p(x)) on R means>a basis for R[x]/(p(x)), regarded as a vector>space over R then I think youfre right and>the solution paper is obviously wrong because>{1, a} is not independent.>There are at least three possibilities:>(i) Ifm wrong>(ii) the solution paper is wrong>(iii) my assumption that basis of ?ld R[x]/(p(x)) on R >means a basis for R[x]/(p(x)), regarded as a vector>space over R is wrong, the phrase actually means>something else here.>Itfs true that {1, x} is a basis (or rather {1 +

, x +

}>is a basis, where

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

, x +

}>is a basis, where

is the ideal generated by p.)>But when I read the statement a exist such that p(a)=0>thus {1,a} is basis I was assuming that a was an element>of R, which is obviously impossible. I suspect you didnft>tell us exactly what the solution paper says: If a is an>element of some _extension_ of R and p(a) = 0 then>{1, a} is a basis as well.>p(a) = 0, then R[x]/(p(x)) is just {0}[*], which isnft a ?ld.Thatfs not so. For example (with R = {reals}, say) let p(x) = x^2 - 4,a = 2. Then p(a) = 0 although R[x]/p(x) is certainly not 0.> So if >R[x]/(p(x)) is a ?ld, and p(a) = 0, then a is not in R._Thatfs_ true (except for the word so) - hence the wordsobviously impossible above!(At least when R is a PID, in particular when R is a ?ld) R[x]/p(x) is a ?ld if and only if p is irreducible; in the example above p has degree 2, so if it has a root in R itfs not irreducible.>[*] Geez, what a headache. Itfs obviously true, I donft see a proof. I >could be talking total nonsense. I guess itfs time for me to leave, I >hope Ifm on the right track and helping.>Jon no FLT proof Miller************************David C. Ullrich === I am a computer science major and i was wondering to myself if there isanythingthat can not be represented in binary format...I mean since any physicalobject can be representedas a set of numbers representing different attributes and since anynumber can berepresented by a binary sequance then that to me suggest there isnothing that can notbe represented in binary format.if the above is true and anything can be represented then is there amathematical proof for it?thanks in advaceGuy === > if the above is true and anything can be represented then is there a> mathematical proof for it?sqrt(2) or pi etc can only be approximated, same with any measurement that comes from real life. Is that what you are looking for?TS === I renig on my comment. I still do not see a difference, the same end resultcomes from the optimized way - just faster.In your example, my method would be the same question (-not differnet-) -only, I get the same end result - only faster with less work.So I still ask:Why bother multiplying by 3 and adding 1 when you can just add one and keepthe problem simple and quick?Mike Curry> Would it not be faster to add +1 to an odd number rather then multiplyby> three and add 1? If I add +1 to an odd number instead, I get faster> results...> If you sere sitting an exam and were asked to multiply 148263 by 237893,> what would you expect to happen if instead of answering the question> as set, you argued that it would be faster to multiply 100000 by 200000> and did that instead? :-)> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen === I renig on my comment. I still do not see a difference, the same end result> comes from the optimized way - just faster.In your example, my method would be the same question (-not differnet-) -> only, I get the same end result - only faster with less work.So I still ask:> Why bother multiplying by 3 and adding 1 when you can just add one and keep> the problem simple and quick?> ************************************************************* *************************What problem are you referring to? The Collatz conjecturehas a de?ite statement. If you change this statement, youare no longer talking about the same problem. _________________________________________________________Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics and StatisticsYoungstown State UniversityOne University PlazaYoungstown, OH 44555-0001330-941-1817 === I renig on my comment. I still do not see a difference, the same end result> comes from the optimized way - just faster.How do you *know* itfs the same? Your method always produces 1, but the 3x+1 method has never been provedto produce 1 every time. So, until you prove that, youare merely making an unsupported assertion that theanswer is the same. Youfve made exactly zero progresstoward the desired goal.In your example, my method would be the same question (-not differnet-) -> only, I get the same end result - only faster with less work.Youfve missed the whole point. Yes, your method is easy,but it tells us nothing we didnft already know about thenatural numbers. That makes it uninteresting. Sorry.So I still ask:> Why bother multiplying by 3 and adding 1 when you can just add one and keep> the problem simple and quick?Why bother? Because itfs an interesting question. If wecan ?ure out how to answer it, that might tell us somethingimportant (perhaps even useful) about numbers. === >I renig on my comment. I still do not see a difference, the same end result>comes from the optimized way - just faster.You seem to be assuming that the end result is always 1. That is notknown - the _problem_ is to _prove_ that it always leads to 1. If youcan prove that the original procedure _always_ leads to a 1 thatwill make you famous.>In your example, my method would be the same question (-not differnet-) ->only, I get the same end result - only faster with less work.Look: _assuming_ that both processes both lead to a 1 eventually,herefs an even _faster_ way to get the same end result:GIven an integer n, let the next integer in the sequence be 1.>So I still ask:>Why bother multiplying by 3 and adding 1 when you can just add one and keep>the problem simple and quick?Youfre still totally missing the point to what the problem _is_.>Mike Curry> Would it not be faster to add +1 to an odd number rather then multiply>by> three and add 1? If I add +1 to an odd number instead, I get faster> results...> If you sere sitting an exam and were asked to multiply 148263 by 237893,> what would you expect to happen if instead of answering the question> as set, you argued that it would be faster to multiply 100000 by 200000> and did that instead? :-)> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen>************************David C. Ullrich === Ok, I am catching on now... (doh!). If the Collatz Conjecture Optimizationis ever proven, it would be interesting to see if this other way could beproven as well:t(x)=x/2 if x is even,t(x)=(x+1)/2 if x is oddMike Curry>I renig on my comment. I still do not see a difference, the same endresult>comes from the optimized way - just faster.> You seem to be assuming that the end result is always 1. That is not> known - the _problem_ is to _prove_ that it always leads to 1. If you> can prove that the original procedure _always_ leads to a 1 that> will make you famous.>In your example, my method would be the same question (-notdiffernet-) ->only, I get the same end result - only faster with less work.> Look: _assuming_ that both processes both lead to a 1 eventually,> herefs an even _faster_ way to get the same end result:> GIven an integer n, let the next integer in the sequence be 1.>So I still ask:>Why bother multiplying by 3 and adding 1 when you can just add one andkeep>the problem simple and quick?> Youfre still totally missing the point to what the problem _is_.>Mike Curry> Would it not be faster to add +1 to an odd number rather thenmultiply>by> three and add 1? If I add +1 to an odd number instead, I get faster> results...> If you sere sitting an exam and were asked to multiply 148263 by237893,> what would you expect to happen if instead of answering the question> as set, you argued that it would be faster to multiply 100000 by 200000> and did that instead? :-)> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen> ************************> David C. Ullrich === Ok, I am catching on now... (doh!). If the Collatz Conjecture OptimizationWhy do you say optimization? This problem is notabout optimization at all. The point is not to tryto make it easier, the point is simply to show thatit is true, as stated.> is ever proven, it would be interesting to see if this other way could be> proven as well:t(x)=x/2 if x is even,> t(x)=(x+1)/2 if x is oddBut that is *easy* to prove (and has been proved). Thatfswhy I said it is uninteresting as a matter of research formathematicians. Your t(x) gives a result that is always*less than* x, for every natural number greater than 1. Andnever zero or negative. It follows that any sequence ofsuch numbers will decrease steadily until it reaches 1.The dif?ult thing about the Collatz conjecture is that3x+1 is *greater* than x. Furthermore, the multiplier is3, which is greater than the divisor for even numbers (2).So, at ?st glance, it looks like the sequence shouldblow up, in general. Yet it does not, for any number thathas been checked. In other words, nobody knows if theconjecture is true or false. That makes it interesting. === > I renig on my comment. I still do not see a difference, the same end result> comes from the optimized way - just faster.In your example, my method would be the same question (-not differnet-) -> only, I get the same end result - only faster with less work.So I still ask:> Why bother multiplying by 3 and adding 1 when you can just add one and keep> the problem simple and quick?As others have pointed out, this is silly. By changing the rules,youfre changing the problem. But there is a legitimate optimizationyou can make. You can remove ALL the factors of 2 in one step insteadof one at a time. So instead of the usual sequence for 7:7 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1you only see the odd numbers:7 11 17 13 5 1This is useful if youfre only interested in how many branches are onthe tree without fundamentally changing the problem. 22_7 34_11 52_17 26 40_13 20 1016_58421becomes _7 _11 _17 _13 _51The even numbers are still accounted for, Ifm just not showing them. 7is still 5 braches away from 1.And as far as this being an optimization, that depends on how youfredoing the math. I use this when I represent binary numbers as textstrings. Since the number of factors of two is equal to the number ofcontiguous least signi?ant 0s, my perl program drops all the 0s inone fell swoop:101 : 510000 : 16 (drop all four 0s in one step)1 : 1But keep in mind that you canft go around claiming that the stoppingtime for 7 is 5, because the usual meaning of stopping time includesthe iterations of x/2. Although some like to use the rules x/2 and(3x+1)/2. Why I donft know. But it means their stopping timestatistics are not apllicable to the proper Collatz Problem.For example, someone spoke of locating all the numbers of stoppingtime 7. He has a nice little chart for ?uring this out that worksreal nice, but itfs worthless to me because the sequence(3x+1)/2 x/2 x/2 x/2 x/2 x/2 x/2would be stopping time 8 under the proper rules. and itfs not justthat the count is different, when he says ALL numbers, hefs includingthe sequences(3x+1)/2 (3x+1)/2 x/2 x/2 x/2 x/2 x/2(3x+1)/2 (3x+1)/2 (3x+1)/2 x/2 x/2 x/2 x/2which would have stopping times of 9 and 10 under the proper rules. Sohefs comparing apples to oranges.In my example of removing all factors of 2 in one step, I donft speakabout stopping time for that reason. I use the term ORDER, whereORDER is the count of odd numbers in the sequence (or the number ofbranches away from 1). I say that 7 is ORDER 5. With this de?ition,it makes no difference how I remove the factors of 2. Whether I removethem all at once or one at a time, the ORDER doesnft change. Thatfswhy it is a legitimate optimization. Your proposal changes the orderof 7 from 5 to 1 and is, therefore, worthless.Mike Curry> Would it not be faster to add +1 to an odd number rather then multiply> by> three and add 1? If I add +1 to an odd number instead, I get faster> results...> If you sere sitting an exam and were asked to multiply 148263 by 237893,> what would you expect to happen if instead of answering the question> as set, you argued that it would be faster to multiply 100000 by 200000> and did that instead? :-)> -- > Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html> The League of Gentlemen === ,> I renig on my comment. I still do not see a difference, the same end result> comes from the optimized way - just faster.In your example, my method would be the same question (-not differnet-) -> only, I get the same end result - only faster with less work.So I still ask:> Why bother multiplying by 3 and adding 1 when you can just add one and keep> the problem simple and quick?Because that makes it an interesting and unsolved problem. === ,> Ok, I am catching on now... (doh!). If the Collatz Conjecture Optimization> is ever proven, it would be interesting to see if this other way could be> proven as well:t(x)=x/2 if x is even,> t(x)=(x+1)/2 if x is oddShouldnft take anyone reading sci.math more than ?e seconds. === De?e circleness (a concept probably already called something else)as being one of these three de?itions:For a (not necessarily?) convex closed-shape, S:alpha(S) = (area a maximum inscribed circle)/(area of S)beta(S) = (area of S)/(area a minimum circumscribed circle)gamma(S) = alpha(S)*beta(S) =(area a maximum inscribed circle)/(area a minimum circumscribedcircle)By maximum inscribed circle I mean the largest circle which can bedrawn so as to be completely contained within S.By minimum circumscribed circle I mean the smallest circle which canbe drawn so that S can be placed completely within it.So, this is really 3 puzzles -- which one puzzle being a result ofwhich de?ition of circleness we are using.Actually, this is really an in?ite number of puzzles, because I amwondering about the solutions for any positive integer n.Question (I am wondering and do not personally know the answer): For a ?ed n:What are the shapes, S(1), S(2),..., S(n), such that the product ofthe circlenesses of these Sfs, and this product multiplied in-turn bythe circleness of super-S (see below), all by one de?ition ofcircleness, is maximized?super-S is any simply-connected and topologically circular (ie. noholes) shape formed by arranging S(1), S(2),..., S(n) so that they areall touching and not overlapping.An example:3 squares with unit-area can be combined as: --- ---! ! ! --- --- ! ! ---(Hopefully, this comes out okay on your browser.)Each square has alpha(S) = pi/4.And alpha(super-S) = x (do not feel like ?uring it out now...)Then the product is:pi^3 *x /64, which I doubt is maximal. Leroy QuetAbuse-Reports-To: durward at hillsboro.net to report improper postings === > De?e circlenessA circle is a polygon that has been stripped of itfs epaulets. === In sci.math, dustbird:> De?e circlenessA circle is a polygon that has been stripped of itfs epaulets.> A polygon is a circle with a few rough edges. :-)-- #191, ewill3@earthlink.netItfs still legal to go .sigless. === >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And thatfs false. I am pretty sure that Dale produced explicit common>factors; but in any case, your claim here is certainly false, since>their product is not coprime to 65.> Ifve proven it true that neither a_1, a_2 nor a_3 have ANY non-unit> factors in common with 5 in the ring of algebraic integers.> And he proves (below) that they do. At this point you have to do more than point at your own proof. You must ?d the error in his.Lemma. Let R be the ring of all algebraic integers, and let a, b, c be>any elements of R. If a and b are coprime to c, then a*b is coprime to>c.>Proof. We use the characterization of coprime valid for commutative>rings with 1: a and b are coprime in R if and only if there exist x>and y in R such that ax+by = 1. > By that de?ition only *one* of the afs is coprime to 5, but none of> them has a factor in common with 5 either.Do you even understand what hefs saying? The application would be: If a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, which means a_1*a_2*a_3 is coprime to 5.The ring of algebraic integers is really screwed up.No, your understanding of them is.For those who donft understand, consider that in the ring of evens,> which does not have 1, you canft use that de?ition of coprime that> Arturo Magidin gives, though it is, interestingly enough, true that in> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is> not in the ring.Why do you make unnecessary side trips to bizarre rings?However, rather than use dueling de?itions or argue about> de?itions I can simply switch to saying that 2 does not share> non-unit factors in the ring of evens with 6.>Since a and c are coprime by assumption, there exist n and m in R such>that an+cm = 1. Since b and c are coprime by assumption, there exist r>and s in R such that br+cs = 1.>Multiplying both together, we have>1 = (an+cm)(br+cs)> = abrn + acns + cbmr + c^2*ms> = ab(rn) + c(ans + bmr + cms).>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and>ab*x + c*y = 1. Therefore, ab and y are coprime. QED>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY>non-unit factors in common with 5 in the ring of algebraic>integers. Then, by the lemma, neither does a1*a_2; and applying the>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which>clearly has 5 as a nonunit common factor with 5. This contradicts the>assumption that none of a_1, a_2, a_3 have common non-unit factors>with 5 in the ring of algebraic integers. Therefore, your assertion is>false.> Well by your de?ition of coprime NONE of the afs have a factor in> common with 5, in the ring of algebraic integers, and you cannot prove> that any of them do.No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. This shows that there is a mistake in your work since you are claiming the product of three algebraic integers coprime to 5 *is* 65.Now if you donft want to call that coprime ?e. It doesnft change> the situation.What I can do is show that with a very quick argument using basic> algebra as Ifve done.You keep posting the same text as if it hasnft been riddled with holes. How about dealing with peoplefs counter-arguments directly instead of posting the same thing over and over?[usual proof deleted]Ifve found the Ring of Objects which includes the ring of algebraic> integers, and does not have this problem, as the bfs are all included> in it.The Ring of Objects is the set of all numbers where -1 and 1 are the> only members that are both a unit, i.e. factor of 1, and an integer,> where no non-unit member is a factor of any two integers that are> coprime.How does the Ring of Objects differ from the algebraic integers?Are you sure your de?ition is consistent?In which ring are you referring to the two integers being coprime?-- Will Twentyman === The problem is that neither a_1, a_2, nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And thatfs false. I am pretty sure that Dale produced explicit common>factors; but in any case, your claim here is certainly false, since>their product is not coprime to 65.> Ifve proven it true that neither a_1, a_2 nor a_3 have ANY non-unit> factors in common with 5 in the ring of algebraic integers.> And he proves (below) that they do. At this point you have to do more > than point at your own proof. You must ?d the error in his.He uses a special de?ition of coprime, and assumes that at least twoof the three numbers is coprime.However, by his de?ition of coprime, only one of the numbers iscoprime. >Lemma. Let R be the ring of all algebraic integers, and let a, b, c be>any elements of R. If a and b are coprime to c, then a*b is coprime to>c.>Proof. We use the characterization of coprime valid for commutative>rings with 1: a and b are coprime in R if and only if there exist x>and y in R such that ax+by = 1. > By that de?ition only *one* of the afs is coprime to 5, but none of> them has a factor in common with 5 either.Do you even understand what hefs saying? The application would be: If > a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, > a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, > which means a_1*a_2*a_3 is coprime to 5.By the de?ition he gave only ONE of the afs is coprime to 5.The others donft have non unit factors in common with 5, in the ringof algebraic integers, but each have a factor that is sqrt(5) in ahigher ring. The ring of algebraic integers is really screwed up.No, your understanding of them is.Nope. The ring of algebraic integers is really screwed up.For those who donft understand, consider that in the ring of evens,> which does not have 1, you canft use that de?ition of coprime that> Arturo Magidin gives, though it is, interestingly enough, true that in> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is> not in the ring.Why do you make unnecessary side trips to bizarre rings?Arturo Magidin uses a special de?ition of coprime which depends oncertain assumptions which fail with the ring of algebraic integers,and I highlight for readers how that de?ition can also fail inanother ring.However, rather than use dueling de?itions or argue about> de?itions I can simply switch to saying that 2 does not share> non-unit factors in the ring of evens with 6.>Since a and c are coprime by assumption, there exist n and m in R such>that an+cm = 1. Since b and c are coprime by assumption, there exist r>and s in R such that br+cs = 1.>Multiplying both together, we have>1 = (an+cm)(br+cs)> = abrn + acns + cbmr + c^2*ms> = ab(rn) + c(ans + bmr + cms).>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and>ab*x + c*y = 1. Therefore, ab and y are coprime. QED>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY>non-unit factors in common with 5 in the ring of algebraic>integers. Then, by the lemma, neither does a1*a_2; and applying the>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which>clearly has 5 as a nonunit common factor with 5. This contradicts the>assumption that none of a_1, a_2, a_3 have common non-unit factors>with 5 in the ring of algebraic integers. Therefore, your assertion is>false.> Well by your de?ition of coprime NONE of the afs have a factor in> common with 5, in the ring of algebraic integers, and you cannot prove> that any of them do.No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then > their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. > This shows that there is a mistake in your work since you are claiming > the product of three algebraic integers coprime to 5 *is* 65.Only one of them is coprime to 5, and the other two should have somenon unit factor in common with 5 but donft because the ring ofalgebraic integers is ?Now if you donft want to call that coprime ?e. It doesnft change> the situation.What I can do is show that with a very quick argument using basic> algebra as Ifve done.You keep posting the same text as if it hasnft been riddled with holes. > How about dealing with peoplefs counter-arguments directly instead of > posting the same thing over and over?[usual proof deleted]Deleting out a mathematical argument does not make it go away.Ifve found the Ring of Objects which includes the ring of algebraic> integers, and does not have this problem, as the bfs are all included> in it.The Ring of Objects is the set of all numbers where -1 and 1 are the> only members that are both a unit, i.e. factor of 1, and an integer,> where no non-unit member is a factor of any two integers that are> coprime.How does the Ring of Objects differ from the algebraic integers?Are you sure your de?ition is consistent?In which ring are you referring to the two integers being coprime?The Ring of Objects is a higher ring than the ring of algebraicintegers, which doesnft have its problems.The two integers are coprime in the Ring of Objects.The de?ition removes the possibility of the fundamentalcontradiction which distinguishes a ?ld from a ring, which is thatin a ?ld coprimeness does not exist, while in the ring of integers,it does.So ?lds are at odds with the ring of integers.Ifve simply highlighted the crucial difference between them.James Harris === This post is classic Harris, marred only by the incoherent mumblingsof others. I remove those mumblings to reveal Jamesf undiluted ef?#####> He uses a special de?ition of coprime, and assumes that at least two> of the three numbers is coprime.However, by his de?ition of coprime, only one of the numbers is> coprime.#######> By the de?ition he gave only ONE of the afs is coprime to 5.The others donft have non unit factors in common with 5, in the ring> of algebraic integers, but each have a factor that is sqrt(5) in a> higher ring.#######> Nope. The ring of algebraic integers is really screwed up.#######> Arturo Magidin uses a special de?ition of coprime which depends on> certain assumptions which fail with the ring of algebraic integers,> and I highlight for readers how that de?ition can also fail in> another ring.#######> Only one of them is coprime to 5, and the other two should have some> non unit factor in common with 5 but donft because the ring of> algebraic integers is ?#######> Deleting out a mathematical argument does not make it go away.#######> The Ring of Objects is a higher ring than the ring of algebraic> integers, which doesnft have its problems.The two integers are coprime in the Ring of Objects.The de?ition removes the possibility of the fundamental> contradiction which distinguishes a ?ld from a ring, which is that> in a ?ld coprimeness does not exist, while in the ring of integers,> it does.So ?lds are at odds with the ring of integers.Ifve simply highlighted the crucial difference between them.> James Harris####### === The problem is that neither a_1, a_2, nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And thatfs false. I am pretty sure that Dale produced explicit common>factors; but in any case, your claim here is certainly false, since>their product is not coprime to 65.> Ifve proven it true that neither a_1, a_2 nor a_3 have ANY non-unit> factors in common with 5 in the ring of algebraic integers.> And he proves (below) that they do. At this point you have to do more > than point at your own proof. You must ?d the error in his.>He uses a special de?ition of coprime, and assumes that at least two>of the three numbers is coprime.No, I use the standard de?ition of coprime: two elements a and b arecoprime in R if and only if the ideals (a) and (b) are coprime, if andonly if there is no prime ideal of R which contains both (a) and (b).It is a THEOREM that in a commutative ring with 1, two elements a andb are coprime if and only if there exist x and y in R such thatax+by=1.What is the de?ition YOU are using?The de?ition you are using is a and b are coprime if and only ifthey have no non-unit common factors.In the ring of all algebraic integers, the two areEQUIVALENT. However, your de?ition is WEAKER than the correctmeaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprimeunder your incorrect usage, but they are not coprime under correct usage.>However, by his de?ition of coprime, only one of the numbers is>coprime.Coprime to ->what<-?And prove it. [.snip.]>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be>any elements of R. If a and b are coprime to c, then a*b is coprime to>c.>Proof. We use the characterization of coprime valid for commutative>rings with 1: a and b are coprime in R if and only if there exist x>and y in R such that ax+by = 1. By that de?ition only *one* of the afs is coprime to 5, but none of> them has a factor in common with 5 either.> Do you even understand what hefs saying? The application would be: If > a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, > a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, > which means a_1*a_2*a_3 is coprime to 5.>By the de?ition he gave only ONE of the afs is coprime to 5.Prove it.>The others donft have non unit factors in common with 5, in the ring>of algebraic integers, but each have a factor that is sqrt(5) in a>higher ring.Which doesnft prove anything. If the others donft have non-unitfactors in common with 5 in the ring of all algebraic integers, thenthey would be coprime under the correct de?ition as well. [.snip.]> For those who donft understand, consider that in the ring of evens,> which does not have 1, you canft use that de?ition of coprime that> Arturo Magidin gives, though it is, interestingly enough, true that in> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is> not in the ring.> Why do you make unnecessary side trips to bizarre rings?>Arturo Magidin uses a special de?ition of coprime which depends on>certain assumptions which fail with the ring of algebraic integers,I use the STANDARD de?ition; it is you who uses a SPECIALone. Please list, explicitly:1. The special de?ition;2. The one you use;3. The certain assumptions on which my supposedly special de?ition depends.4. A proof that they fail in the ring of algebraic integers.It is not enough to say the ring is really screwed up, because thatis supposed to be your conclusion, not your hypothesis. [.snip.] === ===================================== === ==========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===The problem is that neither a_1, a_2, nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And thatfs false. I am pretty sure that Dale produced explicit common>factors; but in any case, your claim here is certainly false, since>their product is not coprime to 65.>Ifve proven it true that neither a_1, a_2 nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And he proves (below) that they do. At this point you have to do more >than point at your own proof. You must ?d the error in his.He uses a special de?ition of coprime, and assumes that at least two> of the three numbers is coprime.What de?ition of coprime would you suggest that he use? What de?ition of coprime are you using?> However, by his de?ition of coprime, only one of the numbers is> coprime.Ifll address this below.>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be>any elements of R. If a and b are coprime to c, then a*b is coprime to>c.>Proof. We use the characterization of coprime valid for commutative>rings with 1: a and b are coprime in R if and only if there exist x>and y in R such that ax+by = 1. >By that de?ition only *one* of the afs is coprime to 5, but none of>them has a factor in common with 5 either.>Do you even understand what hefs saying? The application would be: If >a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, >a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, >which means a_1*a_2*a_3 is coprime to 5.By the de?ition he gave only ONE of the afs is coprime to 5.The others donft have non unit factors in common with 5, in the ring> of algebraic integers, but each have a factor that is sqrt(5) in a> higher ring.Which ring is that? If you mean your Ring of Objects, please explain why sqrt(5) is a factor there and not in the algebraic integers.>For those who donft understand, consider that in the ring of evens,>which does not have 1, you canft use that de?ition of coprime that>Arturo Magidin gives, though it is, interestingly enough, true that in>fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is>not in the ring.>Why do you make unnecessary side trips to bizarre rings?Arturo Magidin uses a special de?ition of coprime which depends on> certain assumptions which fail with the ring of algebraic integers,> and I highlight for readers how that de?ition can also fail in> another ring.What are the assumptions that fail with the ring of algebraic integers? What de?ition *should* he be using if this one is not appropriate? The question becomes: what do you mean by coprime, and why does your de?ition differ from his in the algebraic integers?>However, rather than use dueling de?itions or argue about>de?itions I can simply switch to saying that 2 does not share>non-unit factors in the ring of evens with 6.>Since a and c are coprime by assumption, there exist n and m in R such>that an+cm = 1. Since b and c are coprime by assumption, there exist r>and s in R such that br+cs = 1.>Multiplying both together, we have>1 = (an+cm)(br+cs)> = abrn + acns + cbmr + c^2*ms> = ab(rn) + c(ans + bmr + cms).>Let x = rn, y = ans+bmr+cms. Then x and y are algebraic integrs, and>ab*x + c*y = 1. Therefore, ab and y are coprime. QED>So, assume you were correct and neither a_1, a_2, nor a_3 have ANY>non-unit factors in common with 5 in the ring of algebraic>integers. Then, by the lemma, neither does a1*a_2; and applying the>lemma again, neither does a_1*a_2*a_3. But a_1*a_1*a_3 = 65, which>clearly has 5 as a nonunit common factor with 5. This contradicts the>assumption that none of a_1, a_2, a_3 have common non-unit factors>with 5 in the ring of algebraic integers. Therefore, your assertion is>false.>Well by your de?ition of coprime NONE of the afs have a factor in>common with 5, in the ring of algebraic integers, and you cannot prove>that any of them do.>No, what he has said is that if a_1, a_2, a_3 are coprime to 5, then >their product is as well, which means a_1*a_2*a_3 *cannot* equal 65. >This shows that there is a mistake in your work since you are claiming >the product of three algebraic integers coprime to 5 *is* 65.> Only one of them is coprime to 5, and the other two should have some> non unit factor in common with 5 but donft because the ring of> algebraic integers is ?James, should be donft is the same as donft. What you are saying could be rephrased as They are all coprime to 5, but I want 2 of them not to be. If Ifm not understanding what you are saying here, please explain it. Note: Ifm assuming that no non-unit factor in common with 5 is equivalent to coprime to 5 in the algebraic integers. If you disagree with that, please state why.>Now if you donft want to call that coprime ?e. It doesnft change>the situation.>What I can do is show that with a very quick argument using basic>algebra as Ifve done.>You keep posting the same text as if it hasnft been riddled with holes. > How about dealing with peoplefs counter-arguments directly instead of >posting the same thing over and over?>[usual proof deleted]Deleting out a mathematical argument does not make it go away.No, but you havenft said anything new in a while, and it has nothing to do with what I was talking about.>Ifve found the Ring of Objects which includes the ring of algebraic>integers, and does not have this problem, as the bfs are all included>in it.>The Ring of Objects is the set of all numbers where -1 and 1 are the>only members that are both a unit, i.e. factor of 1, and an integer,>where no non-unit member is a factor of any two integers that are>coprime.>How does the Ring of Objects differ from the algebraic integers?>Are you sure your de?ition is consistent?>In which ring are you referring to the two integers being coprime?> The Ring of Objects is a higher ring than the ring of algebraic> integers, which doesnft have its problems.Aside from the algebraic integers not doing what you think they should do, what problems do they have?The two integers are coprime in the Ring of Objects.Ok, so depending on what is in the Ring of Objects, 2 and 3 may or may not be coprime... how would I check this?> The de?ition removes the possibility of the fundamental> contradiction which distinguishes a ?ld from a ring, which is that> in a ?ld coprimeness does not exist, while in the ring of integers,> it does.Coprimeness exists in a ring, itfs just not very interesting.> So ?lds are at odds with the ring of integers.Considering the ring of integers is *not* a ?ld, this should not be surprising.> Ifve simply highlighted the crucial difference between them.You mean the whole point of this months of discussion has been about the fact that the ring of integers is not a ?ld? Am I missing something?> James Harris-- Will Twentyman === >The problem is that neither a_1, a_2, nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And thatfs false. I am pretty sure that Dale produced explicit common>factors; but in any case, your claim here is certainly false, since>their product is not coprime to 65.>Ifve proven it true that neither a_1, a_2 nor a_3 have ANY non-unit>factors in common with 5 in the ring of algebraic integers.>And he proves (below) that they do. At this point you have to do more >than point at your own proof. You must ?d the error in his.> He uses a special de?ition of coprime, and assumes that at least two> of the three numbers is coprime.>What de?ition of coprime would you suggest that he use? What >de?ition of coprime are you using?I believe that James incorrectly believes that a is coprime to bmeans a has no nonunit common factors with b. That de?ition isequivalent to the standard one in many rings (e.g., in the ring o?tegers, in the ring of all algebraic integers, in any PID), but notin others. The interesting point, of course, is that the de?ition James isusing, if it were not equivalent to the usual one, would be uselessfor his purposes. James needs the coprime property in order to removefactors in a congruence, but that step is not valid if you only havehave no nonunit common factors. You need coprime in the usual sense,with its equivalent statement that there are x and y such thatax+by=1. (If the two are equivalent in the ring in question, ofcourse, there is no problem; but they are not always equivalent).For instance, in Z[sqrt(-5)], the two statements are not equivalent. 2and 1+sqrt(-5) have no common factors other than units, but they arenot coprime; we have that (1+sqrt(-5))*(1-sqrt(-5)) = 0 (mod 2) inZ[sqrt(-5)], but we cannot conclude from the fact that 2 and1+sqrt(-5) have no factors in common that 1-sqrt(-5)=0 (mod 2); itclearly does not (all multiples of 2 are of the form 2a+2b*sqrt(-5), aand b integers).Rather, the usual argument is that if a*b=0 (mod c) and a and c arecoprime, then we know that c|ab. Since there exist elements x,y suchthat ax+cy = 1, multiplying by b we have abx + cby = b, and since cdivides abx and cby, it divides b; thus b=0 (mod c) is a validconclusion.If you want another example that James likes, as he points out(his conclusion is correct, although his reasoning is incorrect), 2and 6 are coprime in 2Z: the ideal (2) consists of all integersdivisible by 4, and is maximal; (6) consists of all integersdivisible by 12; so (6) is contained in (2). The only ideal thatcontains both (2) and (6) is (2); and (2) is not a prime ideal: 2*2lies in (2), but 2 does not. Thus, there is no prime ideal of 2Zcontaining both (2) and (6), so 2 and 6 are coprime (they are not,however, comaximal; the two conditions are equivalent in a ring with1, assuming the Axiom of Choice). However, even though 2 and 6 are coprime, we cannot cancel incongruence in this ring. 12=0 (mod 2), since 12=2*6. However, if wewrite 12=2*6=0 (mod 2) (in 2Z) and argue that 6 is coprime to 2, andtherefore we can cancel from the congruence, we would get that 2=0(mod 2) (in 2Z), and that statement is false: there is no element k in2Z such that 2=0+2k.So, in the end, the de?ition which James is using is not the one heshould be using for his purposes, unless he can prove that in the ringhe is working on it is equivalent to the usual one. Never mind thatthere is no ring of objects as he has de?ed it. [.snip.]>Which ring is that? If you mean your Ring of Objects, please explain >why sqrt(5) is a factor there and not in the algebraic integers.There is no Ring of Objects according to his de?ition. Evenallowing for the possibility that number means complex number oralgebraic number, there is no unique ring determined by theconditions he gives, yet he assumes there is one and only one such ring. [.snip.] === ====================================== === =========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu === He uses a special de?ition of coprime, and assumes that at least two>of the three numbers is coprime.No, I use the standard de?ition of coprime: two elements a and b are> coprime in R if and only if the ideals (a) and (b) are coprime, if and> only if there is no prime ideal of R which contains both (a) and (b).It is a THEOREM that in a commutative ring with 1, two elements a and> b are coprime if and only if there exist x and y in R such that> ax+by=1.Thatfs fascinating. Why donft you give the theorem, as it must be inerror for the ring of algebraic integers.And yes, I would like to see that theorem as Ifd be fascinated to seesome evidence of how deep the weird algebraic integer ring problem hasgone in terms of what mathematicians believe to be true.Still my guess is that it depends on assertions about units, where youbase your belief, once again, on assuming that a number that is not aunit in the ring of algebraic integers must be a non-unit factor inthe ring of algebraic integers.The problem is fascinating indeed, and easily leads to confusion, ifyou arenft careful to be *very* careful with your de?itions andassumptions.That is, you need to follow the math, not what you *think* is thetruth. > What is the de?ition YOU are using?Ifve given the following de?ition for coprime in the proper ring,which is the ring of objects:Two objects are coprime if they donft share a non-unit factor e.g. 2is coprime to 3, while 2 and 4 share 2 as a non-unit factor.> The de?ition you are using is a and b are coprime if and only if> they have no non-unit common factors.Yeah, thatfs basically it.> In the ring of all algebraic integers, the two are> EQUIVALENT. However, your de?ition is WEAKER than the correct> meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime> under your incorrect usage, but they are not coprime under correct usage.They are coprime in the ring Z[sqrt(-5)].Itfs strange to me that youfd miss something so obvious, as you couldalso consider something like the ring of evens, where 2 is coprime to6, in the ring of evens.Apparently, you wish to use the knowledge that in a *higher* ring1+sqrt(-5) and 2 share sqrt(2) as a factor, for the lesser ring.But just as 2 is a factor of 6 in the ring of integers, but is NOT inthe ring of evens, it simply is the case that in the ring Z[sqrt(-5)]1+sqrt(-5) and 2 are coprime, by the logical de?ition.What appears to have happened is that mathematicians consideredvarious rings, and tried to ?d some general way to handle them,rather than simply ?ding the object ring.Itfd be like someone ?dling around with the ring of evens, ratherthan just integers. Sometimes they put in 3, other times they put in7 with integers, when they really just need to put in 1, and be done.>However, by his de?ition of coprime, only one of the numbers is>coprime.Coprime to ->what<-?And prove it.One of the afs is coprime to f, and in particular 5, with f=sqrt(5),m=1.The proof follows.Consider, in the ring of algebraic integers, P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f).Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)where w_1 w_2 w_3 = f, and b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),and at m=0 P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), so two of the bfs must equal 0, which means P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)which is P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)proving that w_1 w_2 must be coprime to 3, if f is coprime to 3, whichleavesb_3 = 3, or a unit multiple of 3.Essentially objections to how f^2 divides off now come down toclaiming that the wfs are functions of m, but consider that w_1 w_2 =1, when m=0, if f is coprime to 3.But that was an arbitrary choice, so let f=3.Now it can be shown that w_1 w_2 is coprime to 3 without regard to m.That is, the wfs can now be shown to all be constant with regard to m,so they have the same value no matter what the value of m is, so theyare also constantwith f coprime to 3.Introducing a_1, a_2, and a_3, that is seen by considering P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)with f=3, as then you have P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 dividesoff as a constant so it is without a dependency on m.Now looking at P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)I can again check at m=0, to see P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),which forces w_3 = 3, or a unit multiple of 3.Therefore, the factorization is P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f = (b_1 x + u)(b_2 x + u)(b_3 x + uf)where youfll notice that the bfs are algebraic integers with m=1,f=sqrt(2), but thatfs a special case as generally they are not, whichshows a problem with the ring of algebraic integers. > [.snip.]>Lemma. Let R be the ring of all algebraic integers, and let a, b, c be>any elements of R. If a and b are coprime to c, then a*b is coprime to>c.>Proof. We use the characterization of coprime valid for commutative>rings with 1: a and b are coprime in R if and only if there exist x>and y in R such that ax+by = 1. > By that de?ition only *one* of the afs is coprime to 5, but none of> them has a factor in common with 5 either.> Do you even understand what hefs saying? The application would be: If > a_1 and a_2 are coprime to 5, then so is a_1*a_2. Extending it, if a_1, > a_2, and a_3 are coprime to 5, then a_1*a_2 and a_3 are coprime to 5, > which means a_1*a_2*a_3 is coprime to 5.>By the de?ition he gave only ONE of the afs is coprime to 5.Prove it.Ifve given the proof above. One of the afs is coprime by thatde?ition, while the others do not have non-unit factors in commonwith 5 in the ring.And in fact your example with Z[sqrt(-5)] outlines how, as similarly,in that ring 1+sqrt(-5) is, by a *logical* de?ition*, coprime to 2.However, by the de?ition youfve been using, it would be that1+sqrt(-5) and 2 do not share non-unit factors in the ringZ[sqrt(-5)]. >The others donft have non unit factors in common with 5, in the ring>of algebraic integers, but each have a factor that is sqrt(5) in a>higher ring.Which doesnft prove anything. If the others donft have non-unit> factors in common with 5 in the ring of all algebraic integers, then> they would be coprime under the correct de?ition as well.Actually, if the ring werenft ?hey *would* have non-unitfactors in common with 5. > [.snip.]> For those who donft understand, consider that in the ring of evens,> which does not have 1, you canft use that de?ition of coprime that> Arturo Magidin gives, though it is, interestingly enough, true that in> fact 2 is coprime to 6 in the ring of evens because 2(3) = 6, and 3 is> not in the ring.> Why do you make unnecessary side trips to bizarre rings?>Arturo Magidin uses a special de?ition of coprime which depends on>certain assumptions which fail with the ring of algebraic integers,I use the STANDARD de?ition; it is you who uses a SPECIAL> one. Please list, explicitly:1. The special de?ition;I quote from above.It is a THEOREM that in a commutative ring with 1, two elements a and b are coprime if and only if there exist x and y in R such that ax+by=1. > 2. The one you use;Ifve given the following de?ition for coprime in the proper ring,which is the ring of objects:Two objects are coprime if they donft share a non-unit factor e.g. 2is coprime to 3, while 2 and 4 share 2 as a non-unit factor. > 3. The certain assumptions on which my supposedly special de?ition> depends.My guess is that you assume that all numbers that *should* be factorsin the ring of algebraic integers are units, when in fact the ring is?in that certain numbers which *should* be units arenft in thering.These numbers are multiplicative inverses of other numbers that *are*in the ring of algebraic integers, but are not units, though they arecoprime to all non-unit algebraic integers.The ring of algebraic integers is ?in a fascinating way. > 4. A proof that they fail in the ring of algebraic integers.Well I just explained it, but Ifll repeat that the problem is thatcertain numbers which *should* be units in the ring, are not, andtheir multiplicative inverses arenft even in the ring, though theyshould be.> It is not enough to say the ring is really screwed up, because that> is supposed to be your conclusion, not your hypothesis.Ifve proven it repeatedly.Rather than work to get to the truth, Ifve seen posters work toconvince, presumably to protect their *beliefs* about mathematics as adiscipline.Their social preoccupation has been of interest to me.My fear is that many mathematicians actually are participants inwhat can be considered to be a dogmatic relgion, where rather thanchallenging what theyfre taught, math students learn to accept fromauthority.However, science is about challenging what is known to make certainthat it ?s with reality, not with giving your trust to particularhuman beings.Trust, but verify, as Ronald Reagan said, in one of the best quotesever.James Harris === >He uses a special de?ition of coprime, and assumes that at least two>of the three numbers is coprime.> No, I use the standard de?ition of coprime: two elements a and b are> coprime in R if and only if the ideals (a) and (b) are coprime, if and> only if there is no prime ideal of R which contains both (a) and (b).> It is a THEOREM that in a commutative ring with 1, two elements a and> b are coprime if and only if there exist x and y in R such that> ax+by=1.>Thatfs fascinating. Why donft you give the theorem, as it must be in>error for the ring of algebraic integers.>And yes, I would like to see that theorem as Ifd be fascinated to see>some evidence of how deep the weird algebraic integer ring problem has>gone in terms of what mathematicians believe to be true.>Still my guess is that it depends on assertions about units, where you>base your belief, once again, on assuming that a number that is not a>unit in the ring of algebraic integers must be a non-unit factor in>the ring of algebraic integers.You are speaking nonsense. A non unit factor of ->what<-?If you again mean, a non-unit factor of itself, then there is nobelief, there is only mathematical truth: by de?ition, x is a factorof y in the commutative ring R if and only there exists an element zof R such that x*z=y. If a is an algebraic integer which is not a unitin the ring of all algebraic integers, then since 1 is an algebraicinteger and a*1=a, it follows, necessarily, that a is a non-unitfactor of itself in the ring of all algebraic integers.Ifll include the de?itions, since I doubt you know them.DEF. Let R be a ring. A subset I of R is an ideal of R if and onlyif: (i) I is nonempty; (ii) If x and y are in I, then so is x-y. (iii) If x is in I and r is any element of R, then rx is in I and xr is in I.Note that if the ring is commutative, it is enough to check that rx isin I.Lemma 0. If I is an ideal of R, then 0 is in I; and if a,b are in I,then a+b is in I.Proof. I is nonempty, so there is an a in I. Since 0 is in R, 0*a=0lies in I by (iii). So 0 is in I.If a,b are in I, then -b=0-b lies in I by (ii). Since a,-b lie in I,then a-(-b)=a+b lies in I by (ii). QED.DEF. Let R be a commutative ring, and let a be an element of R. The principalideal generated by a is the collection { ra : r in R}. It is denotedby (a).Lemma 1. Let R be a commutative ring. Then (a) is an ideal of R. Proof. Since R is a ring, it contains a 0. Therefore, 0*a=0 is in (a), so (a) is nonempty. If ra and sa are in (a), then ra-sa=(r-s)a, which by de?ition is in I. Finally, if ra is in (a), and s is any element of R, then s*(ra) = (sr)a, which lies in (a). QEDDEF. Let R be a ring. An ideal I of R is said to be a prime ideal ifand only if it is not equal to R, and for any two elements x and y ofR, if x*y is in I, then either x is in I or y is in I.DEF. Let R be a commutative ring. Two ideals I and J of R are said tobe coprime (in R) if and only if there is no prime ideal P of R whichcontains both I and J.DEF. Let R be a commutative ring. Two elements a and b of R are saidto be coprime (in R) if and only if the principal ideals (a) and (b)are coprime in R.DEF. Let R be a commutative ring. An ideal I of R is said to be amaximal ideal if and only if I is not equal to R, and there is noideal J of R which properly contains I and is properly contained in R.DEF. Let R be a commutative ring. Two ideals I and J of R are said tobe comaximal if and only if there does not exist a maximal ideal ofR which contains both I and J.Lemma 2. Let I be an ideal of R, and let a be an element of R not inI. Then the set { x + ra : x in I, r in R}is an ideal of R, and moreover, it is the smallest ideal of R whichcontains I and a.Proof. Let S denote the set {x+ra : x in I, r in R}.First, since I is an ideal, it is nonempty. Let x in I. Then x+0*a isin S, so S is nonempty.Assume that b,c are in S. Then b=x+ra for some x in I, r in R; andc=y+sa for some y in I, s in R. Thenb-c = (x+ra) - (y+sa) = (x-y) + (r-s)a.Since I is an ideal, x-y is in I; and r-s is in R. So this is anelement of S; thus, if b and c are in S, then so is b-c.Finally, let y be any element of S and let r in R. Then we may writey=x+sa with x in I and s in R. Thenr*y = r*(x+sa) = (r*x) + r*sa = (rx) + (rs)*a.Since I is an ideal, rx is in I; so this is an element ofS. Therefore, S is an ideal.Now assume that J is any ideal which contains I and contains a. Weprove that it contains S. For let x in I be arbitrary and let r inR. We want to show that x+ra is in J. Since J is an ideal, and a is inJ, it follows that ra is in J. Since J is an ideal, and ra is in J, itfollows that 0=0*ra is in J. Since 0 and ra are in J, it follows that-ra = 0-ra is in J. Since x is in J and -ra is in J, it follows thatx-(-ra)=x+ra is in J. This proves that J contains S, and proves theLemma. QED. THEOREM 3. Let R be a commutative ring with 1. If I is a maximal idealof R, and a is any element of R not in I, then there exists an elementx in I and an element r in R such that x+ra=1.Proof. Let S={x+ra : x in I, r in R}. This set is an ideal, andcontains I. It also contains a, since a=0+1a. Therefore, the setproperly contains I. Since I is maximal, the only ideal which properlycontains I is R itself, so S=R. Since 1 is in R, we conclude that 1 isin S, and therefore, 1 = x+ra for some x in I and r in R, asclaimed. QED.THEOREM 4. Let R be a commutative ring with 1. If I is a maximal idealof R, then it is also a prime ideal of R.Proof. Assume not. Then there exist elements a,b of R, none of them inI, such that ab lies in I.Since a is not in I, by Theorem 3 there exist x in I and r in R suchthat 1 = x+ra. Multiplying by b we haveb = b*1 = b(x+ra) = bx + bra = (bx) + r(ba).Now, x is in I, hence so is bx; likewise, ba is in I, hence so inr*ba. Therefore, bx+r(ba) lies in I. But that means that b is in I,contradicting the choice of a and b. Therefore, I must be a primeideal. QEDTHEOREM 5. (Zornfs Lemma) If C is a nonempty partially ordered set with theproperty that any chain is bounded above, then C has maximal elements.Proof omitted. Known to be equivalent to the Axiom of Choice.Lemma 6. Let R be a commutative ring with 1, and let I be an ideal. IfI contains 1, then I=R.Proof. We know I is contained in R. To prove the reverse inclusion,let x in R. We need to show it is in I. Since 1 in I and x in R, x*1=xmust lie in I, so x is in I. QED.THEOREM 7. Let R be a ring with 1. If I is any ideal of R and is notequal to R, then there exists a maximal ideal M of R which contains I.Proof. Let S= { J ideal of R: J is not equal to R and J contains I}.Order S by inclusion. Then I is in S, so S is nonempty. Let C be achain in S, that is, C is contained in S and C={J_k}, with theproperty that for any indices m,n, either J_m is contained in J_n orJ_n is contained in J_m. If C is empty, I is an upper bound forC. Assume C is nonempty. Let J be the set theoretic union of C. Then Jcontains I, since each element of J contains I. So it is nonempty.If a,b in J, then a in J_k and b in J_n for some k and n. Either J_kis contained in J_n (and so J_n contains both a and b) or J_ncontained in J_k (so J_k contains both a and b). So assume withoutloss of generality that J_n contains both a and b. Since J_n is anelement of S, it is an ideal, so a-b is in J_k, and so a-b is in J.Finally, if a in J, and r in R, then a in J_k for some k, and sincethis is an ideal, ra is in J_k, so ra is in J. Therefore, J is anideal of R which contains S.J contains all elements of C, so it is a bound for C. We just need toshow that it lies in S; the only way in which it would not lie in S isif it violated the condition J is not equal to R. But that wouldmean that J=R, so 1 is an element of J. Since J is the set-theoreticunion of the J_k, so 1 must be in some J_k. But that would imply, byLemma 6, that J_k=R; this contradicts the fact that J_k is an elementof S. The contradiction arose from assuming that R=J, so J is notequal to R, so J lies in S. In conclusion, S satis?s the hypothesis of Zornfs Lemma, and hasmaximal elements. Let M be a maximal element of S.That means that: (1) M is an ideal of R; (2) M is not equal to R; and(3) M contains I [all this by virtue of being in S]. Finally, it alsomeans that there does not exist any ideal J of R, different from R,which contains I, contains M, and is not equal to M [by virtue ofbeing maximal in S]. We need only show that M is maximal to prove the theorem. But if M isnot maximal, then there exists an ideal K of R, different from R,which properly contains M. But then it would necessarily contain I,and so be an element of S larger than M. This contradicts themaximality of M in S, and so we conclude that M is a maximal idealthat contains I. QEDTHEOREM 8. Let R be a commutative ring with 1. Two ideals I and J arecoprime if and only if they are comaximal.Proof. If I and J are coprime, and M is a maximal ideal, then M cannotcontain both I and J, for then M would also be a prime idealcontaining both I and J, which is impossible. So I and J arecomaximal.Conversely, suppose that I and J are comaximal. If P is a prime idealcontaining both I and J, then by Theorem 7 there exists a maximalideal M containing P, and therefore containing I and J, contradictingcomaximality. So I and J are coprime. QEDTHEOREM 9. Let R be a commutative ring with 1, and let a and b be twoelements of R. Then a and b are coprime in R if and only if thereexist x and y in R such that ax+by=1.Proof. First, assume that there exists x and y in R such thatax+by=1. Note that since R has a 1, a lies in (a) (being equal to1*a), and b lies in (b). So if P is any ideal containing both (a)and (b), then it contains both a and b as elements. Since a is in P,then ax must be in P; since b is in P, then b*y is in P. Since ax andby are both in P, we must have that ax+by is in P. But 1=ax+by, so 1is in P. Therefore, P=R. In particular, there is no ideal other than R which contains both (a)and (b); therefore, there is no maximal ideal which contains both (a)and (b); therefore, by de?ition, (a) and (b) are comaximal. ByTheorem 8, they are also coprime. Therefore, a and b are coprime in R.Conversely, assume that a and b are coprime in R. That means that (a)and (b) are coprime, hence comaximal by Theorem 8.Let I = {ra + sb : r,s in R}. We claim that this is an ideal of Rwhich contains both (a) and (b). Indeed, it is an ideal, by the sameargument as Lemma 2: it is equal to the set of all x+sb with x in (a)and s in R. It contains (a) because if ra is in (a), then ra = ra+0*b lies in I; and it contains (b) because if sb is in (b),then sb = 0*a+sb which is in I.Therefore, I is an ideal which contains both (a) and (b). If I is notequal to R, then by Theorem 7 there exists a maximal ideal M of Rwhich contains I; therefore, M contains (a) and (b). This contradictsthe comaximality of (a) and (b). The contradiction arises fromassuming that I is not equal to R, so we must conclude thatI=R. Therefore, 1 is an element of R, so 1 is in {ra+sb : r,s inR}. Therefore, there exist x and y in R such that 1 = xa+yb, asclaimed. QEDDone.>The problem is fascinating indeed, and easily leads to confusion, if>you arenft careful to be *very* careful with your de?itions and>assumptions.Which you are not.>That is, you need to follow the math, not what you *think* is the>truth.> What is the de?ition YOU are using?>Ifve given the following de?ition for coprime in the proper ring,>which is the ring of objects:There is no ring of objects. Your de?ition does not have a referent,as I have pointed out many times.This de?ition is in general weaker than the usual one, but isequivalent to the usual one in the ring of all algebraic integers.DEF. Let R be a commutative ring, a and b elements of R. We say thata is a factor of b (in R) if and only if there exists c in R such thatac=b.DEF. Let R be a commutative ring with 1, and let u in R. Then u is aunit in R if and only if there exists v in R such that u*v=1.DEF. Let R be a commutative ring, a,b,c elements of R. We say that cis a common factor of a and b (in R) if and only if c is a factor ofa in R and c is also a factor of b in R.THEOREM. Let R be a ring commutative ring with 1, and let a and b inR. If a and b are coprime in R, then any common factor of a and b in Ris a unit.Proof. Since a and b are coprime in R, there exist x,y in R such thatax+by=1. Let c in R be a common factor of a and b.Then there exists d in R such that cd=a; and there exists e in R suchthat ce=b. Therefore, cdx=ax, and cey=by. Therefore,1 = ax+by = cdx + cey = c(dx+ey).Since R is a ring, dx and ey are in R, and so is dx+ey. Therefore,dx+ey is an element of R which multiplied by c yields 1. Therefore, cis a unit in R. QEDEXAMPLE. A commutative ring with 1 R in which two elements have nocommon factor other than units, but are not coprime.Proof. Let R = Z[sqrt(-5)], a = 2, b= 1+sqrt(-5).First, we show that a and b are not coprime. For assume that we hadx,y in R such that ax+by = 1. Write x = r+s*sqrt(-5), y =t+u*sqrt(-5) with r,s,t,u integers. Then1 = 2x+(1+sqrt(-5))y = 2r + 2s*sqrt(-5) + t + u*sqrt(-5) -5u +t*sqrt(-5) = (2r + t - 5u) + (2s+u+t)*sqrt(-5).Therefore, 2r+t-5u = 1 2s+u+t = 0Substituting t=-2s-u, we have1 = 2r -2s -u -5u = 2r - 2s - 6u.But since r,s,u are integers, that means that 2r-2s-6u is an eveninteger, and therefore cannot equal 1. Therefore, a and b are notcoprime in R.To prove that they have no common factors, consider the functionN:Z[sqrt(-5)]->Z given byN(a+b*sqrt(-5)) = a^2 + 5b^2, where a and b are integers.Then it is easy to verify by direct calculation that for all x,y in Z[sqrt(-5)]N(xy) = N(x)N(y). Assume that x is a factor of y in Z[sqrt(-5)]. Then there exists z inZ[sqrt(-5)] such that xz=y. Therefore, N(y) = N(xz) =N(x)N(z). Therefore, the integer N(y) is a product of the integersN(x) and N(z), so N(x) is a factor of N(y) IN THE INTEGERS.That is: if x is a factor of y in Z[sqrt(-5)], then N(x) is a factorof N(y) in Z.So, assume that a+b*sqrt(-5) is a common factor of 2 and1+sqrt(-5). Then N(a+bsqrt(-5)) = a^2 + 5b^2 must be a factor, IN THEINTEGERS, of N(2) = 4 and of N(1+sqrt(-5)) = 1+1*5 = 6.The only possibilities are that N(a+b*sqrt(-5)) = 2 or 1 (note N(x) isalways nonzero).But 2 = a^2+5b^2 has no solution with a and b integers, so we musthave N(a+b*sqrt(-5)) = 1. The only solutions to 1 = N(a+b*sqrt(-5)) = a^2 + 5b^2 are b=0, a=1; or b=0, a=-1. So eithera+b*sqrt(-5)=1, hence is a unit; or a+b*sqrt(-5)=-1, hence is aunit. This proves that(1) 2 and 1+sqrt(-5) are not coprime in Z[sqrt(-5)]; but(2) 2 and 1+sqrt(-5) have no non unit common factors in Z[sqrt(-5)]. This proves that in general the two notions are not equivalent; sincecoprime implies no common factors other than units, but the conversedoes not hold, no common factors other than units is a weakerproposition than coprime, in general.The only thing that would remain is to prove that the two notions areequivalent in the ring of all algebraic integers. This is a deeptheorem due to Dedekind and Kronecker, and uses the ?iteness of theclass number. It has been sketched before, but here are the broadstrokes:First, let R be the ring of all algebraic integers. For any a and b inR, there exists c in R such that(a,b) = {xa + yb : x, y in R} = (c)To prove this, let A be the ring of integers of Q(a,b). Then thereexists a positive integer k such that (a,b)^k = (d) for some d in A(?iteness of the class number). Let Af be the ring of integers ofQ(a,b,d^{1/k}), where d^{1/k} is any of the complex k-th roots ofd. In Af, we have (a,b)^k = (d) = (d^{1/k})^k, so by uniquefactorization of ideals in rings of integers, we haveequal, so letting c=d^{1/k} gives c.Let a and b be any two algebraic integers, and assume that a and bhave no common non-unit factors in the ring of all algebraicintegers. Let c be an algebraic integer such that (a,b)=(c). Then cis a common factor of a and b: for a is in (a,b), and so a is in(c)={rc: r in R}; that means, there exists r in R such thatrc=a. Likewise, b is in (a,b), hence in (c), and so there exists s inR such that cs=b. So c is a common factor of both a and b. Since a andb have no common factors other than units, that means that c is a unitin the ring of algebraic integers. That means that there exists v in Rsuch that vc=1. But c is in the ideal (c), and v is in R, so vc=1 isin (c). Since (c) contains 1, it follows that (c)=R.Therefore, (a,b) = {xa+yb : x,y in R} = (c) = R, so 1 is in(a,b). Therefore, there exist x and y in R such that ax+by=1. Therefore, a and b are coprime in R, as claimed. [.rest deleted.]> In the ring of all algebraic integers, the two are> EQUIVALENT. However, your de?ition is WEAKER than the correct> meaning of the term: in Z[sqrt(-5)], 1+sqrt(-5) and 2 are coprime> under your incorrect usage, but they are not coprime under correct usage.>They are coprime in the ring Z[sqrt(-5)].Not under the correct usage.>Itfs strange to me that youfd miss something so obvious, as you could>also consider something like the ring of evens, where 2 is coprime to>6, in the ring of evens.They are coprime under the usual de?ition, and under yourbastardization thereof. However, in the ring Z[sqrt(-5)], they arecoprime under your bastardization, but not under the correct usage. [.snip.]>However, by his de?ition of coprime, only one of the numbers is>coprime.> Coprime to ->what<-?> And prove it.>One of the afs is coprime to f, and in particular 5, with f=sqrt(5),>m=1.>The proof follows.>Consider, in the ring of algebraic integers,> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f).>You have no u in your assertions before; what is u?>Now using b_1, b_2, b_3, w_1, w_2, and w_3, I have the factorization> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)You have not proven that b1, b2, b3, u, w1, w2, w3 are algebraicintegers.>where w_1 w_2 w_3 = f, and > b_1 b_2 b_3 = (m^3 f^4 - 3m^2 f^2 + 3m),>and at m=0> P(0)/f^2 = 3xu^2 + u^3 f = u^2(3x + uf), >so two of the bfs must equal 0, which meansWHEN m = 0.> P(0)/f^2 = w_1 w_2 u^2 (b_3 x + u w_3)>which is> P(0)/f^2 = u^2 (b_3 w_1 w_2 x + u f) = u^2(3x + uf)WHEN m = 0.>proving that w_1 w_2 must be coprime to 3YOu have not proven that w1*w2 is an algebraic integer, so thisstatement is unsupported right now.>if f is coprime to 3, which>leaves>b_3 = 3, or a unit multiple of 3.WHEN m = 0.You have, however, not proven that, under my de?ition, they arecoprime. >Essentially objections to how f^2 divides off now come down to>claiming that the wfs are functions of m, but consider that w_1 w_2 =>1, when m=0, if f is coprime to 3.This is nonsense and is based on your misunderstanding of obejctions.>But that was an arbitrary choice, so let f=3.WHEN m = 0.>Now it can be shown that w_1 w_2 is coprime to 3 without regard to m.Please do so, using the correct de?ition of coprime, since youclaim that under my de?ition you will do so. Be sure to prove thatw_1*w_2 is an algebraic integer while you do so.>That is, the wfs can now be shown to all be constant with regard to m,Please do so.>so they have the same value no matter what the value of m is, so they>are also constant>with f coprime to 3.Please provide a complete proof. See above for what a real proof looks like.>Introducing a_1, a_2, and a_3, that is seen by considering> P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f) => (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)You have not proven that a1, a2, a3 are algebraic integers.>with f=3, as then you have> P(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - > (-1+mf^2 )x u^2 + u^3 => (a_1/3 x + u)(a_2/3 x + uf)(a_3/3 x + u)>so I have a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3 where the 3 divides>off as a constant so it is without a dependency on m.when f=3. You have not shown that a_1/3, a_2/3, a_3/3 are algebraic integers.>Now looking at> P(m)/3^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)>I can again check at m=0, to see> P(0)/3^2 = u^2 (b_3 w_1 w_2 x + 3u) = u^2(3x + 3u),>which forces w_3 = 3, or a unit multiple of 3.When m=0.>Therefore, the factorization is> P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - > 3(-1+mf^2 )x u^2 + u^3 f => (b_1 x + u)(b_2 x + u)(b_3 x + uf)>where youfll notice that the bfs are algebraic integers with m=1,I donft notice that; I assume, by the way, that you screwed up above> a_1 = b_1/3, a_2 = b_2/3, a_3 = b_3/3you meant> b_1 = a_1/3, b_2 = a_2/3, b_3 = a_3/3.>f=sqrt(2), but thatfs a special case as generally they are not, which>shows a problem with the ring of algebraic integers.YOu have analized a number of special cases, and said a lot of things,most of which are not supported. In addition, you CLAIMED that youwould prove that, under my de?ition (i.e., the standardde?ition), the ai would be shown to be coprime to 5 for thepolynomial in question, 65x^3 - 12x + 1.I do not see the values of r_i and s_i such that r_i*a_i + s_i*5 = 1,which is the de?ition you claimed you were going to use.So all you did was waste space and my time. [.snip.]>By the de?ition he gave only ONE of the afs is coprime to 5.> Prove it.>Ifve given the proof above. One of the afs is coprime by that>de?ition, I do not see the elements r and s that make r*a+s5 = 1. Thatfs whatthe de?ition says, after all. PROVE IT.>while the others do not have non-unit factors in common>with 5 in the ring.You have not proven anything.>And in fact your example with Z[sqrt(-5)] outlines how, as similarly,>in that ring 1+sqrt(-5) is, by a *logical* de?ition*, coprime to 2.The STANDARD DEFINITION gives that 1+sqrt(-5) and 2 are NOT coprime inZ[sqrt(-5)]. Ifve proven it.>However, by the de?ition youfve been using, it would be that>1+sqrt(-5) and 2 do not share non-unit factors in the ring>Z[sqrt(-5)].That is true regardless of whether you use my de?ition ofcoprime (i.e., the correct one), or your de?ition of coprime(i.e., the bastardization). The point is that this property does not imply coprimality, using thecorrect de?ition.>The others donft have non unit factors in common with 5, in the ring>of algebraic integers, but each have a factor that is sqrt(5) in a>higher ring.> Which doesnft prove anything. If the others donft have non-unit> factors in common with 5 in the ring of all algebraic integers, then> they would be coprime under the correct de?ition as well.>Actually, if the ring werenft ?they *would* have non-unit>factors in common with 5.You are assuming your conclusion. Thatfs called a ciruclar argument.>Arturo Magidin uses a special de?ition of coprime which depends on>certain assumptions which fail with the ring of algebraic integers,> I use the STANDARD de?ition; it is you who uses a SPECIAL> one. > Please list, explicitly:> 1. The special de?ition;>I quote from above.>It is a THEOREM that in a commutative ring with 1, two elements a and> b are coprime if and only if there exist x and y in R such that> ax+by=1.You see the big capitalized word THEOREM? You did not give the special de?ition I supposedly use. > 2. The one you use;>Ifve given the following de?ition for coprime in the proper ring,>which is the ring of objects:There is no ring of objects. Your de?ition is incoherent as written,and has no referent when ?ed.>Two objects are coprime if they donft share a non-unit factor e.g. 2>is coprime to 3, while 2 and 4 share 2 as a non-unit factor.This is not the correct de?ition of coprime.> 3. The certain assumptions on which my supposedly special de?ition> depends.>My guessYou ->guess<-? But you made a very de?itive ASSERTION above. Are you trying to saythat when you asserted that I was depending on some assertions, youdid not KNOW that to be the case, you were merely ->guessing<-?Is that how you work? [rhetorical question: yes, it is how you work]>is that you assume that all numbers that *should* be factors>in the ring of algebraic integers are units, when in fact the ring is>?in that certain numbers which *should* be units arenft in the>ring.All this should really means numbers which I WOULD REALLY, REALLY,REALLY like to have the properties I want so my proof wouldnft benonsense. [.snip.]>These numbers are multiplicative inverses of other numbers that *are*>in the ring of algebraic integers, but are not units, though they are>coprime to all non-unit algebraic integers.If the multiplicative inverses are NOT algebraic integers, then youcannot talk about them being coprime to algebraic integers in the ringof algebraic integers. If they are multiplicative inverses ofsomething in your ring, then they are units, and therefore coprime toverbiage.>The ring of algebraic integers is ?in a fascinating way.So you claim.> 4. A proof that they fail in the ring of algebraic integers.>Well I just explained it, but Ifll repeat that the problem is that>certain numbers which *should* be units in the ring, are not, and>their multiplicative inverses arenft even in the ring, though they>should be.Translation: numbers that you ->fervently wish<- were there are not,and thatfs a problem. For you.> It is not enough to say the ring is really screwed up, because that> is supposed to be your conclusion, not your hypothesis.>Ifve proven it repeatedly.Youfve given a lot of verbiage, but you have proven nothing. [.snip.] === ===================================== === ==========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===> ...Ifd be fascinated to see...and> ...The problem is fascinating indeed, ...and> ...is ?in a fascinating way...James. Enough with the fascinating. Please drop the Yup., Um., odd, amazing, etc. as well. Ifyour vocabulary needs work, try remedial English.[snip]> Well I just explained it, but Ifll repeat that the problem is that> certain numbers which *should* be units in the ring, are not, and> their multiplicative inverses arenft even in the ring, though they> should be.Fine. Name one. Give us a number which *should* be a unit in the ring of algebraic integers, but which isnot. Just one. Pretty please! For Godfs sake man, name just ONE numeric quantity which supports yourclaim!--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 === Ifm working on an optimization problem where my objective function isthe sum of two functions: a strictly convex function and a strictlyquasiconvex function. Both functions are continuous anddifferentiable.In general, the sum of quasiconvex functions is not necessarilyquasiconvex, but are there any conditions where the sum turns out tobe quasiconvex. Particularly in the case of adding a convex functionand quasiconvex function, under what circumstances could I claim thesum to be quasiconvex?Any pointers/references in this regard would be most appreciated.Karthik === > Ifm working on an optimization problem where my objective function is> the sum of two functions: a strictly convex function and a strictly> quasiconvex function. Both functions are continuous and> differentiable.In general, the sum of quasiconvex functions is not necessarily> quasiconvex, but are there any conditions where the sum turns out to> be quasiconvex. Particularly in the case of adding a convex function> and quasiconvex function, under what circumstances could I claim the> sum to be quasiconvex?Please de?e quasiconvex and strictly quasiconvex. === > ...start with a dodecagon (12-gon) [...] make a path that moves> from vertex to vertex [...] visiting each vertex exactly one time.> And the path returns to the starting-point. But in this puzzle, consecutive vertexes MAY be connected by a segment.> So, I give a list of nonnegative integers below. As the path is drawn> (as opposed to after the path is completed), the n_th segment crosses> a(n) previously drawn segments, where a(n) is the n_th term of the> integer-list.> The path starts at 12. And the ?st segment goes from 12 to 8.> The list {a(n)}: 0, 0, 1, 0, 2, 2, 0, 3, 2, 2, 5, 2> ...> 17 of the 10! paths that start off {12, 8} match that crossing-> count sequence, so solutions seem fairly rare - about 1 per 213000> paths - but on the other hand, there are perhaps O((n-3)!) possible> crossing-count sequences, so 17 could instead be an unusually high> prevalence, and this case no more rare than thousands of others.> I plan to look at this more and ?d out, next weekend.> -jiw> ... > I wonder which sequences (if any), for the 12-gon, produce one> solution, but are interesting.> If you, or anyone, happens to ?d such a sequence, then it would be> interesting to post it to sci.math and rec.puzzles as a challenge for> us all.> ...For the 12-gon there are 893597 different sequences that begin 12,8,> and only 315027 of them belong to unique paths; I donft know if any> of those paths are interesting. :) For example, the> sequential-crossing-counts list 0 0 0 1 1 1 1 0 6 7 8 2 belongs to > a unique path. On the other hand, the list 0 0 0 0 1 2 2 3 4 4 5 4> belongs to 164 different paths, and no list belongs to more than that.paths account for 80% of all sequences. Sequences belonging to 17 > paths (like your original sequence, 0 0 1 0 2 2 0 3 2 2 5 2) are> relatively rare, around 1 in 300. I think my estimate of O((n-3)!) possible crossing-count sequences> was high and that O((n-4)!) is much closer. If I have a chance,> Ifll modify my program to look at all paths starting at 12, rather> than only those that start at 12 and proceed to 8. That may have > a somewhat less arbitrary complexity.Here is a slightly-revised extract from http://pat7.com/jp/counts-r512 :> Occ.Count #Occ.Count Extension An example count sequence> 1 315027 315027 0 0 0 1 1 1 1 0 6 7 8 2> 2 179162 358324 0 0 0 1 1 1 1 0 6 0 7 7> 3 102670 308010 0 0 0 1 1 1 1 0 0 0 1 7> 4 71484 285936 0 0 0 1 1 1 0 0 0 1 1 1> The ?st column is how many paths a pattern belongs to.> The second column is how many patterns belong to that number of> paths. The third column is the product of the ?st two, or> sum of products when the ?st column is a range. The last> group of columns shows an example of a count sequence that > belongs to the number of paths given in column 1.> (In the ?e, there is also an Eg# column that shows program> code numbers of the example count sequences. Also, the ?e > shows similar data for 5, 6, ... 11-gons as well as 12-gons.)> -jiwInteresting. So a puzzle (which *I* could NEVER solve, most probably) forrec.puzzles and sci.math readers is to ?d the solution for the pathfound by Jamesf computer, the *unique* path which follow the rules ofthe original puzzle and has the crossing-count sequence:0 0 0 1 1 1 1 0 6 7 8 2;)(I have no idea personally if this path is at all interesting from apuzzle view-point..)Leroy Quet === Ifm trying to ?ure out this problem, but I canft ?d anyinformation (in my book or on the internet) thatfs related.Ifm asked to solve the following system using the elimination methoddx/dt = 2x - ydy/dt = -x + 2yI appreciate any helpthanks..Boris.. === L1+L2d(x+y)/dt=x+yL1-L2d(x-y)/dt=3(x-y )you get x+y and x-y terminated> Ifm trying to ?ure out this problem, but I canft ?d any> information (in my book or on the internet) thatfs related.> Ifm asked to solve the following system using the elimination method> dx/dt = 2x - y> dy/dt = -x + 2y> I appreciate any help> thanks> ..Boris.. === > L1+L2d(x+y)/dt=x+yL1-L2d(x-y)/dt=3(x-y)you get x+y and x-y terminatedThatfs a nice way to do it. It may not be what the OP meant by the elimination method. The OP may have meant, write the ?st equation in the form y = stuff involving only x and t, then differentiate with respect to t to get dy/dt = other stuff involving only x and t, then in the second of the original equations replace y and dy/dt by the respective stuffs involving only x and t. This procedure eliminates y and leaves a 2nd order equation for x, which OP has presumably already learned how to solve. > Ifm trying to ?ure out this problem, but I canft ?d any> information (in my book or on the internet) thatfs related.> Ifm asked to solve the following system using the elimination method> dx/dt = 2x - y> dy/dt = -x + 2y-- === L1+L2d(x+y)/dt=x+yL1-L2d(x-y)/dt=3(x-y)you get x+y and x-y terminatedThatfs a nice way to do it. It may not be what the OP meant by > the elimination method. The OP may have meant, write the ?st > equation in the form y = stuff involving only x and t, then differentiate with respect to t to get dy/dt = other stuff involving only x and t, then in the second of the original equations replace y and dy/dt > by the respective stuffs involving only x and t. This procedure > eliminates y and leaves a 2nd order equation for x, which OP has > presumably already learned how to solve. Ifve been refreshing myself a bit on ODEs lately. Some texts de?e adifferential operator D, wherebyThe OPfs ?st equationdx/dt = 2x - ycan be writtenDx = 2x - yor(D-2)x + y = 0The second equationdy/dt = -x + 2ysimilarly would be writtenx + (D-2)y = 0The resulting system(D-2)x + y = 0 x + (D-2)y = 0can be readily solved by systematic elimination for x and y.The results will be the characteristic equations of second order linearequations expressed as a linear second order differential operator.Essentially the same method described by Myerson above.I fear I may get pulverized by PDE in the Fall after my long math haitus, soI thought Ifd introduce myself to the group.Andrew> Ifm trying to ?ure out this problem, but I canft ?d any> information (in my book or on the internet) thatfs related.> Ifm asked to solve the following system using the elimination method> dx/dt = 2x - y> dy/dt = -x + 2y === > Point is ASU, USA mathematicians have uni?d all of the known (and> previously unknown) primary laws (equations) of physics for my four> dimensional space time physiker friends, by simple multiplication> and division of the nine (9) primary universal base unit values.> In short, the physiker boneheads could not unify the foundation of> all their published theories, so the ASU, USA mathematicians did.> -- The 9 Primary Universal Base Unit Values...1) mass... (hc/G)^1/2 = 5.4563026(39) x 10^-8 kg> 2) length... (hG/c^3)^1/2 = 4.0507625(38) x 10^-35 m> 3) time... (hG/c^5)^1/2 = 1.3511889(41) x 10^-43 s> 4) current... e/(hG/c^5)^1/2 = 1.1857530(90) x 10^24 A> 5) temperature... b/(hG/c^3)^1/2 = 3.5518626(82) x 10^32 K> 6) substance... M/(hc/G)^1/2 = 1.6605387(31) x 10^-27 kmol> 7) intensity... (hG/c^5)^1/2/sr = 1.9720204(06) x 10^-45 cd> 8) Al Einstein solid angle = 6.8517999(97) x 10^1 sr> 9) Max Planck plane angle = 1.3703599(97) x 10^2 radmultiplied and divided by themselves, up to> and including their fourth (4th) power yield... The Primary Universal Physical Constants001) irradiance constant i. = 4.5211591(52) x 10-122 s^3/kg> 002) radiance constant i.