mm-205 === I have vectors of the form [a1, a2, a3, a4, ...] where a1, a2 etc can be> either +1 or -1. For example, say I have a set S of 16 such vectors for > [a1, a2, a3, a4].I want to find maximum orthogonal vectors from this set S where orthogonality> is defined as a1*b1 + a2*b2 + ... = 0; [a1, a2, ...] and [b1, b2 ...] are> vectors from the set S defined above.Is there a simple algorithm to solve this problem ? Or is there a > similar known problem which I can refer ? > I believe the Gram-Schmidt algorithm for the orthogonalization of a vector basisis what you want, but some have called this the Graham-Schmidt method, so thereappears to be a discrepancy over the name.The algorithm is easy to understand, you pick any 1 vector, then slowly add theother vectors, but eliminate any coefficients from those vectors that create anon-zero dot product (called a projection upon) with the previously determinedvectors, as you go along.There are other methods too, such as QR normalization. I assume that maximumorthogonal vectors means finding all orthogonal vectors, not just a few, thatare actually in the set.See:planetmath.org/encyclopedia/ GramSchmidtOrthogonalization.htmhttp://www.math.ualberta.ca/~ ewoolgar/labs/linalg/Lab14.pdfhttp://www.mathreference.com/la, gram.htmlhttp://ccrma-www.stanford.edu/~jos/mdft/Gram_Schmidt_ orthogonal vectors ?> I believe the Gram-Schmidt algorithm for the orthogonalization of a vector> basis> is what you want, but some have called this the Graham-Schmidt method, so> there> appears to be a discrepancy over the name.But the result will find maximum orthogonal vectors ?x-mimeole: Produced By Microsoft MimeOLE V6.00.2800.1165Try design of experiments or DOEIt is very much used technique there I have vectors of the form [a1, a2, a3, a4, ...] where a1, a2 etc can be> either +1 or -1. For example, say I have a set S of 16 such vectors for> [a1, a2, a3, a4]. I want to find maximum orthogonal vectors from this set S whereorthogonality> is defined as a1*b1 + a2*b2 + ... = 0; [a1, a2, ...] and [b1, b2 ...] are> vectors from the set S defined above. Is there a simple algorithm to solve this problem ? Or is there a> similar known problem which I can refer doing something else, it happened and I managed to prove following equation :pi = lim n*sin(180*n) n->infI dont know, but is this already been proved ?if no, is this equation helpfull to anyone in the world of mathamatics anyhow While doing something else, it happened and I managed to prove following equation : pi = lim n*sin(180*n)> n->inf>I presume range of n is positive integers.If 180n is degrees then lim(n->oo) n.sin 180n = lim n*0 = 0When 180n is radians, I trigonometry and calculusI am extremly sorry for the typo, Pls read the equation as : pi = lim n*sin(180/n) n->infI believe this thing is known and exists.Interestingly, in other words it can be stated as :product of very little ( almost near to zero ) number to large( almost infinite ) number is constant and that constant is pi :-)While doing something else, it happened and I managed to prove following equation : pi = lim n*sin(180*n)> n->inf I presume range of n is positive integers.> If 180n is degrees then> lim(n->oo) n.sin 180n = lim n*0 = 0> When 180n is radians, I doubt limit === exists.Subject: Re: Value of PI using trigonometry and 180/n = 180 lim(n->oo) (sin 180/n)/(180/n) = 180 lim(x->0) (sin x)/x = 180 provided x is in radians> I believe this thing is known and exists.>Basic calculus lim(x->0) (sin x)/x = confusing very fast...I am reading Hersteins Abstract Algebra and teaching myself fieldtheory. I came to the following and it gave me a bit of trouble (I amparaphrasing it for people who dont own that book): If F is a finite field with q elements, then viewing F simply as anabelian group under its addition, +, we have from group theory thatqx=0 for all x in F.Now after a bit of consideration I came to the conclusion that, yes,this is true, if we go by the notation (which Herstein did notexplicitly define) that qx, where q is a counting number and x is anabstract object, means x+x+x+...+x, q times. In terms of fieldsthis takes a little bit of chewing before swallowing since now we havenot one but two very different kinds of multiplication, each of whichuses the same symbol.This is still unambiguous when the elements of our field are abstractthings, or at least anything BUT counting numbers. But if our fieldis in fact counting numbers, it brings up a very gaping problem, atleast as I see things (although, true, I am very much a newbie!) Tobe specific, suppose F is a field of counting numbers, two of whichare a and b. What, then, are we to make of ab? It is ambiguous andhas the three possible meanings:1. a mulitplied by b by the multiplication operation of F2. a+a+a+...+a (b times), or3. b+b+b+...+b (a times)Was Hersteins choice of symbols merely a bad §uke? I certainly hopeso, and that the a+a+a...+a (b times) notation is not actually usedmuch in field theory, else it seems that the study of fields ofintegers would soon be rendered hopelessly ambiguous andcontext-dependant.I am probably just overlooking something Theory} This notation could get very confusing very fast...Z is what is called a universal object in the category of commutativerings. This means given a ring A, there is one and only one morphism from Zto A (this is indeed very easy). This allows us to speak of elements of Z aselements of A without any ambiguity since there is only one way to sendintegers into confusing very fast...> I am reading Hersteins Abstract Algebra and teaching myself field> theory. I came to the following and it gave me a bit of trouble (I am> paraphrasing it for people who dont own that book): If F is a finite field with q elements, then viewing F simply as an> abelian group under its addition, +, we have from group theory that> qx=0 for all x in F.Now after a bit of consideration I came to the conclusion that, yes,> this is true, if we go by the notation (which Herstein did not> explicitly define) that qx, where q is a counting number and x is an> abstract object, means x+x+x+...+x, q times. In terms of fields> this takes a little bit of chewing before swallowing since now we have> not one but two very different kinds of multiplication, each of which> uses the same symbol.This is still unambiguous when the elements of our field are abstract> things, or at least anything BUT counting numbers. But if our field> is in fact counting numbers, it brings up a very gaping problem, at> least as I see things (although, true, I am very much a newbie!) To> be specific, suppose F is a field of counting numbers, two of which> are a and b. What, then, are we to make of ab? It is ambiguous and> has the three possible meanings:> 1. a mulitplied by b by the multiplication operation of F> 2. a+a+a+...+a (b times), or> 3. b+b+b+...+b (a times)It is not a bug. It is a feature.Strictly speaking, you do not consider the integers as a _subset_ of F.But one can define a ring homomorphism f: Z --> F byprescribing f(1)=1_F, where the 1_F is the identity of F.This forces f(n) = 1_F + 1_F + ... 1_F (n times) for n in Z.It also forces f(ab)=f(a)f(b) where the multiplication on the right sideis the one of F.To avoid notational pain, the map f and the subscript in 1_F is usuallyomitted. So e.g. one talks about the integer 2=1+1 in F. Of coursein this notation it can happen that 2=0.Now the Re: {Field Theory} This notation could get very confusing very fast...> This is still unambiguous when the elements of our field are abstract> things, or at least anything BUT counting numbers. But if our field> is in fact counting numbers, it brings up a very gaping problem, at> least as I see things (although, true, I am very much a newbie!) To> be specific, suppose F is a field of counting numbers, two of which> are a and b. What, then, are we to make of ab? It is ambiguous and> has the three possible meanings:It is not usual to take the integers themselves as elements of the field of a prime number of elements, but only as representatives of those field elements, which more commonly are taken as equivalence classes of all those integers which are congruent to that representative modulo the prime in question. E.g., for p as the prime and k as the representative, the equivalence class is the set {k+n*p | n an integer}.Thus in the field modulo 3, 0 represents the set of all integers of form 3*n+0, 1 represents the set of all integers of form 3*n+1, 2 represents the set of all integers of form 3*n+2, for n an integer.One may show that the usual addition and multiplication of integers induces a natural addition and multiplication on the ring of such equivalence classes for an arbitrary nteger modulus > 1, which === is a field when and only when that modulus is a prime.Subject: Re: {Field Theory} This notation could get very confusing very teaching myself field> theory. I came to the following and it gave me a bit of trouble (I am> paraphrasing it for people who dont own that book): If F is a finite field with q elements, then viewing F simply as an> abelian group under its addition, +, we have from group theory that> qx=0 for all x in F. Now after a bit of consideration I came to the conclusion that, yes,> this is true, if we go by the notation (which Herstein did not> explicitly define) that qx, where q is a counting number and x is an> abstract object, means x+x+x+...+x, q times. In terms of fields> this takes a little bit of chewing before swallowing since now we have> not one but two very different kinds of multiplication, each of which> uses the same symbol.>Since q isnt in F, a distinction is implicit.> This is still unambiguous when the elements of our field are abstract> things, or at least anything BUT counting numbers. But if our field> is in fact counting numbers, it brings up a very gaping problem, at> least as I see things (although, true, I am very much a newbie!) To> be specific, suppose F is a field of counting numbers, two of which> are a and b. What, then, are we to make of ab? It is ambiguous and> has the three possible meanings:> 1. a mulitplied by b by the multiplication operation of F> 2. a+a+a+...+a (b times), or> 3. b+b+b+...+b (a times)>As F is a field, it has an multiplicative identity usually notated 1.Thus 1x = x is no problem. If you consider 2 = 1+1, etc., then ï2x = (1+1)x = 1x + 1x = x+x = 2x, etc.Thus youve your situation.> Was Hersteins choice of symbols merely a bad §uke? I certainly hope> so, and that the a+a+a...+a (b times) notation is not actually used> much in field theory, else it seems that the study of fields of> integers would soon be rendered hopelessly ambiguous and> context-dependant.>No, its common usage. Its not much of a {Field Theory} This notation could get very confusing very fast... As F is a field, it has an multiplicative identity usually notated 1.> Thus 1x = x is no problem. If you consider 2 = 1+1, etc.,> then ï2x = (1+1)x = 1x + 1x = x+x = 2x, etc.> Thus youve your situation.