= 3.0978078(26) x 10-120 s^3-sr/kg> 003) radiant volume constant = 1.3554094(15) x 10^-113 m-s^2/kg> 004) measurement volume = 6.6467654(65) x 10^-104 m^3> 005) graviton volume constant = 1.2181812(31) x 10^-96 m^3/kg> 006) luminous ef?acy = 3.7229937(53) x 10^-96 cd-sr-s^3/kg-m^2> 007) electric current volume = 1.3838190(49) x 10^-93 m^2/A> 008) luminous energy = 1.8257115(55) x 10^-86 cd-sr-s> 009) electric charge volume = 4.1485851(42) x 10^-85 m^3/A-s> 010) molar volume = 4.0027765(33) x 10^-77 m^3/kmol> 011) moment of inertia = 8.9530708(38) x 10^-77 kg-m^2> 012) graviton ?y = 1.0031235(26) x 10^-70 m-s/kg> 013) measurement area = 1.6408677(14) x 10^-69 m^2> 014) electric moment = 6.4900363(91) x 10^-54 A-s-m> 015) Euclid capacitance = 5.234567901... x 10^-48 A^2-s^4/kg-m^2> 016) magnetic moment = 1.9456639(62) x 10^-45 A-m^2> 017) luminous intensity = 1.9720204(06) x 10^-45 cd> 018) graviton frequency i. = 1.3511889(41) x 10^-43 s> 019) luminous ?1.3511889(41) x 10^-43 cd-sr> 020) graviton moment = 2.2102186(33) x 10^-42 kg-m> 021) inductance constant = 3.4877980(18) x 10^-39 kg-m^2/A^2-s^2> 022) absorption-emission = 2.4763819(58) x 10^-36 s/kg> 023) graviton wavelength = 4.0507625(38) x 10^-35 m> 024) Planck constant = 6.6260687(65) x 10^-34 kg-m^2/s> 025) relative expansion = 2.8154241(58) x 10^-33 /K> 026) electric resistivity = 1.0456155(41) x 10^-30 kg-m^3/A^2-s^3> 027) uni?d substance = 1.6605387(31) x 10^-27 kmol> 028) kinematic viscosity = 1.2143880(58) x 10^-26 m^2/s> 029) inverse electric current = 8.4334589(42) x 10^-25 /A> 030) Boltzmann constant= 1.3806502(93) x 10^-23 kg-m^2/s^2-K> 031) thermal resistance = 9.7866124(96) x 10^-21 s^3-K/kg-m^2> 032) graviton molality = 3.0433405(93) x 10^-20 kmol/kg> 033) elementary charge = 1.6021764(62) x 10^-19 A-s> 034) primary radiation = 5.9552136(16) x 10^-17 kg-m^4/s^3> 035) speci? heat = 2.5303770(42) x 10^-16 m^2/s^2-K> 036) magnetic ? = 2.0678336(42) x 10^-15 kg-m^2-sr/A-s^2-rad> 037) magnetic ?4.1356672(77) x 10^-15 kg-m^2/A-s^2> 038) electric permittivity = 1.2922425(96) x 10^-13 A^2-s^4/kg-m^3> 039) magnetic exposure = 2.9363775(58) x 10^-12 A-s/kg> 040) electric constant = 8.854187817... x 10^-12 A^2-s^4-sr/kg-m^3> 041) magnetic pole strength = 4.8032041(96) x 10^-11 A-m> 042) Newton constant = 6.6723641(43) x 10^-11 m^3/kg-s^2> 043) density of states = 2.0392015(07) x 10^-10 s^2/kg-m^2> 044) S-B primary constant = 1.3897143(30) x 10^-9 kg/s^3-K^4> 045) radiant distribution constant = 3.335640952... x 10^-9 s/m> 046) graviton mass constant = 5.4563026(39) x 10^-8 kg> 047) molar Planck constant = 3.9903126(87) x 10^-7 kg-m^2/s-kmol> 048) magnetic constant = 1.256637061... x 10^-6 kg-m/A^2-s^2-sr> 049) electric conductance = 3.8740458(43) x 10^-5 A^2-s^3/kg-m^2> 050) conductance q. = 7.7480916(72) x 10^-5 A^2-s^3-rad/kg-m^2-sr> 051) magnetic permeability = 8.6102258(15) x 10^-5 kg-m/A^2-s^2> 052) ?e-structure constant = 7.2973525(36) x 10^-3 /rad> 053) second radiation constant = 1.4387752(29) x 10^-2 m-K> 054) dielectric constant = 1.4594705(05) x 10^-2 /sr> 055) gravitational momentum = 1.6357583(80) x 10^1 kg-m/s> 056) relative permeability = 6.8517999(97) x 10^1 sr> 057) inverse ?e-structure = 1.3703599(97) x 10^2 rad> 058) impedance of vacuum = 3.767303134? x 10^2 kg-m^2/A^2-s^3-sr> 059) molar gas constant = 8.3144720(88) x 10^3 kg-m^2/s^2-kmol-K> 060) spin angle constant = 9.3894326(23) x 10^3 sr-rad> 061) i. conductance q. = 1.2906403(83) x 10^4 kg-m^2-sr/A^2-s^3-rad> 062) von Klitzing constant = 2.5812807(61) x 10^4 kg-m^2/A^2-s^3> 063) inverse gravitational mass = 1.8327429(14) x 10^7 /kg> 064) Faraday constant = 9.6485341(30) x 10^7 A-s/kmol> 065) speed of light in vacuum = 2.99792458 x 10^8 m/s> 066) graviton energy constant = 4.9038802(52) x 10^9 kg-m^2/s^2> 067) Josephson primary = 2.4179894(88) x 10^14 A-s^2/kg-m^2> 068) Josephson q. = 4.8359789(67) x 10^14 A-s^2-rad/kg-m^2-sr> 069) electric displacement = 3.9552465(66) x 10^15 A-s/m> 070) absorbed dose = 8.987551787? x 10^16 m^2/s^2> 071) luminous density = 2.7467669(26) x 10^17 cd-sr-s/m^3> 072) gravity displacement = 4.4930470(15) x 10^18 kg-s/m^2> 073) molar mass constant = 3.2858629(17) x 10^19 kg/kmol> 074) magnetic potential = 1.0209601(87) x 10^20 kg-m/A-s^2> 075) thermal conductance = 1.0218040(21) x 10^20 kg-m^2/s^3-K> 076) electric current constant = 1.1857530(90) x 10^24 A> 077) luminance constant = 1.2018155(94) x 10^24 cd/m^2> 078) luminous ?nsity = 8.2346000(82) x 10^25 cd-sr/m^2> 079) Avogadro constant = 6.0221419(79) x 10^26 /kmol> 080) gravitational ?ld = 1.3469816(08) x 10^27 kg/m> 081) electric potential = 3.0607616(38) x 10^28 kg-m^2/A-s^3> 082) electric conductivity = 9.5637446(19) x 10^29 A^2-s^3/kg-m^3> 083) Celcius temperature = 3.5518626(82) x 10^32 K> 084) graviton wave number = 2.4686709(99) x 10^34 /m> 085) mass ?te constant = 4.0381492(72) x 10^35 kg/s> 086) molar energy = 2.9531863(13) x 10^36 kg-m^2/s^2-kmol> 087) surface concentration = 1.0119881(80) x 10^42 kmol/m^2> 088) graviton frequency = 7.4008894(66) x 10^42 /s> 089) superforce constant = 1.2106066(96) x 10^44 kg-m/s^2> 090) luminous intensity i. = 5.0709414(41) x 10^44 /cd> 091) angular velocity = 1.0141882(87) x 10^45 rad/s> 092) electric ?nsity = 9.7642024(91) x 10^49 A-s/m^2> 093) radiant intensity = 5.2968673(52) x 10^50 kg-m^2/s^3-sr> 094) graviton ?ld strength = 2.2187308(44) x 10^51 m/s^2> 095) superpower constant = 3.6293075(70) x 10^52 kg-m^2/s^3> 096) magnetic ?nsity = 2.5204148(03) x 10^54 kg/A-s^2> 097) thermal conductivity = 2.5224979(53) x 10^54 kg-m/s^3-K> 098) magnetic ?ld strength = 2.9272342(65) x 10^58 A/m> 099) absorbed dose rate = 6.6515877(33) x 10^59 m^2/s^3> 100) graviton surface density = 3.3252544(33) x 10^61 kg/m^2> 101) electric ?ld strength = 7.5560134(90) x 10^62 kg-m/A-s^3> 102) measurement area i. = 6.0943364(99) x 10^68 /m^2> 103) dynamic viscosity = 9.9688619(97) x 10^69 kg/m-s> 104) molar concentration = 2.4982658(70) x 10^76 kmol/m^3> 105) surface tension constant= 2.9885896(41) x 10^78 kg/s^2> 106) electric charge density = 2.4104603(51) x 10^84 A-s/m^3> 107) angular acceleration = 7.5058954(07) x 10^87 rad/s^2> 108) thermal transfer = 6.2272175(40) x 10^88 kg/s^3-K> 109) electric current density = 7.2263783(36) x 10^92 A/m^2> 110) luminous ef?acy i. = 2.6860104(16) x 10^95 kg-m^2/cd-sr-s^3> 111) graviton density constant = 8.2089591(81) x 10^95 kg/m^3> 112) measurement density = 1.5044911(77) x 10^103 /m^3> 113) radiant density constant = 7.3778445(74) x 10^112 kg/m-s^2> 114) radiance constant = 3.2280892(03) x 10^119 kg/s^3-sr> 115) irradiance constant = 2.2118221(59) x 10^121 kg/s^3measured quantities, and 9^9 (387,420,489) more> (less of course the 115 knowns above). The yielded> unknowns are future laws of physics (discoveries), > which also have been veri?d by lab measurement.Weighing the Graviton Primary Equations...E=(hc^5/G)^1/2> 4.9038802 x 10^9 kg-m^2/s^2 = [(6.6260687 x 10^-34 kg-m^2/s)> (2.4216061 x 10^42 m^5/s^5)/(6.6723641 x 10^-11 m^3/kg-s^2)]^1/2> 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2> [rsu 3.9 x 10^-8]E=c^5/Gv> 4.9038802 x 10^9 kg-m^2/s^2 = (2.4216061 x 10^42 m^5/s^5)/> (6.6723641 x 10^-11 m^3/kg-s^2)(7.4008894 x 10^42 /s)> 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2> [rsu 3.9 x 10^-8]E=hc^3/Gm> 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s)> (2.6944002 x 10^25 m^3/s^3)/(6.6723641 x 10^-11 m^3/kg-s^2)> (5.4563026 x 10^-8 kg)> 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2> [rsu 3.9 x 10^-8]E=h(Gd)^1/2> 4.9038802 x 10^9 kg-m^2/s^2 = (6.6260687 x 10^-34 kg-m^2/s)> [(6.6723641 x 10^-11 m^3/kg-s^2)(8.2089591 x 10^95 kg/m^3)]^1/2> 4.9038802 x 10^9 kg-m^2/s^2 = 4.9038802 x 10^9 kg-m^2/s^2> [rsu 3.9 x 10^-8]G=c^3/mv> 6.6723641 x 10^-11 m^3/kg-s^2 = (2.6944002 x 10^25 m^3/s^3)/> (5.4563026 x 10^-8 kg)(7.4008894 x 10^42 /s)> 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2> [rsu 3.9 x 10^-8]G =wc^4/E> 6.6723641 x 10^-11 m^3/kg-s^2 = (4.0507625 x 10^-35 m)> (8.0776087 x 10^33 m^4/s^4)/(4.9038802 x 10^9 kg-m^2/s^2)> 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2> [rsu 3.9 x 10^-8]G=c^4/F> 6.6723641 x 10^-11 m^3/kg-s^2 = (8.0776087 x 10^33 m^4/s^4)/> (1.2106066 x 10^44 kg-m/s^2)> 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2> [rsu 3.9 x 10^-8]G=c^5/P> 6.6723641 x 10^-11 m^3/kg-s^2 = (2.4216061 x 10^42 m^5/s^5)/> (3.6293075 x 10^52 kg-m^2/s^3)> 6.6723641 x 10^-11 m^3/kg-s^2 = 6.6723641 x 10^-11 m^3/kg-s^2> [rsu 3.9 x 10^-8]E is graviton energy constant, h is Planck constant,> c is speed of light in vacuum, G is Newton constant,> v is graviton frequency, m is graviton mass constant,> d is graviton density constant, w is graviton wavelength,> F is superforce constant, P is superpower constant,> e is elementary charge, b is second radiation constant,> M is molar mass constant, i. is inverse, q. is quantumhttp://physics.nist.gov/cuu/Constants/> [UPDATED 1998 CODATA-NIST VALUES]If the math works, whatfs wrong with todayfs physics? Have theysteered themselves down a blind alley? Is Energy really energy, orare all values for E problematic?There is a similar post on sci.physics.electromag at:http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8& group=sci.physics.electromag&selm=30067f9b .0308100956.495d52df%40posting.google.comwhere an equation for Energy does something similar to what is here. The reference link at bottom goes to a page where similar values foreach component of the Axiomatic Equation are worked out. Ifm postingthe original below, should the links not work:-)------------------------------------------------------ ---------------ENERGYfS AXIOMATIC EQUATIONS?1. Energy is the basic common denominator of known physics, aseloquently expressed by Einsteinfs famous equation: E = mc^22. Energy can also be expressed (DeBroglie) as: E = hc/L , where h =Planckfs constant, c = light velocity, L = photon lambda, so that wecan say:E = hc/L = mc^23. Now, if light is an electromagnetic phenomenon, the we can alsosay (per Hyperphysics:http://hyperphysics.phy-astr.gsu.edu/hbase/waves /emwv.html -and- http://hyperphysics.phy-astr.gsu.edu/hbase/electric/ ele?.html), perMaxwellfs equation:Em/Bm = c , and c = 1/(eo mo)^1/2so that magnetic value times lightspeed: Bm * c = Em, its electricvalue equivalent.The question then becomes, can this equation be raised to the level ofEnergy?4. A way to do this is to multiply Em by lightspeed, so that:Em * c = (Bm)c^2, which says that an electron accelerated tolightspeed is energy, which if so gives us:Em * c = (Bm)c^2 = E5. Combining the above into a continuous equation gives us:Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^26. However, this is a still incomplete equation, taking E = mc^2, wethen have:Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^2 = mc^2 = E* * * Now, this last continuous equation gives us a way to interrelateelectric ?ld, magnetic ?ld, Planckfs constant, photon lambda, andmass, all as Energy. This looks like a rather unifying equation forthese forces, except gravity is still missing. The way to solve this(and here we are treading on new and dangerous ground) is to give massa gravity and magnetic value.If we set mass as m = 1, to represent one hydrogen atom, which is anincomplete value, since from it is missing a gravitational constant,which is represented here as negative g = 5.9x10^-39, a negativeremainder from how the atom is formed, so that mass is now expressedas: m (gravity/kilograms) = (1-g)(as per Gravity Force Coupling at:http://hyperphysics.phy-astr.gsu.edu/hbase/forces/ couple.html whichis not the same as Newtonian G)Likewise, if we set mass for magnetic, then the internal atomicpositive and negative magnetic charges equal zero, m = 0, except forthe atomfs magnetic remainder, which here is set as a positive valueat m = (0+Bm). Therefore, m (magnetic potential) = Bm(This can further be illustrated with E/c^2 = m, which is now E/c^2 =Bm, which translates into E = (Bm)c^2, as per #4 above.)7. So, in taking the energy continuous equation to its conclusion,using the new values for mass, we get:Em * c = hc/L = h/L(eo mo)^1/2 = (Bm)c^2 = (1-g)c^2 = Ewhich now represents an Axiomatic Equation for Energy includingelectric ?ld, magnetic ?ld, light speed, Planckfs constant, photonlambda, mass, and gravity. (Please note the gravity constant is notNewtonfs Gravity, for it is a proton to proton value, which needs tobe converted further to become Newtonfs G.) This axiomatic equation,with conversions and SI Base Units values (meters, kilograms, time),was developed more extensively at:http://www.humancafe.com/cgi-bin/discus/show.cgi?84/108. htmlI am posting this equation here to solicit reasonable feedback. Isenergy universal? Can it be expressed as per the 7 equations listedabove? If not, where lies the error? Is is not all Energy?The conclusion (speculative at this point) from how this axiomaticequation is written is that electromagnetic energy and gravity becomeinversely proportional to the energy environment within which they aremeasured. This may mean, a possibility only, that gravity densityaround Pluto will be greater than that of Mercury, which would thenmean that in using the Orbital Equation: v^2*R = GM, yields for us amass reading that may be inaccurate, since we had always assumed G tobe constant. Of course, if it is not constant, then the mass valuesmust represent something else, which may mean that Pluto is perhapsreally a dirty water ice-ball planet, whereas Mercury is nearly allmetallic.The other rami?ation from this equation is that if photon lambda istotally canceled out, gravity g reaches its maximum value (from 0 to1) of g = 1, which means total gravity takes over, mass ceases toexist, light ceases to exist, all is left is gravity... like a blackhole. So these equations may be important, if true, since it wouldgive us a new perspective on universal gravity (not equally constant)and provide a way to solve for various components of the equation inrelation to each other.Happy computing! -Ivan === I want to program some econometric functions in C/C++, VB or SQL for anin-house application. I did see some free source code in Fortran but none inthe any of the above languages.In particular, I am looking at Time Series Analysis, Forecasting, GARCH etc.as applied in the ?ance industry.Does anyone know of a good place from where I can get this - for free or anominal fee (Source Code or Libraries that can be used in customapplications).Jay === That is the method I have used, but there is a ? twisting.(Correct me if I m wrong), the normal vector (principle normal) of acurve always point to the concave side. When the curve changesconcavity, the normal vector reverses direction, and so does thebinormal.For the other question, I was refering to equations likex*sin(x)+y*cos(x)=2*z^2If you restrict x, y, z between -5 to 5, the graph is a forking tube.I m very interested in exploring such equations and theirclassi?ations.Take a smooth curve G, parametrized by r(t) = (x(t),y(t),z(t)). These> vectors can be de?ed easily: rf(t) = (xf(t),yf(t),zf(t)): the velocity vector T(t) = rf(t) / | rf(t) | : unit tangent vector Tf(t) = |rf(t)|^2 K(t) : K = curvature vector N(t) = K(t)/|K(t)| : Principal normal to curve B(t) = T(t) x N(t) : Binormal to curveThen, take a small radius rho, and generate normal circles to the> curve G: C(t, rho, theta) = r(t)> + rho*cos(theta)*N(t)> + rho*sin(theta)*B(t) > If you choose rho suf?iently small (I think smaller than min(1/|K(t)|)> will do the trick), then youfll get a nice embedding of the tube. The> circles form radial ?ribsf along the tube, and tracing out the curves> C(t,rho,theta) for ?ed rho & theta, but with varying t, draws lines> that run ?parallelf [kind of parallel, but not really], to the axis of> the tube.I havenft seen anyone in my math department with explicit interest in> problems like these, or courses that explore this area. Are they more> of a computer science guyfs interest?> I donft know. The formulas seem kind of trivial, so that may be why the> apparent lack of curiosity youfre seeing.Dale. === TWISTING: (Correct me if I m wrong), the normal vector (principlenormal) of a curve always point to the concave side. When the curvechanges concavity, the normal vector reverses direction, and so doesthe binormal. ThisROTATION: If you transform the parametric equation of a circle in the xy planewith the z-axis rotation matrix, and varying theta from 0 to 2pi, youget the equation of a donut ring. There are other ways too, such asusing spherical coordinates.I would very much like mathematical equations instead of computercalculations of the normal, binormal, and using them as a basis.For example: x*sin(x)+y*cos(x)=2*z^2If you restrict x, y, z between -5 to 5, the graph is a forking tube.I m very interested in exploring such equations and theirclassi?ations.> I think you have hit upon the basic method. For example, to construct a> helical tube you might use the following equations:x=a cos(t)> y=a sin(t)> z=b twhere a and b are constants and t varies to give you different points on the> helix. Then, as you said, you would compute the tangent vector and draw a> circle in the plane perpendicular to the tangent. I am not sure what you> meant by rotating circles or problems (twisting).If you are interested in computer graphics there are news groups for that.> One idea then is to just draw lots of spheres centered on points along the> helix.It might help others here to help you if you told us why you want to> describe a tube.> I havenft seen anyone in my math department with explicit interest in> problems like these, or courses that explore this area. Are they more> of a computer science guyfs interest? === > How can I describe a tube (imagine an out of control water hose) in> R3? Or examples of equations that resemble tube like structures?> Preferably no twisting nor self-intersections.An approach that deals with notions of twisting is that of theframed curve.Based on another message of yours, I think you are familiar with theFrenet frame (T,N,B). There are other options for frames; in fact,any time-varying orthonormal basis can be used as a frame. Probablyyou want to restrict to those for which the ?st vector is T; so youwant to consider frames of the form (T(t),e1(t),T(t)xe1(t)). (t isthe curve parameter; x is the cross product.)Off the top of my head, I canft think of any obvious framing that willeliminate twisting; but you can measure twisting by how e1(t)varies, and if you can reduce that then you should be OK.Kevin. <3c65f87.0308080830.c3dd6be@posting.google.com> <3c65f87.0308091410.24e9c1d8@posting.google.com> === > Why is ?Braveheartf worth mentioning in this context?>Both Saving Private Ryan and Braveheart are in genres that have> James,> If you had the *slightest* knowledge of *anything* about William Wallace> or Scottish history you would not describe it as a historical battle> epic, but more as quite a fun ?m, if you totally disregard any facts> but look on it as an amusing sort of Hollywood romp.Ha. Yeah, they did sort of tinker with the details a bit. Guess thatfshow it has to be when you only have 128 pages or so to work with. Ed Howdershelt - Abintra PressScience Fiction and Semi-Fiction http://abintrapress.tripod.com === (x-x/2)(x-x/2+1)... (x-2)(x-1) (x+1)(x+2)...(x+x/2-1)(x+x/2)> ------------------------------------------------------------> x^xNow it should be pretty obvious that the limit of that expression> evaluates to one. There is (x+n)(x-n) in the numerator and x^2 in the> denominator for n=1 to x/2.> Thatfs ridiculous.To get an expression that equals one there what is necessary is II ( (x+n) (x-n) + n^2 )That actually equals II x^2. The expression (x+n)(x-n) is greaterthan (x+n+1)(x-n-1) for positive x and n.Ross === lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1False.Stirlingfs approximation for n! is actually a transformation of aformula thatfs only true in the limit. Thatfs why some use Gosperfsapproximation for n!, itfs more accurate for some values of n, I havealso seen mentioned Burnsidefs formula, and a Lanczos approximation,and Godfreyfs improved Lanczos approximation.Here are some links with information about factorial approximations:http://mathforum.org/library/drmath/view/60692 .htmlhttp://www.google.com/search?q=factorial+ approximationhttp://www.google.com/search?q=%22Gamma+function %22+approximationStirlingfs identity is:lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1An approximation is:n! ~= n^n sqrt(n2Pi) / e^(n-(1/12)n), n > 1The value of n factorial is approximately equal to n to the nfthpower times the square root of n times 2Pi divided by e to the nfthpower, for n much greater than one.What that says is that for given large ?ite values of n, that theresult of the approximation of n! is near n!. The bene? of usingthe other approximations is that they are nearer n!.This thread isnft concerned about four trillion factorial or eighthundred and thirty seven thousand factorial, itfs about the result ofexpressions as the variable diverges to in?ity, n > 4000000000000.Particularly, at this point itfs concerned with setting the value ofone variable of the gamma function to an expression of the othervariable which diverges.lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1Here Ifll call that the Reformulated Gamma expression, RG(x). RG(1)is:lim n->oo ((2n)!) / (n! n^n) = 1If that is not true, then it it is perhaps possible to validly compareit with known results of the value of n! or the other terms in thevariable expresion and to get a known untrue result. That wouldde?itely help establish the truth value of that identity.As we have previously discussed in this thread, there is a knownresult for an expression with n!, (n/2)!, 2^n, and sqrt(n). Stirlingfs formula has terms of n!, n^n, e^n, and sqrt(n). TheDuplication Formula for the Gamma function gives an expression forGamma(2z) in terms of 2^(2z-1/2), Gamma(z), and Gamma(z+1/2).Stirlingfs formula has n! and n^n on opposite sides of the divisor,where RG(1) has them on the same side of the divisor.Stirlingfs formula:1: lim n->oo ( n! e^n ) / ( n^n sqrt(n) sqrt(2P)i ) = 1The expression f(x) = lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1 forx=1:2: lim n->oo (2n!) / ( n! n^n) = 1The common terms of those two expressions are n! and n^n. Put n!alone in the numerator for 1 and 2:3: lim n->oo n! / ( n^n e^(-n) sqrt(n) sqrt (2Pi) ) = 14: lim n->oo n! / ((2n)! n^(-n)) = 1Divide 3 by 4:5: lim n->oo ( (2n)! e^n ) / ( n^(2n) sqrt(n) sqrt(2Pi) ) = 1Put n^n alone in the numerator for 1 and 2:6: lim n->oo n^n / (( n! e^n)/(sqrt(n) sqrt(2Pi))7: lim n->oo n^n / ( (2n)! / n! ) = 1Divide 6 by 7:8: lim n->oo ( (2n)! sqrt(n) sqrt(2Pi) ) / ( n! n! e^n ) = 1So we have two equations, 5 and 8, each is the result of reducingcommon expressions from a product of the two equations 1 and 2. Thatfs about how the l.h.s of each equation is equal to the samevalue, one.How about gathering the (2n)! terms of 5 and 8 and attempting todivide those? Thatfs probably not very informative.9: lim n->oo (2n)! / ( n^(2n) sqrt(n) sqrt(2Pi) e^(-n) ) = 110: lim n->oo (2n)! / (( n! n! e^n )/(sqrt(n) sqrt(2Pi))) = 1Divide 9 by 10.11: lim n->oo n! n! e^n e^n / ( n^(2n) n 2Pi) = 1That is equivalent to:12: lim n->oo n! ^2 e^2n / (n^(2n) n 2Pi) = 1Take the square root of both sides.1: lim n->oo n! e^n / (n^n sqrt(n) sqrt(2Pi)) = 1Well, thatfs just the same thing as Stirlingfs formula. I think itwas derived from Stirlingfs formula and f(x) for x=1 without simplymultiplying Stirling by f(x)/f(x), except in a way it is as f(x)/f(x)= 1 / 1 = 1.Please explain your reasoning. Ifm not trying to be obtuse here, Ifmtrying to be not obtuse here, if itfs invalid I want to know and why.Thatfs because it has the same, true meaning.Why do I say lim n->oo ( (1+1/x)n! ) / ( n! n(n/x) ) = 1? Thatfs notasking why it is so, it is asking why I say it is so. Ifve explainedwhy I think it is so. I think you should address that.Do you have some reason why you say it is not so? You said Falseand Stirlingfs formula. Thatfs not an explanation. I use Falsesometimes, normally in protracted conversations where an explanationis already evident, itfs fun as hell, but itfs not an explanation,just the assertion of the existence of an explanation. It is so that2 n! =/= (2n)!, for n =/=1.So anyways, whatfs your explanation? That is to say, for the bene?of people who donft know what your explanation is, present it.I think it would be great if RG(1) was true. I think it is, for thereasons that I have provided, namely that I think it is a valid use ofEulerfs Gamma function and that it appears as shown here to notcontradict Stirlingfs formula. While that may be true, Ifm not goingto ignore valid objections.So, if you say itfs false, then explain for our bene? why. I donftmind if you reiterate your explanation and answer the questions ofothers about it, I think it would be good.Ross === lim n->oo ((1+1/x)n!) / (n! n^(n/x))Set x=1/n:lim n->oo ((1+n)n)! / (n! n^(n^2))lim n->oo (n^2+n)! / (n! n^(n^2) )Set x=n:lim n->oo ((1+1/n)n)! / (n! n^(1) )lim n->oo (n+1)! / (n! n)lim n->oo (n+1) / (n)The result is equal to one.Set x=n/2:lim ((1+2/n)n!) / (n! n^(2))lim (n+2)! / (n! n^2)lim (n+1)(n+2) / n^2lim (n^2 + 3n + 2) / n^2lim (n^2/n^2) + (3n/n^2) + (2/x^2)That appears to start diverging, and also appears to be equal to one.Set x = 2n:lim ((1+ 1/2n)n)! / (n! n^(1/2))lim (n+1/2)! / (n! n^(1/2))lim Gamma(n+3/2) / (Gamma(n+1) n^(1/2))That appears to diverge. Set x=1:lim (2n)! / (n! n^n)lim (n+1)(n+2)...(2n) / ( n^n)lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n!/n^nThat might also appear to evaluate to one as all but the ?st termwhich evaluates to one evaluate to equal zero, depending on what a, b,... equal. The value of a might be sum n, which is greater than n,with b being the value of the sum of the product of each pair of N, Iwonder if that is n^2. If that were so on and so forth for the otherterms then the value of the expression would be n. If a, b, ... wereeach less than n, n^2, etcetera, that would not be so.lim (n^n/n^n) + (an^(n-1)/n^n) + (bn^(n-2)/n^n) + ... + n! / n^nIf a is sum n, then a is (n+1)(n/2) = n^2/2 + n/2, and an^(n-1) is(n^(n+1) + n^n)/2, and that divided by n^n is n/2+1/2.The limit of the nfth term, n! / n^n, is zero.Letfs see, consider a ?ite case of (x+1)(x+2)...(x+4).(x^2 + 3x + 2)(x+3)(x+4)(x^3 + 6x^2+11x + 6)(x+4)(x^4 + 10x^3 + 35x^2 + 50x + 24)Yep, it appears a = sum n for n= 1 to 4, 5*2, (n+1)/(n/2). The valueof b is the sum of each product of each pair of {1, 2, 3, 4), thevalue of c would be the sum of each product of each triple of 1, 2, 3,4, etcetera, d is the product 1, 2, 3, 4. Normally in the polynomialthe coef?ient a is on the ?st term, this starts with a on thesecond.1*2 + 1*3 + 1*4 +2*3 + 2*4 + 3*4 = 351*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 501*2*3*4 = 24Letfs see, 35 = 1*(2+3+4) + 2*(3+4) + 3*4, the sum for i from 1 to n-1of the product of i and the sum of n from i+1 to n.The third term c is 1*(2*3 + 2*4 + 3*4) + 2*(3*4) ..., a slightly moreconvoluted expression where more terms would be necessary to displayit by example.So, it appears that the expression diverges. Now I want to ?ure outwhat it diverges to, symbolically. Then you could divide thedivergent expression by that and get a convergent expression.Set x=1/2:lim ((1+2)n)! / (n! n^(2n))lim (3n)! / (n! n^(2n))More of the same. lim (n+1)(n+2)...(3n) / (n^(2n)lim n^2n/n^2n + a n^(2n-1) / n^(2n) + b n^(2n-2) / n^(2n) + ...If a is (2n+1)n = 2n^n + n, then the second term would be (n^(2n+1)+n^2n) / n^2n = n+1, where in the above it was (n+1)/2.What are the closed forms for the coef?ients of the polynomial of(x+1)(x+2)...(x+n)? They are {1, sum n, ..., n!), what is the form ofeach of the intervening coef?ients?Going back, set x=1/n:lim n->oo ((1+n)n)! / (n! n^(n^2))lim n->oo (n^2+n)! / (n! n^(n^2) )lim (n+1)(n+2)...(n^2+n) / n^(n^2)lim (n^(n^2))/(n^(n^2)) + ((n^2+n+1)(n^2+n)/2) (n^(n^2-1)) / (n^(n^2))+ ...lim 1 + (n^4 + 2n^3 + n^2 + n) (n^(n^2-1)) / 2(n^(n^2)) + ...lim 1 + (n^(n^2+3) + 2n^(n^2+2) + n^(n^2+1) + n^(n^2))/2(n^(n^2)) +...lim 1 + (n^3)/2 + n^2 + n/2 + n/2 + 1/2 + ...So there appear to be ?e or seven cases, where x is a constantf(x)=c, x is a function of a constant times n, f(x)=cn, for c=1,01, and x is a function of a constant times the reciprocalof n, f(x)=c/n, for c=1, 01.For f(x)= lim n->oo ((1+1/x)n)! / (n! n^(n/x)) = 1, that is true whenx=cn, 0 lim n->oo ((1+1/x)n)! / ( n! n^(n/x) ) = 1> False. > lim n->oo (n! e^n) / (n^n sqrt(n 2Pi)) = 1So, use Stirlingfs formula to obtain the asymptotics(at least the dominant term) of ((1+1/x)n)!/(n! n^(n/x)).-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen <3c6b9c1e.0308091559.485216f4@posting.google.com> <3c6b9c1e.0308100457.73ff5939@posting.google.com> <3c6b9c1e.0308101544.48f969ec@posting.google.com> <3c6b9c1e.0308111623.38637b13@posting.google.com> <3c6b9c1e.0308112243.4fb4e3bc@posting.google.com> === ::That appears to start diverging, and also appears to be equal to one.:and then later.::Bollocks, indeed.::Ross:Speaks for itself. === I want to ?ure out what are the coef?ients. These are 0! = 1, ? =(n+1)(n/2), and basically for each an expression of the sum of theproduct of each triple, quadruple, quintuple, etcetera, of {1, 2, 3,4, ..., n}, which would be the same as for {0, 1, 2, ..., n}.HMF: An in?ite series whose general term is a rational function ofthe index may always be reduced to a ?ite series of psi andpolygamma functions.Letfs see, f(x) = II_i=1^n (n+x)I go to MathWorld and read about polynomial. http://mathworld.wolfram.com/Polynomial.htmlThe function is a univariate polynomial with constant coef?ients. There are n many monomial summands. The order or degree of f is n.Exchanging the coef?ients of a univariate polynomial end-to-endproduces a polynomial whose roots are the reciprocals of the 1/x_i ofthe original roots x_i.The roots x_i of f are -1, -2, -3, ..., their reciprocals 1/x_i are-1, -1/2, -1/3, ....Consider Vietafs Formulas.http://mathworld.wolfram.com/VietasFormulas.htmlI noticed the s1 = r1 + r2 + r3 + r4, it is similar to the 1 + 2 + 3 +4, and so on for the following values of s.Consider Stirling Numbers of the First Kind. The Stirling Number generating functions are de?ed in terms of afalling factorial, where f is de?ed in terms of a rising factorial,almost. The rising factorial or Pochhammer is x(x+1)(x+2)...(x+n-1),f is Pochhammer(x,n) * 2.The function f is easily de?ed as a falling factorial(2n)(2n-1)...(n+1), that is (2n)_n.What is to be done here is to determine the symbolic form of thecoef?ients of the n_1 monomials of the polynomial,(n+1)(x+2)...(n+n), to sum them together to get another polynomial. The ?st coef?ient is 1, the second (n+1)(n/2), and the last n!. There are thus the n-2 other coef?ients that require determination. The coef?ients form another polynomial with each ifth coef?ientmultiplied by n^(n+1-i). Their sum is a univariate polynomial of n. Ifm trying to ?ure out what its degree would be, as the number ofmonomials is a function of the variable, and the coef?ients are eacha function of the variable. Call the polynomial g(n).This polynomial is for the case of RG(1), the expression lim n->oo (2n!) / (n! n^n)is under consideration.lim n->oo (2n)! / (n! n^n g(n)) = 1Basically g(n) = (n+1)(n+2)...(n+n) / n^n = RG(1).In that sense g(n) for RG(x) is RG(x), and when x=cn, 0oo ((1+1/x)n!)/(n! n^(n/x)), it equals onefor certain values of x without explicitly multiplying it by itsreciprocal. Also, it should be possible to determine forms for g(n).Ross === >I cannot think of a counterexample to the following question.>If f is an automorphism of a ?ite group G such that, for every g in G,>the elements g and f(g) are conjugate, this does not necessarily>imply that f is inner, does it? You can have counterexamples to statements (at least to false statements),but what do you mean by a counterexample to a question?OK, so you are looking for an example of a ?ite group G with an outerautomorphism f such that, for every g in G, the elements g and f(g) areconjugate.I do not any such example. Please let me know if you ?d one!Derek Holt. === >NASAfs> hypervelocity guns.> Roughly, they use a gunpowder charge to compress hydrogen which propelsthe> projectile to hypervelocity.> Frank> Yes, I am beginning to see all energy transformations as either obeyingthe> IdealGasLaws which are linear in math form and the EM laws which are> sinusoidal and have cancelling of energies. The linear transformationallows> the exceeding of barriers such as sound barrier. The sinusoidal form suchas> slingshot, crossbow, fusion tokamaks have so much cancelling of energy> that they seldom can ever perform as well as the linear transformation.>this must be the ?st time anyone has directly compared a crossbow to atokamak. amazing> The most common linear transformation of energy is explosion and burning> of fuel and of course air pressure such as airguns. Another linear> transformation> is Fission because the neutrons obey the IdealGasLaws.> We can bifurcate all energy transformations as either direct and linear or> as indirect, cancelling, and sinusoidal.> What I am anxious to prove in the above is that barriers exist such as the> Sound Barrier. And airguns can exceed the Sound Barrier but can bow andarrow> exceed the Sound Barrier or can a slingshot (another form of EM)> ever exceed the Sound Barrier. But the most important of these questions> is whether Fusion can ever exceed 2/3 breakeven and still be under**control**>There is no such thing as the ?sound barrier.fRaz === Consider the sequence where,for m = any positive integer,a(m) = the number of integers j, 1 <= j <= m, where?/j) divides m.Is there an easy direct way to calculate the sequence?The best way I can come up with is:a(m) = sigma(m) - sum{k|m} ?/(1+k)),where sigma is the sum of the positive divisors of m.(I guess my means of calculating this sequence is good enough, if itis a true means to calculate it.)Leroy Quet === > What Ifve endeavored to do to prevent math discussions that go on> endlessly as issues are debated all over the map is focus in on> particular points of contention.Ifd thought Ifd removed all room for debate on the issue of whether or> not, given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)one of the afs must be coprime to f, when f is coprime to 3, without> regard to m, by focusing on P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)with f=3, as Ifd *said* that then each of the wfs must equal 3^{1/3},> but a poster pointed out that in fact I get another factor of 3 that> divides off, and itfs indeterminate as to how it divides off. That> poster continued thinking that shot down my arguments, and presented> their beliefs, which would be that the wfs vary with m.However, Ifll destroy that position using f=3 by considering P(m).Returning to the full expression with f=3, I have P(m) = 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3) = (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u).At issue has been a claim by several posters, whofve spent *months*> arguing with me that f divides off as a variable of m, such that the> afs can have varying factors in common with f, dependent on its value.> They make this argument because I can use m=0 to show that when f is> coprime to 3, one of the afs does not share non-unit factors in common> with f. They claim the m=0 is a special case as they wish for factors> of f to move around in the factorization dependent upon whether or not> it factors over rationals.Yup, if youfve been wondering about all the arguing and all the posts,> a lot of it has been about that one point of contention, which Ifm> going to shoot down here.Now then, if these other posters were correct, then a variable factor> in common with 3 should divide off through each of the afs, but> looking at P(m), 3 itself must divide through AS A CONSTANT without> regard to the value of m.Thatfs because P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)proves that *each* of the afs must have a factor that is 3 without> regard to m, so m=0 doesnft matter.> What you have been claiming is about the form of thefactorization of P(m)/f^2. In terms of the factorizationthat you gave above,> P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3),you want to show (direct quote from your text above): each of the wfs must equal 3^{1/3}.What you REALLY need to show is that the analog of this istrue for f = prime > 3, and for some reason you think it will be suf?ient to show it only for f = 3. But even withf = 3 you have a problem. What you have shown here is notabout P(m)/f^2. You have instead considered P(m)/f^3.The reasons are (1) you know that I have already shown thatfor f = 3, the wfs are indeterminate in the factorization ofP(m)/f^2, and (2) you have lost track of what you were doing.Bottom line: what you have done here regarding P(m)/f^3 forf = 3 is not what you set out to do, and it does not implywhat you wanted to do either. I actually donft disagree with the result you noted for the a_ifs above, but it is inthe so-what category: it is not what you need. You want to focus on the factorization of P(m)/f^2.Second bottom line: my examples showing that there are in?itelymany choices for the wfs when you are considering what youreally intended - P(m)/f^2 - are all still completely valid.What you have done here is irrelevant to that also. The implication of these examples is: you CANfT do what youwanted to do when f = 3.Third bottom line: What you REALLY need is a demonstration that what you (incorrectly) think holds for f = 3, also holds forf = 5, 7, 11, 13, .... This you have not even touched. AsI noted in the other thread, you need to prove something about the trees deep in the forest (f = prime > 3). You CANNOT (and very de?itely have not!) done that by consideringthe underbrush out on the edge of the forest (f = 3). Reasoningthat a pattern on the boundary must extend into the interiortakes proving also. And what you have done here doesnft evenestablish the pattern out on the boundary (f = 3).Fourth and ?al bottom line: After all this work, you are back whereyou started: no proof at all that the pattern of factorization youhave observed for m = 0 extends to m <> 0. Nora B.> BUT, these posters believe that the afs have a *variable* function of> f, and my having f=3 wouldnft change that fact, so if true, itfd force> some of the afs, depending on the value of m, to not have a factor> that is 3. [propaganda deleted]> I await attempts at further debate from posters, but I warn them that> my job is to focus on the issues, and take away their ability to> endlessly argue, expecially through distraction.> James Harris === > |BUT, these posters believe that the afs have a *variable* function of> |f, and my having f=3 wouldnft change that fact, so if true, itfd force> |some of the afs, depending on the value of m, to not have a factor> |that is 3.> You *say* having f=3 wouldnft change it, but why? The rest of> the posting doesnft provide an argument. Doesnft it bother you> to be stuck repeating such a claim without any way to show> that itfs actually true besides telling us how weird it would be> if it werenft?> Keith RamsayPosters repeatedly delete out the math in replies.Here the pertinent expressions are P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)where if f divides off the afs as a function of m, then you get a> contradiction, as in fact, itfs clear that f divides off as a constant> independent of m.Of course, here f=3, but whatfs amazing is that posters would try to> convince anyone that having f change its value to being coprime to 3,> could *introduce* a dependency on m.And that explains why they delete out pertinent expressions as they> want to convince readers.Or does it?Ifm curious about what readers think.Why do YOU think posters like Keith Ramsay keep deleting out the math?There is a much more general question. Why does anyone read these JamesHarris threads at all?However, I will address your ?st questions as a casual reader. Theremoval of the constantly repeated algebra is a good idea since theremust be hundreds of repetitions of the stuff at hand. Same old stuffover and over again. Never the slightest attempt to clarify. Just thesame tired stuff endlessly.Now to my more general question. In my case, Ifm learning from thevarious posters who have posted refutations to your proof. These postersmostly explain very well their steps and provide de?itions and otheruseful information. They are more focused on the mathematics than youare. They seem to have an excellent grasp on the subject. Indeed, theyare rminding me of material I studied 30 years ago and never used.Chuck-- ... The times have been, That, when the brains were out, the man would die. ... Macbeth Chuck Simmons chrlsim@earthlink.net === > Ifm curious about what readers think.> Why do YOU think posters like Keith Ramsay keep deleting out the math?I think if YOU deleted all the math, wefd be better off. === > What Ifve endeavored to do to prevent math discussions that go on> endlessly as issues are debated all over the map is focus in on> particular points of contention.Ifd thought Ifd removed all room for debate on the issue of whether or> not, given P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)one of the afs must be coprime to f, when f is coprime to 3, without> regard to m, by focusing on P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3)with f=3, as Ifd *said* that then each of the wfs must equal 3^{1/3},> but a poster pointed out that in fact I get another factor of 3 that> divides off, and itfs indeterminate as to how it divides off. That> poster continued thinking that shot down my arguments, and presented> their beliefs, which would be that the wfs vary with m.That is, I made a mistake, which Ifve acknowledged.Intriguingly, Nora Baron appears to be trying to convince you thereader of something else.See below... > However, Ifll destroy that position using f=3 by considering P(m).Returning to the full expression with f=3, I have P(m) = 3^2((m^3 3^4 - 3m^2 3^2 + 3m) x^3 - 3(-1+m3^2 )x u^2 + 3u^3) = (a_1 x + 3u)(a_2 x + 3u)(a_3 x + 3u).At issue has been a claim by several posters, whofve spent *months*> arguing with me that f divides off as a variable of m, such that the> afs can have varying factors in common with f, dependent on its value.> They make this argument because I can use m=0 to show that when f is> coprime to 3, one of the afs does not share non-unit factors in common> with f. They claim the m=0 is a special case as they wish for factors> of f to move around in the factorization dependent upon whether or not> it factors over rationals.Yup, if youfve been wondering about all the arguing and all the posts,> a lot of it has been about that one point of contention, which Ifm> going to shoot down here.Now then, if these other posters were correct, then a variable factor> in common with 3 should divide off through each of the afs, but> looking at P(m), 3 itself must divide through AS A CONSTANT without> regard to the value of m.Thatfs because P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)proves that *each* of the afs must have a factor that is 3 without> regard to m, so m=0 doesnft matter. What you have been claiming is about the form of the> factorization of P(m)/f^2. In terms of the factorization> that you gave above, P(m)/f^2 = (b_1 x + u w_1)(b_2 x + u w_2)(b_3 x + u w_3),you want to show (direct quote from your text above): each of the wfs must equal 3^{1/3}.See what I mean? The poster Nora Baron posted as if in fact Ididnft myself acknowledge the mistake above!!!In fact, two of the wfs are coprime to 3, while the remaining one hasa factor that is 3. > What you REALLY need to show is that the analog of this is> true for f = prime > 3, and for some reason you think it > will be suf?ient to show it only for f = 3. But even with> f = 3 you have a problem. What you have shown here is not> about P(m)/f^2. You have instead considered P(m)/f^3.> The reasons are (1) you know that I have already shown that> for f = 3, the wfs are indeterminate in the factorization of> P(m)/f^2, and (2) you have lost track of what you were doing.But they are not indeterminate, which can be seen by consideringP(m)/3^3 = (m^3 3^3 - 3m^2 (3) + m) x^3 - (-1+mf^2 )x u^2 + u^3 = (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)as thatfs the only way the 3^3 can divide off when f=3.There is NO other way, so itfs wrong for Nora Baron to claim itfsindeterminate.> Bottom line: what you have done here regarding P(m)/f^3 for> f = 3 is not what you set out to do, and it does not imply> what you wanted to do either. I actually donft disagree > with the result you noted for the a_ifs above, but it is in> the so-what category: it is not what you need. You want to > focus on the factorization of P(m)/f^2.Actually, it blows away your assertions that f^2 divides off indifferent ways dependent on the value of m.Look at P(m)/f^2 = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 = (a_1/f x + u)(a_2/f x + u)(a_3 x + uf)where here now, if f is coprime to 3, that *last* f is blocked by a_3.Now Ifve proven that factorization, while Nora Baron is working toconvince you that the ffs divide out differently dependent on m.But when f=3 thatfs easily shown to be just wrong.So that leaves Nora Baron trying to convince you that the equationscheck to see if f is coprime to 3.Why would the poster ?ht so hard?Because accepting the truth means accepting that therefs a problemwith the ring of algebraic integers.> Second bottom line: my examples showing that there are in?itely> many choices for the wfs when you are considering what you> really intended - P(m)/f^2 - are all still completely valid.> What you have done here is irrelevant to that also. The > implication of these examples is: you CANfT do what you> wanted to do when f = 3.Third bottom line: What you REALLY need is a demonstration > that what you (incorrectly) think holds for f = 3, also holds for> f = 5, 7, 11, 13, .... This you have not even touched. As> I noted in the other thread, you need to prove something > about the trees deep in the forest (f = prime > 3). You > CANNOT (and very de?itely have not!) done that by considering> the underbrush out on the edge of the forest (f = 3). Reasoning> that a pattern on the boundary must extend into the interior> takes proving also. And what you have done here doesnft even> establish the pattern out on the boundary (f = 3).Actually, what I demonstrate with f=3 is that f^2 does NOT divide offas dependent on m.The f^2 divides off one way from P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf). > Fourth and ?al bottom line: After all this work, you are back where> you started: no proof at all that the pattern of factorization you> have observed for m = 0 extends to m <> 0. > Nora B.> BUT, these posters believe that the afs have a *variable* function of> f, and my having f=3 wouldnft change that fact, so if true, itfd force> some of the afs, depending on the value of m, to not have a factor> that is 3. [propaganda deleted]And notice how much effort the poster Nora Baron goes to in order totry and convince you the reader.Here youfre told there was propaganda but it was deleted, so if youwant to know what was deleted, you have to go back to the earlier postand ?d it.