> When you say 2 = 1+1, by your + are you talking aboutnormal addition, or the fields addition? Note that it is quitepossible to have a unit identity which is an integer but is NOT infact the 1 of normal arithmetic. So when faced with a statementlike 2 = 1+1 over a field of integers i am faced with threedilemmas, the first being whether the + is normal arithmetic, thesecond begin whether the 1 is the normal one or just a symbol forthe fields identity (which is not necessarily the normal one), andfinally whether the 2 is the normal two or a symbol for 1+1(whatever THAT is!). In a very well behaved field, like the integersmodulo p, then yes the notation is fine... but in a field of integerswhose addition and multiplication have nothing to do with normaladdition and multiplication, it becomes ambiguous. Sorry for being so newbieish over this, I am sure that I am stilljust overlooking something and I hope you will share your superiorknowledge with Heres a demonstration of how you can get a problem, using xy=2, where: x and y are algebraic integers. Consider x=2a, and y = b, so b is an: algebraic integer, but ïa is not, so x does not technically have 2 as: a factor in the ring of algebraic integers, as that requires *both*: factors be algebraic integers, while ïa is not one. So b *should* be: a unit, but because ab=1, it is NOT a unit, because the unit: definition would require that both ïa and b be algebraic integers.I dont understand how you are deciding what should and shouldnot be in the algebraic integers. : Now Ive given that example before and someone sent me an email with: a=1/2, so Ive also included that ïa is a member of the object ring,: where the definition follows.: The object ring is a commutative ring that includes all numbers such: that -1 and 1 are the only members that are both a unit and an: integer, where no non-unit member is a factor of any two integers that: are coprime.I dont understand this definition. Could you specify a numericalvalue for a, to make the example more concrete?: Since a=1/2 would make 2 a unit, it is excluded by the definition.: Now it turns out that ïa exists and it is the root of a non-monic: polynomial with integer coefficients, but its excluded arbitrarily by: the definition used for algebraic integers.: And you can see several problems that popped up by that exclusion as b: is not a unit in the ring of algebraic integers, when it should be,: and x does NOT have 2 as a factor in the ring, though x=2a.So, what is the problem? I havent seen anything that looks like a contradiction, only an assertion that something is not a unitwhich should be according Core error, FEAR is a natural response> : Heres a demonstration of how you can get a problem, using xy=2, where> : x and y are algebraic integers. Consider x=2a, and y = b, so b is an> : algebraic integer, but ïa is not, so x does not technically have 2 as> : a factor in the ring of algebraic integers, as that requires *both*> : factors be algebraic integers, while ïa is not one. So b *should* be> : a unit, but because ab=1, it is NOT a unit, because the unit> : definition would require that both ïa and b be algebraic integers.I dont understand how you are deciding what should and should> not be in the algebraic integers. Heres a simpler example, consider 2 and 6 in the ring of evens, but 2is NOT a factor of 6 because 3 is not even, understand?Similarly, because ïa above is not an algebraic integer, then thoughb should be a unit, its not because the *definition* for unitrequires that both ïa and b be in the ring of algebraic integers, sob is not a unit, which means it is a non-unit factor of 2, which givesa false implication, since x = 2a, like with 6, 2(3) = 6. > : Now Ive given that example before and someone sent me an email with> : a=1/2, so Ive also included that ïa is a member of the object ring,> : where the definition follows.> > : The object ring is a commutative ring that includes all numbers such> : that -1 and 1 are the only members that are both a unit and an> : integer, where no non-unit member is a factor of any two integers that> : are coprime.I dont understand this definition. Could you specify a numerical> value for a, to make the example more concrete?Well, I could, but I dont think its worth the effort, and itwouldnt really show you anything.As for the definition it suffices to note that it doesnt allow ïa tobe 1/2, as that would mean that 2 is a unit, since 2(1/2) = 1.> : Since a=1/2 would make 2 a unit, it is excluded by the definition.> > : Now it turns out that ïa exists and it is the root of a non-monic> : polynomial with integer coefficients, but its excluded arbitrarily by> : the definition used for algebraic integers. : And you can see several problems that popped up by that exclusion as b> : is not a unit in the ring of algebraic integers, when it should be,> : and x does NOT have 2 as a factor in the ring, though x=2a.So, what is the problem? I havent seen anything that looks like > a contradiction, only an assertion that something is not a unit> which should be according to some unstated principle. Whats the> principle?Given that xy=2, with x=2a, y=b, with b an algebraic integer, x analgebraic integer, while ïa is not, it *should* be that y does notshare non-unit factors with 2, since x *should* have 2 as a factor,but in the ring of algebraic integers, because the definitionarbitrarily excludes ïa, neither of those is the case.That is, a proper definition would include ïa, but ïa happens to bethe root of a polynomial that doesnt have algebraic integercoefficients.An example of such a polynomial isb^3+...+ 3(-1+mf^2)b^2+...- (m^3 f^4 - 3m^2 f^2 + 3m)where for certain values of m and f, the inexpressible terms arealgebraic integers and are are expressible.Its actually fascinating, but not surprisingly, potentially difficultto understand, which is why mathematical logic--like §ying a plane byinstruments--is required.Your intuition isnt worth is a natural response> : Heres a demonstration of how you can get a problem, using xy=2, where> : x and y are algebraic integers. Consider x=2a, and y = b, so b is an> : algebraic integer, but ïa is not, so x does not technically have 2 as> : a factor in the ring of algebraic integers, as that requires *both*> : factors be algebraic integers, while ïa is not one. So b *should* be> : a unit, but because ab=1, it is NOT a unit, because the unit> : definition would require that both ïa and b be algebraic integers.I dont understand how you are deciding what should and should> not be in the algebraic integers. Heres a simpler example, consider 2 and 6 in the ring of evens, but 2> is NOT a factor of 6 because 3 is not even, understand?Im going to join with the others in asking what pointyou meant to make here regarding what should be inthe ring of algebraic integers.You have the situation that 2a = 6. Yet a is not in thering of evens. Clearly 2 is a factor of 6, but 2 is nota factor of 6 in the ring of evens. Are you sayingthat a should be in the ring of evens, that theresan error in the definition of the ring, that the ringis error, FEAR is a natural responsegood one; its this level of dyscoursethat gives me hope for mathematicians. on the other hand,Universe is expanding faster & faster,youd know that NASA will be officially interred, alongwith all of those nice engineers over in Huntsville. now,it followed a rather recondite (a-hem) solutionto Einsteins equations. it included a mentionof Eric Lerners _The Big Bang Never Happenned_,which is a work-a-day treatment of Alfvens stuff(I only recently learned taht he was dead,which may explain the peculiar look Id getfrom former UCSD students .-) PS: just as Ive been saying (for years,in hte larger aspect), Captain Arnie has immediately proposed complete deregof energy, along with green supplies, no offshore drilling, andnot happy with the President on the Clean Air Act. at least,he didnt come out against nuclear power. Wheres Warren? > Heres a simpler example, consider 2 and 6 in the ring of evens, but 2> is NOT a factor of 6 because 3 is not even, understand? > You have the situation that 2a = 6. Yet a is not in the> ring of evens. Clearly 2 is a factor of 6, but 2 is not> a factor of 6 in the ring of evens. Are you saying> that a should be in the ring of evens, that theres> an error in the definition of the ring, that the ring> is incomplete?What is the should test?--les ducs de Buffet;vote NONE OF THE BELOWon Trickier Dick Cheneys Re: Core error, FEAR is a natural response:> :> I dont understand how you are deciding what should and should:> not be in the algebraic integers. : Heres a simpler example, consider 2 and 6 in the ring of evens, but 2: is NOT a factor of 6 because 3 is not even, understand?: Similarly, because ïa above is not an algebraic integer, then though: b should be a unit, its not because the *definition* for unit: requires that both ïa and b be in the ring of algebraic integers, so: b is not a unit, which means it is a non-unit factor of 2, which gives: a false implication, since x = 2a, like with 6, 2(3) = 6.What is the false implication? :> : And you can see several problems that popped up by that exclusion as b:> : is not a unit in the ring of algebraic integers, when it should be,:> : and x does NOT have 2 as a factor in the ring, though x=2a.:> :> So, what is the problem? I havent seen anything that looks like :> a contradiction, only an assertion that something is not a unit:> which should be according to some unstated principle. Whats the:> principle?: Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an: algebraic integer, while ïa is not, it *should* be that y does not: share non-unit factors with 2, since x *should* have 2 as a factor,: but in the ring of algebraic integers, because the definition: arbitrarily excludes ïa, neither of those is the case.What is behind the shoulds, i.e. what are the bad consequences of excluding ïa? Before you were talking about Re: Core error, FEAR is a natural response Heres a demonstration of how you can get a problem, using xy=2, where : x and y are algebraic integers. Consider x=2a, and y = b, so b is an : algebraic integer, but ïa is not, so x does not technically have 2 as : a factor in the ring of algebraic integers, as that requires *both* : factors be algebraic integers, while ïa is not one. So b *should* be : a unit, but because ab=1, it is NOT a unit, because the unit : definition would require that both ïa and b be algebraic integers. I dont understand how you are deciding what should and should not be in the algebraic integers. Heres a simpler example, consider 2 and 6 in the ring of evens, but 2> is NOT a factor of 6 because 3 is not even, understand?How does this answer the question?Are you saying that 2 should be a factor of 6 in the even numbers?-- Im talking about mathematics--hard, brutal, extreme ... pushing yourmind beyond the limits to understand what no one else can becausetheyre afraid to risk it all, to lose their freaking worthless mindsin the push to know. --James Harris, for the Nike responseNntp-Posting-Host: hera.cwi.nl > : Heres a demonstration of how you can get a problem, using xy=2, where > : x and y are algebraic integers. Consider x=2a, and y = b, so b is an > : algebraic integer, but ïa is not, so x does not technically have 2 as > : a factor in the ring of algebraic integers, as that requires *both* > : factors be algebraic integers, while ïa is not one. So b *should* be > : a unit, but because ab=1, it is NOT a unit, because the unit > : definition would require that both ïa and b be algebraic integers. > > I dont understand how you are deciding what should and should > not be in the algebraic integers. > > Heres a simpler example, consider 2 and 6 in the ring of evens, but 2 > is NOT a factor of 6 because 3 is not even, understand?So you think the ring of evens is a §awed ring because 3 should be in it? > Similarly, because ïa above is not an algebraic integer, then though > b should be a unit,*Why* should b be a unit? *Why* should a = 1/b be an algebraic integer? > its not because the *definition* for unit > requires that both ïa and b be in the ring of algebraic integers, so > b is not a unit, which means it is a non-unit factor of 2, which gives > a false implication, since x = 2a, like with 6, 2(3) = 6.What false implication? You are of the opinion that a should be analgebraic integer? For *all* b? In that case you are of the opinionthat 1/2 should be an algebraic integer (take b = 2). > So, what is the problem? I havent seen anything that looks like > a contradiction, only