> I await attempts at further debate from posters, but I warn them that> my job is to focus on the issues, and take away their ability to> endlessly argue, expecially through distraction.But theyfre still trying, as theyfve made it their job to hide thetruth, and thereby hide the embarrassment to mathematicians.James Harris === |Posters repeatedly delete out the math in replies.Let me remind you that you were the one who kept saying thatthe postings we were writing to each other should keep gettingshorter and more to the point. Thatfs not possible if I keeprequoting the same stuff of yours over and over.|Here the pertinent expressions are || P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - || (-1+m 3^2 )x u^2 + u^3 = || (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)||where if f divides off the afs as a function of m, then you get a|contradiction, as in fact, itfs clear that f divides off as a constant|independent of m.A contradiction is when the same premisses logically implyboth A and A is false. This is not a contradiction. This isjust a situation that looks wrong to you.|Of course, here f=3, but whatfs amazing is that posters would try to|convince anyone that having f change its value to being coprime to 3,|could *introduce* a dependency on m.You ?d the possibility amazing. This is apparently all you haveto go on. There are plenty of mathematical claims where it would beamazing if they were false, but nobody knows how to prove them yet.For instance, nobody can prove yet that the decimal expansion of pi hasin?itely many 3fs in it. The idea that they should appear on averageevery tenth digit out to the tens of billions of places and then disappearwould be amazing. Thatfs not a proof.Itfs not the case that every function of m and f which is independentof m for f having common factors with 3 is also independent of mfor other values of f. So to show that it doesnft happen in this case,youfd have to tell us something speci? about these common divisorsthat makes it not just amazing to you but not actually possible.|And that explains why they delete out pertinent expressions as they|want to convince readers.||Or does it?||Ifm curious about what readers think. ||Why do YOU think posters like Keith Ramsay keep deleting out the math?It seems to me that you present a rather insulting picture of thelurkers. You describe them as if they were such mindless sheep thatthey canft read what youfre saying and understand it, unless I keeprequoting it to them for you. Not just a few times either, but over andover again. Just who do you think they are?Even supposing they were intimidated somehow, donft you supposeright, there really is a contradiction in the de?ition of algebraicinteger, how can we convince the world? I think itfs safe to sayyoufre failing to attract support like that because you do NOT havea clear and convincing case. Even supposing for the sake ofargument that wefre all terribly confused and youfre actually rightin your ?al conclusions, canft you see that it would be unclearthat youfre correct?Keith Ramsay === [snip bottomless pit of demented arguments]> ... theyfve made it their job to hide the> truth, and thereby hide the embarrassment to mathematicians.Speaking of hiding things, you forgot to post one of the numbers you discovered which support your claim that thering of algebraic integers is ? You claim that there are numbers which *should be* algebraic integers, butwhich have been *left out* of the ring of algebraic integers.Name one. Just one will do. (You wouldnft be trying to ?hide the truth, and thereby hide the embarrassmentf wouldyou?)--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > |Posters repeatedly delete out the math in replies.Let me remind you that you were the one who kept saying that> the postings we were writing to each other should keep getting> shorter and more to the point. Thatfs not possible if I keep> requoting the same stuff of yours over and over.Yes it is. Your conclusion does not follow logically.Given a mathematical argument there are typically points not indispute, so you can delete those out until you reach the *last* pointyou donft disagree with, which you should leave in for context, andthen you consider the *next* point you do disagree with, and state whyyou believe it doesnft follow logically.Now that would typically lead to replies shorter than the entire mathargument, unless possibly the error took a really long time toexplain.Remember, a math proof begins with a truth and proceeds by logicalsteps to a conclusion which then must be true.If you think someone has a false claim of discovering a proof, thenall you need do is either show their argument does not begin with atruth, or show a break in the logical chain. > |Here the pertinent expressions are > |> | P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - > |> | (-1+m 3^2 )x u^2 + u^3 = > |> | (a_1/3 x + u)(a_2/3 x + u)(a_3/3 x + u)> |> |where if f divides off the afs as a function of m, then you get a> |contradiction, as in fact, itfs clear that f divides off as a constant> |independent of m.A contradiction is when the same premisses logically imply> both A and A is false. This is not a contradiction. This is> just a situation that looks wrong to you.Actually your claim here is just wrong Keith Ramsay.The question is whether or not f^2 divides off as a function of m,which would suppose that it varies dependent on m.However, I show that when f=3, not only f^2 but an additional factorof f, divides off, showing that it divides off as a constant.Maybe this will help you see P(m)/3^3 = (a_1/w_1(m) x + u)(a_2/w_2(m) x + u)(a_3/w_3(m) x + u)where Ifve introduced wfs as a function of m, where w_1(m) w_2(m) w_3(m) = 3^3.However, *any* variation of the wfs that leaves any of them without afactor that is 3, will force non-unit factors in common with 3 into P(m)/3^3 = (m^3 3^3 - 3^2 m^2 + m) x^3 - (-1+m 3^2 )x u^2 + u^3which would be a contradiction. > |Of course, here f=3, but whatfs amazing is that posters would try to> |convince anyone that having f change its value to being coprime to 3,> |could *introduce* a dependency on m.You ?d the possibility amazing. This is apparently all you have> to go on. There are plenty of mathematical claims where it would be> amazing if they were false, but nobody knows how to prove them yet.> For instance, nobody can prove yet that the decimal expansion of pi has> in?itely many 3fs in it. The idea that they should appear on average> every tenth digit out to the tens of billions of places and then disappear> would be amazing. Thatfs not a proof.So are you then claiming that you consider it possible that factors off divide off with a dependency on m?> Itfs not the case that every function of m and f which is independent> of m for f having common factors with 3 is also independent of m> for other values of f. So to show that it doesnft happen in this case,> youfd have to tell us something speci? about these common divisors> that makes it not just amazing to you but not actually possible.Itfs simple enough, as P(m) = f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f)where the focus is on m, and the question is how f^2 divides off.Now it might help to see the underlying P(m) without so manyvariables, so let x=1, u=1, f=7, so you have P(m) = 7^2((m^3 7^4 - 3m^2 7^2 + 3m) - 3(-1+m 7^2 ) + 7), which is P(m) = 49(2401 m^3 - 147 m^2 + 3m + 3 - 147m + 7), which is P(m) = 49(2401 m^3 - 147 m^2 -144 m + 10).Now the assertion by posters is that the 49 might divide off of P(m)as a function of m, because they attack my use of P(0), which in thiscase gives P(0) = 49(7)as a special case.And yes readers, you *can* stick in numbers to help yourselfunderstand, and then you may also have the dubious pleasure ofcatching posters, who have advanced math training, like from BerkeleyUniversityfs Ph.D program, lying to you.What I use is the non-polynomial factorization P(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf), which in this case is P(m) = (a_1 + 7)(a_2 + 7)(a_3 + 7) = 49(2401 m^3 - 147 m^2 -144 m + 10)with a special expression I call an uber-polynomial, which again is f^2((m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 f).So, yes, posters have been working to convince you that m=0 is aspecial case and that they can *tell* you how 49 divides off of P(m) = 49(2401 m^3 - 147 m^2 -144 m + 10)where they claim reducibility over rationals constrains how 49 dividesthrough (a_1 + 7)(a_2 + 7)(a_3 + 7).The reality is that these methods are something beyond whatmathematicians have used before, and with their own techniques, theycanft say much about how 49 divides off of P(m), and they canftconstrain it as a variable dependent on m.However, they ?ht the math because they must not believe inmathematics itself as a body of truths. Instead it is a *social*function to them, possibly just a way to pay the bills. So they ?htletting society know of an error that has sat in mathematics for overa hundred years.They are working to convince you the reader.The reality is that these people have no way to force 49 to ? intoany particular factorization given 49(2401 m^3 - 147 m^2 -144 m + 10)because they donft have advanced factorization techniques.> |And that explains why they delete out pertinent expressions as they> |want to convince readers.> |> |Or does it?> |> |Ifm curious about what readers think. > |> |Why do YOU think posters like Keith Ramsay keep deleting out the math?It seems to me that you present a rather insulting picture of the> lurkers. You describe them as if they were such mindless sheep that> they canft read what youfre saying and understand it, unless I keep> requoting it to them for you. Not just a few times either, but over and> over again. Just who do you think they are?My audience is a very particular one. > Even supposing they were intimidated somehow, donft you suppose> right, there really is a contradiction in the de?ition of algebraic> integer, how can we convince the world? I think itfs safe to say> youfre failing to attract support like that because you do NOT have> a clear and convincing case. Even supposing for the sake of> argument that wefre all terribly confused and youfre actually right> in your ?al conclusions, canft you see that it would be unclear> that youfre correct?Keith RamsayGrow up.James Harris === > There is a much more general question. Why does anyone read these James> Harris threads at all?>Good question. Maybe to see if hefs ever going to bet back on his meds!? === James,Looking over recent threads I donft seem to be able to ?d where you posted one of the numbers whichyou claim *should be* in the ring of algebraic integers, but which is *left out*. Itfs notconceivable that you have overlooked this, since it is the fundamental basis for your claim the ringof algebraic integers is ? and the motivation for your notion of a ring of objects, whateverthat is.Please take advantage of this wonderful opportunity to support your claim and post one of thesenumeric values (there must be millions of them). Of course, if you have something to hide, thatfsdifferent --dare I say fascinating?--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === > There is a much more general question. Why does anyone read these James> Harris threads at all?> Good question. Maybe to see if hefs ever going to bet back on his meds!?My main reason is for the comedy. It usually happens like this:People spend weeks and weeks (or months and months) trying to pound XYZinto Jamesf head. He yells and screams about liars, dueling proofs,military to punish mathematicians, etc. Then one day (and this is thedelicious part) he suddenly exclaims, Oh no! Ifve just realized thatXYZ may be true! If it is, Ifll have to retract my proof! I need tothink about this for a while. Then comes a period of anywhere from afew hours to a few days of James getting drunk, singing to the walls,struggling with despair and self-doubt, etc. Finally he emerges with acry of victory: Eureka! XYZ is true, but it doesnft matter! Ifve justrealized that Z is an uber-super-duper-poly-want-a-cracker-nomial, somy proof is right after all! Now we need to launch a Congressionalinvestigation into why mathematicians have conspired to keepuber-super-duper-poly-want-a-cracker-nomials secret from me! And thewhole mess starts all over again.You just canft buy entertainment like that.-- Wayne Brown | When your tailfs in a crack, you improvisefwbrown@bellsouth.net | if youfre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- Euler | -- John Myers Myers, Silverlock === > Has this unusual piece been shot down yet?> http://www.geocities.com/dharwadker/> It doesnft display properly for me in Netscape, Opera, Mozilla> or Konqueror ....> I can see it in Netscape Communicator 4.8.I can see it in Nutscrape 4.78 (indeed in all of these)> but from Section II onward visual gibberish starts to appear.Unfortunately I am unquali?d to tell you if it is still mathematicalgibberish even when it isnft visual gibberish.-- G.C. === > I can see it in Nutscrape 4.78 (indeed in all of these)> but from Section II onward visual gibberish starts to appear.The problem lies in the use of the Symbol font.There is a semi-of?ial scripthttp://hutchinson.belmont.ma.us/tth/Xfonts.?which you can apply if you are using X-windows,after which the document becomes readable,although not, in my case, comprehensible.-- Timothy Murphy tel: +353-86-233 6090 === > http://www.geocities.com/dharwadker/...> (Paul) Let me know if you manage to get through all the jargon and seewhether it> is a valid proof.As far as I can see, there is some wishful thinking in the paper. N issomewhere in the set {4,5,6}, true. There is a Steiner triple (n+1,2n,6n)for n=4 but not for n=5 or n=6, true. But how does that relate tocolorations of planar graphs? Itfs none too clear even how he uses planarityor colorations. He does know all about the Witt pattern (5,8,24); but heseems to be too sanguine about its signi?ance.The algebra about split extensions of groups seems to be okay; but most ofthose extensions are trivial, i.e. just direct products.LH === Please see http://www.tln.net/~reriker/fermat.html for the details. === >Please see http://www.tln.net/~reriker/fermat.html for the details.You start by takinga^q + b^q = c^qand transforming it into a pythagorean triple by writing(a^{q/2})^2 + (b^{q/2})^2 = (c^{q/2})^2and then proceed to use the well known characterization of rationalpythagorean triples.Now, tell me, if q is not a multiple of 2, then how do you guaranteethat a^{q/2}, b^{q/2}, and c^{q/2} are rational numbers?For instance, say q=7. Then you need to take the 7th power of thesquare root of an integer. What makes you think that a, b, and c areperfect squares to begin with? === ======================================= === ========Itfs not denial. Ifm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) === ====================================== === =========Arturo Magidinmagidin@math.berkeley.edu===> What is your de?tion of program then?A formal de?ition would be: A Base of Computing (aka programminglanguage) is a mapping from a recursively enumerable set onto the setof recursive functions. A program, then, would be any element inthe domain of that mapping (Base of Computing.)But to capture the spirit of a program, we need to be less abstractand less formal, and just point out that:1. A program is composed of constructs (whose semantics are) such asvariable assignment, branching and conditionalized execution.2. Some programs do not terminate on some inputs.A single expression isnft really a program. (Multiple expressionsthat allow recursion can be.) The problem is that it is too limited. You seem to be inputting and outputting only expressions (couched inthe syntax of a LISP program.) That is, the program that yougenerate is limited to those that consist of a single expression,limiting your capabilities tremendously. It is not general programsynthesis.Furthermore, I still donft see where you actually do anything with theexpression, other than to input it and then output it again with LISPsyntax surrounding it but I would like to learn more. I think thatif you can generate programs for the 5 problems (specs) below, thenthat would go a long ways towards explaining, and demonstrating, howgeneral your system is.The problem here is that I happened to give an example (determining ifone number is a factor of another) for which there is in fact a LISPexpression. In an attempt to determine if what you have in mind isvery general (as general as the Predicate Calculus, which I use), hereare some useful program speci?ations that I describe in my papersand the corresponding Predicate Calculus wff. (See my papers for theaxioms and formal de?itions of Number Theory. The 8 rules areuniversal for all domains. I will copy all of that here if peoplewant, though it is all in my papers.) What would be the LISPexpression for each?Again, my papers are at:http://www.mathpreprints.com/math/Preprint/ CharlieVolkstorf/20021008.1/1andhttp://www.arxiv.org/html/ cs.lo/00030711. List all prime numbers between 1000 and 1005.PRIME(x)^BETW(1000,x,1005)2. Is a given number prime?PRIME(I)3. Determine the largest proper factor of a given number.FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A)4. Determine the largest prime factor of a given number.FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A)5. Determine the smallest prime factor of a given number.FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x)If you can express these as LISP expressions, then we have somethingto talk about. Otherwise, you have a very limited system that isreally just dealing with expressions as opposed to full-?programs, and you canft express, much less generate, hardly anything.> You do not want a program generator, you want a computerized human programmer.With all due respect, I think youfre missing the point.1. We can easily imagine a situation in which there is a need for aperson to see the programs and subjectively choose his favorite. Butthis is not the point.2. The system needs to CONSIDER multiple algorithms when generatingthe program, to differentiate it from the mere translation from oneprogramming language into another. In the latter case, the user hasto conceptualize the algorithm, then express it in one language to betranslated into another language. In the former (which my systemdoes), there is no algorithm being expressed by the user, and thesystem constructs (considers, outputs, whatever) the algorithms, thusproducing multiple programs using different algorithms.For example, in the factor checking example, my system constructs 2algorithms, which I have copied directly from my papers below. TheEnglish lines are added by me to explain what the program is doing. The other lines are the input (spec) entered by the user and theoutput (program) generated by the system:User Input: FAC(I,J)English: Is one given number a factor of another ?Program 1English: Search for an integer A such that A*I=J.for A=1:1:J set B=A*I if B=J write (true)write (false)Program 2English: Check the remainder of J/I.set A=rem(J,I) if A=0 write (true)write (false)> What I am doing is showing> that in fact, it is not much different from other systems, that *you* do not> accept as input speci?ation. I would.My system is different from your system in that I can express andgenerate a wide range of programs, whereas you have yet to demonstratethat a single LISP expression can express that much. Here is yourchance. Express the above 5 requirements. (My system is differentfrom other systems in that they give no examples at all or are justprogramming language translators.)> As I said above, it quali?s as a programming language.No, Predicate Calculus has no assignment, branching or conditionalizedexecution. It canft get into loops. It is non-procedural. Again,how would you specify what the program is supposed to do? Show that aLISP expression can represent the above 5 problems, and then we havesomething to discuss.> While authors of bogus papers refuse to answer criticisms such as why> it never sees more than one algorithm, I will gladly respond to any> and all criticisms of my paper and the system that it describes.I am glad you are open to criticism. Please address mine concerning 1/ the> fact that PC quali?s as a programming language as much as Lisp does, and 2/> the pointlessmess of generating many algorithms when a computer can generate> the best one from the start.See above for both 1 and 2. But ?st show how a LISP expression canrepresent the above 5 useful programming tasks.Charlie VolkstorfCambridge, MA> Sam === > A single expression isnft really a program. Then (1 + 1) is not, contrary to what you said in your previous post.> The problem is that it is too limited.> You seem to be inputting and outputting only expressions (couched in> the syntax of a LISP program.) That is, the program that you> generate is limited to those that consist of a single expression,> limiting your capabilities tremendously. It is not general program> synthesis.I did not say it was. You asked me to write a program that generates a programthat checks if a number is a factor of another. I did (actually I generalizedthat a bit, since it checks the truth of any expression).> Furthermore, I still donft see where you actually do anything with the> expression, other than to input it and then output it again with LISP> syntax surrounding it but I would like to learn more. I check for its validity. Thatfs what needed to achieve what you asked. Noteit in 30 seconds. My intent was to prove that Lisp an be used to generateprograms.> The problem here is that I happened to give an example (determining if> one number is a factor of another) for which there is in fact a LISP> expression. Lisp is Turing complete. Thus anything you can specify using a Turing machine,you can specify with Lisp.> In an attempt to determine if what you have in mind is> very general (as general as the Predicate Calculus, which I use), here> are some useful program speci?ations that I describe in my papers.> What would be the LISP expression for each?Note that this would be more appropriate in comp.lang.lisp. I will be givingways to *specify* your jobs. The actual implementation is irrelevant (albeittrivial in these cases), just as Predicate Calculus is only a syntax, whoseimplementation *is* procedural (partly given in your paper, in the form ofAPC, which is very akin to pseudo code).> 1. List all prime numbers between 1000 and 1005.> PRIME(x)^BETW(1000,x,1005)(format t ~s (mapcar ?primef (list)))Where list contains all the elements.(prime x) evaluates to x if x is prime, to nil otherwise.> 2. Is a given number prime?> PRIME(I)(primep x)> 3. Determine the largest proper factor of a given number.> FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A)(max (proper-factors x))> 4. Determine the largest prime factor of a given number.> FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A)(max (prime-factors x))> 5. Determine the smallest prime factor of a given number.> FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x)(min (prime-factors x))> If you can express these as LISP expressions, then we have something> to talk about. Otherwise, you have a very limited system that is> really just dealing with expressions as opposed to full-?> programs, and you canft express, much less generate, hardly anything.Refer to my comment above. Lisp is Turing complete.> 2. The system needs to CONSIDER multiple algorithms when generating> the program, to differentiate it from the mere translation from one> programming language into another.Again, you are applying the human way of solving problems to a computer.A well-designed system needs not differentiate [the actual program] from themere translation from a programming language into another. It does not havethe notion of such difference. It is designed to take a certain input thatspeci?s a task to be done and return a program that performs the speci?dtask.> As I said above, it quali?s as a programming language.No, Predicate Calculus has no assignment, branching or conditionalized> execution. It canft get into loops. It is non-procedural.It still quali?s as a programming language, albeit not a Turing complete one(just as a Maple-like language without the procedural constructs would be)Sam-- Fear is the path to the dark side. Fear leads to anger, anger leads to hatred, hatred leads to suffering. I sense much fear in you. === A single expression isnft really a program. Then (1 + 1) is not, contrary to what you said in your previous post.Single expressions are a proper subset of programs. You can talkabout a program that consists of a single expression, but that is notprogramming in general. A system that only creates single expressionsis not much of a Program Synthesis system because very few programsconsist of just one expression. It is very limited. (And Ifm not sosure that your system does even that.) See my 5 examples below.> Furthermore, I still donft see where you actually do anything with the> expression, other than to input it and then output it again with LISP> syntax surrounding it but I would like to learn more. I check for its validity. Thatfs what needed to achieve what you asked.So you just have a program syntax checker?> Note that I am not saying that my program is a general program generator.And I am saying that in fact it is not. So we agree? Well, byProgram Synthesis, I mean a general program generator.> Lisp is Turing complete. Thus anything you can specify using a Turing machine,> you can specify with Lisp.With a single expression? Without sepcifying or knowing the algorithmthat will be used? Program Synthesis means that the system constructsthe algorithms.> 1. List all prime numbers between 1000 and 1005.> PRIME(x)^BETW(1000,x,1005)(format t ~s (mapcar ?primef (list)))Where list contains all the elements.You donft even refer to 1000 and 1005! How did list get to containthe prime numbers between 1000 and 1005? > (prime x) evaluates to x if x is prime, to nil otherwise.So prime is a predicate supplied in LISP?> 2. Is a given number prime?> PRIME(I)(primep x)So primep is a predicate supplied by LISP? Whatfs the differencebetween prime and primep? > 3. Determine the largest proper factor of a given number.> FAC(x,I)^LT(x,I)^~(exists A)FAC(A,I)^LT(A,I)^LT(x,A)(max (proper-factors x))So proper-factors is a construct supplied by LISP?> 4. Determine the largest prime factor of a given number.> FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(x,A)(max (prime-factors x))So prime-factors is a primitive supplied by LISP?> 5. Determine the smallest prime factor of a given number.> FAC(x,I)^PRIME(x)^~(exists A)FAC(A,I)^PRIME(A)^LT(A,x)(min (prime-factors x))> 2. The system needs to CONSIDER multiple algorithms when generating> the program, to differentiate it from the mere translation from one> programming language into another.Again, you are applying the human way of solving problems to a computer.So? Anything a computer can do a human can do. You are disqualifyingevery possible program.> A well-designed system needs not differentiate [the actual program] from the> mere translation from a programming language into another. It does not have> the notion of such difference.It doesnft. I am saying that if it is not a mere translator from oneprogramming language to another, then multiple programs (usingdifferent algorithms) would be generated from one program requirement.> As I said above, it quali?s as a programming language.No, Predicate Calculus has no assignment, branching or conditionalized> execution. It canft get into loops. It is non-procedural.It still quali?s as a programming language, albeit not a Turing complete oneThat is a self-contradiction. By de?ition, a programming languageis Turing complete. The point is that it doesnft use programminglanguage comnstructs and is non-procedural. What better way to beable to specify what a program is supposed to do?SUMMARYAre prime, primep, proper-factor and prime-factor all primitivessupplied by LISP? If not, then how can you use them? If so, then atsome point you will come across a program requirement that is not aprimitive supplied by LISP. What do you do then?You seem to be merely declaring that each requirement is a primitivesupplied by LISP and displaying trivial expressions using thoseprimitives. I donft see any signi?ance to that at all. You are notdescribing the general case where the programming requirement is not aLISP primitive. What would you do then?It sounds like you are talking about a programming system in which themerely compiles them into one program. Thatfs just a compiler and notnon-procedural input.Charlie VolkstorfCambridge, MA> Sam === > Furthermore, I still donft see where you actually do anything with the> expression, other than to input it and then output it again with LISP> syntax surrounding it but I would like to learn more.> I check for its validity. Thatfs what needed to achieve what you asked.So you just have a program syntax checker?I donft check the syntax. I check the truth of the expression.> 1. List all prime numbers between 1000 and 1005.> PRIME(x)^BETW(1000,x,1005)> (format t ~s (mapcar ?primef (list)))> Where list contains all the elements.You donft even refer to 1000 and 1005! How did list get to contain> the prime numbers between 1000 and 1005?I said Ifm not giving the whole implementation. This is not about Lispprogramming, this is about specifying requirements. Itfs trivial to constructa list that contains all integers between 1000 and 1005.> (prime x) evaluates to x if x is prime, to nil otherwise.So prime is a predicate supplied in LISP?Irrelevant. If itfs not supplied, supply it. (indeed it is not part of the CLstandard). That applies to the following functions too. If youfre arguingthat PC is better because it provides a prime checking predicate by default,then youfre agreeing that PC is merely a higher level language.Less important, prime is *not* a predicate. A predicate returns either t ornil (true or false). (prime x) returns x if x is prim, nil otherwise. primep(that follows) is a predicate. > Again, you are applying the human way of solving problems to a computer.So? Anything a computer can do a human can do. You are disqualifying> every possible program.You misunderstood my critic. Ifm not saying itfs not possible to do so (thoughit would be hard), Ifm saying itfs not the most ef?ient way.> A well-designed system needs not differentiate [the actual program] from> the mere translation from a programming language into another. It does not> have the notion of such difference.It doesnft. I am saying that if it is not a mere translator from one> programming language to another, then multiple programs (using> different algorithms) would be generated from one program requirement.Show it! That is, write an implementation of your system. For now, you onlyhave a paper that describes your ideal system. It does not describe how toactually implement that system. Lisp has implementations.> By de?ition, a programming language is Turing complete.That is false. A programming language needs not be Turing complete. Aprogramming language is a language that allows you to specify a task to bedone by the computer. It needs not be able to specify *any* task the computercan do.> The point is that it doesnft use programming> language comnstructs and is non-procedural. Simply because you provided (in your design, since there is no implementation)enough high level feaures so that we do not need branching to solve theproblems you are addressing. Note that it does not provide features troadress other kinds of problems. Thus the following:Using PC, please provide a non procedural construct that veri?s if a givenset with 2 laws is a vector space. PC does not provide any facility to verifyif a given object is a member of a given set, neither does it provide a wayto construct objects or laws, so I doubt this is even adressable.Note that this is not even program generation. If you manage to do that, go onto the following:Using PC, provide a program that generates 2 algorithms that verify if a givensubset of a vector space, with the vector space laws, is a sub-vector space.And last, provide a program, using PC, that generates algorithms to check if agiven ?e on a Unix system can be opened by the current user.> SUMMARY[summary snipped]I have adressed these above in this post.> It sounds like you are talking about a programming system in which the> merely compiles them into one program. Thatfs just a compiler and not> non-procedural input.That is so, to a certain extent. But PC is not different, apart from the factthat it is not Turing complete and does provide higher level functions notde?ed in the Common Lisp standard (but that could be part of a mathlibrary). PC is a mathematical non-procedural language. The sole fact that itis purely mathematical makes it unsuitable for program generation outside thecontext of mathematics.Sam-- Creativity can be a social contribution, but only in so far as society is free to use the results. - Richard Stallman === >It is often said that the Lorentz transformation can be derived from>the homogeneity and isotropy of space alone. Ifm looking for a>concrete counterexample to this claim.They canft be. Homogeneity (isnft isotropy redundant with homogeneity?) and the assumption of no special frame (principle of relativity) gets you, for a boost in x, xf = a(x - vt) yf = y zf = z tf = bt + cxThese are Galilean if a=1, b=1, c=0. To derive the Lorentz transforms you can suppose when the origin of two frames S and Sf coincide a ?f light expands out in a sphere. Throw in the additional postulate of the invariance of the speed of light and you get equations in the two frames describing the sphere, (1) x^2 + y^2 + z^2 = c^2 t^2 (2) xf^2 + yf^2 + zf^2 = c^2 tf^2Note that equation (2) explicitly has the speed of light unchanged and the light shell spherical in Sf. Insert the primed quantities de?ed above into equation (2), rearrange terms, and ?d which values of a, b, c are needed to reproduce equation (1).The Lorentz transforms can be derived from the principle of relativity and Maxwellfs equations. That is, since no experiment seems able to identify an aether or aether frame, rather than taking Lorentzfs course of assuming the aether to interact with matter such that it remains undetectable, take Einsteinfs course and assume there really is no special frame, that the principle of relativity applies to electromagnetism as well as to mechanics. Maxwellfs equations remain invariant under Lorentz transformations, so you can do an exercise similar to the above but with more dif?ult math.Either way, you need something more than just homogeneity. Maybe what youfve often seen said just didnft make it clear that Maxwellfs equations were the additional input.-- A good plan executed right now is far better than a perfect planexecuted next week. -Gen. George S. Patton === >It is often said that the Lorentz transformation can be derived from>the homogeneity and isotropy of space alone. Ifm looking for a>concrete counterexample to this claim.> They canft be. Homogeneity (isnft isotropy redundant with homogeneity?) No. Homogeneity means the manifold in question looks the sameeverywhere; isotropy means it looks the same no matter which directionone looks. An anisotropic manifold can be homogeneous, as long as itlooks the same everywhere (i.e. some special direction is everywherethe same). E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2 is homogeneous but not isotropic. Of course a simple change of coordinates can make it isotropic -- that is not always the case: Minkowski spacetime is homogeneous but is not isotropic in 4-d, only in its spatial 3-d slices.A manifold that is isotropic everywhere is necessarily homogeneous.> and the assumption of no special frame (principle of relativity) gets you, > for a boost in x,> xf = a(x - vt)> yf = y> zf = z> tf = bt + cxYou can further constrain a,b,c by requiring these transforms form agroup (i.e. that the composition of two transforms of this type resultsin another transform of this type). One ?ds there are three possibilities: 1) the Galileo group 2) the Lorentz group 3) the Euclid group(3) is unphysical (x,y,z,t all look the same, and the composition o?enough boosts along +x results in a boost along -x).> These are Galilean if a=1, b=1, c=0. To derive the Lorentz transforms you > can suppose when the origin of two frames S and Sf coincide a ?f > light expands out in a sphere. Throw in the additional postulate of the > invariance of the speed of light [...]The Lorentz transforms can be derived from the principle of relativity and > Maxwellfs equations. [...]Given the constraints imposed by the group property, one merely needs toselect from the three possible groups. A postulate like there is anupper bound on speeds suf?es, but one needs a measurement toestablish that this upper bound is the speed of light. A postulate likepion beams exist serves double duty (and is manifestly true)....> Either way, you need something more than just homogeneity. Yes.Tom Roberts tjroberts@lucent.com === >It is often said that the Lorentz transformation can be derived from>the homogeneity and isotropy of space alone. Ifm looking for a>concrete counterexample to this claim.> They canft be. Homogeneity (isnft isotropy redundant with homogeneity?) >No. Homogeneity means the manifold in question looks the same>everywhere; isotropy means it looks the same no matter which direction>one looks. An anisotropic manifold can be homogeneous, as long as it>looks the same everywhere (i.e. some special direction is everywhere>the same).> E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2> is homogeneous but not isotropic. Of course a simple change> of coordinates can make it isotropic -- that is not always> the case: Minkowski spacetime is homogeneous but is not> isotropic in 4-d, only in its spatial 3-d slices.>A manifold that is isotropic everywhere is necessarily homogeneous.Ehh... I guess Ifd buy that, as long as homogeneity is de?ed as looks the same everywhere along one particular direction. I was thinking of homogeneity as looks the same everywhere, no quali?rs.> and the assumption of no special frame (principle of relativity) gets you, > for a boost in x,> xf = a(x - vt)> yf = y> zf = z> tf = bt + cx>You can further constrain a,b,c by requiring these transforms form a>group (i.e. that the composition of two transforms of this type results>in another transform of this type). One ?ds there are three possibilities:> 1) the Galileo group> 2) the Lorentz group> 3) the Euclid groupIfm not familiar with this argument. Could you clue me in?-- A good plan executed right now is far better than a perfect planexecuted next week. -Gen. George S. Patton === >It is often said that the Lorentz transformation can be derived from>the homogeneity and isotropy of space alone. Ifm looking for a>concrete counterexample to this claim.> They canft be. Homogeneity (isnft isotropy redundant with homogeneity?) I tend to agree. To me a point having the same properties as everyother point de?es homogeneity. What does homogeneity mean withoutisotropy?> No. Homogeneity means the manifold in question looks the same> everywhere; isotropy means it looks the same no matter which direction> one looks. An anisotropic manifold can be homogeneous, as long as it> looks the same everywhere (i.e. some special direction is everywhere> the same).I understand the distinction youfre making technically and wonder if aconcrete illustration of an anisotropic, homogeneous manifold exists.> E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2> is homogeneous but not isotropic. Please pardon my ignorance. I had differential geometry from TedFrankel. We never covered any practical examples. What are themathematical de?itions of homogeneity and isotropy in this case?Please explain the essential features of this geometry and the specialdirection. Why are 3 dimensions required? Why doesnft your examplework with just ds^2 = 2dx^2 + 3dy^2 ?Eugene Shuberthttp://www.everythingimportant.org === Perfectly Innocent skrev i melding>It is often said that the Lorentz transformation can be derived from>the homogeneity and isotropy of space alone. Ifm looking for a>concrete counterexample to this claim.> They canft be. Homogeneity (isnft isotropy redundant with homogeneity?)> I tend to agree. To me a point having the same properties as every> other point de?es homogeneity. What does homogeneity mean without> isotropy?> No. Homogeneity means the manifold in question looks the same> everywhere; isotropy means it looks the same no matter which direction> one looks. An anisotropic manifold can be homogeneous, as long as it> looks the same everywhere (i.e. some special direction is everywhere> the same).> I understand the distinction youfre making technically and wonder if a> concrete illustration of an anisotropic, homogeneous manifold exists.Many crystalline substances (e.g. quartz) are homogenous, but not isotropic.The em-properties, including the anisotropy, are the same everywherein the crystal.Think of it.Wouldnft it be rather meaningless to say that a crystal whichmacroscopically is the same everywhere isnft homogenous?Paul === What was I thinking? The cylinder SxR is obviously homogeneous but not isotropic!Eugene Shuberthttp://www.everythingimportant.org/relativity/ simultaneity.htm === > To me a point having the same properties as every> other point de?es homogeneity. What does homogeneity mean without> isotropy?As applied to ?lds on a manifold, homogeneity means the ?ld isindependent of position, and isotropy means the ?ld is independent oforientation. No nonzero vector ?ld can be isotropic except possibly atisolated points, but can certainly be homogeneous if it is constanteverywhere. For the manifold itself, it is common to say the manifold isisotropic or homogeneous iff the metric is. For instance, a 3-dspherically-symmetric manifold is isotropic around its center, butunless it is isometric to E^3 it is not homogeneous, and is notisotropic at any other location. The surface of a sphere S^2 (with theusual metric) is both homogeneous and isotropic. Note I use orientation in the more general sense than the binary choice of orientation for orientable manifolds. I believe there is a technical term for this, but forget what it is.> I understand the distinction youfre making technically and wonder if a> concrete illustration of an anisotropic, homogeneous manifold exists.Minkowski spacetime in 4 dimensions. Timelike orientations are DIFFERENTfrom spacelike orientations, yet the metric is independent of position.The surface of a cylinder RxS (with the usual metric) is isotropic andhomogeneous by the above de?ition. Topologically it certainly has aspecial orientation, but isotropy is a geometrical property, not atopological one.> E.g. the 3-d manifold with metric ds^2 = 2dx^2 + 3dy^2 + 4dz^2> is homogeneous but not isotropic. I misspoke. The manifold and its metric are indeed isotropic, but theCOORDINATES are not. Oops. This manifold is isometric to E^3 (up totopological considerations which arenft mentioned).Tom Roberts tjroberts@lucent.com === > Be careful with what you mean by one-to-one. Goedelfs method is a> method for setting up a one-to-one correspondence between the> expressions in the calculus and a _certain subset_ GC of the integers.