an assertion that something is not a unit > which should be according to some unstated principle. Whats the > principle? > > Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an > algebraic integer, while ïa is not, it *should* be that y does not > share non-unit factors with 2, since x *should* have 2 as a factor,*Why*? But let us elaborate. b is an algebraic integer that is alsoa factor of 2. Now 1/b (by your reasoning) should also be an algebraicinteger. Now b is either a unit (a root of a monic polynomial withconstant term +1 or -1) or b is a root of a monic irreducible polynomialwith constant term +2 or -2. Those are all the factors of 2 in thealgebraic integers. Lets have such a polynomial: x^3 - 2. It hasthree roots: r1 = cbrt(2) r2 = cbrt(2) * (-1 + sqrt(-3))/2 r3 = cbrt(2) * (-1 - sqrt(-3))/2If we add r1 and its inverse to the object ring we have immediatelythat 2 is a unit in that object ring. So r1 can not be in thatobject ring. If we add r2 and its inverse we get also r3 becauser3 = -2cbrt(2) - r2 and the two operands of the subtraction arein the object ring. But 1/r3 can not be in the object ring, otherwise2 would become a unit. So why must we chose 1/r2 in the object ringand 1/r3 not? Note that when this is the case, x *has* 2 as a factorif b = r2, but *not* if b = r1 or r3. Why is that not a problem?I think that it is the case that for two complex conjugate factors oftwo you must have one a unit in the object ring and the other not.Appears to be pretty arbitrary in my eyes. > Given that xy=2, with x=2a, y=b, with b an algebraic integer, x an > algebraic integer, while ïa is not, it *should* be that y does not > share non-unit factors with 2, since x *should* have 2 as a factor, > but in the ring of algebraic integers, because the definition > arbitrarily excludes ïa, neither of those is the case.This is also the case with the ring extended as outlined above...But you use *should* twice without elaboration *why* that should be so.It is the case that x=2a *has* 2 as a factor in the algebraic integers,if and only if a is an algebraic integer. And that is how it should be.Similarly x=2a *has* 2 as a factor in the integers, if and only if a isan integer. That is how having a factor is defined.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn amsterdam, nederland; natural responseIn sci.physics, Arturo Magidin:Typo patrol> THEOREM. An algebraic integer r is a unit if and only if it is the rootof a monic polynomial with integer coefficients, whose constant termis equal to 1 or -1.Proof. Suppose first that it is a root of such a polynomial,f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0with a_0 = 0 or -1. That should be a_0 = 1 or -1. [.snip.]Arturo Magidin, sans .sigHmm...I was *wondering* how to prove the converse. :-)That was simpler than I thought.As it is, JSH has yet to answer my little question regardinghow to resolve the issue of3 = sqrt(3) * sqrt(3).Or, for that matter, in *regular* integers, how theequationa_nx^n + a_{n-1}x^{n-1} + ... + {a_1}x + a_0 = 0can be proven to have two roots (real or non-real) thathave factors of 3 if a_0 is divisible by 9, when aclear counterexample is available:x^3 + 9*x^2 - x - 9 = 0 = (x+9)*(x+1)*(x-1)And then there are issues, switching to algebraic integersagain, such asx^3 - 9 = 0where there are three roots, none of which are divisible [*]by 3, but all of them algebraic integers.That feeling Im getting isnt fear. Its amusement. :-)Or maybe more like ye gods, what an execrably bad errorto make.[*] one can extend the concept here to indicate that 3 evenly divides an algebraic integer a when the quotient a/3 is another algebraic integer. Since the roots of x^3 - 9 are divisible by 3^(2/3), but not 3, in this sense, Ive clearly got another counterexample here.-- #191, ewill3@earthlink.netIts still legal to go responseNntp-Posting-Host: apps.cwi.nl >The definition of algebraic integers excludes ALL roots of non-monic >polynomials with integer coefficients, when the polynomial is >irreducible over Q. > Only if you can prove there exists a root of a monic polynomial which > is also the root of a non-monic polynomial irreducible over Q.that excludes roots of primitive irreduciable non-monic polynomials, thatis a theorem.-- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn Core error, FEAR is a natural response Typo patrolTHEOREM. An algebraic integer r is a unit if and only if it is the root>of a monic polynomial with integer coefficients, whose constant term>is equal to 1 or -1.Proof. Suppose first that it is a root of such a polynomial,f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0with a_0 = 0 or -1. That should be a_0 = 1 or -1. [.snip.] further on this, but evidently, since thehighest order term (the monic term) and least order term (constant term) are1, replacing the ïx by its reciprocal, setting the resulting polynomial tozero, and clearing fractions by multiplying by x^n, always results in a new,monic polynomial with integer coefficients. This new polynomial has rootswhich are the reciprocals of the original. Hence, each root of the originalis an algebraic integer and its reciprocal is an algebraic integer.Therefore those roots are units since r_j * 1/(r_j) equals 1.This sketch is not a rigorous proof, such as a mathematician might prefer,and which you provided, but is only a conceptual description of the process.(I am not a mathematician, but a fry cook with a national burger is a natural response Visiting Assistant Professor at the University of Montana.> Typo patrolTHEOREM. An algebraic integer r is a unit if and only if it is the rootof a monic polynomial with integer coefficients, whose constant termis equal to 1 or -1.Proof. Suppose first that it is a root of such a polynomial,f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1x + a_0with a_0 = 0 or -1. That should be a_0 = 1 or -1. time further on this, but evidently, since the>highest order term (the monic term) and least order term (constant term) are>1, replacing the ïx by its reciprocal, setting the resulting polynomial to>zero, and clearing fractions by multiplying by x^n, always results in a new,>monic polynomial with integer coefficients. This new polynomial has roots>which are the reciprocals of the original. Hence, each root of the original>is an algebraic integer and its reciprocal is an algebraic integer.>Therefore those roots are units since r_j * 1/(r_j) equals 1.This sketch is not a rigorous proof, such as a mathematician might prefer,>and which you provided, but is only a conceptual description of the process.>(I am not a mathematician, but a fry cook with a national burger chain.)Yes, that proves that if it is a root of a polynomial with constantterm 1, then it is a unit (or constant term -1, adjusting by a sign atthe end). It is the converse which is a bit more difficult (that if itis a unit, then it must be the root of such a polynomial), and thatrelies on the theorem that an algebraic number is an algebraic integerif and only if the monic minimal polynomial over Q has integercoefficients. (Which is something James contested for months on end trying failedcounterexample after failed counterexample, until he finally agreedgrudgingly to look at a proof and discovered it was reasoner as Mr. Smith? I answer as a deceased friend of mine used to answer on like occasions - A mans capacity is no measure of his power to do mischief. Mr. Smith has untiring energy, which does something; self-evident honesty of conviction, which does more; and a long purse, which does most of all. He has made at least ten publications, full of figures few readers can critize. A great many people are staggered to this extend, that they imagine there must be the indefinite something in the mysterious all this. They are brought to the point of suspicion that the mathematicians ought not to treat all this with such undisguised contempt, at least. -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de response> reasoner as> Mr. Smith? I answer as a deceased friend of mine used to answer> on like occasions - A mans capacity is no measure of his power> to do mischief. Mr. Smith has untiring energy, which does > something; self-evident honesty of conviction, which does more;> and a long purse, which does most of all. He has made at least> ten publications, full of figures few readers can critize. A great> many people are staggered to this extend, that they imagine there> must be the indefinite something in the mysterious all this.> They are brought to the point of suspicion that the mathematicians> ought not to treat all this with such undisguised contempt,> at least.> -- A Budget of Paradoxes, Vol. 2 p. 129 by Augustus de Morgan> consecration ofsocietal power in the upper classes of any age]. Their scarlet robes,the ermine in which they swaddle themselves, the palaces where theypreside, all this august apparel was most necessary; and if doctorshad no cassocks and scholars no four-cornered caps, no spreading,four-pieced robes, they would never have fooled the world, whichcannot resist such authentic display. Only men of war have notdisguised themselves in this way, because their rule is indeed moreessential: they establish themselves by force, the others with airsand graces.--Pascal, Pensees You cant eat a math proof, so for many of you ideas are nothing,> unless society TELLS you theyre ok.Youve been spouting that cant eat a math proof nonsense quite abit lately. Why? Are you waiting for someone to tell you how cleverit is? Well, youll have quite a long wait, because it sounds *idiotic*-- just like most of the things you say.-- Wayne Brown (HPCC #1104) | When your tails in a crack, you improvisefwbrown@bellsouth.net | if youre good enough. Otherwise you give | your pelt to the trapper.e^(i*pi) = -1 -- error, FEAR is a natural response> >The definition of algebraic integers excludes ALL roots of non-monic>polynomials with integer coefficients, when the polynomial is>irreducible over Q.Its that arbitrary exclusion that leads to a way to *appear* to>prove two different and contradictory things.Only if you can prove there exists a root of a monic polynomial which> is also the root of a non-monic polynomial irreducible over Q.I dont follow you here.There would not be a problem if a root r of a monic polynomial(with integer coefficients) is also the root of a non-monic(primitive) polynomial (with integer coefficients) irreducible over Q.This doesnt happen, but whether it does or doesnt isnt relevant.For, if it would happen, r would still be an algebraic integersince there is some monic polynomial (with integer coefficients)of which it is a root, and thus meets the requirements of thedefinition to be an algebraic integer.I am using the definition that r is an algebraic integer if itis a root of a monic polynomial (with integer coefficients). Youare apparently using that definition Teller having proposed, as an explanation for the divergencebetween quantum statistics and classical statistics, that ïquanta> >lack haecceities, I posted to sci.physics.relativity what I took> >to be the import of Tellers position for the logic of identity.> >On this topic a fair amount has been written by philosophers of> >science who, unlike myself, are knowledgeable about the formal> >details of quantum mechanics. What these people have had to> >say on the topic may be of interest to some onsci.physics.relativity...> > The content of your thread contains no physics. My replies to one> of those (who, I cant recall) also posting on this thread, indicates> that whatever logic is being argued, totally neglects any physical> meaning to the quantum mechanics.Either you missed the following; 1 I. The Problem, and the Problem with the Problem,> 2 of Identical Articles and Quantum Statistics> 3> 4 Suppose we have a box with two qualitatively> 6 bouncing around inside. We think of the box> 7 as having a left (*l*) and a right (*r*) side.> 9 at random without interacting, so that their> 10 motions are independent; in particular we> 12 that we may neglect collisions. What are the> 13 chances for finding one or both on one side> 14 or the other?> 15> 16 Many find the following reasoning persuasive.> 19 in *r*, and 2 in *l* and 1 in *r*. These should> 20 be equally likely, so that each has a probability> 21 of 1/4, or a probability of 1/4 for two in *l*,> 22 1/4 for two in *r*, and 1/2 for one on each side.> 23> 24 This stylized example is a simple mock-up for a> 25 kind of situation that can occur with quantum entities> 26 and properties. For many of these situations the> 27 probabilities are in fact found to be 1/3 for each> 28 of the three cases: two in *l*, two in *r* and one> 29 on each side. Many interpreters have found this> 30 fact utterly astonishing.> 31> 32 But on the face of it, there is a very simple> 33 resolution of the puzzle: give up supposing> 34 that there are two qualitatively identical but> 36 there are two *quanta*, as Ill put it, to which> 37 the notion of being numerically distinct does not> 38 apply. . . (p. 114)or youre unaware that interpreters of quantum mechanics largelyagree that classical concepts do not apply without alterationor restriction to quantum objects. (Michael Redhead and Paul Teller,Quantum Mechanics, Brit. J. Phil. Sci. 43 (1992), p. crossposting this to sci.physics.relativity> Actually John Jones did it. (*SNIP*) (...Starblade Riven Darksquall...)i do Immortalist: Stop crossposting this to sci.physics.relativity> Actually John Jones did it.(*SNIP*)(...Starblade Riven for Dave Rusins> http://mathforum.org/discuss/sci.math/m/230743/484252> To any q-sided die (which will roll to each side with equal probability)> with integer labels a_i we associate> the polynomial P = sum (1/q) x^{a_i}, a polynomial with P(1) = will roll to each side with equal probability)> with integer labels a_i we associate> the polynomial P = sum (1/q) x^{a_i}, a polynomial with P(1) = 1. What is x?What are we summing roll to each side with equal probability):> with integer labels a_i we associate:> the polynomial P = sum (1/q) x^{a_i}, a polynomial with P(1) = 1.:>:> What is x?: What are we summing over?We are summing over i from 1 to q, the number of sides on the die, and xis the variable of the polynomial (we could write P(x) = .... instead).a_i is the label of the impossibility arguments concerning probabilities> of certain rolls of two dice, invoking a cyclotomic> polynomial and given here some time ago.> Does anyone have a reference?You are probably looking for Dave Rusinshttp://mathforum.org/discuss/sci.math/m/230743/ Keplers EquationJohn Schutkeker> Does anybody else agree with me that a mathematical solution to Keplers> Equation, in closed fom, is a sufficiently important unsolved mathematical> problem to merit inclusion in the Clay Foundations list of million dollar> prizes for mathematics?I guess it would depend on whos paying :) Would _you_ stump up US$1 millionfor this problem?The equation in question isx = y - a sin ywhich we would like to solve for y in terms of x. I cant swear to it, but Ithink it has been shown that the solution in closed form <>sSHfTy;{Dhe&:+?b`9fUj5A~$gIYlYT0/$-asR-K~3S3[]q.R3YSmpR|$- GiZp>UN2a}!Fmw+%h}YL`!h_XXr5Q>_nGsY2_> Does anybody else agree with me that a mathematical solution to Keplers> Equation, in closed fom, is a sufficiently important unsolved mathematical> problem to merit inclusion in the Clay Foundations list of million dollar> prizes for mathematics?> [...]> The equation in question is> x = y - a sin y> which we would like to solve for y in terms of x. I cant swear to it, but I> think it has been shown that the solution in closed form (which needs> clarification) is impossible.> LarryTrue. It is not unsolved in the sense that the Clay problems are.It has been proved that the solution is not an elementary function.-- Re: Million Dollar Prize for Solving Keplers Equation>John Schutkeker> >Does anybody else agree with me that a mathematical solution to KeplersEquation, in closed fom, is a sufficiently important unsolved mathematicalproblem to merit inclusion in the Clay Foundations list of million dollarprizes for mathematics? >I guess it would depend on whos paying :) Would _you_ stump up US$1 million>for this problem?>The equation in question is>x = y - a sin y>which we would like to solve for y in terms of x. I cant swear to it, but I>think it has been shown that the solution in closed form (which needs>clarification) is impossible.>I dont see much interest in that problem. For two-body problems, the possible orbit types are well-known: parabolic, hyperbolic, elliptical/circular, degenerate, and with the computational power at lowcost available today, I cant think of anyone who would gain much from a closed-formsolution to Keplers Solving Keplers Equation> Does anybody else agree with me that a mathematical solution to Keplers > Equation, in closed fom, is a sufficiently important unsolved mathematical > problem to merit inclusion in the Clay Foundations list of million dollar > prizes for mathematics?Yeah, because I came up with what I thought was an excellent proof,over five years ago.James Equation> Does anybody else agree with me that a mathematical solution to Keplers > Equation, in closed fom, is a sufficiently important unsolved mathematical > problem to merit inclusion in the Clay Foundations list of million dollar > prizes for mathematics?Yeah, because I came up with what I thought was an excellent proof,> over five years ago.> James HarrisYuck! I thought he was talking about spherical packing. I shouldhave read past Kepler. Um, so no, I didnt find a Million Dollar Prize for Solving Keplers EquationDoes anybody else agree with me that a mathematical solution to Keplers Equation, in closed fom, is a sufficiently important unsolved mathematical problem to merit inclusion in the Clay Foundations list of million dollar prizes for someone please explain in lay terms what al-Kharkis method is? TIAI couldnt find it. Could you tell us something about al-Kharki> Larry Steinberg Can someone please explain in lay terms what al-Kharkis method is? TIA> I couldnt find it. Could you tell us something about the context of the> term, i.e. what its about?> LHI have never heard about an al-Kharki method. Might it be that you meanKarmarka method? This is a method for solving linear programmingproblems by starting from an interior point and not going around thecorners of the an uncountable set with the discrete topology. >2) Prove that the one-point compactification X cannot >be embedded into the plane RxR. (R are the reals)Exercise. Show a 2nd countable space cant have an uncountable discrete subspace.Conclude === neither X nor X are embeddable into RxR.----Subject: Re: uncountable set with the discrete topology. 1) Determine all compact subsets of X.> This is trivial. 2) Prove that the one-point compactification X cannot be embedded into the> plane RxR. (R are the reals) Heres my problem: Im not sure what an embedding is; to me, its a map: f:> X -> RxR such that:> g: X -> f(X) is an homeomorphism. Am I right?>Yes, that is correct.g:X -> Y embeds X into Y when, g is a bi-continuous injection.or equivalently g:X -> f(X) is a Embeddingbmbrth$kch$1@news-reader2.wanadoo.fr...> Hello. Heres a problem I cant solve: Let X be an uncountable set with the discrete topology. 1) Determine all compact subsets of X.> This is trivial. 2) Prove that the one-point compactification X cannot be embedded intothe> plane RxR. (R are the reals) Heres my problem: Im not sure what an embedding is; to me, its a map:f:> X -> RxR such that:> g: X -> f(X) is an homeomorphism. Am I right?>I would say an embedding is an injective continuous map, which would notrequire the inverse to be continuous.Anyway, this notion implies the one you gave, and if there was an embeddingof X in RxR in my sense, the image of X would be an uncountable compactset with only one accumulation point (for the image of X{a}, aneq {00} isa compact set), which happens to be problem I cant solve:Let X be an uncountable set with the discrete topology.1) Determine all compact subsets of X.This is trivial.2) Prove that the one-point compactification X cannot be embedded into theplane RxR. (R are the reals)Heres my problem: Im not sure what an embedding is; to me, its a map: f:X -> RxR such that:g: X -> f(X) is an homeomorphism. or two, and was wonderingif someone could give a hint or two in the right direction? :)The first one asks that you prove the Fundamental Theorem of Algebra (thateach complex polynomial of degree >= 1 has at least one complex zero)using the Maximum principal. The maximum principle states that if aharmonic function f(z) has bounded modulus on the boundary of a boundedclosed domain D, then it has bounded modulus inside the domain. That is,if |f(z)| <= M of the boundary of D, then |f(z)| <= M for all z in D.The book (Complex Analysis by T.W. Gamelin) states that I should considerusing the maximum principle on the function f(z) = 1/p(z) on a disc o§arge radius, where p(z) is the polynomial of degree n>=1.However, Im not too sure on how to proceed. If I assume that p(z) has nozero, then 1/p(z) is well defined on all of C. Also, we can find an Rlarge enough so that |p(z)| > (1/2)|a_n|R^n, if |z| > R, and then 1/|p(z)|< 2/(|a_n|R^n), if |z| > R. But that is outside the disc of radius R andis not bounded, so Im not sure if this leads anywhere. I guess thequestion Im asking is, what does a large R give me so I can reach acontradiction? Am I wrong to assume that for really large R, |p(z)| with|z| = R is also really large? If so, then for any bound M we can findsuch an R such that 1/|p(z)| < M. Then it is bounded on the disc ofradius R, so we can use the maximum principal to say that 1/|p(z)| < M forall z in D. Then |p(z)| > 1/M; but this says there is no zero.Argh, and Im back to where I started. have a problem that has stumped me for a day or two, and was wondering> if someone could give a hint or two in the right direction? :)The first one asks that you prove the Fundamental Theorem of Algebra (that> each complex polynomial of degree >= 1 has at least one complex zero)> using the Maximum principal. The maximum principle states that if ahedging you bets? Its principle.> harmonic function f(z) has bounded modulus on the boundary of a bounded> closed domain D, then it has bounded modulus inside the domain. That is,> if |f(z)| <= M of the boundary of D, then |f(z)| <= M for all z in D.The book (Complex Analysis by T.W. Gamelin) states that I should consider> using the maximum principle on the function f(z) = 1/p(z) on a disc of> large radius, where p(z) is the polynomial of degree n>=1.However, Im not too sure on how to proceed. If I assume that p(z) has no> zero, then 1/p(z) is well defined on all of C. OK> Also, we can find an R> large enough so that |p(z)| > (1/2)|a_n|R^n, if |z| > R, and then 1/|p(z)|> < 2/(|a_n|R^n), if |z| > R. Hmm. Better if you can prove that |p(z)| > (1/2)|a_n|R^n on the circle{|z| = R} for sufficiently large R. The point is that theresan R with |p(z)| > |p(0)| for z on {|z|=R}. Then |1/p(z)| < |1/p(0)|for |z| = R, so that the maximum of 1/p on {|z| <= R} is NOT on theboundary of this disc.-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 trying to compose a definition of L-series for one of the>volunteer-written online websites (www.planetmath.org). Maybe in the end,>L-series are not a structure but only a formalism, like generating ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^>functions. I may need to resort to arm-waving or (what is much the same>thing) category theory:)Im not sure what you mean exactly, but isntF(x)=Sum a_n u_n(x)for a given sequence {u_n} of functions a general enough definition ofgenerating function?And isnt what youve just written a perfect example of a formalism?