> In particular, not every integer is a Goedel number. Consider the number> 100 = 2^2*5^2. Looking at this prime factorization it becomes> immediatelly obvious that the number 100 cannot be a Goedel codon, since> powers of the prime 3 are skipped, TOGETHER with the fact that powers of> 5 appear. This is by de?ition not allowed.> On the other hand, given G in GC, you can easily decompose G into its> prime number factorization and assuming the decomposition follows proper> codi?ation rules (i.e. no numbers like 100) therefrom derive a unique> sequence of exponents of the ?st n primes, which will lead you back to> a sequence of fundamental codons, which can be used to recover the> calculus expression uniquely.> The above paragraph can be used slightly modi?d to prove what you> want.> number we get to a unique sequence of exponents and, given a codi?ation,> to a statement in PM. My question is: how is it proved that there exists a> complete codi?ation between statements in PM and these exponent> (basically natural numbers). In other word how does Godel prove that all> statements in PM can be mapped into the natural numbers? Intuitively what is> the Godel number of the class of all classes?>The essense is that the statements are countable. The Godel numbering showsthat any sequence of alphabet symbols is countable. Are you implying thatformula may exist that are not sequences of alphabet symbols?Herc === > Be careful with what you mean by one-to-one. Goedelfs method is a> method for setting up a one-to-one correspondence between the> expressions in the calculus and a _certain subset_ GC of the integers.> In particular, not every integer is a Goedel number. Consider thenumber> 100 = 2^2*5^2. Looking at this prime factorization it becomes> immediatelly obvious that the number 100 cannot be a Goedel codon,since> powers of the prime 3 are skipped, TOGETHER with the fact that powersof> 5 appear. This is by de?ition not allowed.> On the other hand, given G in GC, you can easily decompose G into its> prime number factorization and assuming the decomposition followsproper> codi?ation rules (i.e. no numbers like 100) therefrom derive aunique> sequence of exponents of the ?st n primes, which will lead you backto> a sequence of fundamental codons, which can be used to recover the> calculus expression uniquely.> The above paragraph can be used slightly modi?d to prove what you> want.>a> number we get to a unique sequence of exponents and, given acodi?ation,> to a statement in PM. My question is: how is it proved that there existsa> complete codi?ation between statements in PM and these exponent> (basically natural numbers). In other word how does Godel prove thatall> statements in PM can be mapped into the natural numbers? Intuitivelywhat is> the Godel number of the class of all classes?> The essense is that the statements are countable. The Godel numberingshows> that any sequence of alphabet symbols is countable. Are you implying that> formula may exist that are not sequences of alphabet symbols?> Herc>What I am asking is: how is it proved that the symbols from which thesestatements are derived is countable. For example, these symbols cannotinclude irrational numbers. Taking it a step further does this mean thatthere are no statements about irrational numbers in PM? === >number we get to a unique sequence of exponents and, given a codi?ation,>to a statement in PM. My question is: how is it proved that there exists a>complete codi?ation between statements in PM and these exponent>(basically natural numbers). In other word how does Godel prove that all>statements in PM can be mapped into the natural numbers?Given a speci?ation of the language PM (in particular the alphabet it uses), this is very easy.> Intuitively what is>the Godel number of the class of all classes?It is not mathematical objects, but their descriptions in the language, that have Godel numbers. The same object could have many different descriptions in the language, and each will have a different Godel number. On the other hand, some objects (e.g. almost all real numbers) can not be completely described in the language. I donft know how youfd describe the class of all classes in PM.If you were encoding English, with letters correspondingto their ASCII codes, you could give the class of all classes theGodel number 2^116 * 3^104 * 5^101 * ... * 89^115, where 116, 104, 101, ... are the ASCII codes of t, h, e, ....Robert Israel israel@math.ubc.caDepartment of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === architechture. Unlike JSH, there is nothing groundbreaking orearthshattering about it. Are there any journals that look for papers onthe GR speci?ally (like Fibonnaci Quarterly for example) or any otherjournals that would look for a paper on this topic. === 8.9 x 14.4 ?===Out of curiosity. Why on earth do sci.mathfers - which otherwise seemsto be quite intelligent people - even consider replying to JamesHarrisf posts?First of all, he is the crank fatale of our century. Second, he isprobably the kind of crank who rolls on the ?aughing his ass offeach time someone replies.It appears that replying to Harris is like watching a gold ?h in abowl. You canft stop, yet you know itfs not going anywhere but aroundand around, ad in?itum. === > Out of curiosity. Why on earth do sci.mathfers - which otherwise seems> to be quite intelligent people - even consider replying to James> Harrisf posts?Another Harris basher. What a surprise.> First of all, he is the crank fatale of our century. Second, he is> probably the kind of crank who rolls on the ?aughing his ass off> each time someone replies.That would make him a troll, not a crank. If you must insult him, atleast get your insults right.> It appears that replying to Harris is like watching a gold ?h in a> bowl. You canft stop, yet you know itfs not going anywhere but around> and around, ad in?itum.It does seem to be going around and around, doesnft it? And herefs why:*** James Harris has proved Fermatfs Last Theorem ***but people are either unwilling or unable to accept it. I admit that Ipersonally cannot understand his proof: itfs beyond my comprehension.James says it should be easy for everyone to understand, but (pleaseforgive my impudence, James, in offering you some constructive criticism)perhaps he has failed to take into account that he is so much smarter thanthe rest of us that what seems easy to him may not be so easy for us.The reason that his threads go around and around is that people keep havingthe same misunderstandings, and he keeps trying to enlighten them with thetruth . . . but to no avail. I wonder whether people are as thick as theyseem, or if they have some sort of mental block against the mere possibilityof a simple proof of FLT (at least by a non-mathematician), or, and this iswhat I fear, if they *do* understand that itfs correct, but are attemptingto sti?out of some perverted form of self-interest.But maybe Ifm just being paranoid. The reaction to James could be explainedby sheer stupidity. Therefs not necessarily a conspiracy. I myself, untilrecently, did not accept Harrisfs proof because . . . (drumroll please) . . .I did not understand it. For example, take his use of objects. I can sensethat what hefs doing here is profound and radical. In contrast to the rampantabstraction of conventional mathematics, James is working with concrete things,things that have an actual existence, not just some member of a set, which wasconstructed under the ZFC axioms, with such and so operators de?ed to havethese various properties, yadda yadda yadda. James doesnft need all that crap!He just gets to the core of the matter in one leap! I wish I could follow him.I donft even understand what is and is not an object. But unlike certainabusive Harris harrassers, I have the guts to admit that the fault lies with mefor not understanding. How cowardly to blame James for onefs incomprehension!Shame on you people!The real crime here is not that FLT is being ignored -- sometimes it is onlygenerations later that genius is understood -- but that sci.math is wasting somuch of Jamesfs time with our nitpicking. James has proved FLT. Itfs over anddone with. And sure, that would be enough for one lifetime, but think about howmuch more he has to offer!James has already begun work on the theory of prime numbers. I for one wouldlike to see him bring this work to fruition in a proof of the Riemann Hypothesis.Then maybe we can talk about this whole mathematical messiah thing without thedisruption of rude snickerings. === Can anyone help me identify the sequence of polynomials de?ed by therecurrence relation:p_n (x) = x p_{n-1} (x) + p_{n-2} (x)This leads to the following few terms:p_0 = 0p_1 = 1p_2 = xp_3 = 1 + x^2p_4 = 2x + x^3p_5 = 1 + 3x^2 + x^4p_6 = 3x + 4x^3 + x^5p_7 = 1 + 6x^2 + 5x^4 + x^6-Lotofun === > Can anyone help me identify the sequence of polynomials de?ed by the> recurrence relation:p_n (x) = x p_{n-1} (x) + p_{n-2} (x)This leads to the following few terms:p_0 = 0> p_1 = 1> p_2 = x> p_3 = 1 + x^2> p_4 = 2x + x^3> p_5 = 1 + 3x^2 + x^4> p_6 = 3x + 4x^3 + x^5> p_7 = 1 + 6x^2 + 5x^4 + x^6If you arrange the coef?ients carefully, you getPascalfs triangle.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html The League of Gentlemen === Can anyone help me identify the sequence of polynomials de?ed by the> recurrence relation:p_n (x) = x p_{n-1} (x) + p_{n-2} (x)This leads to the following few terms:p_0 = 0> p_1 = 1> p_2 = x> p_3 = 1 + x^2> p_4 = 2x + x^3> p_5 = 1 + 3x^2 + x^4> p_6 = 3x + 4x^3 + x^5> p_7 = 1 + 6x^2 + 5x^4 + x^6If you arrange the coef?ients carefully, you get> Pascalfs triangle.p_{2m} (x) = sum_{k=0}^m C(m+k,2k) x^{2k}andp_{2m+1} (x) = sum_{k=0}^m C(m+k+1,2k+1) x^{2k+1}where C(n,j) is the binomial coef?ient. === > Can anyone help me identify the sequence of polynomials de?ed by the> recurrence relation:p_n (x) = x p_{n-1} (x) + p_{n-2} (x)This leads to the following few terms:p_0 = 0> p_1 = 1> p_2 = xLet C = sqrt(x^2/4 + 1), A = x/2 + C, B = x/2 - Cthen p_n = (A^n - B^n)/(2C) === > |There have been NO demonstrations to the contrary. If you have one,> |give it.> Ifve lost count of how many times theyfve presented a non-unit common> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3> as usual are algebraic integers satisfying> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,> and likewise for a_2 and a_3 by substituting them for a_1. Wefd be happy> to post the _calculation_ again if we thought youfd read it.>Huh? Are you sure youfre claiming that the non-monic primitive>irreducible over Q supports your case?You are just saying words. 8a_1^2 + 4a_1 - 45 is not a non-monicprimitive irreducible over Q. Itfs not even a polynomial. Itfs analgebraic integer, just like 8*(4)^2 + 4(4) - 45 is an integer.>Oh, I know. Youfre probably talking about that argument that Arturo>Magidin tossed at me in a post that I just refuted.>Basically any such claims depend on the assertion that an algebraic>integer ?af that is not a unit, must have a non-unit algebraic integer>factor.If a is not a unit, then a itself is a non-unit algebraic integerfactor of itself. Thatfs because this ring has a 1.If thatfs hard for you, think about integers: any integer other than1, -1, is a factor of itself.>Thatfs a sneaky little argument as itfs getting the reader to assume>something not proven, which is that ?af actually has a non-unit>algebraic integer factor because itfs not a unit.> Your argument, on the other hand, always relies at the crucial point> on the reader sharing your feeling that the kind of weird behavior> the common divisors have as f and m vary just couldnft possibly> be so.> Keith Ramsay>Well if youfre telling the truth, ?l in the gap Ifve pointed out in>YOUR argument.>The gap is the assertion that given an algebraic integer ?af which is>not a unit, that ?af must have some non-unit factor *in the ring of>algebraic integers*.This is nonsense.>To protect readers from thinking itfs simple, I remind them of the>ring of evens, that is the set of even numbers, where you have 2 not a>factor of 6, and not a unit. But 2 and 6 donft have factors in the>ring.>Thatfs because the ring doesnft have 1.Are you claiming that the ring of algebraic integers does not have a1? === ========================================= === ======Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu=== [.snip.]>If thatfs hard for you, think about integers: any integer other than>1, -1, is a factor of itself.That, of course, should be non-unit factor or nontrivial factor. === |There have been NO demonstrations to the contrary. If you have one,>|give it.>Ifve lost count of how many times theyfve presented a non-unit common>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3>as usual are algebraic integers satisfying> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,>and likewise for a_2 and a_3 by substituting them for a_1. Wefd be happy>to post the _calculation_ again if we thought youfd read it.> Huh? Are you sure youfre claiming that the non-monic primitive> irreducible over Q supports your case?Oh, I know. Youfre probably talking about that argument that Arturo> Magidin tossed at me in a post that I just refuted.Basically any such claims depend on the assertion that an algebraic> integer ?af that is not a unit, must have a non-unit algebraic integer> factor.> The factor that has been provided, r(a) = 8 a^2 - 4 a - 45is> (1) an algebraic integer for any algebraic integer a,> (2) a divisor of both a and 5 for> a = -(any of the aifs in the above> factorization of 65x^3 - 12x + 1)The factorizations have already been given, but since youfre so adept> at ignoring the evidence, I will provide them once again: q(a) = 8 a^2 - 76 a - 185> r(a) = 8 a^2 - 4 a - 45> s(a) = 4 a^2 - 37 a - 104Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative*> of any of the aifs of the above factorization of 65x^3 - 12x + 1), the> following factorizations hold: q(a) r(a) = 5> r(a) s(a) = a.Hmmm...the poster has tried to dupe *you* the reader, and it all hasto do with the word unit. See below... > In fact, the minimal polynomial of this number (r(a) for -a = any of> the above aifs) is given as: MinPoly(r) = x^3 - 969 x^2 + 315 x + 5The above facts prove that this a has a non-unit algebraic integer> as a factor.No it doesnft. What you can prove is that if ?af is coprime to 5 itcanft be a unit in the ring of algebraic integers, but as Ifve saidthe ring of algebraic integers is screwed up.Your algebraic manipulations will just keep running into that over andover again.> Thatfs a sneaky little argument as itfs getting the reader to assume> something not proven, which is that ?af actually has a non-unit> algebraic integer factor because itfs not a unit.> This is actually *much* easier to prove: the ring of algebraic> integers is closed under the extraction of roots of monic> polynomials with algebraic integer coef?ients. In particular,> every equation: x^n - a = 0Thatfs not the equation you have.What you have is x^3 - 969 x^2 + 315 x + 5and what you want to do is convince readers that none of its roots canbe coprime to 5.But you cannot do that, and instead have maintained that proving thatnone of the roots can be a unit proves that one of them must have anon-unit factor in common with 5 in the ring of algebraic integers.Ifve just shot down that little trick, but youfre squirming.> has algebraic integer roots, whenever a is an algebraic integer,> and n is a natural number.Thus, if a is an algebraic integer, and n is any natural number,> the number a^(1/n)is an algebraic integer, irrespective of which of the n roots you take.Duh, but youfre not talking about such a simple expression as x^n - a = 0.Youfre talking about x^3 - 969 x^2 + 315 x + 5 = 0and *repeatedly* posters like you try to fool readers on thenewsgroups as you work to *convince* rather than get to themathematical truth.The truth is that youfre relying on the negative, which is that NONEof the roots of that expression can be units, to try and prove thepositive, which is that then they all have a non-unit factor in commonwith 5 in the ring of algebraic integers.But thatfs beyond bogus as my point is that the ring of algebraicintegers is screwed up to the extent that a root of that expressioncan be coprime to 5, and yet not be a unit.Your argument, on the other hand, always relies at the crucial point>on the reader sharing your feeling that the kind of weird behavior>the common divisors have as f and m vary just couldnft possibly>be so.>Keith Ramsay> Well if youfre telling the truth, ?l in the gap Ifve pointed out in> YOUR argument.> No gap.Thatfs what you want to convince readers.However, simply asserting something that is not true does not ?l thegap.Again, what you have is the negative--certain numbers arenftunits--and you want the positive--that they all have to have non-unitfactors in common with 5.But you havenft proven the positive.Itfs a gap.Denying it wonft ?l it, so quit being lazy and try to ?l it.> The gap is the assertion that given an algebraic integer ?af which is> not a unit, that ?af must have some non-unit factor *in the ring of> algebraic integers*.> As I mentioned above, the factor r that I provided *is* an algebraic> integer, and in the case that a is (-1) times any of the aifs in the> factorization cited, I have provided its minimal polynomial. It is> simple to verify from that minimal polynomial that itfs an algebraic> integer and that itfs not a unit in the ring of algebraic integers.See readers? You can catch the trick here as notice the poster saidnot a unit, and what Ifm telling you is that these posters areworking hard to convince YOU the reader.> To protect readers from thinking itfs simple, I remind them of the> ring of evens, that is the set of even numbers, where you have 2 not a> factor of 6, and not a unit. But 2 and 6 donft have factors in the> ring.Thatfs because the ring doesnft have 1.You have to go to a higher ring, in this case the ring of integers, to> escape that interesting little situation.Now the problem is different in the ring of algebraic integers, but> the ring is screwed up, and the ? is again a higher ring.What I want you to consider is the *possibility* that weird things can> happen with rings, when youfre not careful enough, and that> mathematicians werenft careful enough with the ring of algebraic> integers.> The problem is with your ignorance. The ring of algebraic integers has> none of the shortcomings you seem to want it to have. It is your method> which is at fault, since it forces you to believe a falsehood, namely> that the aifs in that above factorization are coprime to 5.Now this poster is the one who tried to fool all of you, and now thatIfve caught him, he decides to hurl insults.These posters are intellectually lazy, and theyfre rude.But theyfre not lazy when it comes to posting as they *keep* postingnow donft they?Ifm curious about what you, the reader, thinks is the reason for theirpersistence.Why do YOU think they keep posting?James Harris === > |There have been NO demonstrations to the contrary. If you have one,> |give it.> Ifve lost count of how many times theyfve presented a non-unit common> divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3> as usual are algebraic integers satisfying> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).> Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,> and likewise for a_2 and a_3 by substituting them for a_1. Wefd be happy> to post the _calculation_ again if we thought youfd read it.>Huh? Are you sure youfre claiming that the non-monic primitive>irreducible over Q supports your case?>Oh, I know. Youfre probably talking about that argument that Arturo>Magidin tossed at me in a post that I just refuted.>Basically any such claims depend on the assertion that an algebraic>integer ?af that is not a unit, must have a non-unit algebraic integer>factor.>Thatfs a sneaky little argument as itfs getting the reader to assume>something not proven, which is that ?af actually has a non-unit>algebraic integer factor because itfs not a unit.If thatfs really what is behind it, then itfs easy to prove: if a isnot a unit in a ring with 1, then a is a nonunit factor of a. Again, remember:DEFINITION. Let R be a commutative ring, and let a and b be elementsof R. Then a is a factor of b (in R) if and only if there exists anelement c in R such that a*c = b.Since a=1*a, then a is a factor of a. Since a is not a unit, a is anon-unit factor of a.Surely you agree with ->that<-.As to the refutation, you sure talked a lot, but you did notrefute anything. Herefs why you did not refute it:You are demanding a proof that if a1, a2, a3 are algebraic integerschosen so that 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1), theneach of a1, a2, a3 have a non unit factor in common with 5.That is, we need to ?d algebraic integers r1,s1,q1,r2,s2,q2,r3,s3,q3such that(1) r1*s1 = a1; r1*q1 = 5 (so r1 is a factor in common of a1 and 5);(2) r2*s2 = a2; r2*q2 = 5 (so r2 is a factor in common of a2 and 5);(3) r3*s3 = a3; r3*q3 = 5 (so r3 is a factor in common of a3 and 5);(4) none of r1, r2, r3 are units in the ring of algebraic integers. Surely, you agree, that would necessarily show that a1 has a non unitfactor in common with 5 (namely r1); that a2 has a nonunit factor incommon with 5 (namely r2); and that a3 has a nonunit factori n commonwith 5 (namely r3). The values were given explicitly: Note that-a1^3 - 12*a1^2 + 65 = 0-a2^3 - 12*a2^2 + 65 = 0-a3^3 - 12*a3^2 + 65 = 0.Let r1 = 8*a1^2 + 4*a1 - 54 q1 = 8*a1^2 + 76a1 - 185 s1 = -(4a1^2 + 37a1 - 104) r2 = 8*a2^2 + 4*a2 - 54 q1 = 8*a2^2 + 76a2 - 185 s1 = -(4a2^2 + 37a2 - 104) r3 = 8*a3^2 + 4*a2 - 54 q1 = 8*a3^2 + 76a3 - 185 s1 = -(4a3^2 + 37a3 - 104)Then a simple calculation using the fact that -a1^3 - 12*a1^2 + 65 = 0;-a2^3 - 12*a2^2 + 65 = 0;-a3^3 - 12*a3^2 + 65 = 0;shows that r1*q1 = r2*q2 = r3*q3 = 5.Another simple calculation shows thatr1*s1 = a1r2*s2 = a2r3*s3 = a3.(The explicit calculations may be found either in Dalefs originalpost,http://groups.google.com/groups?selm= 3F148963.6000800%40farir.comor my more recent almost-verbatim-quote inhttp://groups.google.com/groups?selm=bh6666%24a7e%241% 40agate.berkeley.edu)Note also that all of r1, r2, r3, s1, s2, s3, q1, q2, a3 are algebraicintegers. This simply because they are obtained from algebraicintegers by addition and multiplication by integers and/or algebraicintegers.So it is CERTAINLY the case that r1 is an algebraic integer which is acommon factor, IN THE RING OF ALGEBRAIC INTEGERS, of a1 and 5; that r2is an algebraic integer which is a common factor IN THE RING OFALGEBRAIC INTEGERS, of a2 and 5; and that r3 is an algebraic integerwhich is a common factor IN THE RING OF ALGEBRAIC INTEGERS, of a3 and5. We have not, in any way, left or been pushed out of the ring ofalgebraic integers. Finally, we just need to verify that none of r1, r2, or r3 are unitsin the ring of algebraic integers. If that is indeed the case, WE AREDONE. And we are done, because we already know they are commonfactors, we only need to know they are not units in this ring.The word unit is shorthand. It means nothing more and nothing lessthan:DEF. Let R be a ring with 1. An element a in R is a unit (in R) if andonly if there exists an element b IN R such that a*b=b*a = 1.In a commutative ring, of course, it is enough to check a*b=1.Then we have the following easy theorem, consequence of the theoremyou accepted last December after 10 months of denying it:THEOREM. Let a<>0 be an algebraic integer, and let f(x) be a monicpolynomial with integer coef?ients, irreducible over Q, which has aas a root. Then a is a unit in the ring of all algebraic integers ifand only if the constant term of f(x) is either 1 or -1.Proof. Let b be the unique complex number such that a*b=1. That is,b=1/a. By de?ition, a is a unit in the ring of all algebraicintegers if, and only if, b is an algebraic integer.Let f(x) = x^n + ... + a_1*x + a_0.Then f(a) = 0. Multiplying through by b^n, we have0 = b^nf(a) = b^n*a^n + ... + a_1*a*b^{n} + a_0*b^n = 1 + ... = a_1*b^{n-1} + a_0*b^n.Let g(x) = a_0*x^n + ... + a_{n-1}*x + 1.That means that g(b) = 0. Since any factorization of g(x) will producea factorization of f(x) by substituting x^{-1} for x and clearingdenominators, we see that g(x) is an irreducible polynomial withinteger coef?ients. Since the last term equals 1, it is thereforeprimitive.Therefore, b is an algebraic integer if and only if the leading termof g(x) is either 1 or -1. Since the leading term of g(x) is a_0, thatsays that b is an algebraic integer if and only if the constant termof f(x) is 1 or -1. IN summary:a is an algebraic integer unit if and only if b is an algebraicinteger; b is an algebraic integer if and only if the constant term off(x) is 1 or -1. Therefore, a is an algebraic integer unit if and onlyif the constant term of f(x) is 1 or -1.QEDSo one way to prove that none of r1, r2, r3 are units in the ring ofall algebraic integers is to exhibit a monic polynomial with integercoef?ients, irreducible over Q, with constant term different from 1and -1, which has r1, r2, r3 as roots. (Not the only way: we could try?ding a different one for each ri, but it so happens they are allroots of the same polynomial).And indeed, as Dale noted, if we letf(x) = x^3 - 969x^2 + 315x + 5, it so happens that f(r1)=f(r2)=f(r3)=0. The irreducibility over Q off(x) may be checked by the rational root theorem: the only possibleroots are 1, -1, 5, and -5, none of them are roots, and so we aredone.Now, you can dance around the issue all you want, James. You canproduce paragraph upon paragraph, post upon post, saying all sorts ofthings about problems and sneakiness and how it is impossible to showsomething or other.But you cannot get around one very simple issue: we have EXPLICITLYshown the factors, we have EXPLICITLY shown they are factors, we haveEXPLICITLY shown they are not units, we have EXPLICITLY shown they arecommon factors. We have EXPLICITLY produced the factors you claim cannot exist.Anything else is sophistry.(Though I say we, the credit goes of course to Dale who actuallytook the time to make the calculations and initial veri?ations). === ================================= === ==============Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === ========>|There have been NO demonstrations to the contrary. If you have one,>|give it.>Ifve lost count of how many times theyfve presented a non-unit common>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3>as usual are algebraic integers satisfying> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,>and likewise for a_2 and a_3 by substituting them for a_1. Wefd be happy>to post the _calculation_ again if we thought youfd read it.> Huh? Are you sure youfre claiming that the non-monic primitive> irreducible over Q supports your case?> Oh, I know. Youfre probably talking about that argument that Arturo> Magidin tossed at me in a post that I just refuted.> Basically any such claims depend on the assertion that an algebraic> integer ?af that is not a unit, must have a non-unit algebraic integer> factor.> The factor that has been provided,> r(a) = 8 a^2 - 4 a - 45 is> (1) an algebraic integer for any algebraic integer a,> (2) a divisor of both a and 5 for> a = -(any of the aifs in the above> factorization of 65x^3 - 12x + 1)> The factorizations have already been given, but since youfre so adept> at ignoring the evidence, I will provide them once again:> q(a) = 8 a^2 - 76 a - 185> r(a) = 8 a^2 - 4 a - 45> s(a) = 4 a^2 - 37 a - 104> Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative*> of any of the aifs of the above factorization of 65x^3 - 12x + 1), the> following factorizations hold:> q(a) r(a) = 5> r(a) s(a) = a.>Hmmm...the poster has tried to dupe *you* the reader, and it all has>to do with the word unit. See below... In fact, the minimal polynomial of this number (r(a) for -a = any of> the above aifs) is given as:> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5> The above facts prove that this a has a non-unit algebraic integer> as a factor.>No it doesnft. Why not? There is an algebraic integer, called here r(a), with thefollowing three properties: (1) It is not a unit in the ring of all algebraic integers; you apparently agree with that conclusion, but then engage in sophistry about a supposed problem with it. (2) There is an algebraic integer, called s(a), such that r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic integers. (3) There is an algebraic integer, called q(a), such that r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic integers.So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5.Why is it not the case that r(a) is a non-unit algebraic integer whichis a common factor of a and 5? [.snip.]>What you have is> x^3 - 969 x^2 + 315 x + 5>and what you want to do is convince readers that none of its roots can>be coprime to 5.There is no convincing that needs to be done. NONE Of the roots areunits, and ALL the roots are factors of 5. Youfre done.>But you cannot do that,Correction: James Harris cannot do that because he does not know whatcoprime means.> and instead have maintained that proving that>none of the roots can be a unit proves that one of them must have a>non-unit factor in common with 5 in the ring of algebraic integers.They are each FACTORS of 5, and EACH non-units.Do you agree or disagree with (all happening in a commutative ringWITH 1): (a) If a is a factor of b, then a and b have a common factor (namely, a). (b) If a is a factor of b and is not a unit, then a is a non-unit common factor of a and b. (c) If is a factor of b and is not a unit, then a and b have a non-unit common factor (namely, a).>Ifve just shot down that little trick, but youfre squirming.I do see a lot of squirming, but itfs not coming from Dale. [.snip.]>The truth is that youfre relying on the negative, which is that NONE>of the roots of that expression can be units, to try and prove the>positive, which is that then they all have a non-unit factor in common>with 5 in the ring of algebraic integers.You are completely, totally, utterly lost and confused.The roots themselves are factors of 5. Each root is a factor o?self, and a factor of 5. Therefore, each root is a common factor o?self and 5. And since each of them is not a unit, then we haveEXHIBITED a non-unit common factor of each with 5. The roots are notthe original numbers we were interested in (the afs), they are thecommon factors that have been produced. [.snip.]> As I mentioned above, the factor r that I provided *is* an algebraic> integer, and in the case that a is (-1) times any of the aifs in the> factorization cited, I have provided its minimal polynomial. It is> simple to verify from that minimal polynomial that itfs an algebraic> integer and that itfs not a unit in the ring of algebraic integers.>See readers? You can catch the trick here as notice the poster said>not a unit, and what Ifm telling you is that these posters are>working hard to convince YOU the reader.Are you claiming that non-unit and not a unit are not the samething? [.snip.] === ===================================== === ==========Why do you take so much trouble to expose such a reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A manfs capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of ?ures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the inde?ite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan === ======================================= === =========Arturo Magidinmagidin@math.berkeley.edu===|There have been NO demonstrations to the contrary. If you have one,>|give it.>Ifve lost count of how many times theyfve presented a non-unit common>divisor between each of a_1, a_2, and a_3 with 5, where a_1, a_2, and a_3>as usual are algebraic integers satisfying> 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).>Namely, 8 a_1^2 + 4 a_1 - 45 is a non-unit common factor of a_1 with 5,>and likewise for a_2 and a_3 by substituting them for a_1. Wefd be happy>to post the _calculation_ again if we thought youfd read it.>Huh? Are you sure youfre claiming that the non-monic primitive>irreducible over Q supports your case?>Oh, I know. Youfre probably talking about that argument that Arturo>Magidin tossed at me in a post that I just refuted.>Basically any such claims depend on the assertion that an algebraic>integer ?af that is not a unit, must have a non-unit algebraic integer>factor.>The factor that has been provided,> r(a) = 8 a^2 - 4 a - 45>is> (1) an algebraic integer for any algebraic integer a,> (2) a divisor of both a and 5 for> a = -(any of the aifs in the above> factorization of 65x^3 - 12x + 1)>The factorizations have already been given, but since youfre so adept>at ignoring the evidence, I will provide them once again:> q(a) = 8 a^2 - 76 a - 185> r(a) = 8 a^2 - 4 a - 45> s(a) = 4 a^2 - 37 a - 104>Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative*>of any of the aifs of the above factorization of 65x^3 - 12x + 1), the>following factorizations hold:> q(a) r(a) = 5> r(a) s(a) = a.> Hmmm...the poster has tried to dupe *you* the reader, and it all has> to do with the word unit. See below...>In fact, the minimal polynomial of this number (r(a) for -a = any of>the above aifs) is given as:> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5>The above facts prove that this a has a non-unit algebraic integer>as a factor.> No it doesnft. What you can prove is that if ?af is coprime to 5 it> canft be a unit in the ring of algebraic integers, but as Ifve said> the ring of algebraic integers is screwed up.> No, no one can prove that. After all, a unit *is* coprime to everynumber!Let u be a unit algebraic integer, and let p be any algebraic integer.Then there is an algebraic integer v for which uv = 1, and we get: u v + 0 p = 1,where 0 is zero.This establishes the fact that a UNIT algebraic integer is coprime toevery algebraic integer, in the ring of algebraic integers. A veryslight rewording proves the result for a unit of any commutativering.No wonder you think the ring of algebraic integers is screwed up!Letfs get the language straight. I have been clear in this that ?af refers to (-1) times any one of the coef?ients ai in the factorization 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1).I claim that *each* of those numbers a has a non-unit algebraic integerfactor in common with 5.PROOF: The above expressions q(a),r(a),s(a) all yield algebraic integersfor any algebraic integer a. When, in addition, a is a root of thepolynomial x^3 - 12 x^2 + 65then the products are as I mentioned above: q(a) r(a) = 5 r(a) s(a) = a.Your claim: > No it doesnft. What you can prove is that if ?af is > coprime to 5 it canft be a unit in the ring of algebraic > integers, but as Ifve said the ring of algebraic integers > is screwed up.amounts to this:PARAPHRASE: It doesnft matter whether anyone provides algebraic integers q,r, and s for which 5 = q*r a = r*s it doesnft prove that a and 5 are not coprime.However, it *does* prove that. Assume otherwise, that a is coprime to 5. Then there exist algebraic integers u and v for which a u + 5 v = 1and the above factorizations show this: rs u + qr v = 1so r( su + qv ) = 1,and we deduce that r is a unit. However, itfs not one of your fancifulUnit with no inverse hallucinations. In fact, the above equationprovides us with an inverse for r. Based on its construction, it*must be* an algebraic integer: u and v are speci?d to be algebraicintegers, and q,r,s were already speci?d as algebraic integers.Thus, r must be an algebraic integer unit, in the ordinary sense: an algebraic integer whose inverse is *also* an algebraic integer.However, I show below that the minimal polynomial for r precludes itbeing a unit.> Your algebraic manipulations will just keep running into that over and> over again.> Ifve given the proof before, and itfs only cut ?nf paste, so here it is,once again:Given: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104A quick bit of arithmetic will show the following: q(x)*r(x) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8325 r(x)*s(x) = 32 x^4 - 312 x^3 - 864 x^2 + 2081 x + 4680and: (64 x + 128)*(x^3 - 12 x^2 + 65) = 64 x^4 - 640 x^3 - 1536 x^2 + 4160 x + 8320 (32 x + 72)*(x^3 - 12 x^2 + 65) = 32 x^4 - 312 x^3 - 864 x^2 + 2080 x + 4680and so, if you care to compare: q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x,where p(x) = x^3 - 12 x^2 + 65. Note that this immediately showsthat for z = any root of p(x), q(z)*r(z) = 5, r(z)*s(z) = z.To derive the minimal polynomial: > MinPoly(r) = x^3 - 969 x^2 + 315 x + 5let mpr(x) = x^3 - 969 x^2 + 315 x + 5.First, expand the polynomial mpr(r(x)): mpr(r(x)) = (r(x))^3 - 969 (r(x))^2 + 315 (r(x)) + 5 = (8 x^2 - 4 x - 45)^3 - 969 (8 x^2 - 4 x - 45)^2 + 315 (8 x^2 - 4 x - 45) + 5 = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520Next, multiply these two polynomials: p(x) = x^3 - 12 x^2 + 65,and w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808to get this: p(x)*w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520Notice the equality mpr(r(x)) = p(x)*w(x).That means for every value of x, the polynomial you get by computingr(x), then evaluating mpr(x) at that value, is equal to the product ofp(x) and w(x).In short, if x is a root of p(x) (in particular, this is true forx = -ai for any of your aifs), mpr(r(x)) = 0.Thus, the rfs are roots of the polynomial mpr(x) = x^3 - 969 x^2 + 315 x + 5This cubic polynomial has integer coef?ients, is monic,and has no integer roots, thus itfs irreducible. Its rootscannot be units in the ordinary sense (i.e., with inverseswithin the ring of algebraic integers), because theconstant term is a non-unit.So, what can I conclude at this point? 1. each of the afs has an algebraic integer factor in common with 5: 5 = q(a)*r(a) a = r(a)*s(a). where q(a) = 8 a^2 - 76 a - 185 r(a) = 8 a^2 - 4 a - 45 s(a) = 4 a^2 - 37 a - 104 These values q(a),r(a),s(a) are all algebraic integers. 2. If a and 5 were coprime, then there would be an inverse *in the ring of algebraic integers* for r(a). 3. The minimal polynomial mpr(x) for r(a) is: mpr(x) = x^3 - 969 x^2 + 315 x + 5Thatfs a sneaky little argument as itfs getting the reader to assume>something not proven, which is that ?af actually has a non-unit>algebraic integer factor because itfs not a unit.>This is actually *much* easier to prove: the ring of algebraic>integers is closed under the extraction of roots of monic>polynomials with algebraic integer coef?ients. In particular,>every equation:> x^n - a = 0> Thatfs not the equation you have.> Idiot! Ifm showing that your alleged something not proven, as Iquote: ... itfs getting the reader to assume something not proven, which is that ?af actually has a non-unit algebraic integer factor because itfs not a unit.is hogwash! EVERY NON-UNIT ALGEBRAIC INTEGER HAS NON-UNIT FACTORS AMONG THE ALGEBRAIC INTEGERS!Ifll repeat that: Take any algebraic integer K that is NOT a unit. Then take the equation x^n - K = 0. and solve it. The roots are algebraic integers, and they are ALSO non-unit factors of K.What was it you claimed was unproven: that ?af actually has a non-unit algebraic integer factor because itfs not a unit.What did I just ?d: a non-unit algebraic integer factor K^(1/n), for an ARBITRARY non-unit algebraic integer K.Does that address this tiny little point?Can we get off that one then? Youfre wrong, Ifm right.> What you have is x^3 - 969 x^2 + 315 x + 5and what you want to do is convince readers that none of its roots can> be coprime to 5.> This is ridiculous. What was to be shown was this: NONE OF THE COEFFICIENTS ai IN THE FACTORIZATION 65x^3 - 12x + 1 = (a1*x + 1)(a2*x+1)(a3*x+1). IS COPRIME TO 5.I showed, for each a, two factorizations: 5 = q(a) r(a) a = r(a) s(a).The common factor r has minimal polynomial x^3 - 969 x^2 + 315 x + 5I donft have to show *ANYTHING* about that polynomial: its rootsare by de?ition algebraic integers, and the inverses of thoseroots have minimal polynomial: 5 x^3 + 315 x^2 - 969 x + 1and so none of them is a unit. > But you cannot do that, and instead have maintained that proving that> none of the roots can be a unit proves that one of them must have a> non-unit factor in common with 5 in the ring of algebraic integers.> Idiot. I have displayed the factor r(a), that you have claimed does not exist. That later polynomial is the minimal polynomial for the common factor r(a), which shows that r(a) is not a unit.Geez, itfs just high school algebra, whatfs the confusion?I have just exhibited the common factor. I have shown such a factorfor *each* of those roots. I have also shown the minimal polynomialfor those common factors, and used it to show that the common factoris not a unit algebraic integer.> Ifve just shot down that little trick, but youfre squirming.> Please show me an error. Youfve had probably three months of lookingat the derivation, in one form or another, and *THAT* is all you cancome up with? I see that you could be confused, since all this isjust simple numbers, and anyone could be confused with numbers.>has algebraic integer roots, whenever a is an algebraic integer,>and n is a natural number.>Thus, if a is an algebraic integer, and n is any natural number,>the number> a^(1/n)>is an algebraic integer, irrespective of which of the n roots you take.> Duh, but youfre not talking about such a simple expression as x^n - a = 0.> No, I was showing you that ANY non-unit algebraic integer hasnon-unit algebraic integer factors, in fact will have in?itelymany of them.> Youfre talking about x^3 - 969 x^2 + 315 x + 5 = 0and *repeatedly* posters like you try to fool readers on the> newsgroups as you work to *convince* rather than get to the> mathematical truth.> I have the truth up there. You are now acting to deny it.Thatfs OK, since we have been through YEARS of seeing you behave likea little boy who has pooped his britches. No, I donft have poop inmy britches, um, I grew a tail!> The truth is that youfre relying on the negative, which is that NONE> of the roots of that expression can be units, to try and prove the> positive, which is that then they all have a non-unit factor in common> with 5 in the ring of algebraic integers.> Hey, WHAT negative: 5 = q(a)*r(a) a = r(a)*s(a).Wherefs the negative? This is what Ifve shown, and you still canft seethrough your own hatred of mathematicians to recognize that it plainlyshows your error.> But thatfs beyond bogus as my point is that the ring of algebraic> integers is screwed up to the extent that a root of that expression> can be coprime to 5, and yet not be a unit.> Where did I claim somethingfs not a unit? Not in the above pair ofexpressions, in which the root is shown NOT to be coprime to 5.No, it was in ANOTHER expression altogether: x^3 - 969 x^2 + 315 x + 5where I argue that none of the roots of THIS polynomial can be a unit. ... stuff deleted ...>No gap.> Thatfs what you want to convince readers.However, simply asserting something that is not true does not ?l the> gap.