>Wheres the structure?Then Im *sure* I dont know what he meant exactly... the problem*seemed* (to me!) to be that the OP couldnt find a general enoughdefinition encompassing all L-series and said that it is the same forgenerating functions.After all isnt the difference more philosophical than mathematical?That is, on the one hand, a structure is a formalism itself, and onthe other hand it seems to me that generating functions have a welldefinite structure: that of a sum (as of my definition), whateverthis might be/mean...Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on L-series?Keith Ramsay:> It would be interesting if there were a definition of L-series ratherthan> just definitions of individual kinds of L-series. As far as I know, there> may be one. But I never did see one, in spite of having seen a number of> talks and lectures about specific kinds of L-series. It seems plausible to> me, then, that what has happened is simply that as each kind of L-series> has been defined (Artin L-series, etc.), its gotten termed another kindof> L-series because of its family resemblanceI agree.> to the other kinds of L-series> previously defined. As far as I know they are all naturally expressed as> Dirichlet series, sum_{n} too.Yes they are all Dirichlet series (AFAIK). But some things spoken of asL-series have Euler products involving a quadratic term (in contrast withe.g. Dirichlet L-functions) such asprod_{p prime}(1 - a_p p^{-s} + p^-{-2s})^{-1}.(That example comes from what little I know about the Ramanujan tauproblem.)Now the factors all split in C, but the product of them all doesnt yield amultiplicative sequence of numerators in the Dirichlet series in an_obvious_ way (as does an Euler product that comes from a modular character,say).The best prototype of an L-series that Ive seen yet is Langsgeneralization of Artins, which can be seen in the last exercise of thelast chapter of Langs Algebra (1st edition). A zeta function and anL-series pop out of a fairly general representation statement, about acategory that includes Artins case of finite groups. And Langs prototypeis in formalism class. Ill go with that one for now.LarryP.S. If the reader thinks Im out of my depth on this question, hes function>log(zeta(x)) = sum_p -log(1-p^{-x})[...]> = sum_k (1/k) * sum_p p^{-kx}This expansion and summation switch is valid for x > 1, or even Re(x) 1. Notice that your sum is the part with k = 1. Solog(zeta(x)) = sum_p p^{-x} + (Other Terms With k > 1).[...]>Anyway, you should be able to find this and more in any basic textbook>on analytic number theory. The trick of taking the logarithm of an>L-series or zeta function with an Euler product in order to get a sum>over primes is quite standard.First, thank you for the supplied information. Second, I am aware ofthe trick you mentioned; however, getting a sum over primes is notthe same thing as getting *that* particular sum.In other words, it *seems to me* that Other Terms With k > 1 are anunavoidable feature and while log(zeta(x)) is informative to getasymptotic results of the kind of those you gave, *as a whole* myseries cannot be expressed, say, in terms of algebraic combinationsof zetas, composition with some function or derivation. Am I right?Please forgive if my wording is not precise, I couldnt find anybetter one for the moment...Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on generating functionThe Riemann zeta function iszeta(x) = sum_n n^{-x} = prod_p (1 - p^{-x})^{-1}Taking logs and expanding using the formula for log(1-t) giveslog(zeta(x)) = sum_p -log(1-p^{-x}) = sum_p sum_k p^{-kx}/k = sum_k (1/k) * sum_p p^{-kx}This expansion and summation switch is valid for x > 1, or even Re(x)> 1. Notice that your sum is the part with k = 1. Solog(zeta(x)) = sum_p p^{-x} + (Other Terms With k > 1).Its not hard to show that the Other Terms converge at x = 1, sousing the easy fact that zeta(x) = 1/(x-1) + O(1) as x approaches 1from above, youll get the useful formula sum_p p^{-x} = -log(x-1) + O(1) as x --> 1+.So this gives something a little stronger than the fact that the sumof the reciprocal of the primes diverges.Anyway, you should be able to find this and more in any basic textbookon analytic number theory. The trick of taking the logarithm of anL-series or zeta function with an Euler product in order to get a sumover primes is quite standard.> Im reading some introductory material about Dirichlets generating> functions of arithmetical functions, but could not find anything about> the obvious generating functionSum_p p^{-x},where the sum is extended over all primes. Has it been studied? Is it> an independent function or can it be expressed in terms Riemanns> zeta? Anything else Generating a congruence class in the polynomial ring GF(2^m)[x]?Suppose I have a factor ring F[x]/P(x), where F[x] is a polynomialring where the polynomials have coefficients in the finite fieldGF(2^m), and P(x) is the polynomial X^n + 1. (Thus, I consider onlypolynomials of degree n-1 and less in F[x])Further, say I have a polynomial g(x), and we have that g(x) is afactor of P(x). (i.e. g(x)|P(x)).How can I generate the different congruence classes of g(x) forF[x]/P(x)?The congruence class [0] is quite easy to generate, as it consists ofall polynomials multiplied by g(x) => this have the form D(x)g(x) =>it can be proved that this class is an ideal => .How can I generate any other congruence class? Will this classes alsohave generators? Or will all the polynomials in a congruence class berelative prime to each other? If they are relative prime to eachother, will there be a way to construct a congruence class?Your time, effort and suggestions of planning an elementary mathematical proof> Hello. I am just a self-studying math hobbyist; I like to help out> calculus students who post questions on the Live Math website.> Recently a young lady there has posted a question that I find> intriguing; trying to solve it, I realize that maybe what I lack is an> understanding of how real mathematicians go about analyzing the> requirements for a prospective proof.Anyway: given that a > 0, b > 0, p > 1, q < infinity, and (1/p) +> (1/q) = 1,prove thatab <= (a^p)/p + (b^q)/q.This is used in the proof of Holders inequality.A cunning trick one can use to prove this is to consicder the graph Gof y = x^{p-1} in the first quadrant. This is also the graph ofx = y^{q-1} (why?). We consider three regions: A bounded by G, the x-axisand x = a, B bounded by G. the y-axis and y = b, and C the rectanglebounded by the axes, the lines x = a and y = b. Then A has area a^p/p(why?) and similarly B has area b^q/q. Obviously C has area ab, but Cis contained in the union of A and B ....-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 mathematical proof> Anyway: given that a > 0, b > 0, p > 1, q < infinity, and (1/p) +> (1/q) = 1,prove thatab <= (a^p)/p + (b^q)/q.Of course, the stipulation that q < infinity also struck me as odd> (isnt it true that any real number is less than infinity?).Ive been able to ascertain that pq = p + q, and that q > 1.How does a real mathematician go about planning his/her strategy for> proving a conjecture like this (assuming it isnt already known)? I,> myself, have only advanced as far as about chapter 7 (Techniques of> Integration) in my favorite calculus text, but I suspect that the> proof of this proposition doesnt require calculus at all.I dont know if you will call this elementary, it uses some facts thatare usually taught in basic calculus courses.as you noted, p and q are postive numbers greater than 1.Suppose p and q are both rational numbers.Then there is some positive integer K such thatP = Kp and Q = Kq are both positive integers.as you noticed, p + q = pq so K(P + Q) = PQ Let T = a^p/p + b^q/qnow T = (q*a^p + p*b^q)/pq = (Q*a^p + P*a^q)/(P + Q)Since P and Q are positive integers, we can rewrite T as(a^p + a^p + ... Q times + b^q + b^q + ... P times)/(P + Q)applying the arithmetic mean >= geometric mean inequality, we getT >= (a^(pQ)* b^(qP)) ^ (1/(P+Q)) = abSo, if p and q are both rational numbers then T >= abSo far the proof has used elementary concepts.We still have to prove this for irrational numbers.To do that, we use the following fact (which some may not considerelementary)if p is any real number, there exists a sequence of rational numbersr_n such that r_n tends to p as n tends to infinity.suppose r_n is a sequence that converges to p.since p > 1, we can ensure that r_n > 1 for all n.for every postive integer k, let s_k satisfy 1/r_k + 1/s_k = 1clearly s_k is a rational number > 1 and s_n tends to q as n tends toinfinity.Now considerT_n = (a^r_n)/r_n + (b^s_n)/s_n Since r_n and s_n are positive rational numbers satisfying 1/r_n +1/s_n = 1we get, from part 1 (p, q rational) thatT_n >= abNow T_n tends to T = a^p/p + b^q/q as n tends to infinityso we have that T >= ab. This completes the proof.The definition of a^p where a and p are real numbers is usuallytaught/formalized in a basic calculus course. So I guess the aboveproof might be Technique of planning an elementary mathematical proof> Anyway: given that a > 0, b > 0, p > 1, q < infinity, and (1/p) +> (1/q) = 1,prove thatab <= (a^p)/p + (b^q)/q.Of course, the stipulation that q < infinity also struck me as odd> (isnt it true that any real number is less than infinity?).Ive been able to ascertain that pq = p + q, and that q > 1.How does a real mathematician go about planning his/her strategy for> proving a conjecture like this (assuming it isnt already known)? I,> myself, have only advanced as far as about chapter 7 (Techniques of> Integration) in my favorite calculus text, but I suspect that the> proof of this proposition doesnt require calculus at all.I dont know if you will call this elementary, it uses some facts thatare usually taught in basic calculus courses.as you noted, p and q are postive numbers greater than 1.Suppose p and q are both rational numbers.Then there is some positive integer K such thatP = Kp and Q = Kq are both positive integers.as you noticed, p + q = pq so K(P + Q) = PQ Let T = a^p/p + b^q/qnow T = (q*a^p + p*b^q)/pq = (Q*a^p + P*a^q)/(P + Q)Since P and Q are positive integers, we can rewrite T as(a^p + a^p + ... Q times + b^q + b^q + ... P times)/(P + Q)applying the arithmetic mean >= geometric mean inequality, we getT >= (a^(pQ)* b^(qP)) ^ (1/(P+Q)) = abSo, if p and q are both rational numbers then T >= abSo far the proof has used elementary concepts.We still have to prove this for irrational numbers.To do that, we use the following fact (which some may not considerelementary)if p is any real number, there exists a sequence of rational numbersr_n such that r_n tends to p as n tends to infinity.suppose r_n is a sequence that converges to p.since p > 1, we can ensure that r_n > 1 for all n.for every postive integer k, let s_k satisfy 1/r_k + 1/s_k = 1clearly s_k is a rational number > 1 and s_n tends to q as n tends toinfinity.Now considerT_n = (a^r_n)/r_n + (b^s_n)/s_n Since r_n and s_n are positive rational numbers satisfying 1/r_n +1/s_n = 1we get, from part 1 (p, q rational) thatT_n >= abNow T_n tends to T = a^p/p + b^q/q as n tends to infinityso we have that T >= ab. This completes the proof.The definition of a^p where a and p are real numbers is usuallytaught/formalized in a basic calculus course. So I guess the aboveproof might be considered ïelementary in that convexity I assume you meant:> show that the second derivative is always positive for x > 0, hence> the value of f(x) where f ï (x) is zero will be the absolute minimum> etc.No, thats for completing the proof that I sketched in the first paragraph. What I meant was the following: If g(x) is concave down on an interval, and P, Q are points on the graph of g, then the graph of g from P to Q will lie above the line segment PQ. This seemingly simple property implies a host of important results, among which is the inequality you asked function.