> I made the following assertions. The proofs, up to veri?ation ofa few polynomial expansions, can be seen above, but the interestedreader can produce these products explicitly, without any form o?? on my part.ASSERTION 1:Given: q(x) = 8 x^2 - 76 x - 185 r(x) = 8 x^2 - 4 x - 45 s(x) = 4 x^2 - 37 x - 104we have q(x)*r(x) = (64 x + 128)*p(x) + 5 r(x)*s(x) = (32 x + 72)*p(x) + x,where p(x) = x^3 - 12 x^2 + 65.Corollary: q(a)r(a) = 5, r(a)s(a) = aTherefore a and 5 have the common factor r(a) in the algebraic integers.ASSERTION 2:Given the polynomials mpr(x) = x^3 - 969 x^2 + 315 x + 5 r(x) = 8 x^2 - 4 x - 45 p(x) = x^3 - 12 x^2 + 65. w(x) = 512 x^3 + 5376 x^2 - 5760 x - 31808we have: mpr(r(x)) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520and p(x)w(x) = 512 x^6 - 768 x^5 - 70272 x^4 + 70592 x^3 + 731136 x^2 - 374400 x - 2067520Therefore, for a = any root of p(x), mpr(r(a)) = 0,so mpr(x) is a polynomial for which r(a) is a root. mpr(x) is easilyseen to be irreducible, and so is the minimal polynomial of r(a).Corollary: NONE OF THE FACTORS r(a) is a unit in the ring of algebraicintegers.> Again, what you have is the negative--certain numbers arenft> units--and you want the positive--that they all have to have non-unit> factors in common with 5.> I have shown as much, but you refuse to look at it: a and 5 have common factors in the ring of algebraic integers at least one of those common factors (r(a)) is not a unitSo, youfre confusing what has to be shown non coprime to 5, with whathas to be shown not to be a unit: a is not coprime to 5 r(a) is not a unit.> But you havenft proven the positive.> Of course I have, but you got your ears all plugged upand refuse to listen.> Itfs a gap.Denying it wonft ?l it, so quit being lazy and try to ?l it.> You should say that to yourself once in a while.>The gap is the assertion that given an algebraic integer ?af which is>not a unit, that ?af must have some non-unit factor *in the ring of>algebraic integers*.>As I mentioned above, the factor r that I provided *is* an algebraic>integer, and in the case that a is (-1) times any of the aifs in the>factorization cited, I have provided its minimal polynomial. It is>simple to verify from that minimal polynomial that itfs an algebraic>integer and that itfs not a unit in the ring of algebraic integers.> See readers? You can catch the trick here as notice the poster said> not a unit, and what Ifm telling you is that these posters are> working hard to convince YOU the reader.> What? Now youfre claiming that I have attempted to fool some unnamedclass of readers by asserting somethingfs not a unit?Anyone can scroll back to verify *exactly* what I am claiming. Ifvelabelled it clearly.>To protect readers from thinking itfs simple, I remind them of the>ring of evens, that is the set of even numbers, where you have 2 not a>factor of 6, and not a unit. But 2 and 6 donft have factors in the>ring.>Thatfs because the ring doesnft have 1.>You have to go to a higher ring, in this case the ring of integers, to>escape that interesting little situation.>Now the problem is different in the ring of algebraic integers, but>the ring is screwed up, and the ? is again a higher ring.>What I want you to consider is the *possibility* that weird things can>happen with rings, when youfre not careful enough, and that>mathematicians werenft careful enough with the ring of algebraic>integers.>The problem is with your ignorance. The ring of algebraic integers has>none of the shortcomings you seem to want it to have. It is your method>which is at fault, since it forces you to believe a falsehood, namely>that the aifs in that above factorization are coprime to 5.> Now this poster is the one who tried to fool all of you, and now that> Ifve caught him, he decides to hurl insults.> Please stop mischaracterizing a proof as an attempt at fraud.Just because YOU couldnft follow the fairly simple arithmetic Ifveprovided, and just because it showed that everything youfve beendoing for, what is it, 8 years now(?), has been for no good reasonat all?You have perpetually misunderstood the mathematics youfre attemptingto work with, and have, to date, NEVER accepted even the smallestbit of responsibility for knowing what it is youfre talking about.It is not an insult to say that the problem with all this is that youare ignorant of the subject. That is the simple truth.Have I insulted you?How?By saying youfre ignorant of algebra?Can you demonstrate that you know some nontrivial algebraic topic, sayGalois theory?What about any of the [other] standard theorems about ?lds and theirextensions?To claim that you have caught me, you should at least provide somestatement that you take issue with. All I saw above was your claim thatI was making the argument: If ?af isnft a unit, then ?af canft be coprime to 5.where my argument was this: A. Here are two products: a,q,r,s are all algebraic integers: 5 = q*r a = r*s Then, unless 5 is a unit, then a and 5 have a non-unit common factor in the ring of algebraic integers B. Here is the minimal polynomial for r(a). mpr(x) = x^3 - 969 x^2 + 315 x + 5 Thus, r(a) is not a unit C. Thus, a and 5 share a non-unit algebraic integer factor.> These posters are intellectually lazy, and theyfre rude.> Um, intellectually lazy? I think you just have a little ?e of all theepithets that have come your way, and choose one that seems to ?.Unfortunately, I have no grounds for claiming not to have beenintellectually lazy. Oh, I did ?d those factors you said wereimpossible, didnft I? And that other cubic a couple of weeks ago,where you said that no one could provide the monic polynomial forwhich something was a root. I gave you that polynomial.So, I guess that Ifm intellectually lazy, yet can do what mankindcannot possibly do, by your own estimation!Ifm SUPERMAN!!!! WOO HOOOO!!!Mere mortals, step aside! I can solve your problems, even thosethat mankind cannot possibly do, and not break a mental sweat.SOOOO PPERRRRRR...MAAAAAAANNNNNNN!!!!!!Yesssssssssssssssssss!!!!!!! !!!!I gotta get one of those cape thingies, some blue tights, with thered codpiece, er, maybe I can be Clark Kent? I never did likethose superhero gymnastsf costumes, capes, tiaras, big S on theshirt. It always seemed a bit, er, sissy for my tastes.Well, I have time to re?n that one for awhile.I have crushed your argument. It has nowhere to go, now that we canall see how it predicts stuff that is plain wrong. Itfs ?e forsymbols, you can push little afs and bfs and ufs and vfs aroundon your paper all day long, and once you come to plugging realvalues in for the variables it says that the afs are coprime to5. Thatfs got to be embarassing.Yeah, therefs no insult on your side. At least Ifve been factuallycorrect: you ARE ignorant, and parade that ignorance around day afterday. In addition, I havenft been running around proclaiming to haveshown something to be true, all the while ignoring a direct calculationthat shows it to be false.Who has been doing that?Now that youfre pretending to address the issue (since it wonft goaway), who is it who is continuing to mis-label every thing, out ofsome vain belief that if it fools the old JSHter, itfll fool someoneelse on earth.I think for these purposes, you will just have to settle on being THE MOST GULLIBLE MAN ON EARTH.No one else will ever believe you, and you will end up at the age of50 a broken little man who couldnft let it go when it was clear hewas done.> But theyfre not lazy when it comes to posting as they *keep* posting> now donft they?> Letfs do a post count for me vs. you WDH: 693 JSH: 4180if I count JSH-related posts, I have 322. About halfof what I contribute to sci.math goes to JSH nonsense.Already Ifm ashamed of myself. Boy, Ifm just a-burninfup that internet, donft you think?Next baseless accusation?> Ifm curious about what you, the reader, thinks is the reason for their> persistence.> Why do YOU think I keep posting?Ifve already given my reasons: you are a punk you leech of the help of others and treat your benefactors shamefully. you have an in?sense of your own worth, originality, talent, intelligence you behave as though yours is the only thought thatfs worth having you bully others you clog the newsgroup with ill will, and lead other discussions into the ground you despise what it is that I have spent the better part of my life doing. you cast aspersions at honorable people, and make idle threats you are a cynical abuser of the very concept of intellectual debate, you are a crybaby, your useage of the English language is substandard, even for a southerner. you pretend to lecture those who are your betters (in terms of comprehension, intuition, and experience).> Why do YOU think they keep posting?> James HarrisDale === [snip hysterical rant]Is your nose starting to grow? If you had any credibility left, which you donft, your own posts over thelast few days would have gone a long way toward demolishing it. You continue to ignore those posts fromothers which disprove your claim that the ring of algebraic integers is ? The ? entirelybetween your ears.Just to recap (for the record): Every root of a monic polynomial with integer coef?ients divides theconstant term within the ring of algebraic integers. That is, the quotient obtained by dividing theconstant term by any root is an algebraic integer.The above is easy to prove, and has been shown repeatedly by more than one poster, but so far seemsbeyond your comprehension. Furthermore, you have not shown a single example of a number which *should be*in the ring of algebraic integers, but which is *left out*. Tsk, tsk. That doesnft look good for you,James. You have claimed repeatedly that such numbers exist, but you have never produced a single one!At this point, what you have claimed is a ?linchpin of my FLT prooff has become a ?nail in your FLTprooffs cof?f.--There are two things you must never attempt to prove: the unprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === The factor that has been provided,> r(a) = 8 a^2 - 4 a - 45> is> (1) an algebraic integer for any algebraic integer a,> (2) a divisor of both a and 5 for> a = -(any of the aifs in the above> factorization of 65x^3 - 12x + 1)> The factorizations have already been given, but since youfre so adept> at ignoring the evidence, I will provide them once again:> q(a) = 8 a^2 - 76 a - 185> r(a) = 8 a^2 - 4 a - 45> s(a) = 4 a^2 - 37 a - 104> Whenever a is a root of x^3 - 12 x^2 + 65 (that is, a is the *negative*> of any of the aifs of the above factorization of 65x^3 - 12x + 1), the> following factorizations hold:> q(a) r(a) = 5> r(a) s(a) = a.>Hmmm...the poster has tried to dupe *you* the reader, and it all has>to do with the word unit. See below... In fact, the minimal polynomial of this number (r(a) for -a = any of> the above aifs) is given as:> MinPoly(r) = x^3 - 969 x^2 + 315 x + 5> The above facts prove that this a has a non-unit algebraic integer> as a factor.>No it doesnft. Why not? There is an algebraic integer, called here r(a), with the> following three properties: (1) It is not a unit in the ring of all algebraic integers; you> apparently agree with that conclusion, but then engage in> sophistry about a supposed problem with it.It is true that r(a) is not a unit in the ring of algebraic integers.Now what are the actual expressions?They are q(a) r(a) = 5 r(a) s(a) = a.where I simply note that r(a) may not have non-unit factors in commonwith 5, which *should* make q(a) have a factor of 5 ***in the ring ofalgebraic integers***, but the ring of algebraic integers is screwedup.In trying to prove that r(a) must have a non-unit factor in commonwith 5, you rely on the fact that it canft be a unit ***in the ring ofalgebraic integers***!!!The trick is to act as if proving the negative, proves that it musthave some non-unit factor in common with 5, but therefs a fascinatingerror with the ring of algebraic integers. (2) There is an algebraic integer, called s(a), such that> r(a)*s(a)=a. So r(a) is a factor of a in the ring of algebraic> integers.Well, what if ?af doesnft share non-unit factors in common with 5? Then both r(a) and s(a) donft either.That *should* leave q(a) with a factor of 5, but the ring is screwedup.In order to try and prove that it does have a factor of 5, you have totry and claim that it must because otherwise r(a) would be a unit.But the ring of algebraic integers is screwed up, which is my point.> (3) There is an algebraic integer, called q(a), such that> r(a)*q(a)=5. So r(a) is a factor of 5 in the ring of algebraic> integers.However, how do you know that r(a) must share non-unit factors incommon with 5?Ultimately your claim must be that itfs because itfs not a unit in thering of algebraic integers!!! > So: r(a) is (1) not a unit; (2) a factor of a; and (3) a factor of 5.> Why is it not the case that r(a) is a non-unit algebraic integer which> is a common factor of a and 5?For readers, these posters are working to *convince* not get to thebottom of the problem.If they cared about the truth, then they only need consider my proofof the problem with algebraic integers.Proofs donft duel. If theyfre correct, then theyfd be able to ?d anerror with my argument, but it has no error as the ring of algebraicintegers IS screwed up, so instead they work to convince YOU thereader. > [.snip.]>What you have is> x^3 - 969 x^2 + 315 x + 5>and what you want to do is convince readers that none of its roots can>be coprime to 5.There is no convincing that needs to be done. NONE Of the roots are> units, and ALL the roots are factors of 5. Youfre done.No, you have a gap. The gap is that now you need to prove that theroots share non-unit factors in common with 5 in the ring of algebraicintegers.>But you cannot do that,Correction: James Harris cannot do that because he does not know what> coprime means.Which is the semantic argument.> and instead have maintained that proving that>none of the roots can be a unit proves that one of them must have a>non-unit factor in common with 5 in the ring of algebraic integers.They are each FACTORS of 5, and EACH non-units.Which does NOT prove that they each have a non-unit factor in commonwith 5, because the ring of algebraic integers is ?> Do you agree or disagree with (all happening in a commutative ring> WITH 1): (a) If a is a factor of b, then a and b have a common factor> (namely, a).> (b) If a is a factor of b and is not a unit, then a is a non-unit> common factor of a and b. (c) If is a factor of b and is not a unit, then a and b have a> non-unit common factor (namely, a).Hmmm...I can see that youfre dedicated at working to convince theaudience, so Ifll give a demonstration to show them *how* youfreworking.Consider abc = 5, where a = sqrt(5), b=sqrt(5)(-1+sqrt(-3))/2,c=(-1-sqrt(-3))/2, but imagine that youfre talking to someone whodoesnft know about radicals, as all they know about are integers.Now neither ?af nor ?bf is an integer, but this person is using d=ab,and they now see cd = 5and tell you that ?cf and ?df must be factors of 5 ***in the ring o?tegers***.You say nope.Well they come back, and argue, and point out that neither is a unit***in the ring of integers***.You say, yup, that is correct.Then they come back and say, well, cd = 5, neither is a unit in thering, so they must be factors of 5.And you say, nope, they are not.But the ring of algebraic integers is VERY screwed up, so that onlygives you an idea.>Ifve just shot down that little trick, but youfre squirming.I do see a lot of squirming, but itfs not coming from Dale.What mathematicians can do, because itfs such an odd error, is keepcasting doubt, and running away from the proof of the error.Probably for most of you the idea that you could have abc=5 whereneither ?af, ?bf, nor ?cf is a factor of 5, in the ring of algebraicintegers, seems nonsensical.However, itfs an esoteric problem in an esoteric branch ofmathematics, which has been there for over a *hundred* years.If mathematicians want to confuse most of you about it, they can.> [.snip.]>The truth is that youfre relying on the negative, which is that NONE>of the roots of that expression can be units, to try and prove the>positive, which is that then they all have a non-unit factor in common>with 5 in the ring of algebraic integers.You are completely, totally, utterly lost and confused.The roots themselves are factors of 5. Each root is a factor of> itself, and a factor of 5. Therefore, each root is a common factor of> itself and 5. And since each of them is not a unit, then we have> EXHIBITED a non-unit common factor of each with 5. The roots are not> the original numbers we were interested in (the afs), they are the> common factors that have been produced.> [.snip.]And again the trick is that use of the word unit as in fact thefactors are NOT units ***in the ring of algebraic integers***.In a higher, more complete ring, one of them IS a unit, while two ofthem have a factor that is sqrt(5).> As I mentioned above, the factor r that I provided *is* an algebraic> integer, and in the case that a is (-1) times any of the aifs in the> factorization cited, I have provided its minimal polynomial. It is> simple to verify from that minimal polynomial that itfs an algebraic> integer and that itfs not a unit in the ring of algebraic integers.>See readers? You can catch the trick here as notice the poster said>not a unit, and what Ifm telling you is that these posters are>working hard to convince YOU the reader.Are you claiming that non-unit and not a unit are not the same> thing? [.snip.]Nope.For readers who donft realize how this ring business can be confusing,consider my example with the ring of evens. That ring doesnft have 1in it, as 1 is odd, so 2 and 6 do not share factors with each other.However, you go to the higher ring--integers--and they do.Mathematicians have this problem where they didnft think there wasone, until I pushed it into the open.Rather than be fascinated by it, and accept the mathematics, you cansee posters like Arturo Magidin instead working to convince YOU.Why does he have to keep replying to me in posts?Because Ifm right. If he goes away, and isnft around to confusereaders, he probably realizes that eventually I might convince some ofyou to follow the mathematical logic.That explains why posters like Arturo Magidin are so dedicated.They need to hang around to keep YOU confused.James Harris === |Well if youfre telling the truth, ?l in the gap Ifve pointed out in|YOUR argument.||The gap is the assertion that given an algebraic integer ?af which is|not a unit, that ?af must have some non-unit factor *in the ring of|algebraic integers*.One obvious answer is that ?af is a factor of itself.If you mean nontrivial factor, then the square roots of ?af arenon-unit factors of ?af. If ?af is an algebraic integer, then itfsa root of a monic polynomial with integer coef?ients x^n + c_1*x^{n-1} + ... + c_n = 0.If b^2=a, then b is a root of the monic polynomial with integercoef?ients x^{2n} + c_1*x^{2n-2} + ... + c_n = 0.We know b is not a unit in the algebraic integers, because if bwere a unit then by de?ition there would be some algebraicinteger c such that bc=1. But then we would have 1=(bc)^2=b^2c^2 = ac^2. Because c is an algebraic integer, then so isc^2. That would imply a is a unit, but we assumed to start withthat a was not a unit.If you want an explicit proof that if c is an algebraic integer thenso is c^2: suppose c is the root of a polynomial x^m + d_1*x^{m-1} + ... + d_m = 0with integer coef?ients. Then we can group the even terms on oneside and the odd terms on the other: d_m + d_{m-2}*x^2 + ... = -(d_{m-1} + d_{m-3}*x^2 + ...)*x.Square both sides: (d_m+d_{m-2}*x^2+...)^2 = (d_{m-1}+d_{m-3}*x^2+...)*x^2.This is now a polynomial in x^2. Substituting y for x^2 we geta monic polynomial with integer with integer coef?ients whichhas c^2 as a root.|To protect readers from thinking itfs simple, I remind them of the|ring of evens, that is the set of even numbers, where you have 2 not a|factor of 6, and not a unit. But 2 and 6 donft have factors in the|ring.||Thatfs because the ring doesnft have 1.Before reading this past yearfs discussion, I hadnft realized whatthe de?ition of coprime was in such cases. So you sometimeslearn something new. (Not that this de?ition was ever importantto anything I had to do.)|You have to go to a higher ring, in this case the ring of integers, to|escape that interesting little situation.||Now the problem is different in the ring of algebraic integers, but|the ring is screwed up, and the ? is again a higher ring.||What I want you to consider is the *possibility* that weird things can|happen with rings, when youfre not careful enough, and that|mathematicians werenft careful enough with the ring of algebraic|integers.Ifll do that if you agree to do something much easier. Consider thepossibility that weird things are causing the common factors whichyou believe to be independent of m to depend upon m, but thatyou werenft careful enough.Keith Ramsay <3c65f87.0308110729.3f68620@posting.google.com> <3F37D13C.40008@farir.com> <3c65f87.0308111350.19f0910e@posting.google.com> <3F38305E.4050008@farir.com> === > your useage of the English language is> substandard, even for a southerner.Donft be an ass. At least, not in the same sentence in which you misspell usage andfail to capitalize Southerner.(Now, wherefs my requisite typo in this spelling ?-- Jesse Hughes[I]tfs the damndest thing. Therefs something wrong with every lastone of you, and I *never* thought that was a possibility. But now Ifeel itfs the only reasonable conclusion. --JSH sees some sorta light === >The integration problem>int p Beta(p|a,b) dp>where a and b are the parameters for the beta distribution, has the>exceedingly nice solution>a / (a + b).>Is there a somewhat nice (tho certainly not AS nice!) solution>available to the related integration problem:>int log(p) Beta(p|a,b)dp>If so, it would be very useful for something Ifm working on. Ifd be>very grateful for any help!It is Psi(a) - Psi(a+b). Psi is the derivative of the logarithm of Gamma. Just differentiate theequationint p^{a-1}(1-p)^{b-1}dp = Gamma(a)*Gamma(b)/Gamma(a+b)with respect to a.-- This address is for information only. I do not claim that these viewsare those of the Statistics Department or of Purdue University.Herman Rubin, Department of Statistics, Purdue University === >The integration problem int p Beta(p|a,b) dp >where a and b are the parameters for the beta distribution, has the> exceedingly nice solution [....] >[Herman Rubin=.. It is Psi(a) - Psi(a+b). Psi is the derivative of > the logarithm of Gamma. Just differentiate the> equation > int p^{a-1}(1-p)^{b-1}dp = Gamma(a)*Gamma(b)/Gamma(a+b)> with respect to a.(a)_k=a(a+1)...(a+k-1) . k=1,2,... , (a)_0:=1 -(Pochhammer Symbol) F_{2,1}(a,b;c; z)=F(a,b;c; z)==SUM_{k=0 to k=infty}(a)_k*(b)_k*z^k/((c)_k*k!) -(Gauss Hypergeometric Series) F_{3,2}(a,b,c;p,q; z)= =SUM_{k=0 to k=infty}(a)_k*(b)_k*(c)_k*z^k/((p)_k*(q)_k*k!) -(a generalization of Gauss Hypergeometric Series) G(z)=Gamma Function , B(p,q)=Beta Function ,Psi(z)=d(G(z))/dz =Gf(z)/G(z)= the Logarithmic Derivative of Gamma Function,and for a>0 , b>0 (1) T(a,b;f;z)==(1/B(a,b))*Integral_{t=0 to t=1}t^{b-1}*(1-t)^{a-1}*f(zt) dt (Beta Transform of f)GAry need t a ,,closed form for (#) Z:= T(a,b; ln(.); 1) = ??For instance, let us note that(2) ln(z) =-(1-z)*F(1;1;2;1-z) , |z-1|< 1 ,and (3) F_{3,2}(a,a_1,a_2;a+b,b_1; z)= T(a,b;H;z) with H(x)=F(a_1,a_2;b_1; x) .STEP 1; select in (3) a_1=a_2=1 , b_1=2 and z=1 .According to (1)-(2) one ?ds for a>0 , b>0 (3f) F_{3,2}(a,1,1;a+b,2; 1) = = -(1/B(a,b))*Integral_{t=0 to t=1}t^{b-1}*(1-t)^{a-1}*(1-t)*ln(1-t) dt == -(1/B(a,b))*Integral_{t=0 to t=1}t^a*(1-t)^{b-1}*ln(t) dt == - (a/(a+b))* T(a+1,b;ln(.) ; 1) .Therefore (see (#)) Z=T(a,b;ln(.);1)= - ((a+b-1)/(a-1))* F_{3,2}(a-1,1,1;a+b-1,2; 1) , a>1,b>0 .STEP 2; Use the fact that for A=/=1 , C-A > 0 , we have (5) F_{3,2}(A,1,1; C,2; 1)=(C-1)/(A-1))*( Psi(C-1) - Psi(C-A) ) . If we choose in (5) A=a-1 , C=a+b-1 and then using (4), we give === ======================================== === ============Z =((a+b-1)*(a+b-2)/(a-1)(a-2))*( Psi(b)-Psi(a+b-1) ) , a>1 ,a=/=2, b>0 . === ========================================== === ===========Note that Psi(1)=-gamma=-(Euler constant) and Psi(n+1)= 1+1/2+...+1/n - gamma , Psi(1/2)=-gamma -2*ln(2) , Psi (z+n)-Psi(z)= 1/z + 1/(z+1) + ... +1/(z+n-1) ,n being a positive integer. I hope that there are no many misprints. However, Herman solution is nice === =========Many thanks for your responses.My setup (based on the python language running on OS X) doesnft seemto have a readily-available psi function. There is one in SciPy, but Ihavenft been able to get SciPy to compile on OS X.I wonder if therefs another way to calculate it that uses less-exoticfunctions. I may have to run a simulation to get it...Many thanks again -- hopefully Ifll be able to get access to a psifunction if therefs no other way.Garyhttp://www.garyrobinson.net === wow,the internet is really an awesome tool. thanks to both you fellows very much.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBKFmp204548; === Do you still have this book?- J. J. Barton, L. R. Nackman, Scienti? and Engineering C++: An Introduction with AdvancedTechniques and Examples, Addison Wesley, 1994, $25 {Book Condition: New}.If so, what is you lowest price?Antonio Ferrao NetoX-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBKLX4328498; === idoit===quatium bomb @ multidemtional website @ http://www.speedypc.20m.com === I am trying to solve poisson equation using bunemanfs algorithm on a single machine (no need of any parallelization). I have read buzbee, golub and nielson; and stoer and bulirsch regarding this. While solving poisson equation I have to use the nine-point formula.For the usual 5-point formula the offdiagonal block matrix is just identity matrix. Buzbeefs paper and stoerfs book discuss only this case.I could not ?d any reference when the off-diagonal terms are non-identity matrices. Could someone point me to the right direction?will the same stability conditions hold for 9-point formula also? what are the necessary changes to be done...?I have derived the equations by myself, but would like to check with some one/paper before I start coding this stuff.thanksraju === There are ?ready to usef formulas in Abramowitz & Stegun,limited in the numbers of points used for the Lagrangemethod.Are there any readings [on the web?] using polynomialsinstead of the concrete solutions?Moreover i would like to ?d something for non-equalspaced points assuming my functions are very smooth[like 2nd derivative ~ exp(polynomial)]. Any good hintsin the sense above?--- === There are ?ready to usef formulas in Abramowitz & Stegun,> limited in the numbers of points used for the Lagrange> method.Are there any readings [on the web?] using polynomials> instead of the concrete solutions?Moreover i would like to ?d something for non-equal> spaced points assuming my functions are very smooth> [like 2nd derivative ~ exp(polynomial)]. Any good hints> in the sense above?---Use Gram polynomials to ? either equally-spaced or unequallyspace data in the least-squares sense. This IS a ?tering method,of course, if you use polynomials of (much) lower degree than thenumber of data.Then use a standard algorithm to compute the derivative of the?ting polynomial. (See, e.g. F.S. Acton, Numerical Methodsthat (Usually) Work, AMS pub.)-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === > There are ?ready to usef formulas in Abramowitz & Stegun,> limited in the numbers of points used for the Lagrange> method.> Are there any readings [on the web?] using polynomials> instead of the concrete solutions?> Moreover i would like to ?d something for non-equal> spaced points assuming my functions are very smooth> [like 2nd derivative ~ exp(polynomial)]. Any good hints> in the sense above?You might want to investigate Savitzky-Golay ?ters whichnot only support taking derivatives but smoothing (leastsquares) as well. For unequally spaced points you may have tobe content with ?ting splines (or polynomials) and takingderivatives from them.--There are two things you must never attempt to prove: theunprovable -- and the obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === Dear all,At the moment I use NAG routines to solve sets of non-linear equations.Untill now I was able to solve sets of at maximum 5000 equations. However, Ihave to go to at least 30000-50000. Does anybody has a good tip for a solverable to handle such systems? Henk === > Dear all,At the moment I use NAG routines to solve sets of non-linear equations.> Untill now I was able to solve sets of at maximum 5000 equations. However, I> have to go to at least 30000-50000. Does anybody has a good tip for a solver> able to handle such systems?> Henkhttp://www-unix.mcs.anl.gov/petsc/petsc-2/Hans Mittelmann === > I have prepared> http://informatik.uibk.ac.at/users/c703211/solok.jpg> as an example for the GSN-SOR, residual ~ 1e-5 after 18 iterations.Compare this to> http://informatik.uibk.ac.at/users/c703211/solnok.jpg> You can recognize the small oscillations as the moire pattern.> This was made with Jacobi, residual ~ 25 after 17 iterations, changing > hardly thereafter.Which damping factor do you use? The standard 5-point stencil needs dampedJacobi for smoothing (a factor 0.5 works well).Nicolas. === > Strange. Jacobi is very local, so if anything it should be good at> precisely losing the oscillatory components of the error, while really> slow at ?ding the low frequencies.Indeed. (If it is correctly damped.)Nicolas. === hy all,ifm searching a good (=VERY FAST) algorithm to compute eigenvalue of amatrix in skyline format in order to study it and reimplement it(donft ask me why, please... :)Any help is appreciated.Very many thanks,Attilio Gelosa. === There always seem to be interesting discussions here but very little mentionof application (presumably to avoid divulging what your company isscheming).So, to satisfy my curiosity, I would like to ask what kind of work you alldo that requires knowledge of num-analysis? I.E. why are you reading thisnewsgroup?Personally, Ifm an engineering student with an interest in this kind ofstuff. While I have had a reasonable amount of exposure to engineering work,I have never come across anyone requiring num-analysis knowledge outside ofacademia.So, letfs hear it.CM === You have had very little experienceIn Engineering we use numerical methods for almost anything.You may notice the large engineering companies like say Boeing have humongous libraries of proprietrary routines.But there is a lot of info out on the net and at various site that hold fortran and other codes.So there is really tonnes of stuff,Paul> There always seem to be interesting discussions here but very little mention> of application (presumably to avoid divulging what your company is> scheming).So, to satisfy my curiosity, I would like to ask what kind of work you all> do that requires knowledge of num-analysis? I.E. why are you reading this> newsgroup?Personally, Ifm an engineering student with an interest in this kind of> stuff. While I have had a reasonable amount of exposure to engineering work,> I have never come across anyone requiring num-analysis knowledge outside of> academia.So, letfs hear it.CM> === Chris,> There always seem to be interesting discussions here but very> little mention of application...So, to satisfy my curiosity,> I would like to ask what kind of work you all do that requires> knowledge of num-analysis? I.E. why are you reading this> newsgroup?I write a circuit analysis program used for analog circuit design.Itfs used both in-house(wefre an IC manufacturer) and it is givenaway to customers as a general purpose SPICE program to virtuallydemo the companyfs IC products. I recently asked here forpointers on symbolic dimensional analysis of expressions (i.e.,to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K*time)*time as Joule-seconds). Julian N. gave some useful pointersand the code is now release as http://LTspice.linear-tech.com/software/swcadiii.exe. (BTW, Julian, I see your pointabout Forth being ideal for recursively reducing the symbolicexpressions, but C++ returning String objects on the stack wasfortunately just as effective and just as easy to handle thememory management of all those strings.)--Mike === Chris,There always seem to be interesting discussions here but very> little mention of application...So, to satisfy my curiosity,> I would like to ask what kind of work you all do that requires> knowledge of num-analysis? I.E. why are you reading this> newsgroup?I write a circuit analysis program used for analog circuit design.> Itfs used both in-house(wefre an IC manufacturer) and it is given> away to customers as a general purpose SPICE program to virtually> demo the companyfs IC products. I recently asked here for> pointers on symbolic dimensional analysis of expressions (i.e.,> to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K*> time)*time as Joule-seconds). Julian N. gave some useful pointers> and the code is now release as http://LTspice.linear-> tech.com/software/swcadiii.exe. (BTW, Julian, I see your point> about Forth being ideal for recursively reducing the symbolic> expressions, but C++ returning String objects on the stack was> fortunately just as effective and just as easy to handle the> memory management of all those strings.)--MikeFirst, thanks for the acknowledgement of my advice. I should send abill if it was that useful ;-)Seriously, my only objection to using C++ is the HUMUNGUOUS sizeof the resulting code. I guarantee that, had you used a Windoze-based Forth such as Win32Forth or SwiftForth, the code size wouldhave been way smaller than 4+ Mb, probably would have allowed usermodi?ation (although companies rarely want to do this even forstuff they give away for free), and just as much GUI functionalityor more. Maybe using OpenGL. Win32Forth was written by a guy whogot tired of sweating the M$ PDK for C++ apps. And I am sure itwould have run about as fast. Certainly SwiftForth would.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ Science knows only one commandment: contribute to science. -- Bertolt Brecht, Galileo. === Julian,> There always seem to be interesting discussions here but very> little mention of application...So, to satisfy my curiosity,> I would like to ask what kind of work you all do that requires> knowledge of num-analysis? I.E. why are you reading this> newsgroup?> I write a circuit analysis program used for analog circuit design.> Itfs used both in-house(wefre an IC manufacturer) and it is given> away to customers as a general purpose SPICE program to virtually> demo the companyfs IC products. I recently asked here for> pointers on symbolic dimensional analysis of expressions (i.e.,> to identify the units of 10+2*pi*V(n002)*I(R1)*sine(2*pi*1K*> time)*time as Joule-seconds). Julian N. gave some useful pointers> and the code is now release as http://LTspice.linear-> tech.com/software/swcadiii.exe. (BTW, Julian, I see your point> about Forth being ideal for recursively reducing the symbolic> expressions, but C++ returning String objects on the stack was> fortunately just as effective and just as easy to handle the> memory management of all those strings.)> First, thanks for the acknowledgement of my advice. I should send a> bill if it was that useful ;-)> Seriously, my only objection to using C++ is the HUMUNGUOUS size> of the resulting code. I guarantee that, had you used a Windoze-> based Forth such as Win32Forth or SwiftForth, the code size would> have been way smaller than 4+ Mb, probably would have allowed user> modi?ation (although companies rarely want to do this even for> stuff they give away for free), and just as much GUI functionality> or more. Maybe using OpenGL. Win32Forth was written by a guy who> got tired of sweating the M$ PDK for C++ apps. And I am sure it> would have run about as fast. Certainly SwiftForth would.C++ can also be used to make ef?ient use of objectcode size, but it isnft a trend after which manypeople aspire. I certainly do in the interest ofesthetics and code execution speed. BTW, I foundthat the MS compiler can make the smallestcomplete executable of all PC-based compilers Itried.While LTspice is now 4.65MB(slightly larger ifcompiled to use P4fs 128 bit width data types tosimultaneously do two FLOPS on two 64bit doubleprecision numbers), years ago LTspice was justover 1.4MB, as I recall. In that 1.4MB there was acomplete SPICE engine, a waveform viewer(that haditfs own 64bit virtual address space for waveformdata) a schematic capture program, a SMPSsynthesizer, about 30 small compilers(circuitsimulation is ultimately a complier problem thesedays as I see it) and MFC statically linked in fora multi-threaded GUI. Yep, people were surprisedto see the small execution size. Todayfs 4.65MBversion bloated in that it has now contains twocomplete SPICE solvers(one uses a different SPARSEmatrix package that was incompatible with the normalsolver) and very large MOSFET device evaluationcode for various of experimental MOSFETfs devices.However, the only penalty for the size is thedownload time and by industry standards for this?ld, itfs still a lean, thin, and mean executable.Itfs distributed as a 5MB self-extracting gzipfedexecutable that contains 1500+ ?es in addition tothe executable. My guess is that it wound not bepractical to implement such a project in Forth.--Mike === > I have never come across anyone requiring num-analysis knowledge outside of> academia.So, letfs hear it.CM> Two words.Cleve Moler. === CenterSpace Software, a leading provider of enterprise class numericalcomponent libraries for the .NET platform, today announced the releaseof NMath Stats. NMath Stats is part of CenterSpace Softwarefs NMathproduct suite, which provides object-oriented components formathematical, engineering, scienti?, and ?ancial applications onthe .NET platform.NMath Stats provides functions for statistical computation, includingdescriptive statistics, probability distributions, combinatorialfunctions, multiple linear regression, hypothesis testing, andanalysis of variance (ANOVA).Fully compliant with the Microsoft Common Language Speci?ation, allNMath Stats routines are callable from any .NET language, includingC#, Visual Basic.NET, and Managed C++.A free 14-day evaluation version is available for download from theCenterSpace website at http://www.centerspace.net, along with completeuser documentation, performance benchmarks, a whitepaper, and codeexamples. === hi, ALL,I am wondering whether AMG likes banded matrixthan general sparse matrix. Two similar linear systems. (generated frompoisson equations)A * x = bA_* x = bThey share the same RHS, and exact solution, but different matrix (A, and A_).In fact, x = { 1, 1, 1,......,1 }(Trans)A is a banded matrix. A_ is very similar to A, except several entries out of the band. If we combine thoseentries to diagonal elements, we will get A. It seems that AMG working on A_ gives better resultthan working on A.Could anyone tell me which kind of matrix AMG like to see?lin === > [...] > Right now, I am interested in doing the following: setup the> model equations using Maple and convert them into C or MATLAB routines> and solve them. Final aim is to be able to develop detailed models (a> highly nonlinear system of PDEs) quickly using Maple, but solve them> in MATLAB. Can somebody please let me know whether this is practical> and if possible can you provide me some Maple code as an example to> start with.See the ?CodeGeneration,Matlab help page. Some more examples are inhttp://www.mapleapps.com/categories/maple9/html/ CodeGeneration.html.-- Thomas RichardMaple SupportScienti? Computers GmbHhttp://www.scienti?.de === I have recently discovered generalizations of the complex-conjugatefor a number of algebras.http://library.wolfram.com/infocenter/MathSource/ 4894 contains aMathematica notebook, Hoopalgebras.ma (designed to be comprehensibleto non-Mathematica users) that investigates many algebras and includes47 conjugate formulations. Brie?ops are all the divisionalgebras (including C, H, O, Clifford algebras, etc.) that conserveone or more size on multiplication and division. The sizes are thereal factors of the inverse of the multiplication table determinant,so Det[A] * Det[B] = Det[A*B] where the Dets are calculated from thismultiplication table after mapping with the vectors. Multiplying a vector by its conjugate is usually employed to ?d thesquared length of the vector. It also gives a vector with elementsthat de?e these sizes. Complex algebra provides the archetype.Multiply the complex number {a,b} (==a + b i) by its complex-conjugate(a,-b} (==a - b i) using complex arithmetic and the result is{a^2+b^2, 0}; the real component of the product is both the squaredlength and the conserved size; the complex component is zero. The sizes are always divisors of the inverse, e.g. the complexinverse of {a,b} is {a/(a^2+b^2, -b/(a^2+b^2)}. The inverse goes toin?ity (division by zero) when a size goes to zero. Quaternion &octonion conjugate products also have a single non-zero element (thesum of the squared elements), which is the (scalar) conserved propertyand the divisor in the inverse. R, C, H, & O are the only real algebras without divisors of zerobecause their sizes are the sum of the squared elements, which (withreal coef?ients) can only be zero in the trivial case. (The onlyother operations with sums of squared elements as divisors are thehigher Cayley - Dickson algebras such Sedenions, but they have lostany form of associativity.) I know of no other algebras that conservethe sum of the squares of their elements, but a few have a signed sumof their elements as a size (and so have real divisors of zero -non-trivial sets of elements can give a zeroed size). Their conjugateproducts also have a single non-zero element. E.g.:- (1) The Pauli-sigma-matrix algebra product of {a,b,c,d} &{a,-b,-c,-d} is {a^2 -b^2 - c^2 - d^2, 0,0,0}, with size a^2 - b^2 -c^2 - d^2. (2) The Clifford(2) product of {a,b,c,d} & {a,-b,-c,-d} is {a^2 - b^2+ c^2 - d^2, 0,0,0}, with size a^2 - b^2 + c^2 - d^2. I use {a,b,...} as the initial vector in the following examples. Some algebras have quartic sizes that split into compact expressionsas signed sums of squared quadratics. I have found several suchcompact expressions by searching for a conjugate and con?ming thatthe sums of squares of the product elements de?e conserved sizes.E.g. the Clifford(3) conjugate {a, -b, -c, -d, -e, -f, -g, h} gives aproduct with scalar element a^2 -b^2 -c^2 +d^2 -e^2 +f^2 +g^2 -h^2 &trivector element -2d e +2c f -2b g +2a h, the other elements beingzero. The (single, quartic) conserved factor is the sum of the squaresof these two terms. I have found three other Clifford(3) conjugates,each with four non-zero terms; signed sums of their squares providesdifferent formulations of the same conserved factor. Most algebras have several sizes, so their inverses split intopartial fractions. All the above algebras have a single size; they aredegenerate because their inverses do not split into partial fractions. I have found plex-conjugates for 40 other algebras; these usuallygive a product containing many zero coef?ients, with the fewnon-zero elements being related to quadratic or quartic conservedsizes, though the relationship is not always obvious. In algebras with ternary symmetry the conjugate has reversed pairs ofelements (instead of negated elements), and the product has repeated(instead of zero) elements. The C3 group de?es a division algebra inwhich the product of {a,b,c} and {a,c,b} is {a^2+b^2+c^2, a b +a c +bc, a b +a c +b c} whilst the sizes are {a+b+c, a^2 +b^2 +c^2 -(a b +ac +b c)}. The quadratic size is the sum of the unrepeated and one ofthe repeated terms. Linear sizes (a+b+c in this case) obviously cannotbe given by conjugate multiplication. Plex-conjugates appear to be a new concept that raises someinteresting research problems. Do all hoops have them? I have failedto ?d them for C5, D3, C7, D4, D3C2 & A4 and a few smallsigned-table algebras; I have only looked at larger algebras where Iexpected an obvious conjugate. Is there an extension to tripleproducts to generate cubic factors? A4 is the smallest group with acubic size. Complex conjugation is a special case of a more general property ofsome algebras, which gives a squared length for a vector because ofthe simple form of the sizes of C, H, & O. Roger Beresford.Facts do not cease to exist because they are ignored. T. H. Huxley.