[Unfortunately, this question was posted by tgs _separately_ toseveral different math newsgroups.]Of course, if one were to take the signum function to be -1 if x < 0sgn(x) = undefined if x = 0 1 if x > 0and take 0/0 to be undefined,then an algebraic formula would be sgn(x) = Sqrt(x^2)/xwhere Sqrt denotes the principal (i.e., nonnegative) square root.But the signum function is most commonly defined so that sgn(0) = 0.Presumably that cannot be achieved by what tgs would call an algebraicformula unless one can find a way to handle the definition at thediscontinuity. The simplest such way, IMO, would be to take 0/0 to be 0.Then Sqrt(x^2)/x would give sgn(x), as most commonly defined, for all realx. However, there is another (less controversial) way available. Considerthe curious function 0^x. Many mathematicians accept 0^0 = 1. Furthermore,for x < 0, 0^x = oo if one allows arithmetic in the extended reals. Soheres what tgs might call an algebraic formula for sgn(x) asmost commonly defined (and using extended real arithmetic, rather thanresorting to taking 0/0 to be 0): sgn(x) = 2/(1 + 0^x) - the sin function. [Unfortunately, this question was posted by tgs _separately_ to> several different math newsgroups.] Of course, if one were to take the signum function to be -1 if x < 0> sgn(x) = undefined if x = 0> 1 if x > 0 and take 0/0 to be undefined, then an algebraic formula would be sgn(x) = Sqrt(x^2)/x where Sqrt denotes the principal (i.e., nonnegative) square root. But the signum function is most commonly defined so that sgn(0) = 0.> Presumably that cannot be achieved by what tgs would call an algebraic> formula unless one can find a way to handle the definition at the> discontinuity. The simplest such way, IMO, would be to take 0/0 to be 0.> Then Sqrt(x^2)/x would give sgn(x), as most commonly defined, for all real> x. However, there is another (less controversial) way available. Consider> the curious function 0^x. Many mathematicians accept 0^0 = 1. Furthermore,> for x < 0, 0^x = oo if one allows arithmetic in the extended reals. So> heres what tgs might call an algebraic formula for sgn(x) as> most commonly defined (and using extended real arithmetic, rather than> resorting to taking 0/0 to be 0): sgn(x) = 2/(1 + 0^x) - 1. David CantrellAnother expression for sgn(x) which has been suggested is:lim tanh(nx)n->inftybut this has similar problems when x = 0.--There are two things you must never attempt to prove: the unprovable -- andthe obvious.--Democracy: The triumph of Re: sign function> function, NOT the sin function. Of course, if one were to take the signum function to be -1 if x < 0> sgn(x) = undefined if x = 0> 1 if x > 0 and take 0/0 to be undefined, then an algebraic formula would be sgn(x) = Sqrt(x^2)/x where Sqrt denotes the principal (i.e., nonnegative) square root. But the signum function is most commonly defined so that sgn(0) = 0.> Presumably that cannot be achieved by what tgs would call an algebraic> formula unless one can find a way to handle the definition at the> discontinuity. The simplest such way, IMO, would be to take 0/0 to be> 0. Then Sqrt(x^2)/x would give sgn(x), as most commonly defined, for> all real x. However, there is another (less controversial) way> available. Consider the curious function 0^x. Many mathematicians> accept 0^0 = 1. Furthermore, for x < 0, 0^x = oo if one allows> arithmetic in the extended reals. So heres what tgs might call an> algebraic formula for sgn(x) as most commonly defined (and using> extended real arithmetic, rather than resorting to taking 0/0 to be 0): sgn(x) = 2/(1 + 0^x) - 1. Another expression for sgn(x) which has been suggested is: lim tanh(nx)> n->infty but this has similar problems when x = 0.Actually, due to the limit, it presents no problem whatsoever when x = 0.[OTOH, if we avoid a limit process, something which would have presented asimilar problem when x = 0 is sgn(x) = tanh(oo * x). Of course, thatproblem could be resolved, I would suggest, by taking oo * 0 to be 0.]The problem with the limit expression is that Id doubt that the OPwould consider it to be an algebraic formula, especially content with 0 not having a sign, then sgn x = x/sqr in advance...>If youre content with 0 not having a sign, then you mean something only involving addition, multiplication and constant exponentiation, I think youre out of luck. If dont mean that then maybe you mean this: sign(x) = x/|x|. Of course algebraic formula you mean something only involving> addition, multiplication and constant exponentiation, I think youre> out of luck. If dont mean that then maybe you mean this: sign(x)> = x/|x|. Of course this doesnt work for 0.There is a neat expression if one accepts Iversons notation asrecommended by Knuth: sign(x) = [x > 0] - [x < 0].Handles zero. Is easy to change sign(0) be 1 if that is wanted.The basic meaning of the notation is that [S] is 1 for any truestatement S and 0 for any false statement S. Sometimes it is alsouseful to regard [S] as functionWell, you mean the signum function?Anyway, hope the following useful to you.The real function sgn is defined as below: -1, if x < 0,sgn(x) = 0, if x = 0, 1, if x > 0.Michael Leung .b9.a6.b9g.97.a6l.97sD:Emtib.97$UQ.69@newssvr25. mean the signum function? Anyway, hope the following useful to you. The real function sgn is defined as below:> -1, if x < 0,> sgn(x) = 0, if x = 0,> 1, if x > 0. Michael LeungOrsgn(x) = x/|x|, if x =/= 0, sgn(0) = 0-- > .b9.a6.b9g.97.a6l.97sD Emtib.97$UQ.69@newssvr25.news.prodigy.com... function, NOT solution of Cubic equations.For ax^2 + bx + c = 0,x = (-b +- sqrt(b^2 - 4ac))/2a.Could someone be kind enough to tell me, for ax^3 + bx^2 + cx + d = 0,what the corresponding general General solution of Cubic equations.> For ax^2 + bx + c = 0,> x = (-b +- sqrt(b^2 - 4ac))/2a.Could someone be kind enough to tell me, for ax^3 + bx^2 + cx + d = 0,> what the corresponding http://mathworld.wolfram.com/CubicEquation.htmlHave a says we have this form: a ^ x mod n = yEg. 5 ^ 2 mod 4 = 1It will be easy to find y if the a, x, and n is small.In case if I have a VERY large number of a, x, and n, how should Icompute it?The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.Even using Fast Exponential method ... Im not able to solve it because thea is too large when powered with 2 form: a ^ x mod n = yEg. 5 ^ 2 mod 4 = 1It will be easy to find y if the a, x, and n is small.> In case if I have a VERY large number of a, x, and n, how should> I compute it?> The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.> Even using Fast Exponential method ... Im not able to solve it> because the a is too large when powered with 2 (3549123456 ^ 2 is> too large) > Two ways to make the problem smaller:1. Factor 3549123456 into prime powers. Compute each p^78901 mod n,then multiply the results together.Thats probably still not small enough for your calculator, so2. Factor the modulus into prime powers 67891283121=p*q*r....Do step one for each prime power factor of the modulus, so that youhave 3549123456 ^ 78901 mod p etc. Then you can apply the Re: Large number modulo> Lets says we have this form: a ^ x mod n = yEg. 5 ^ 2 mod 4 = 1It will be easy to find y if the a, x, and n is small.> In case if I have a VERY large number of a, x, and n, how should I> compute it?> The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.> Even using Fast Exponential method ... Im not able to solve it because the> a is too large when powered with 2 (3549123456 ^ 2 is too large)In Pari/GP:(16:10) gp > lift(Mod(3549123456,67891283121)^78901)%1 = 11755218675In calc:> pmod(3549123456,78901,67891283121) 11755218675In dc:3549123456 78901 67891283121|p11755218675Or use Python, Mathematica, Perl (with bignum library), ...Google for number moduloX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>Lets says we have this form: a ^ x mod n = yEg. 5 ^ 2 mod 4 = 1It will be easy to find y if the a, x, and n is small.>In case if I have a VERY large number of a, x, and n, how should I>compute it?>The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.>Even using Fast Exponential method ... Im not able to solve it because the>a is too large when powered with 2 (3549123456 ^ 2 is too large)I believe the number youre looking for is 11755218675. Someonemight confirm thats right.First you need something that handles large integers, like Pythonor something else. Then you need to (i) do all the calculationsmod 67891283121, (ii) use binary exponentiation, which maybe what youre calling the Fast Exponential. (Calculate x^(2^n)by repeated squaring and get x^N by multiplying the appropriatex^(2^n)s.)That 67891283121 took less than a second in Python (usingcode Id written previously that does (i) and (ii) above.)(The documentation says that the builtin function callpow(3549123456L, 78901, 67891283121L) should dothe same thing, but at least in my version of Pythonthat doesnt seem === CCH************************Subject: Re: Large number mod 4 = 1It will be easy to find y if the a, x, and n is small.>In case if I have a VERY large number of a, x, and n, how should I>compute it?>The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.>Even using Fast Exponential method ... Im not able to solve it becausethe>a is too large when powered with 2 (3549123456 ^ 2 is too large) I believe the number youre looking for is 11755218675. Someone> might confirm thats right. First you need something that handles large integers, like Python> or something else. Then you need to (i) do all the calculations> mod 67891283121, (ii) use binary exponentiation, which may> be what youre calling the Fast Exponential. (Calculate x^(2^n)> by repeated squaring and get x^N by multiplying the appropriate> x^(2^n)s.) That 67891283121 took less than a second in Python (using> code Id written previously that does (i) and (ii) above.)>I really appreciate your help, but, actually the question should be donewith a paper and scientific calculator.(My model - Casio fx 570MS). This is the question that my lecturer ask us onHow to solve such question.He said, with small number, we can do it easily, How if he give us a largenumber?Then how should the question being solved? (We were asked to find out thesolution)Actually what I need is the algorithm.(If the solution / algorithm is complex and not convenience for you to writeit down, please provide me some URL of page that teach the documentation says that the builtin function call> pow(3549123456L, 78901, 67891283121L) should do> the same thing, but at least in my version of Python> that doesnt seem moduloX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>Lets says we have this form: a ^ x mod n = yEg. 5 ^ 2 mod 4 = 1It will be easy to find y if the a, x, and n is small.In case if I have a VERY large number of a, x, and n, how should Icompute it?The calculator cannot support until that large number.Eg. 3549123456 ^ 78901 mod 67891283121This number would be too large.Even using Fast Exponential method ... Im not able to solve it because>thea is too large when powered with 2 (3549123456 ^ 2 is too large) I believe the number youre looking for is 11755218675. Someone might confirm thats right. First you need something that handles large integers, like Python or something else. Then you need to (i) do all the calculations mod 67891283121, (ii) use binary exponentiation, which may be what youre calling the Fast Exponential. (Calculate x^(2^n) by repeated squaring and get x^N by multiplying the appropriate x^(2^n)s.) That 67891283121 took less than a second in Python (using code Id written previously that does (i) and (ii) above.)I really appreciate your help, but, actually the question should be done>with a paper and scientific calculator.