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBMBIP319625; === Can anyone suggest a book (or internet source) that describes indetail how to construct (and read in) a Maple library? I have dozensof Maple procedures in a *.txt ?e that I read into Maple every timeI begin a session. I want to somehow construct a library for theseprocedures and also make documentation ?es that could be accessedthrough the Maple help menus. I believe I am really asking twoseparate questions here:(1) How to build a library (given the procedures are alreadyconstructed and currently sit in a large *.txt ?e).(2) How to build your own documentation Help menu -- just as Maplehas Help menus for its various functions.Diane EvansMathematics Assistant Professor atRose-Hulman Institute of Technology === Can anyone suggest a book (or internet source) that describes in> detail how to construct (and read in) a Maple library? I have dozens> of Maple procedures in a *.txt ?e that I read into Maple every time> I begin a session. I want to somehow construct a library for these> procedures and also make documentation ?es that could be accessed> through the Maple help menus. I believe I am really asking two> separate questions here:(1) How to build a library (given the procedures are already> constructed and currently sit in a large *.txt ?e).(2) How to build your own documentation Help menu -- just as Maple> has Help menus for its various functions.> Diane Evans> Mathematics Assistant Professor at> Rose-Hulman Institute of TechnologyI forwarded this to comp.soft-sys.math.maple in hopes a maventhere could answer.-- Julian V. NobleProfessor Emeritus of Physicsjvn@lessspamformother.virginia.edu ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything, and thereby take away our free wiil and that share of glory that rightfully belongs to ourselves. -- N. Machiavelli, The Prince.X-Received: (from approve@localhost) by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id hBMGtSf10747; === I have a proof for Goldbachfs Conjecture that is prominently posted indirectories throughout the web. It is most notably found at FloridaStatefs virtual library under: www.math.fsu.edu/Science/Specialized asIn Defense of Mr. Fermat. === On Wed, 4 Feb 1998 10:01:12 -0500 (EST) in Geometry-Research,But Bucky Fuller, though a wonderful architect, was close to crazy, as his books clearly show. He regularly used phrases (such as the fundamental structure of the plane is hexagonal) that sound wonderful but have no meaning. In fact itfs a mere matter of convenience whether we use orthogonal or hexagonal coordinates - neither is intrinsically better than the other; itfs just that some coordinate-systems are better suited to some problems than others. Ifve used dozens of different coordinate-systems in my life, as have most other professional mathematicians, and I prefer not to waste time by muttering meaningless mumbo-jumbo to show how one is somehow more moral than the others.---I have read some things that say the rect in rectilinear means correct and it implies rectilinear Cartesian coordinates are somehow righteous. Bucky Fuller insisted that academia, all over the globe, does not agree with John Conway about the merits of non-orthogonal coordinate systems, and that is what Ifve seen in everything Ifve read in the last forty years except Bucky Fullerfs writing. Could it be that the top people in academia donft know what is being taught to the public and the students?I was asked to show examples of how the Synergetics coordinate system can make things clearer or easier than Cartesian coordinates. Thatfs a job for mathematicians (teachers) who agree that some coordinate-systems are better suited to some problems than others.If you pull string through equal length soda straws, the triangle holds its shape and the square does not. The tetrahedron with four triangular faces holds its shape and the cube with six square faces does not. Why?The only author Ifve read who has written about the importance of the stability of the triangle is Buckminster Fuller. He is very persuasive that the stability of the triangle is very important; it is a primary fact of existence, it comes before other facts of geometry. He is very persuasive that if there is no triangulation there is no structure in anything.De?ition mathematics:Math`e*matics, n. [F. math[?e]matiques, pl., L. mathematica, sing., Gr. ? (sc. ?) science. See Mathematic, and -ics.] That science, or class of sciences, which treats of the exact relations existing between quantities or magnitudes, and of the methods by which, in accordance with these relations, quantities sought are deducible from other quantities known or supposed; the science of spatial and quantitative relations.Note: Mathematics embraces three departments, namely: 1. Arithmetic. 2.Geometry, including Trigonometry and Conic Sections. 3. Analysis, in which letters are used, including Algebra, Analytical Geometry, and Calculus. Each of these divisions is divided into pure or abstract, which considers magnitude or quantity abstractly, without relation to matter; and mixed or applied, which treats of magnitude as subsisting in material bodies, and is consequently interwoven with physical considerations.Source: Websterfs Revised Unabridged Dictionary, 1996, 1998 MICRA, In---The Cartesian coordinate system should be considered part of pure or abstract mathematics, and the Synergetics coordinate system part of mixed or applied mathematics, even though the Synergetics coordinate system was developed in pure principle.R. Buckminster Fuller thought that the mistake most people make is to leave out relevant parameters because they want to be brief; they want leave out relevant parameters but was not redundant, and, he is the only writer Ifve read who has even tried to do that.A small summary of Synergetics Coordinates would be misleading because it would leave out relevant parameters. Here are a few URLs to prepare you to read RBFfs books Synergetics 1 and 2.A brief description of Synergetics coordinates at:http://mathworld.wolfram.com/ SynergeticsCoordinates.htmlThe Mathematica notebook SynergeticsApplication7 at:http://library.wolfram.com/infocenter/MathSource/600/or the SynergeticsApplication7 notebook as html at:http://users.adelphia.net/~cnelson9/R. Buckminster Fullerfs Synergetics 1 and 2 at:http://www.rwgrayprojects.com/synergetics/synergetics.html Cliff Nelson === > I just looked in a second algebra book on my shelf,> Birkhoff and MacLane, A Brief Survey of Modern Algebra.> Page 17. In fact it say> of the two possible g.c.d.fs +-d for a and b, the> positive one is often denoted by the symbol .> (a , b)...And here is another place..> http://mathworld.wolfram.com/GreatestCommonDivisor.html> Ifm a mathematician, and again I can insure you I never> saw this notation before. Which means for me a couple> of (integers or polynomials). Maybe itfs a local US notation.Its not US speci?. I am a mathematician too, and (a,b) is the standard notion in mathematics for the gcd of two integers. Pick up any book on elementary number theory and it willcontain this notation. eg. I just picked Niven, Zuckerman andMontgomery of my bookshelf, and there on page 7 is thenotation (a,b) for gcd. graham === > So to throw some wood in this ?e, I would say that Java (for me)> is a very good language to develop CAS software.This depends on what you mean by good language. Java has somestrengths -- garbage collection, runs everywhere, JIT compilers, etc. Butto me it also has some weaknesses (esp for symbolic code): noalgebraic data types, very weak type system, OO is a stupid idea (thisone is a personal peeve).YMMVgraham === > The most popular CAS seems to have a C kernel and> a proprietary extension language that has an algebraic syntax.> If the designers of these systems would design them today, it> would be interesting to know what they would do...They could answer that question here if they are reading this!> My guess is that they would write in C because their programmers know> C. Ok I will bite. I was one of the designers of Magma. It has a C kerneland a proprietary extension language.If I could do it again what wouldI do? Before answering let me be clear that these are my personalopinions and not those of the Magma group (I no longer work for them).I would do one of two things :-a) start with Lisp, Caml, or (S)ML, and build my CA system as a set o?ies on top of such languages.b) (more ambitiously) start totally from scratch, and rethink everythingin the design of CA systems. I suspect the end result of this would bea language similar to Lisp or SML, but with some area speci? addons.I would not repeat the C kernel plus proprietary language design.graham === > The most popular CAS seems to have a C kernel and> a proprietary extension language that has an algebraic syntax.> If the designers of these systems would design them today, it> would be interesting to know what they would do...> They could answer that question here if they are reading this!> My guess is that they would write in C because their programmers know> C.> Ok I will bite. I was one of the designers of Magma. It has a C kernel> and a proprietary extension language.If I could do it again what would> I do? Before answering let me be clear that these are my personal> opinions and not those of the Magma group (I no longer work for them).> I would do one of two things :-> a) start with Lisp, Caml, or (S)ML, and build my CA system as a set of> libraries on top of such languages.I wouldnft do that.> b) (more ambitiously) start totally from scratch, and rethink everything> in the design of CA systems. I suspect the end result of this would be> a language similar to Lisp or SML, but with some area speci? add> ons.> I would not repeat the C kernel plus proprietary language design.I would program the kernel in assembly meaning 64-bit assembly because of itbecoming a near future for all of us. Lisp, C and SML seem to be somewhatoutdated at this time. Interface for Windows can be also designed inthey just canft provide a normal performance now and certainly wonft be ableto do that in the future.Dr. Alec Mihailovshttp://webpages.shepherd.edu/amihailo/ === mixed up the columns in my notes, and I was differentiating when I wassupposed to be integrating. Ifm very sorry.B. === Ifm trying to prove mathematically that:For all primes p, 2^p - 1 is prime. True or False.I know that the proposition is false, because 2^11 - 1 = 2047 which isdivisible by 23 & 89.What I cant ?ure out is how to go about proving it.So far I have tried by Induction and now Disproof, but I have to admitthat my algebra isnft that good after 25 years away from high school.Ifve been trying to prove that if p = x*y (i.e. p is composite) that2^p-1 is composite. (provided p>2)It follows that if p is not composite (i.e. prime) then 2^p-1 is alsonot composite (i.e. prime). Problem is, that that logic proves theoriginal proposition as True ... which I know to wrong.Can someone give me some guidance or URLs?-- Luke------Q: What does FAQ stand for?A: We are Frequently Asked this Question, and we have no idea.------ === Ifm trying to prove mathematically that:For all primes p, 2^p - 1 is prime. True or False.I know that the proposition is false, because 2^11 - 1 = 2047 which is> divisible by 23 & 89.> What I cant ?ure out is how to go about proving it.You just did! One counterexample, such as p = 11, is enough to falsify the general statement. === Ifm trying to prove mathematically that:For all primes p, 2^p - 1 is prime. True or False.I know that the proposition is false, because 2^11 - 1 = 2047 which is> divisible by 23 & 89.> What I cant ?ure out is how to go about proving it.You just proved itfs false. The proposition said for ALL primes p...; you only have to ?d a single counterexample to prove that the proposition is false.End of story. You donft need to go any further.So far I have tried by Induction and now Disproof, but I have to admit> that my algebra isnft that good after 25 years away from high school.Ifve been trying to prove that if p = x*y (i.e. p is composite) that> 2^p-1 is composite. (provided p>2)It follows that if p is not composite (i.e. prime) then 2^p-1 is also> not composite (i.e. prime). Problem is, that that logic proves the> original proposition as True ... which I know to wrong.Can someone give me some guidance or URLs? === This has been asked as an assignment question for Discrete Maths that Iam doing over the x-mas break (I use the word break loosely!)I know that p=11 is a counter example, but is simply stating thatadequate proof from an academic point of view? My experience to datesays they want a more professional/academic explanation.Perhaps Ifm barking up the wrong tree insofar as there isnft a way ofproving it mathematically?Itfs been many years (over 2 decades) since Ifve done any seriousalgebra,and Ifm a little surprised at how rusty I am.Any further guidance ... ideas?Luke.> Ifm trying to prove mathematically that:> For all primes p, 2^p - 1 is prime. True or False.> I know that the proposition is false, because 2^11 - 1 = 2047 whichis> divisible by 23 & 89.> What I cant ?ure out is how to go about proving it.> You just proved itfs false. The proposition said for ALL primes p...;> you only have to ?d a single counterexample to prove that the> proposition is false.> End of story. You donft need to go any further. === > I know that p=11 is a counter example, but is simply stating that> adequate proof from an academic point of view? Yes.> My experience to date> says they want a more professional/academic explanation.Perhaps Ifm barking up the wrong tree insofar as there isnft a way of> proving it mathematically?You (and conceivably they) may be confusing different sorts of propositions. Can you please reproduce EXACTLY what the question says? The precise wording may be crucial. === Everett,> While LTspice is now 4.65MB(slightly larger if> compiled to use P4fs 128 bit width data types to> simultaneously do two FLOPS on two 64bit double> precision numbers), years ago LTspice was just> over 1.4MB, as I recall. In that 1.4MB there was a> complete SPICE engine, a waveform viewer(that had> itfs own 64bit virtual address space for waveform> data) a schematic capture program, a SMPS> synthesizer, about 30 small compilers(circuit> simulation is ultimately a complier problem these> days as I see it) and MFC statically linked in for> a multi-threaded GUI. Yep, people were surprised> to see the small execution size. Todayfs 4.65MB> version bloated in that it has now contains two> complete SPICE solvers(one uses a different SPARSE> matrix package that was incompatible with the normal> solver) and very large MOSFET device evaluation> code for various of experimental MOSFETfs devices.> However, the only penalty for the size is the> download time and by industry standards for this> ?ld, itfs still a lean, thin, and mean executable.> Itfs distributed as a 5MB self-extracting gzipfed> executable that contains 1500+ ?es in addition to> the executable. My guess is that it wound not be> practical to implement such a project in Forth.> We put men on the moon with probably less code than> that in all the computers involved, ground and mobile.> I sure hope so, itfs about half a million lines of densely> written C++ code. The simulator is far beyond 60fs> technology. It does full chip, transistor level> simulation of ICfs much larger than were available during> the moon shots. Itfs routinely used for ICfs with 10,000> transistors solved in a single matrix 100,000 elements on> a side. Technology moves on, even if it feels like all> the good engineering has already been done.> Ifm just trying to fathom how it takes so much code to> simulate transistors. One line of code per electron?Eh? I already described what else besides transitorevaluation is in that 4.6MB two posts back. The size ofthe transitor evalution code is probably less than a 1MB.It handles 14 different types of transitors, each withobject code sizes ranging from less than 20K for a uni?dbipolar device to 80K for lastest Berkeley release ofBSIM4.--Mike === In Von Neumannfs stability analysis, each element of the solution, u_j^nis written as: u_j^n = A^n(k) e^{i k j dx} eq(1)I really donft understand this.What I *DO* understand is that we can expand the solution u(x,t) in aFourier series: u(x,t) = sum_k A^n(k) e^{i k x} eq(2) where, at least numerically, x = j dr and t = n dt.What I *THINK* I understand is that we can play some orthogonality games(borrowing Diracfs notation and surpressing the time index for clarity): |u> = sum_k A(k) |e_k> = sum_k A(k) = A(j) eq(3)Although eq(3) looks like eq(1), is not really u_j. At least, Idonft think it is.Can someone explain how and why you get eq(1) from eq(2)? Ifm lookingfor something intuitive if possible. Anything to help understand why wewrite down eq(1) to begin Von Neumannfs stability analysis.Pete === Merry Christmas!I have some questions to the statistic application amos. I have some speci? questions about modelling latent variables. Which one would the right News Group be? === Data encryption 360 degrees rotation document 90 degrees andencryptionon every angel then change it two binary code and fold it over like apiece of paper then having the onefs and zero cancel each other out.if you written a very long letter and then change it two binary codeit would look like this0101010101010101010101010010101010101010101010010101010010 1010101001000010101000101010101010 10010101010100101010101 would equal = 01010101010101000011001010100101010101010101011111110111001101 0101010100101010101010101010101010101010101010101010101if you took the piece of paper and folded it and folded it and foldedit the 0 and 1 would cancel each out and if you keep folding the pieceof paper too the smallest you would have 4 numbers left if 1+1 =nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 01 now if the key new the folding times you could send 2 bytes overthe internet and unzip a100 zetabyte program you computer could store all the programs everwritten but just need the key to unzip then you could us this for SETIfor signals or can you imagine a computer processor that would be 1.8Hz but run like 100 million zeta hz you could use the new 64 bitprocess second side to unzip while the front side processes. or usethis for the matrix or quantum computing or supercomputer. 64 bit. 1+1= nothing and 0 + 0 = nothing 1+0=1 and 0+1+0 dont forrget to use signes 0>1+1<0HTTP://WWW.SPEEDYPC.20M.COM === > please send me any source code for pid controllerCompanies that want to sell embedded controller chips have such things:http://www.microchip.com/1010/suppdoc/appnote/all/ an532/index.htm === please send me any source code for pid controllerWould its transfer fcn, y/u = Kp + Ki/s + Kd*shelp you roll your own? === Has anyone been able to compile the BLAS and LAPACK libraries with Intel Fortan 8.0?. What were the compilation ?hat you used?If this has been discussed before, could you direct me to a link/discussion thread ? -Nyarlathotep === > Has anyone been able to compile the BLAS and LAPACK libraries with Intel> Fortan 8.0?. What were the compilation ?hat you used?Instead of blas you should really use www.netlib.org/atlas, which givesa much higher performance. Regardless your compiler.V.-- homepage: cs utk edu tilde lastname === > given a 2-d pdf with P = 0 for x<0 OR y<0 OR x>1 OR y>1.> i need to ?d an algorithm which divides this domain [(0, 0) - (1,> 1)] into n areas (e.g. polygons) of equal probabilities, i.e. each> areafs probability will be 1/n.I wouldnft be surprise if this is a known problem; maybe thiscan be considered a variation on the Voronoi tesselation. Trysearching for weighted tessellation or something.Considering this abstractly, there are n equations (area_1 = 1/n, ..., area_n = 1/n) and the number of variables isequal to the twice the number of vertices. In general the numberof variables will be greater than the number of equations, soyou have an underdetermined system. That should help you ?d an appropriate solution. It seems like there should be a suitablemodi?ation of Newtonfs method for undetermined systems; Ifvecross-posted to sci.math.num-analysis -- any ideas there?If Ifm not mistaken, derivatives of the mass of the pdf over a polygon with respect to its vertices do exist, assuming the pd? at least continuous. I believe the derivative of the mass withrespect to the x coordinate of one vertex, a, is something like 1/2 dx int_0^1 pdf(xy_1(t)) dt + 1/2 dx int_0^1 pdf(xy_2(t)) dtwith xy_1(t) = (1-t) a + t b, and xy_2(t) = (1-t) a + t c,where b and c are the other two vertices which share an edge with a. Hmm, I might have goofed that up, youfll want to double check. In any event, the point is that the derivativescan be computed without too much trouble -- even if the integrationin the derivative canft be computed symbolically, it is (hopefully) not dif?ult to compute it numerically; itfs onlyone dimensional.> actually, ifve done this with a genetic algorithm but itfs not> ef?ient at all. i want to know if there exists a better (perhaps> more elegant) solution to this problem.The advantage of randomized algorithms such as genetic algorithmsor simulated annealing is that theyfre applicable to problems,such as combinatorial problems, which have no notion of which wayto go. If you can tell which way to go, you are usually betteroff using a method that can exploit that information.For what itfs worth,Robert Dodier--If I have not seen as far as others, it is because giants were standing on my shoulders. -- Hal Abelson === > I canft prove that this equation : y^2=x^3 + 7> has no integer solutions.> Can you help me?Itfs fairly easy to show that x must be odd. By considering theequivalent form y^2 + 1 = (x + 2)*(x^2 - 2*x + 4) you can show thaty^2 + 1 must be divisible by a prime congruent to 3 mod 4, and fromthat derive a contradiction.Geoff.----------------------------------------- ------------------------------------Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --ftww@maths.usyd.edu.au | Gameplayer by vocation.---------------------------------------------------- ------------------------- === Prime numbers are actually very important in todays world. Those and thefact that you cant easily factor big numbers that is! Thats how encryptionworks. If I get get two really big prime numbers, nd multiply them togetherthen I have a whopping great number that I can use in a formula that noonewill be able to break (hopefully) since they wont be able to ?ure out whatthe initial two numbers were.For a really good explanation of this goto the sci.crypt newsgroup and readthe FAQ there.For practical applications, the ATM you get your money from uses encryption.Online transactions use encrypted data, your communication sattelites, yourISP. Anywhere where there is a password your prolly gonna ?d encryption.Hope this helps - its pretty basic.Mathew B> As a person of *extremely* limited mathematical education could someone> please explain to me the signi?ance and importance of prime numbers?> How are they employed in real-world applications?> big H> === Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y.But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I canassume that the expression is a difference of squares.Dennis === >Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y.>But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I can>assume that the expression is a difference of squares.???Ifm sorry, but I donft understand how you get what you said in your second paragraph. What operations did you perform?a = 2; b = (x+y); a^2 - b^2 = 4 - (x+y)^2.Factors are a+b = 2+x+y and b = 2-x-y.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comModern cyberspace is a deadly festering swamp, teeming withdangerous programs such as viruses, worms, Trojan horses,and licensed Microsoft software that can take over yourcomputer and render it useless. --Dave Barry === You almost answer your own question...4-(x+y)^2 = (2 + x + y)(2 - x - y)Which can perhaps better be seen as:(2 + (x + y))(2 - (x + y))I hope that this helps - if it does, then please visit:www.brainbashers.com?postcardKev === > Can anyone help alleviate some dif?ulty of mine in understanding the> details of the real projective plane and a few of its homeomorphic> spaces? What is the signi?ance of the addition of the in?ite line> to the projection? One way to think about it is that the addition of the in?ite line and the in?ite point removes some special cases from the geometry. For example, if you donft have the in?ite point, then two lines intersect in one of zero points, depending on whether they are parallel. If you add the in?ite point, then every pair of lines intersects in exactly one point -- the in?ite point is the intersection of all pairs of parallel lines. Likewise, once you add the in?ite point, if you want to maintain the property that for every line and every point there is a line that passes through that point thatfs parallel to the line, you need to add the in?ite line. > I would assume that a better understanding of the> idea of the thing would assist me in understanding certain> homeomorphisms. For instance, its being homeomorphic to a sphere with> antipodal points identi?d. If you call each great circle of the sphere a line then each pair of lines intersects in two antipodal points of the sphere. Now you call that pair a point and suddenly you have a geometry which satis?s the same constraints: every pair of lines gives you a point, one pair of points gives you a line, etc.> Or a disc with boundary points identi?d. Or, in what seems the most confusing analogy, the set of> all lines in R^3. How can that be the case? Ifm just at a total loss> here.For lines in R^3, you call each line a point and each plane a line and then you again have the same property (you do have to choose a suitable way to de?e the line that passes through two points that represent skew lines). might be useful.hthmeeroh-- If this message helped you, consider buying an itemfrom my wish list: === confuse everyone with this message:> ... So what is the status of automatic theorem> proving? Have any great breakthroughs been made using this approach?>Herefs a place at least to begin looking for the answers to your>questions:> http://gtps.math.cmu.edu/tps.html>Maybe someone knows how to compile this thing with GNU CLisp for Windows?--|E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15, === ============;|E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--||E1M3_:43__E1M7_:14__E3M1_:43__E4M2 _:52__END__:37; === =========[4*72]=== ... So what is the status of automatic theorem> proving? Have any great breakthroughs been made using this approach?Herefs a place at least to begin looking for the answers to your> questions: http://gtps.math.cmu.edu/tps.htmlthanks === All messages from thread Message 1 in thread (a) Show that I_n = (4n -2)*q^2*I_n-1 - p^2*q^2*I_n-2 for n>= 2 wherepi = p/q (b) Deduce that I_n is an integer(c) show that 0 < I_n < (p/q)*(p/2)^(2n) /n! n = 0, 1, 2, ....(d) If (p/q)*(p/2)^(2n) /n! < 1, deduce that pi is irrational.I am stuck at (c) at the moment, (d) seems to be quite easy once you gotI try to present some hints. Following notation is used : I[f]= Integral_{t=a to t=x}f(t) dt E(t)|_{a,x}=E(b)-E(x) W^{(p)}= p-th derivative of W (if exists) C(z,k)= z(z-1)(z-2)...(z-k+1)/k! M(n,k)=C(n,k)*(2n-k)!/(2n)! === =========================================== === J_n= Integral {from -pi/2 to pi/2}(((pi^2)/4 - t^2)^n)*cos(t) dt. === =========================================== === Suppose that p is a positive integer and W,F are in C^{p}[a,b] and x is in [a,b].The repeated integration by parts formula (so called ,,Green-Lagrange rule ) is I[W*F^{(p)}]=(-1)^p*I[W^{(p)}*F]+ (1) +SUM_{k=0 to k=p-1}(-1)^{p-1-k}W^{(p-1-k)}(t)F^{(k)}(t)|_{a,x} .STEP I. Select in (1) the following input data : W(t)=W_1(t)= (t-a)^n*(x-t)^n , F= f in C^{2n+1}[a,b] . You ?d(2) f(x)= f(a)+ + SUM_{k=1 to k=n}M(n,k)(x-a)^k*(f^{(k)}(a)-(-1)^k*f^{(k)}(x))+ r_n(f;x)where the ,,remainder-term r_n(f;x) is r_n(f;a,x)=((-1)^n/(2n)!)* I(W*f^{(2n+1)})= =(see Mean-value theorem for Riemann integrals)=(2f) =(-1)^n*(x-a)^{2n+1}f^{(2n+1)}(z)/(C(2n,n)*(2n+1)!)for some z=z(a,x,f) in [a,x] . Identity (2f) may be called ,,HERMITE FORMULA (1878), see [3]. This formula has a long history. After me, he was rediscovered in 1915 by K. Petr [11]1940 by Nikola Obresckhoff [10] (in a more general form)1947 by I. Niven [9]1948 by P.M. Hummel and C.L.Seebeck (as in [10])1957 by D.V.Ionescu (as in [10]).For application see [8].STEP II. In (2) let x= pi/2 , a=-pi/2, i.e. [a,x]=[-pi/2,pi/2] and W(t)=W_2(t)=((pi^2)/4 - x^2)^n , f(t)= sin(t) .r_n(sin;-pi/2,pi/2)=(1/(2n)!)*J_n ==pi^{2n+1}*cos(z)/(C(2n,n)*(2n+1)!), that integral J_n veri?s(3) J_n=pi^{2n+1}*(n!)^2*cos(z_1)/(2n+1)! for some z_1 in [-pi/2,pi/2].In fact it may be justi?d that z_1 is in (-pi/2,pi/2). Suppose that pi=p/q . According to (3) (4) J_n= p^{2n+1}*(n!)^2*cos(z_1)/((2n+1)!*q^{2n+1}). === ============Further use ....(2). Perhaps you need also the remark that === ============(5) sqrt(n*pi)/4^n < (n!)^2/(2n)! = 1/C(2n,n) < sqrt(n*pi+pi/2)/4^n for n=1,2,... . Inequalities (5) may be called ,,WALLIS INEQUALITIES. Possible literature about this subject : REFERENCES [1] AIGNER M., ZIEGLER G.M., Proofs from THE BOOK. Springer-Verlag , Berlin,1998. [2] GROSSWALD E., Bessel Polynomials. Lecture Notes in Mathematics- 698,Springer Verlag, Berlin, 1978. [3] HERMITE Ch., Sur la formule dfinterpolation de Lagrange. Journ.f.Reine Angew. Mathematik 84(1878), 70-79. [4] HUMMEL P.M., SEEBECK C.L., A generalization of Taylorfs expansion. Amer.Math.Monthly , 56(1949) 243-247. [5] IONESCU D.V., Numerical Quadratures , (Romanian) Editura tehnica, 1957. [6] IWAMOTO Y., A proof that pi^2 is irrational. J.Osaka Institute of Science and Technology , 1, (1949) 147-1148. [7] KOKSMA J.F., On Nivenfs proof that pi is irrational. Nieuw Archiv Wiskunde (2) , 23, (1949) 39 . [8] ...., Computation of some elementary functions (I), (Romanian), Gazeta Matematica (seria A), VII, 1, (1986) 15-26. [9] NIVEN I., A simple proof that pi is irrational. Bulletin Amer.Math.Soc., 53, (1947), 509.[10] OBRESCHKOFF N., Neue Quadraturformeln. Abhandlugen der Preussischen Akademie der Wissenschaften, 4, 1940, 1-20.[11] PETR K., O jedne formuli pro numerichy vypost urcitych integralu, Casopis pro pertovani Matematiks 44 (1915) 454-455. === Let f:[0,1] X [0,1] - > R be de?ed by f(x,y) = 0 if x is irrational = 0 if x is rational, y is irrational = 1/q if x is rational, y = p/q (in lowest terms)Prove that the function f is Riemann integrable, and that int [f,[0,1] X [0,1]] = 0(Spivak, Calculus on Manifolds, problem 3-7.)According to the text (as well as standard de?tions/theorems), thefunction is integrable iff for any epsilon > 0 there exists apartition P of [0,1] X [0,1] such that U(f,P) - L(f,P) < epsilon. Itis easy to see that for any partition of [0,1] X [0,1] all Lower SumsL(f,P) = 0 . The problem I am having is with the proper setup of thepartition so that the upper sums U(f,P) can be shown to be arbitrarilysmall. We know for example that for say 1, 1/2, 1/3, ..., 1/k thefunction will take on all of these values (and beyond) so we will needto construct the rectangles of the partition P small enough so thatU(f,P) will be arbitrarily small.I have a preliminary attempt which constructs two sums:U(f,P) = 1*v_1 + (1/2)*v_2 + (1/3)*v_3 + ... + (1/k)*v_kand v_1 + v_2 + ... + v_k = 1.The v_i each represent the area of each rectangle in the partition sothat there sum must be equal to 1. Each term in the ?st equationrepresents the supremum of the function on the rectangle whose area isv_i times that area (i.e. (1/r)*v_r).The goal is to make the value of U(f,P) arbitrarily close to 0 whilemaintaining the second relation (that the sum of all the v_i is 1).Does anyone have a speci? construction that would solve this? === Let epsilon>0 be given.Look at this set: Y={y in [0,1] : y is rational and y=p/q, where q <2/epsilon}.Notice that this is a ?ite set!!! Write Y={y_1,...,y_n}.Choose a number delta > 0 such that: delta < epsilon/(4n) and the sets[y_1-delta,y_1+delta],...,[y_n-delta,y_n+delta] do not overlab. (Thiscan be done since Y is ?ite.)For k=1,...,n let P_k=[0,1] X [y_k-delta,y_k+delta].P_0=[0,1] X [0,1] union(P_1,...,P_k). ( means setminus.)Now P_0,....,P_n is a partion of [0,1] X [0,1].If (x,y) is in P_0 and y=a/b then b>=2/epsilon, otherwise y would been Y.So 1/b < epsilon/2.Therefore U(f,P_0)Which of the following is a factor of 4-(x+y)^2? The answer is 2+x+y.>But I get at best only 4+4x-4y+x^2+y^2 or 4(1+x-y)+x^2+y^2 if I can>assume that the expression is a difference of squares.???Ifm sorry, but I donft understand how you get what you said in your second paragraph. What operations did you perform?a = 2; b = (x+y); a^2 - b^2 = 4 - (x+y)^2.Factors are a+b = 2+x+y and b = 2-x-y.-- Stan Brown, Oak Road Systems, Cortland County, New York, USA http://OakRoadSystems.comModern cyberspace is a deadly festering swamp, teeming withdangerous programs such as viruses, worms, Trojan horses,and licensed Microsoft software that can take over yourcomputer and render it useless. --Dave Barry === You almost answer your own question...4-(x+y)^2 = (2 + x + y)(2 - x - y)Which can perhaps better be seen as:(2 + (x + y))(2 - (x + y))I hope that this helps - if it does, then please visit:www.brainbashers.com?postcardKev === > Can anyone help alleviate some dif?ulty of mine in understanding the> details of the real projective plane and a few of its homeomorphic> spaces? What is the signi?ance of the addition of the in?ite line> to the projection? One way to think about it is that the addition of the in?ite line and the in?ite point removes some special cases from the geometry. For example, if you donft have the in?ite point, then two lines intersect in one of zero points, depending on whether they are parallel. If you add the in?ite point, then every pair of lines intersects in exactly one point -- the in?ite point is the intersection of all pairs of parallel lines. Likewise, once you add the in?ite point, if you want to maintain the property that for every line and every point there is a line that passes through that point thatfs parallel to the line, you need to add the in?ite line. > I would assume that a better understanding of the> idea of the thing would assist me in understanding certain> homeomorphisms. For instance, its being homeomorphic to a sphere with> antipodal points identi?d. If you call each great circle of the sphere a line then each pair of lines intersects in two antipodal points of the sphere. Now you call that pair a point and suddenly you have a geometry which satis?s the same constraints: every pair of lines gives you a point, one pair of points gives you a line, etc.> Or a disc with boundary points identi?d. Or, in what seems the most confusing analogy, the set of> all lines in R^3. How can that be the case? Ifm just at a total loss> here.For lines in R^3, you call each line a point and each plane a line and then you again have the same property (you do have to choose a suitable way to de?e the line that passes through two points that represent skew lines). might be useful.hthmeeroh-- If this message helped you, consider buying an itemfrom my wish list: === confuse everyone with this message:> ... So what is the status of automatic theorem> proving? Have any great breakthroughs been made using this approach?>Herefs a place at least to begin looking for the answers to your>questions:> http://gtps.math.cmu.edu/tps.html>Maybe someone knows how to compile this thing with GNU CLisp for Windows?--|E1M1 :29 E1M5 :19 E2M1 :10 E3M2 :22 E4M5 :15, === ============;|E1M2 :36 E1M6 :11 E2M3 :27 E4M1 :30 E4M6 :24|->grue3.tripod.com<--||E1M3_:43__E1M7_:14__E3M1_:43__E4M2 _:52__END__:37; === =========[4*72]=== ... So what is the status of automatic theorem> proving? Have any great breakthroughs been made using this approach?Herefs a place at least to begin looking for the answers to your> questions: http://gtps.math.cmu.edu/tps.htmlthanks === All messages from thread Message 1 in thread (a) Show that I_n = (4n -2)*q^2*I_n-1 - p^2*q^2*I_n-2 for n>= 2 wherepi = p/q (b) Deduce that I_n is an integer(c) show that 0 < I_n < (p/q)*(p/2)^(2n) /n! n = 0, 1, 2, ....(d) If (p/q)*(p/2)^(2n) /n! < 1, deduce that pi is irrational.I am stuck at (c) at the moment, (d) seems to be quite easy once you gotI try to present some hints. Following notation is used : I[f]= Integral_{t=a to t=x}f(t) dt E(t)|_{a,x}=E(b)-E(x) W^{(p)}= p-th derivative of W (if exists) C(z,k)= z(z-1)(z-2)...(z-k+1)/k! M(n,k)=C(n,k)*(2n-k)!/(2n)! === =========================================== === J_n= Integral {from -pi/2 to pi/2}(((pi^2)/4 - t^2)^n)*cos(t) dt. === =========================================== === Suppose that p is a positive integer and W,F are in C^{p}[a,b] and x is in [a,b].The repeated integration by parts formula (so called ,,Green-Lagrange rule ) is I[W*F^{(p)}]=(-1)^p*I[W^{(p)}*F]+ (1) +SUM_{k=0 to k=p-1}(-1)^{p-1-k}W^{(p-1-k)}(t)F^{(k)}(t)|_{a,x} .STEP I. Select in (1) the following input data : W(t)=W_1(t)= (t-a)^n*(x-t)^n , F= f in C^{2n+1}[a,b] . You ?d(2) f(x)= f(a)+ + SUM_{k=1 to k=n}M(n,k)(x-a)^k*(f^{(k)}(a)-(-1)^k*f^{(k)}(x))+ r_n(f;x)where the ,,remainder-term r_n(f;x) is r_n(f;a,x)=((-1)^n/(2n)!)* I(W*f^{(2n+1)})= =(see Mean-value theorem for Riemann integrals)=(2f) =(-1)^n*(x-a)^{2n+1}f^{(2n+1)}(z)/(C(2n,n)*(2n+1)!)for some z=z(a,x,f) in [a,x] . Identity (2f) may be called ,,HERMITE FORMULA (1878), see [3]. This formula has a long history. After me, he was rediscovered in 1915 by K. Petr [11]1940 by Nikola Obresckhoff [10] (in a more general form)1947 by I. Niven [9]1948 by P.M. Hummel and C.L.Seebeck (as in [10])1957 by D.V.Ionescu (as in [10]).For application see [8].STEP II. In (2) let x= pi/2 , a=-pi/2, i.e. [a,x]=[-pi/2,pi/2] and W(t)=W_2(t)=((pi^2)/4 - x^2)^n , f(t)= sin(t) .r_n(sin;-pi/2,pi/2)=(1/(2n)!)*J_n ==pi^{2n+1}*cos(z)/(C(2n,n)*(2n+1)!), that integral J_n veri?s(3) J_n=pi^{2n+1}*(n!)^2*cos(z_1)/(2n+1)! for some z_1 in [-pi/2,pi/2].In fact it may be justi?d that z_1 is in (-pi/2,pi/2). Suppose that pi=p/q . According to (3) (4) J_n= p^{2n+1}*(n!)^2*cos(z_1)/((2n+1)!*q^{2n+1}). === ============Further use ....(2). Perhaps you need also the remark that === ============(5) sqrt(n*pi)/4^n < (n!)^2/(2n)! = 1/C(2n,n) < sqrt(n*pi+pi/2)/4^n for n=1,2,... . Inequalities (5) may be called ,,WALLIS INEQUALITIES. Possible literature about this subject : REFERENCES [1] AIGNER M., ZIEGLER G.M., Proofs from THE BOOK. Springer-Verlag , Berlin,1998. [2] GROSSWALD E., Bessel Polynomials. Lecture Notes in Mathematics- 698,Springer Verlag, Berlin, 1978. [3] HERMITE Ch., Sur la formule dfinterpolation de Lagrange. Journ.f.Reine Angew. Mathematik 84(1878), 70-79. [4] HUMMEL P.M., SEEBECK C.L., A generalization of Taylorfs expansion. Amer.Math.Monthly , 56(1949) 243-247. [5] IONESCU D.V., Numerical Quadratures , (Romanian) Editura tehnica, 1957. [6] IWAMOTO Y., A proof that pi^2 is irrational. J.Osaka Institute of Science and Technology , 1, (1949) 147-1148. [7] KOKSMA J.F., On Nivenfs proof that pi is irrational. Nieuw Archiv Wiskunde (2) , 23, (1949) 39 . [8] ...., Computation of some elementary functions (I), (Romanian), Gazeta Matematica (seria A), VII, 1, (1986) 15-26. [9] NIVEN I., A simple proof that pi is irrational. Bulletin Amer.Math.Soc., 53, (1947), 509.[10] OBRESCHKOFF N., Neue Quadraturformeln. Abhandlugen der Preussischen Akademie der Wissenschaften, 4, 1940, 1-20.[11] PETR K., O jedne formuli pro numerichy vypost urcitych integralu, Casopis pro pertovani Matematiks 44 (1915) 454-455. === Let f:[0,1] X [0,1] - > R be de?ed by f(x,y) = 0 if x is irrational = 0 if x is rational, y is irrational = 1/q if x is rational, y = p/q (in lowest terms)Prove that the function f is Riemann integrable, and that int [f,[0,1] X [0,1]] = 0(Spivak, Calculus on Manifolds, problem 3-7.)According to the text (as well as standard de?tions/theorems), thefunction is integrable iff for any epsilon > 0 there exists apartition P of [0,1] X [0,1] such that U(f,P) - L(f,P) < epsilon. Itis easy to see that for any partition of [0,1] X [0,1] all Lower SumsL(f,P) = 0 . The problem I am having is with the proper setup of thepartition so that the upper sums U(f,P) can be shown to be arbitrarilysmall. We know for example that for say 1, 1/2, 1/3, ..., 1/k thefunction will take on all of these values (and beyond) so we will needto construct the rectangles of the partition P small enough so thatU(f,P) will be arbitrarily small.I have a preliminary attempt which constructs two sums:U(f,P) = 1*v_1 + (1/2)*v_2 + (1/3)*v_3 + ... + (1/k)*v_kand v_1 + v_2 + ... + v_k = 1.The v_i each represent the area of each rectangle in the partition sothat there sum must be equal to 1. Each term in the ?st equationrepresents the supremum of the function on the rectangle whose area isv_i times that area (i.e. (1/r)*v_r).The goal is to make the value of U(f,P) arbitrarily close to 0 whilemaintaining the second relation (that the sum of all the v_i is 1).Does anyone have a speci? construction that would solve this? === Let epsilon>0 be given.Look at this set: Y={y in [0,1] : y is rational and y=p/q, where q <2/epsilon}.Notice that this is a ?ite set!!! Write Y={y_1,...,y_n}.Choose a number delta > 0 such that: delta < epsilon/(4n) and the sets[y_1-delta,y_1+delta],...,[y_n-delta,y_n+delta] do not overlab. (Thiscan be done since Y is ?ite.)For k=1,...,n let P_k=[0,1] X [y_k-delta,y_k+delta].P_0=[0,1] X [0,1] union(P_1,...,P_k). ( means setminus.)Now P_0,....,P_n is a partion of [0,1] X [0,1].If (x,y) is in P_0 and y=a/b then b>=2/epsilon, otherwise y would been Y.So 1/b < epsilon/2.Therefore U(f,P_0) Will you help me in solving the following ODE k-th order non omog.> differential equation?> let u denote a real function of one real variable , say x.> u^(k) (x) = 1/x> Pheraps i need some iterative formula...>u^(k-1) = log x + cu^(k-2) = x log x - x + cx + d = x log x + cx + d, where new c = c-1etc.Youfll want integral x^n log x dx = [x^(n+1) log x]/(n+1) - x^(n+1) / (n+1)^2and some elboe grease for the ?al result. === could u plz integrate this for me? 2/(4x^2+4x+5)^0.5the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc)i dont get wher the ln come from?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > could u plz integrate this for me?> 2/(4x^2+4x+5)^0.5> the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc)> i dont get wher the ln come from?>First, complete the square on the denominator4x^2+4x+54(x^2+x)+5 1/4 -14(x+1/2)^2+4Let u=x+1/2, du=dx2/(4u^2+4)Integral(1/u^2+1) duArctan(u)+CArctan(x+1/2)+C or ArcSinh(x+1/2)+C (The latter was what I got when I ran itthrough Mathematica).David Moran-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === The ln came from the fact that:If f(x) = 1/x then the integral of f(x) = ln(x)> could u plz integrate this for me?