>(My model - Casio fx 570MS). Well then you do (i) and (ii) above (you followed what (i) and (ii)meant?), but you do the calculations with paper and a scientificcalculator.You recall the algorithms for doing multiplication and divisionby hand? Do that, except if you calculator can handle numberswith 2N digits then regard your numbers as written in base10^N. So you use the calculator to find the product of twodigits when you need to, write down the answers as inlong multiplication and add them up.>This is the question that my lecturer ask us on>How to solve such question.>He said, with small number, we can do it easily, How if he give us a large>number?>Then how should the question being solved? (We were asked to find out the>solution)Actually what I need is the algorithm.(If the solution / algorithm is complex and not convenience for you to write>it down, please provide me some URL of page that teach the algorithm [in>case the builtin function call pow(3549123456L, 78901, 67891283121L) should do the same thing, but at least in my helping.Peter CCH ************************ moduloPeter Chan escribi.97 en el 2 mod 4 = 1 It will be easy to find y if the a, x, and n is small.> In case if I have a VERY large number of a, x, and n, how> should I compute it?> The calculator cannot support until that large number. Eg. 3549123456 ^ 78901 mod 67891283121 This number would be too large.> Even using Fast Exponential method ... Im not able to solve it> because the a is too large when helping.>MOD(3549123456^78901, 67891283121) = 11755218675Instantanealy with DERIVE.If you dont wanto to use a Symbolic Calculator program, you can use alibrarie for C++ o well UBASIC. With all of them you can solve that. Visithttp://primes.utm.edu/links/programs/large_arithmetic/-- equationF(ln(x))/x where F is an odd function> Which continous Functional equationX-DMCA-Notifications: http://www.giganews.com/info/dmca.html>Which continous function(s) satisfy:[*] f(x) + 1/x^2 * f(1/x) = 0 ?Presumably for x > 0. There are _many_ functionsthat do this: If g(x) is any continuous functionon the interval [1,infinity) such that g(1) = 0then you can extend g to a function f definedon (0,infinity) that satisfies (*). (Because ofsome symmetry: Its clear you can extend gso as to satisfy (*) for x in (0,1], and then thefact that f satisfies (*) for x in [1,infinity) related to artinian group> # The puzzle is:> # > # Suppose n people sit around a table and n-1 cards are dealt to them.> # There is no asumption on the number of cards a player receive. In each> # round, all players with 2 or more cards pass one card to the left and> # one card to the right. Prove that eventually, all players but one have> # exactly one card.> # > # I do not have the solution, so any hint would be highly appreciated.> # Please restrain from posting messages saying this is obvious, and> # the like, because what is required here is a proof.> Your puzzle is a special case of the so-called chip-firing games> in A Bjoerner, L Lovasz, PW Shor> Chip-firing games on graphs> European Journal of Combinatorics 12, 1991, pp 283-291Bjoerner, Lovasz, and Shor prove that for every chip-firing game,> the initial configuration determines whether or not the game will> continue forever. If the game eventually terminates, then the final> configuration is fully determined by the initial configuration.For your game, it is easy to prove that there always exists> SOME sequence of moves for which the game terminates.> [BLS] then yields that your game ALWAYS terminates.--GerhardI think you are taking the rules to be that in each round the players withtwo or more cards make their moves one after the other (in some unspecifiedorder). Your analysis of that non-deterministic game is very interesting.But I dont think that solves the problem on my reading of the rules, whichis that the players with more than two cards make their moves in parallel.This makes the game deterministic and harder to analyse, because a playerwhose neighbours both have two or more cards cannot decrease his or ______________________________________________________________ ____> Gerhard J. Woeginger puzzle related to artinian group# #> # #> # Suppose n people sit around a table and n-1 cards are dealt to them.#> # There is no asumption on the number of cards a player receive. In each#> # round, all players with 2 or more cards pass one card to the left and#> # one card to the right. Prove that eventually, all players but one have#> # exactly one card.#> #> Your puzzle is a special case of the so-called chip-firing games#> in#> #> A Bjoerner, L Lovasz, PW Shor#> Chip-firing games on graphs#> European Journal of Combinatorics 12, 1991, pp 283-291#> #> Bjoerner, Lovasz, and Shor prove that for every chip-firing game,#> the initial configuration determines whether or not the game will#> continue forever. If the game eventually terminates, then the final#> configuration is fully determined by the initial configuration.#> #> For your game, it is easy to prove that there always exists#> SOME sequence of moves for which the game terminates.#> [BLS] then yields that your game ALWAYS terminates.#> # # I think you are taking the rules to be that in each round the players with# two or more cards make their moves one after the other (in some unspecified# order). Yes, I do.The analysis of [BLS] applies to games, where the players move one by one. In one move, some player with at least two cards passes one card to the left and one card to the right. My statements above all refer to thissituation.The puzzle of Yannick is a very special case of the [BLS] games:Yannick compresses many moves into a single super-move, and he makesall players with at least two cards move SIMULTANEOUSLY. However,this does not change the combinatorics of the game. If you decomposea super-move into its one-player moves, you will have the same effecton the distribution of cards.Hence: If there exists SOME sequence of one-player moves for whichthe game terminates, then also the game with super-moves must terminate.--Gerhard# Your analysis of that non-deterministic game is very interesting.# # # But I dont think that solves the problem on my reading of the rules, which# is that the players with more than two cards make their moves in parallel.# This makes the game deterministic and harder to analyse, because a player# whose neighbours both have two or more cards cannot decrease his or her# holding related to artinian group# The puzzle is:# # Suppose n people sit around a table and n-1 cards are dealt to them.# There is no asumption on the number of cards a player receive. In each# round, all players with 2 or more cards pass one card to the left and# one card to the right. Prove that eventually, all players but one have# exactly one card.# # I do not have the solution, so any hint would be highly appreciated.# Please restrain from posting messages saying this is obvious, and# the like, because what is required here is a proof.Your puzzle is a special case of the so-called chip-firing gamesin A Bjoerner, L Lovasz, PW Shor Chip-firing games on graphs European Journal of Combinatorics 12, 1991, pp 283-291Bjoerner, Lovasz, and Shor prove that for every chip-firing game,the initial configuration determines whether or not the game willcontinue forever. If the game eventually terminates, then the final configuration is fully determined by the initial configuration.For your game, it is easy to prove that there always existsSOME sequence of moves for which the game terminates.[BLS] then yields that your game ALWAYS terminates.--Gerhard__________________________________________ ________________________Gerhard J. Woeginger you very much, James. You are so kind. I am sorry to reply you solate. The method is very skillful. I believe it has a excellent efficiency.Thaks again. ^_^Best wishesLeng of a few circles?> Make a list of the N*(N-1) circle intersection points and the 2N> points p_i with x +/- r that are not in the interior of another> circle, and sort into ascending order by x coordinate, then sum the> areas of vertical zones bounded by these critical points. There> are no arc intersections within a zone. A zone may contain disjoint> segments but each segment is bounded above and below by an arc of a> circle, and left and right by straight lines. For example, if we have 3 circles of radius 5 and centers at (5,12),> (8,9), and (9,5), the event-points list for the plane sweep is:> 0.0 12.0> 3.3 7.3> 4.0 5.0> 4.1 5.9> 9.7 13.7> 12.9 and Randy. To my understanding, James> plane-sweep method is similar to the numerical integration. The sunk> parts that I mentioned is the disjoint segments in the reply from James.> I am still in the trouble to determine which arcs should be used to bound> the above and below in my simulation programme, especially in the cases of> the disjoint segments, since the ascending order by x coordinate will be> disturbed. Could you please show me the detail of how to deal with this> problem? In James example, how to identify which arcs are the above bound> and the below bound between (3.3, 7.3) and (4.0, 5.0), as well as between> (4.0, 5.0) and (4.1, 5.9)?I dont have access to Edelsbrunners paper that David Eppsteinmentioned, but feel certain that he (Herb Edelsbrunner) worked out allthe picky details and special cases that arise, and probably presenteda method using near-optimal data structures and algorithms. So keepin mind that the following may be far less efficient than possible,may overlook various special cases or optimizations, etc.BTW, this is not a numerical integration method in that it is exact(as long as you use proper formula for segment areas, exactarithmetic, and epsilon=0). Numerical integration commonly refers toapproximations whose accuracy depends on a chosen delta-x. See, forexample, http://mathworld.wolfram.com/TrapezoidalRule.htmlI see no reason to use numerical integration for this problem,given the simplicity and ease of an exact method like the following.I would sort the circles into x order before making the list ofcritical points (to decrease average complexity via simpleintersection pre-checks) and would add circle-number fields andevent-type fields to the event object. [For example, (3.3, 7.3) is anintersection of circles 1 and 2 so the event entry could be (3.3 7.3 12 I). (0,12) is the left edge of circle 1 and (14,5) is the rightedge of circle 3 so they would have event entries like (0 12 1 L) and(14 5 3 R).] After making list Events, make an empty list calledActive that tells which circles are active. The first entry inEvents is an L item. Add its circle to Active and set xright to its xvalue. Then, do steps 1-6 for each remaining event:1. Set xleft to xright, and set xright to the current events x value.2. If Event is an I event, add its circles to Active. 3. If xright-xleft < epsilon go to step 5. [f]4. Integrate between xleft and xright. [a]5. If Event is an R event, remove its circle from Active. [b]6. If Event is an L event, add its circle to Active.[a] Integration process: Make a list Arcs as follows: For each Activecircle, add its arcs between xleft and xright [f] to Arcs, along with§ags for circle number. See example at [d]. Sort Arcs [e] so bothends are in non-decreasing y-order (which is always possible becauseof the Event list construction). In one pass, §ag arcs that areinside other circles, and in another pass, remove the §agged arcs.At this point, each even/odd pair of entries in Arcs gives bounds of acontiguous area, bounded below by the first arc, above by the secondarc, and left and right by xleft and xright. Add its area to sum.