> 2/(4x^2+4x+5)^0.5> the answer is ln (|4x^2+4x+5+(2x+1)|) (from my casio FX2.0 calc)> i dont get wher the ln come from?>-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === HELP! I have been searching for a software program that would providediagnostic information on multiple math strands. I am a remedial mathteacher. I am looking for a program to help me quickly assess theknowledge and skill a student has in math, as well as give me somesort of narrative report. We are currently using the KeyMath Test --but would like lomething a little more current. Any suggestions?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === hi-We are testing a small interactive learning lab so students can experiment with the quadratic equation. It is for a pre-algebra type curriculum. It runs on Mac OS X v10.3. If you would like to run it and give any feedback Ifd be happy to provide a download URL and license key for it.TIA--lanceLance BlandVVI888-VVI-PLOThttp://www.vvi.com-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I need to determine a percentage for a SALES application.$168,000.00. They grew in sales by a total of $28,231.00. Do I takesales) that gives me .8319 and say that they grew by 17%????When I take the lowest number and multply by 20% I get $27,953.00.How does the business world determine a true percentage for this typeof scenario??-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Has anyone used Qwizdom? It looks like a great way to get thestudents involved with learning. Using the remotes!! Before I try tourge my school to buy this software I wanted to know if anyone hasused it and how they feel about it?!!-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I have been wondering on whether there exist methods to ?d the eigenvaluesof a matrix other than solving the characteristic polynomial equation. Dothere exist elementary row operation methods that will ?d the realeigenvalues of a matrix?I have been using a site that will ?d the eigenvalues of a matrix and Ifmwondering how it does it because Ifm pretty sure its not doing it by solvingthe characteristic equation.Jeremy === > I have been wondering on whether there exist methods to ?d the eigenvalues> of a matrix other than solving the characteristic polynomial equation. Do> there exist elementary row operation methods that will ?d the real> eigenvalues of a matrix?Sort of - see below.> I have been using a site that will ?d the eigenvalues of a matrix and Ifm> wondering how it does it because Ifm pretty sure its not doing it by solving> the characteristic equation.That is correct. There are ways to ?d eigenvalues (approximately)without ?st getting the characteristic polynomial. Google QRdecomposition to see one way. is a start.It is perhaps counter-intuitive but Math packages, AFAIK, ?d thecharacteristic polynomial by ?st ?ding the eienvalues.-- Paul SperryColumbia, SC (USA) === What I like about adding to Rick Deckerfs example is that it makes iteasy.In case you missed it, Rick Decker is a professor at Hamilton College(do a web search to ?d out about his school) who posted a*quadratic* in an attempt to refute some of my conclusions.Well, I found his example grabbed me, for various reasons, and ?allyfound that I could modify it slightly to suit my purposes, and nowcon?ent in your understanding--even as undergrads--of independencebetween variables, I feel I can ?ally show you how I have beendefending mathematics against people working to undermine it!Herefs the quadratic that Decker gave:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the afs are roots of a^2 - (x - 1)a + 7(x^2 + x).And my modi?ation was just to add in the variable y, to get(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the afs are roots of a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I havenft relatedthem to each other in any way.That second polynomiala^2 - (x - y)a + 7(x^2 + xy)is the polynomial being analyzed by considering(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).You ?d it from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where whatfs on the right side is just a regrouping of terms fromwhatfs on the left, as you can ?d out by simplifying the right side.Notice that with(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on theleft side, but that contradicts with 7(2y^2) on the right!So what gives?Well the problem is that x is in the way, and a simple technique inanalysis is to clear out one variable by setting it to 0.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly.Obviously, now the coef?ients of the terms that have y but donfthave x as a factor are revealed, and not surprisingly, 7y(2y) matchesup with what you see on the right side with 7(2y^2).So if x does not equal y then it follows that(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2but if x-y=0, then you havea^2 + 7(x^2 + xy), de?ing the afs, and you can directly calculate that a = +/- sqrt(14)y, and pickinga_1(y,y) = sqrt(14)y, you have(5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2).Now considering when x does not equal y itfs easy enough to see thatregardless of anyonefs feelings on the matter, itfs not possible forthe terms *independent* of x, to actually have a dependency on x.So, with(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)if (5b_2(x,y) + 2y) had any factors in common with 7, then thosefactors would have to be a factor of 2y, as that term is both visibleand independent of x.So you can see my conclusion about what happens when you divide bothsides by 7, when x does not equal y, follows from some basic algebra.What I like here is that quadratics are simpler than cubics to showthe result, the example I modi?d was presented ?st by Rick Decker,a professor at Hamilton College, and my conclusion follows fromindependence between independent variables.That is to doubt me here you need to doubt that concept in mathematicsitself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, is thatif you supposed that you were in the ring of algebraic integers(notice I made no mention of a ring before now) then you run into aproblem speci? to that ring, where dividing both sides by 7 pushesyou out of the ring of algebraic integers.It might seem esoteric and distant as a problem, but my discovery ofthat problem is actually a fascinating story, and now you can see howmathematicians *worldwide* answer the question:What do you do when your paradigm is forced to shift by a discovererwho thought outside of the box?Want more?Then read my other threads on sci.math or check out my blog archives:James Harris === > A while back I found these algebraic tools for ?ding out about roots> of a polynomial using another polynomial, but the problem is that> apparently (from the arguing) mathematicians had never found my> technique before, and not having had it, theyfd come up with some> false assumptions about roots of polynomials.> So not surprisingly, worldwide, mathematicians when contacted about my> work have tried to run away from it. Ifve tried to explain it before,> but it is just complicated enough that it seems to shoot past people,> but recently a Rick Decker, a professor at Hamilton College came up> with a *quadratic* example to try and refute my position, but I found> that by modifying his example, I could use it to explain, knowing that> quadratic are simple enough that therefs less room for confusion.> Consider> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > where the afs are roots of > a^2 - (x - y)a + 7(x^2 + xy).> Notice that x and y are independent variables, as I havenft related> them to each other in any way.> That second polynomial> a^2 - (x - y)a + 7(x^2 + xy)> is the polynomial being analyzed by considering> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).> You ?d it from> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2> where whatfs on the right side is just a regrouping of terms from> whatfs on the left, as you can ?d out by simplifying the right side. Now letting x=0, gives > a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have> a_2(x,y) = b_2(x,y) - y, so> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > CORRECTION:Oh hey, now that I have another variable, I can set *it* to 0 to seewhatfs left over as well, so letting y=0, givesa^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,so you can actually separate out further, where Ifll introducec_1(x,y) and c_2(x,y) to get(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Want more advanced polynomial factorization? Then check out my blog archives: James Harris === > A while back I found these algebraic tools for ?ding out about roots> of a polynomial using another polynomial, but the problem is that> apparently (from the arguing) mathematicians had never found my> technique before, and not having had it, theyfd come up with some> false assumptions about roots of polynomials.> So not surprisingly, worldwide, mathematicians when contacted about my> work have tried to run away from it. Ifve tried to explain it before,> but it is just complicated enough that it seems to shoot past people,> but recently a Rick Decker, a professor at Hamilton College came up> with a *quadratic* example to try and refute my position, but I found> that by modifying his example, I could use it to explain, knowing that> quadratic are simple enough that therefs less room for confusion.> Consider> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > where the afs are roots of > a^2 - (x - y)a + 7(x^2 + xy). Notice that x and y are independent variables, as I havenft related> them to each other in any way.> That second polynomial> a^2 - (x - y)a + 7(x^2 + xy)> is the polynomial being analyzed by considering> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).> You ?d it from> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2> where whatfs on the right side is just a regrouping of terms from> whatfs on the left, as you can ?d out by simplifying the right side.> Now letting x=0, gives > a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have> a_2(x,y) = b_2(x,y) - y, so> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Oh hey, now that I have another variable, I can set *it* to 0 to seewhatfs left over as well, so letting y=0, givesa^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,so you can actually separate out further, where Ifll introducec_1(x,y) and c_2(x,y) to get(5(1+sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1-sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Want more advanced polynomial factorization? Then check out my blog archives: James Harris === [cut]> Want more advanced polynomial factorization?[cut]> James HarrisSince you asked, I would like to see your factorization of x^3-x+8 === >[...]>So why would mathematicians argue against such a simple result?>Pride? Fear?What if you were simply _wrong_? I mean of course you canft be wrong, anyone can seefrom the record that while you make occaisional slips like anyone else, when you insist for months that somethingfs trueit< turns out to be true, once the dust has settled.Of course youfre not wrong, but think about it for a second:if you _were_ wrong that would be another possible reasonwhy mathematicians would say you were wrong. >Want more advanced polynomial factorization?>Then check out my blog archives:>James Harris************************David C. Ullrich === > A while back I found these algebraic tools for ?ding out about roots> of a polynomial using another polynomial, but the problem is that> apparently (from the arguing) mathematicians had never found my> technique before, and not having had it, theyfd come up with some> false assumptions about roots of polynomials.> So not surprisingly, worldwide, mathematicians when contacted about my> work have tried to run away from it. Ifve tried to explain it before,> but it is just complicated enough that it seems to shoot past people,> but recently a Rick Decker, a professor at Hamilton College came up> with a *quadratic* example to try and refute my position, but I found> that by modifying his example, I could use it to explain, knowing that> quadratic are simple enough that therefs less room for confusion.> Consider> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > where the afs are roots of > a^2 - (x - y)a + 7(x^2 + xy).> Notice that x and y are independent variables, as I havenft related> them to each other in any way.> That second polynomial> a^2 - (x - y)a + 7(x^2 + xy)> is the polynomial being analyzed by considering> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).> You ?d it from> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2> where whatfs on the right side is just a regrouping of terms from> whatfs on the left, as you can ?d out by simplifying the right side. Now letting x=0, gives > a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have> a_2(x,y) = b_2(x,y) - y, so> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > Obviously, now the coef?ients of the terms that have y but donft> have x as a factor are revealed, and not surprisingly, 7y(2y) matches> up with what you see on the right side with 7(2y^2).> Thatfs the simple result which mathematicians canft handle.> So instead posters ?hting the mathematical reality have made claims> that here would mean that that 7 divides as some kind of factor of x,> which is to say that depending on what value x has, 7 divides through> whatfs on the left hand side in different ways.> Mathematicians, you see, believe that if the quadratic> a^2 - (x - y)a + 7(x^2 + xy)> doesnft have *integer* roots, then *both* roots would have to share> factors with 7.> Correct. Thatfs what we believe.> If you ask them why youfll probably get a lot of complicated> explanations in return, and theyfre very defensive on the issue from> what Ifve gathered, as well as being wrong.> Let the roots be a1 and a2. Suppose, as Harris claims, that not both of them share factors with 7. That means essentially that oneof them has no factors in common with 7, which implies that the other one must be *divisible* by 7. Say a1 is divisible by 7. That means that b1 = a1/7 is an algebraic integer. Of course a1 = 7*b1. Since a1 is a root of the polynomial,so is 7*b1. Therefore 7^2*b1^2 - (x - y)*7*b1 + 7*(x^2 + x*y) = 0. Divide through by 7: 7*b1^2 - (x - y)*b1 + (x^2 + x*y) = 0.Now, in general this polynomial in b1 is irreducible, non-monic,and primitive. These properties are all true for most choices of x and y. For example, x = 5 and y = 3; x = 1, y = 0;x = 1, y = -3; etc., etc.. This means (by a basic theorem in algebraic number theory) that b1 in general cannot be an algebraic integer. This contradicts the statement that a1 is divisible by 7,and therefore it also contradicts Harrisfs statement that one of the roots must have no factors in common with 7. Now: is that a complicated explanation? Is it defensive?Is it *wrong* ?> Here, for physicists itfs easy to see that their wrong, as x and y are> INDEPENDENT of each other, so therefs no way with> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > that (5b_2(x,y) + 2y) can in general have any factor of 7.> Itfs just not mathematically possible, as if it did that factor would> show up multiplied times y, along with 2. Notice you *do* see that> factor showing up with (5a_1(x,y) + 7y).> Just proving what we have said repeatedly: the only way you have ever learned to factor is by inspection. Good old high-school math.> Importantly, there is a special case here when x=y, as then the> quadratic becomes> a^2 - (x - y)a + 7(x^2 + xy) = a^2 + 14y^2.> At that value the equation belongs to a different family as the> solution then is identical to that from> a^2 + 7(x^2 + xy)> when x=y.> This is not a special case. This is just an example of whatwas described above, where the polynomial is irreducible.> So why would mathematicians argue against such a simple result?> Pride? Fear?> We argue against your claim for one very simple reason: It is wrong.> Want more advanced polynomial factorization?> Good idea, though what I give below is not exactly advanced. Assume, as is usually the case, that x^2 + x*y is coprime to 7. Here is the right way to factor analyze divisibility of the roots of a^2 - (x - y)*a + 7 *(x^2 + x*y)with respect to 7: Let w1 = GCD(a1, 7) and w2 = GCD(a2, 7). Then both a1 and 7 are divisible by w1, and both a2 and 7are divisible by w2. By the argument above, neither w1 norw2 can be equal to 7 and neither can be a unit in the algebraicintegers. Therefore, in direct contrast to what Harris claimedabove, both of the roots a1 and a2 have nonunit factors in common with 7. Do you see any errors in this ? Nora B.> Then check out my blog archives:> James Harris === [snip]> Mathematicians, you see, believe that if the quadratic> a^2 - (x - y)a + 7(x^2 + xy)> doesnft have *integer* roots, then *both* roots would have to share> factors with 7.If I wanted to know what mathematicians believe, I would ask amathematician and not take your word for it. You have a prior history ofmisrepresenting the position that other posters, including mathematicians,have taken. Generally you engage in these misrepresentations maliciouslyand dishonestly, as is revealed by your posting record.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of popularity over principle.--http://www.crbond.com === What I like about adding to Rick Deckerfs example is that it makes iteasy.In case you missed it, Rick Decker is a professor at Hamilton College(do a web search to ?d out about his school) who posted a*quadratic* in an attempt to refute some of my conclusions.Well, I found his example grabbed me, for various reasons, and ?allyfound that I could modify it slightly to suit my purposes, and nowcon?ent in your understanding--even as undergrads--of independencebetween variables, I feel I can ?ally show you how I have beendefending mathematics against people working to undermine it!Herefs the quadratic that Decker gave:(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) where the afs are roots of a^2 - (x - 1)a + 7(x^2 + x).And my modi?ation was just to add in the variable y, to get(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) where the afs are roots of a^2 - (x - y)a + 7(x^2 + xy).Notice that x and y are *independent* variables, as I havenft relatedthem to each other in any way.That second polynomiala^2 - (x - y)a + 7(x^2 + xy)is the polynomial being analyzed by considering(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).You ?d it from7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2where whatfs on the right side is just a regrouping of terms fromwhatfs on the left, as you can ?d out by simplifying the right side.Notice that with(5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) it *looks* like you have 7y and 7y as terms independent of x on theleft side, but that contradicts with 7(2y^2) on the right!So what gives?Well the problem is that x is in the way, and a simple technique inanalysis is to clear out one variable by setting it to 0.Now letting x=0, gives a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you havea_2(x,y) = b_2(x,y) - y, so(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). And now everything matches up correctly.Obviously, now the coef?ients of the terms that have y but donfthave x as a factor are revealed, and not surprisingly, 7y(2y) matchesup with what you see on the right side with 7(2y^2).So if x does not equal y then it follows that(5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2but if x-y=0, then you havea^2 + 7(x^2 + xy), de?ing the afs, and you can directly calculate that a = +/- sqrt(14)y, and pickinga_1(y,y) = sqrt(14)y, you have(5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2).Now considering when x does not equal y itfs easy enough to see thatregardless of anyonefs feelings on the matter, itfs not possible forthe terms *independent* of x, to actually have a dependency on x.So, with(5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)if (5b_2(x,y) + 2y) had any factors in common with 7, then thosefactors would have to be a factor of 2y, as that term is both visibleand independent of x.So you can see my conclusion about what happens when you divide bothsides by 7, when x does not equal y, follows from some basic algebra.Whatfs fascinating is that you can also consider what happens when y=0to separate out further, so letting y=0, I havea^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). Whatfs fascinating here is that completely broken is *any* idea thattherefs a variable factor in common with 7 that divides differentlydependent on the value of x, as you have the lead coef?ients1-sqrt(-27))/2 and 1+sqrt(-27))/2 which are, of course, constant.What I like here is that quadratics are simpler than cubics to showthe result, the example I modi?d was presented ?st by Rick Decker,a professor at Hamilton College, and my conclusion follows fromindependence between independent variables.That is to doubt me here you need to doubt that concept in mathematicsitself--the concept of independence between variables.Of course, the problem for the old view on algebraic integers, is thatif you supposed that you were in the ring of algebraic integers(notice I made no mention of a ring before now) then you run into aproblem speci? to that ring, where dividing both sides by 7 pushesyou out of the ring of algebraic integers.It might seem esoteric and distant as a problem, but my discovery ofthat problem is actually a fascinating story, and now you can see howmathematicians *worldwide* answer the question:What do you do when your paradigm is forced to shift by a discovererwho thought outside of the box?Want more?Then read my other threads on sci.math or check out my blog archives:James Harris === [propaganda clipped]> Herefs the quadratic that Decker gave:> (5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30xy + 2y^2) > where the afs are roots of > a^2 - (x - 1)a + 7(x^2 + x).> And my modi?ation was just to add in the variable y, to get> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > where the afs are roots of > a^2 - (x - y)a + 7(x^2 + xy).> Notice that x and y are *independent* variables, as I havenft related> them to each other in any way.> That second polynomial> a^2 - (x - y)a + 7(x^2 + xy)> is the polynomial being analyzed by considering> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2).> You ?d it from> 7(25x^2 + 30xy + 2y^2) = 7(x^2 + xy)(5^2) + 7y(x - y)(5) + 7^2 y^2> where whatfs on the right side is just a regrouping of terms from> whatfs on the left, as you can ?d out by simplifying the right side. Notice that with> (5a_1(x,y) + 7y)(5a_2(x,y) + 7y) = 7(25x^2 + 30xy + 2y^2) > it *looks* like you have 7y and 7y as terms independent of x on the> left side, but that contradicts with 7(2y^2) on the right!> So what gives?> Well the problem is that x is in the way, and a simple technique in> analysis is to clear out one variable by setting it to 0.> Now letting x=0, gives > a(a + y) = 0, so a = 0, or -y, and letting a_2(0,y) = -y, you have> a_2(x,y) = b_2(x,y) - y, so> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2). > And now everything matches up correctly.> Obviously, now the coef?ients of the terms that have y but donft> have x as a factor are revealed, and not surprisingly, 7y(2y) matches> up with what you see on the right side with 7(2y^2).> So if x does not equal y then it follows that> (5a_1(x,y)/7 + y)(5b_2(x,y) + 2y) = 25x^2 + 30xy + 2y^2> but if x-y=0, then you have> a^2 + 7(x^2 + xy), de?ing the afs, > and you can directly calculate that a = +/- sqrt(14)y, and picking> a_1(y,y) = sqrt(14)y, you haveThe zeros of a in a^2 + 7(x^2 + xy), with x = y, are NOT +/- sqrt(14)y, they are +/- sqrt(-14)y.> (5 sqrt(14)y + 7y)(5 sqrt(14)y + 7y) = 7(25y^2 + 30 y^2 + 2y^2).> Now considering when x does not equal y itfs easy enough to see that> regardless of anyonefs feelings on the matter, itfs not possible for> the terms *independent* of x, to actually have a dependency on x.And whom are you alleging has made such a claim? Or is this just another of your straw men?> So, with> (5a_1(x,y) + 7y)(5b_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)> if (5b_2(x,y) + 2y) had any factors in common with 7, then those> factors would have to be a factor of 2y, as that term is both visible> and independent of x.It is quite possible in the ring of algebraic integers for 7 and 2y, for y an integer, always to have a common non-unit factor, at least until you or someone else proves it to be impossible. Note that the common factor may be different for different values of y.> So you can see my conclusion about what happens when you divide both> sides by 7, when x does not equal y, follows from some basic algebra.So basic, that it is wrong in the more advanced context!> Whatfs fascinating is that you can also consider what happens when y=0> to separate out further, so letting y=0, I have> a^2 - xa + 7x^2 = 0,> so> a = (1+/-sqrt(-27))x/2,> and introducing c_1(x,y) and c_2(x,y) I then have> (5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y)> = 7(25x^2 + 30xy + 2y^2). > Whatfs fascinating here is that completely broken is *any* idea that> therefs a variable factor in common with 7 that divides differently> dependent on the value of x, as you have the lead coef?ients> 1-sqrt(-27))/2 and 1+sqrt(-27))/2 > which are, of course, constant.Same fault as above. No one but JSH is insisting that the common factors of 7 and a_1 and of 7 and a_2 be independent of either x or y. > [more propaganda snipped] === Both of you gave truly wonderful posts - greatly appreciated. === hi all,I assume that a bezier patch has 16 control points and 4 corners controlpoints are attached onto the surface. It implies that there are 4 beziercurves on the edges of the patch. My question is how can the middle 4control points change the shape of the surface?newbie... === > I assume that a bezier patch has 16 control points and 4 corners control> points are attached onto the surface. It implies that there are 4 bezier> curves on the edges of the patch. My question is how can the middle 4> control points change the shape of the surface?I donft understand your question. They change the shape because thatfs how Bezier patches are de?ed. Can you try to elaborate on what exactly is confusing you?meeroh-- If this message helped you, consider buying an itemfrom my wish list: === It looks like there was more to adding in that y as a variable thanwas realized as I saw the usual suspects replying--yet again--tryingto attack the mathematics, attacking algebra, as if they could stillconvince all of you.However, the other shoe fell as I set y=0 givinga^2 - xa + 7x^2 = 0,soa = (1+/-sqrt(-27))x/2,and introducing c_1(x,y) and c_2(x,y) I then have(5(1-sqrt(-27))x/2 + c_1(x,y) + 7y)(5(1+sqrt(-27))x/2 + c_2(x,y) + 2y) = 7(25x^2 + 30xy + 2y^2)from my modi?ation of Rick Deckerfs example.Now I guess that Decker is a good enough mathematician to recognizethat now itfs completely over, but I wonder about the others.Will they *dare* step out now despite the mathematics clearly showingthat Ifve been right all along?Letfs see, as I will admit that Ifm now curious about what thesepeople will do, and how many of you will still trust and believe them.That is, Ifm curious about you, and what mathematics means to you, byhow you *act* not what you say.James Harris === > When are we ?ally going to see the Tablet PC operating systems and> software give us math persons the holy grail?> Itfs where we write out everything on a tablet, and then push a couple> of buttons and the software transforms the entire document into a form> that looks like we went through all the trouble of typesetting it when> we didnft! Better yet, the document would be compatible and> transformable into and from LaTeX! (I understand that MathType and> LaTeX are now convertible.) The software companies have already given the language persons this> capability (write it out, push some buttons and it looks like it was> typed). So what about us math persons?Math typesetting has historically been much more dif?ult than plaintext. The math formulae have a two-dimensional structure thatrequires more careful placement, and spacing and positioning carrymore of the meaning than in plain text.The market for math typesetting is also much smaller than the marketfor plain text, so there has been less incentive for industry todevelop tools. (Consider that the software that does the besttypesetting for math, TeX, has been in the public domain for over 20years, but the software industry is still putting out third-rate mathtypesetting tools when they could freely use the algorithms and codein the TeX program.)-- Kevin Karplus karplus@soe.ucsc.edu http://www.soe.ucsc.edu/~karpluslife member (LAB, Adventure Cycling, American Youth Hostels)Effective Cycling Instructor #218-ck (lapsed)Professor of Computer Engineering, University of California, Santa CruzUndergraduate and Graduate Director, BioinformaticsAf?iations for identi?ation only.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > Math typesetting has historically been much more dif?ult than plain> text. The math formulae have a two-dimensional structure that> requires more careful placement, and spacing and positioning carry> more of the meaning than in plain text.> The market for math typesetting is also much smaller than the market> for plain text, so there has been less incentive for industry to> develop tools. (Consider that the software that does the best> typesetting for math, TeX, has been in the public domain for over 20> years, but the software industry is still putting out third-rate math> typesetting tools when they could freely use the algorithms and code> in the TeX program.)> Ifd say that the industry underestimates how many math knowledgeworkers there are. Have they included the very many teachers andespecially students world-wide who teach and learn any math at all, atany level at all? Maybe the market is much smaller percentage-wise,but what businessperson would knowingly want to walk away from allthose potential customers?Paul-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > Itfs not about how often we state it. Itfs about how often we apply it> and even about how often we can apply it. For any expression involving> both operations, this axiom is lurking in the background as the main> tool for creating equivalent expressions. Havenft you ever wondered> why polynomials are so common and why sums of products are far more> useful throughout math than products of sums? I donft know about the relative usefulness of products of sums vs.> sums of products. After so much work in arithmetic of polyfs (adding> them, multiplying them, combining like terms) we come to factoring,> where the goal is to write P(x) as (x-r1)(x-r2)...(x-rn) in place of> the usual x^n+...+a0. Those factorizations can be pretty useful. I was just quoting and agreeing with Michael Hamm. Giving precedence to multiplication over addition in notation is> giving deference to this lurker.> What about set union (U) and intersection (^)? We have two> distributive laws:> 1) (a^b)U c = (a U c)^(b U c)> 2) (a U b)^c = (a^c) U (b^c)> Now there is no preference. So, a^b U c must be decided by convention?> In a ?ld there is no distribution of addition over multiplication.> The analogy you create therefore doesnft hold. The fact that there is> distributivity in a ?ld in only one way in my view opens the door> for a bias towards a certain way of doing things in terms of operation> precedence.The natural numbers donft constitute a ?ld either. Nor do they forma ring. But their algebraic structure is very similar to that for thepower set of a given set X, with the operations of union andintersection. The operations are closed, associative, and commutative.The main difference is that there are two distributive laws instead ofone. But, I have seen in several books that intersection precedesunion in the order of operations. Itfs arbitrary. If we were inclinedto write all polynomials (over an appropriate ring) in factored form,we would perhaps choose to perform additions ?st.> Imagine if there were only one type of distributivity in logic and set> theory. What do you think would happen in terms of operation> precedence?> Paul-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > The natural numbers donft constitute a ?ld either. Nor do they form> a ring. But their algebraic structure is very similar to that for the> power set of a given set X, with the operations of union and> intersection. The operations are closed, associative, and commutative.> The main difference is that there are two distributive laws instead of> one. But, I have seen in several books that intersection precedes> union in the order of operations. Itfs arbitrary. If we were inclined> to write all polynomials (over an appropriate ring) in factored form,> we would perhaps choose to perform additions ?st.Letfs stick to the natural numbers if you wish, and so examine theargument given by Ron. (I wanted to start with the ?ld structure, hewith the natural numbers. Maybe he was right to do so. According toKronecker, the natural numbers are the work of God, but everythingelse is the work of Humanity.)I do not see how a decision is arbitrary when one choice repeatedlyleads to a contradiction and the other never does.Let multiplication in the natural numbers be understood as repeatedaddition of the same number:a + a + . . . + a (for n instances of a)= (a*1) + (a*1) + . . . + (a*1) (for n instances of a*1)= a*(1 + 1 + . . . + 1) (for n instances of 1) = a*nThen giving default precedence to addition repeatedly yieldscontradiction, but giving default precedence to multiplication neverdoes. For example:3 + 5 + 5 = 13 3 + 5 * 2 = 16 (addition gets default precedence) 3 + 5 * 2 = 13 (multiplication gets default precedence)To anticipate:Of course we can give precedence to addition if we want, such as in (3+ 5) * 2 = 16, but this is about default precedence. On expressionsinvolving both addition and multiplication with no parenthesesdenoting precedence, this is about having a default precedence grammarthat avoids contradiction.Cordially, Paul-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Depends on the question of course, but most ask what is the probability ofsomething happening?So the answer would be:The probability of X happening is Y.Funny you should ask, I just had my ?al exam in prob and stat...I got an Ain the class!! woohoo!!John> How do you explain your answer to a probability question>-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I am an elementary PE teacher that has been asked to be on a committeeto come up with ideas for a Math Week to be held in Feb 2004. We wouldlike for each day to have a theme. I am basically at a loss right nowas this is not really my area of expertise. I would love any input-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === >I am an elementary PE teacher that has been asked to be on a committee>to come up with ideas for a Math Week to be held in Feb 2004. We would>like for each day to have a theme. I am basically at a loss right now>as this is not really my area of expertise. I would love any inputHerefs a few topics that might be fun to expand:Prime days. Special badges for those who are 5, 7, 11, or 13.Tangram contests. Lots of interesting math facts come out of tangrams.Probability games. Guess the number of M&Ms in a Halloween-size bag,guess the color distribution, plot the distribution as bar graphs,simple equations where you add the total per color to get the overalltotal.Tetrahedral kites. Ifve taught that to 6th graders, and all 120 studentsin the sixth grade were able to build their kites in one hour, even witha short introduction on the history of the tetrahedron and tetrahedralkites. It took a ratio of one adult per 10 students to get enough helpto keep the kids on track, and some of the special needs kids neededindividual help. Jill Britton has her pre-service students work ingroups of two or three to take it into classrooms in BC.For a much better selection of ideas, pick up one of Jill Brittonfs booksor look at her website:http://camosun.bc.ca/~jbritton/Home.htm(The page is not responding as I type, but the pages for each book havetons of links to activities associated with each chapter of the book.) -- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Does anyone know of a university offering a 5-8 Math Educationcourse (preferrably on-line)? I live in south-central Minnesota--andneed that one course in order to complete my 5-12 Math Teachingprogram. -- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === I need help on my hmwk. but it is to hard to understand-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === in part:> I need help on my hmwk. but it is to hard to understandThe answer to number 3 is 54. After all, it doesnft matter what aorder you take the partials in.Hope this helps,AM, Math, Wash. U. St. Louis Ifve been erasing too much UBE.msh210@math.wustl.edu Of a reply, then, if you have been cheated,-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > I need help on my hmwk. but it is to hard to understand>Well, if it were easy youfd probably not need help.Could you be more speci? about: 1. Which problem(s) youfre having trouble with. NB: you usually getbetter responses if you con?e it to no more than two or three at a time. Isuspect not many readers are interested in doing your homework for you -- but quite a few are willing to explain mathematical concepts and problemsolving techniques. 2. What you have done to solve the problems or at least your thoughtsare on how you think they should be approached.- Sox-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === In a ?ancial arrangement you are promised $1000 the ?st day andeach day after that you will recieve 35% of the previous days amount.When one days amount drops below $1, you stop getting paid from thatday on. What day is the day you would receive no payment and what isyour total income? Use a formula for the nth partial sum of ageometric sequence. -- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === if i reviewed 30 charts and 13 were compliant what would be thepercent. Please show me how to calculate.-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === > if i reviewed 30 charts and 13 were compliant what would be the> percent. Please show me how to calculate.Percent means out of 100. If you had 100 charts, how many of them would be compliant?-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === >if i reviewed 30 charts and 13 were compliant what would be the>percent. Please show me how to calculate.>100 times the quantity of objects which qualify divided by total quantity ofobjects.If want percent compliant, use 100* (number compliant /number of total)If want percent non-compliant, use 100*(number noncompliant/number of total)G C-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Has anyone tried qualitytutors.com? I have heard that teachers makegood money tutoring there.Sam-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === My son is currently in Algebra 2 and he canft ?ure out the followingproblem (nor can I):sqrt(14)/(sqrt(6) + sqrt(7) - 1)-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === Below Ifll point out a link between number theory and music theory,speci?ally modular arithmetic modulo 12 and the 12-tone musicalsystem. It also links to the standard 12-hour clock if you want areference.On the piano, starting in any key you want, wefll move up or down thekeyboard a number of notes at a time, including both black and whitekeys. Here are some options:1) Move up one note, this interval called a semitone, then another,etc.2) Move down one note, this interval called a semitone, then another,etc.3) Move up a perfect fourth (5 notes), then another perfect fourth (5more notes), etc.4) Move down a perfect fourth, then another perfect fourth, etc.5) Move up a perfect ?th (7 notes), then another perfect ?th (7more notes), etc.6) Move down a perfect ?th, then another perfect ?th, etc.7) Move up a major seventh (11 notes), then another major seventh (11more notes), etc.8) Move down a major seventh, then another major seventh, etc.In each case, youfll cycle through all twelve keys exactly once. Doesthis remind you of a certain something? Itfs modular arithmetic. In modulo 12, itfs sometimes called clockarithmetic.In almost all Western music, there are 12 keys as in tones. On thepiano, if we start on any key, say a middle C, and play every note,one at a time upwards, the next C we hit is the 12th note up. Thisinterval is an octave. Another 12 notes up, another C. Another octave.This occurs regardless of which of the 12 keys we start in. We thuslabel the keys so that we can have modular arithmetic, modulo 12. The12 keys, appropriately labeled, function as the 12 congruence orresidue classes mod 12.To apply case 1) above:Letfs choose C, and label it 0 mod 12. A perfect 4th up from any noteis 5 notes up. A perfect 4th up from C is F, which we label 5 mod 12:12.The intervals mentioned above are generators of this additive group,because repeated addition cycles through all of the keys (congruenceclasses mod 12). Each of the intervals mentioned above has order 12.The interval of a minor sixth, 8 notes (or units) up (or down), hasorder 3, because the greatest common divisor of 8 and 12 is 4 and 12/4= 3. This means that going up (or down) a minor sixth, an intervalcovering 8 notes (or units), at a time cycles through only 3 keys(congruence classes mod 12). Starting at C, these would be C (labeled0 mod 12), A-?# or G-sharp) (labeled 8 mod 12), and E (labeled 4mod 12). (And then of course C again, but that is back to 0 mod 12).A minor third, which is a 3-note interval, is of order 4. All fournotes obtained by repeatedly going up a minor third played togetherform a diminished chord. Wagner LOVED this chord as a so-calledpassing chord. Properly used, it creates rich, dark sonorities in themusic.Now you get the idea. Show this to interested students. Number theoryand music theory. Have fun!Cordially,Paul-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html === symmetry/ half-turn symmetry and the af?e equivalence of cubicpolynomials that Ifve recently redone a little for re-publication.Both results use transformation geometry and algebraic functionnotation and can be nicely illustrated with Sketchpad as it allowsstudents to see how any cubic polynomial can be translated, shearedand/or stretched. Though probably more suited as enrichment, theresults should be accessible to most high school students and can bedownloaded directly from: http://mzone.mweb.co.za/residents/profmd/cubic.pdfhttp:// mzone.mweb.co.za/residents/profmd/cubeaf?e.pdfA zipped Sketchpad sketch (4 KB) that could be used to demonstrate orallow students to investigate these results by themselves can bedownloaded from: http://mzone.mweb.co.za/residents/profmd/cubic.zipIfd be very interested & happy to hear from anyone regarding anyexperiences you may have with students investigating these properties.Note that the above Sketchpad sketches certainly need some adaptationand improvement for direct use in a class.Michael de Villiers-- newsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: http://www.thinkspot.net/k12math/charter.html