[b] So, right edges inside other circles and otherwise irrelevant tointegration, would need to be on the event list. This problem can beameliorated by coding them as B events and not integrating or changingxleft or xright at B events.[d] For example, the Active-circle arcs between events (3.3 7.3 0 1 I)and (4 5 2 L) are: 0 3.2984 16.7016 4.0000 16.8990 0 3.2984 7.2984 4.0000 7.1010 1 3.2984 10.7016 4.0000 12.0000 1 3.2984 7.2984 4.0000 6.0000which sorts to 1 3.2984 7.2984 4.0000 6.0000 0 3.2984 7.2984 4.0000 7.1010 1 3.2984 10.7016 4.0000 12.0000 0 3.2984 16.7016 4.0000 16.8990and §ags to 1 3.2984 7.2984 4.0000 6.0000f 0 3.2984 7.2984 4.0000 7.1010f 1 3.2984 10.7016 4.0000 12.0000 0 3.2984 16.7016 4.0000 16.8990and reduces to 1 3.2984 7.2984 4.0000 6.0000 0 3.2984 16.7016 4.0000 16.8990which is a contiguous zone of area about 7.12.[e] Maintaining Active in some y-sorted order or as a priority heapmight reduce the computational complexity somewhat. The given method(with an Arcs sort) is O(n^3 log n) worst case for circles in generalposition (and I think could achieve that with n circles having centersall on one vertical line) but your conditions of common r and at leasthalf a disc of overlap will result in much better average performance.[f] Consider 1/(r*n*10^6) for epsilon to limit error to at most to do a Applied Maths major but at my university there isnormally not enough people wanting to do them to be tough as face to facesubjects. But I am able to take them as reading courses or even go to another university to do them.But I dont really know what I want to do. For my major I need to do 4subjects(each subject lasts one semester). They will count any pure orapplied type subject and they will also let me pick something I want to dothen try to find a person on staff that is willing to run the readingcourse.I am just wanting to do something that is interesting and that is juststarting to take off. I have been told I might be able to do things likewavelets, high performance computing and cryptography. But I would reallylike to see what is really out there.I am already doing Stats subjects as I am doing a major in that as wellAny information that you might be able to give me will be most be honest, the distinction between applied and pure math doesntreally become a reality until you get beyond the undergraduate level,or at least it shouldnt. I say this because, if you remove yourselffrom what might be called pure math before youve gone through astandard undergraduate mathematics curriculum, youre going to end upwith a lot of applications and absolutely no way to prove any of thetheory.For instance, lets say you come into college where most math majorsIm familiar with did. You understand most 3d calculus and some o§inear algebra, but have never done long, rigorous proofs, and youreonly familiar with continuity and differentiability when it comes toanalysis, not much more.So you jump into an pure applied math course, say, numericalanalysis. Immediately youll be hit by theorems and vocabulary thatyoure not familiar with; you might also not have the sophisticationon paper to write down the information you need the teacher to knowyou know. Furthermore, if the class is anything like the one I took,itll look a _lot_ like a pure math course for a while, while you laydown the theory before actually programming. Or, if you decide to gointo cryptography, youll have to learn a large chunk of algebrabefore youre able to go much further than RSA.My suggestion, then, is to spend at least two of those classes ongetting a good grounding in that pure, theoretical math. Id startwith an analysis course, to get a good handle on the real numbers andfunctions. Youll want to understand things like the Fourier transformand what they really mean before you look at wavelets, and that willrequire some knowledge of integration beyond what most people know inhigh school. In fact, a (depending on your level of expertise,partial) differential equations class can introduce you to some of thebigger ideas in analysis while teaching you some interestingapplications of it, but Id say again, only after an analysis course.If, at that point, you really like analysis, stick with it. Somepeople take to those ideas like a fish to water, while some preferalgebra, which brings up my next point. If they offer an abstractalgebra course where you are and you find that the analysis coursedidnt exactly spark any §ames, you might consider talking to theprofessor to find out if its for you. After thats over, much ofcryptography will simply fall into place.Id say it should only be at this point that you consider looking atan application. I mentioned numerical analysis before; you and theteacher might sit down and discuss if this is for you, too. A lot ofit is tedious, hard work, but in the end it gives a lot of powerful,applicable tools that we really cant do without. At this point, youcould look at numerical solutions to differential equations, which arealways interesting. Cryptography, too. If youre looking for a teacherthat can supervise a reading course, the person who taught you algebrashould be able, if not willing.I guess my suggestions boil down to this: give yourself a strongtheoretical background before going off and trying to apply. Onceyouve got that, however, the possibilities are endless. Id say,though, that you should see which you like better, the algebra or theanalysis, because that can tell you to a certain degree of accuracyhow much youll enjoy their Re: Science is a human activity (was: Python syntax in Lisp and Scheme)>Its certainly true that mathematicians do not _write_>proofs in formal languages. But all the proofs that Im>aware of _could_ be formalized quite easily. Are you>aware of any counterexamples to this? Things that>mathematicians accept as correct proofs which are>not clearly formalizable in, say, ZFC?I am not claiming that it is a counterexample, but Ive always metwith some difficulties imagining how the usual proof of Eulerstheorem about the number of corners, sides and faces of a polihedron(correct terminology, BTW?) could be formalized. Also, however thatcould be done, I feel an unsatisfactory feeling about how complex itwould be if compared to the conceptual simplicity of the proof itself.Just a thought,Michele-- > Comments should say _why_ something is being done.Oh? My comments always say what _really_ should have happened. :)- Tore Aursand on (was: Python syntax in Lisp and Scheme)Well, since you crossposted this to sci.math you must be hoping> for replies from that direction:> Alex Martelli:> would you kindly set right the guys (such as your> namesake) who (on c.l.lisp with copy to my mailbox but not to here) are> currently attacking me because, and I quote,> > Software is a department of mathematics.> And anyone who doesnt think mathematics has its ownculture with ideas and even mistaken preferences for whatis right and wrong should readThe Mystery of the Aleph: Mathematics, the Kabbalah, and the Human Mindto see how Cantors ideas of transfinite numbers (and other ideas,as I recall, like showing there are functions which are everywherecontinuous and nowhere differentiable) were solidly rejected bymost other mathematicians of his time.Mathematicians are people as well.> .> .> .>And let no one assume that these are mere foibles of the>past that we moderns have overcome; mathematics remains>stunningly incoherent in whats labeled foundations.>Theres a wide, wide divergence between the intuitionism>working mathematicians practice, Actually inuitionism has a certain technical meaning,> and actual intuitionism is not what most mathematicians> practice. But never mind, I believe I know what you meant.> >and the formalism they>profess.Far be it from me to insist weve overcome the foibles> of the past. But:Its certainly true that mathematicians do not _write_> proofs in formal languages. But all the proofs that Im> aware of _could_ be formalized quite easily. Are you> aware of any counterexamples to this? Things that> mathematicians accept as correct proofs which are> not clearly formalizable in, say, ZFC?How about the following?C3 EyAx[x in y <-> Et(x in t) & A] (with y not free in A) Classification C4 AyAx[Az(z in y <-> z in x) -> {(set y & set x) <-> y=x}] (Equi-membered classes are identical iff these are sets.) 1) EyAx(x in y <-> Et(x in t) & ~(x in x)) C3 Hence 2) Ax(x in r <-> Et(x in t) & ~(x in x)) 1,EI and 3) r in r <-> (Et(r in t) & ~(r in r)) 2,UI so that 4) ~Et(r in t) 3 and 5) ~(set r) 4 and 6) ~(r=r) 5,C4 so that 7) Ex~(x=x) 6,EG >Good thing, too; our age enjoys the blessing of superb>mathematicians, and Im relieved that philosophical in->consistencies dont (appear to) slow them down.Whats an actual example of one of those philosophical> inconsistencies that luckily doesnt slow us down?************************--John^^^^^^^^^^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^The most successful ideological effects are those which haveno need for words, and ask no more than complicitous silence.--Pierre news.mailgate.org; posting-host=gatekeeper.tripos.com; posting-account=35661; posting-date=1066055371X-URL: http://mygate.mailgate.org/mynews/sci/sci.math/ 760fbf184d01790019233c862779216d.35661% 40mygate.mailgate.orgIm reading (trying to read) the book Mathematical Methods of QuantumMechanics by G. Fano and he introduces measure theory through Jordanmeasure based on generalised intervals - so he ends up with two sumscorresponding with inner and outer Jordan measure. The outer measure isM_i which is the GLB of a set of sums based on intervals R_i, and M_ jis the LUB of the set of all possible sums based on the intervalsR(hat)_ j. Then he says that delta is the greatest diagonal of thegeneralised intervals R_1, R_2......R(hat)_1, R(hat)_2.....etc. Then hestarts talking about limits of delta.But what Id like to know is what is this delta? What are thesediagonals exactly? If anyone can explain in very simple language thatwould be need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the plane, but in 2d, the plane and all its lines are the whole thing implement a collision detection algorithm >and cant figure out how Id be able to do this in 2D. What would you >suggest. point and time ofa moving particule, and a wall given by the line segment in your 1stpost. You dont need a normal to the wall to do that (and the wall isnot a ïplane, its just confusing).So this problem is basically the intersection of a line with a linesegment. Equation of a particule moving at constant velocity:Px = Px0 + t*VxPy = Py0 + t*Vywhere (Px0,Py0) is the position of the particule at time t=0, and(Vx,Vy) is the constant velocity, and t is the time which is theunknown variable.Equation of the line segment (wall):Wx = Wx1 + u*(Wx2-Wx1)Wy = Wy1 + u*(Wy2-Wy1)where (Wx1,Wy1) and (wx2,Wy2) are the 2 endpoints of the segment, andu is the variable to solve for.For the collision solve the system:Px0 + t*Vx = Wx1 + u*(Wx2-Wx1)Py0 + t*Vy = Wy1 + u*(Wy2-Wy1)These are 2 equations with 2 unknowns t and u, so its solvable.If there is a collision then t must be >0 because t represents thetime of the collision(which must be in the future), and u must be0<=u<=1 because the collision must be on the line segment and notoutside of it, so if t<=0 or u<0 or u>1 then there is no collision.If you dont understand the parametric equations of lines in 2d thenread this http://www.geocities.com/SiliconValley/2151/math2d.htmlAfter determining if there is a collision and where it will be on theline segment, you might want to calculate the rebound of the movingparticule. Then tq2f7vo050nc1pvhdhp2olmkcbhr2vm828%404ax.comSee also section 1 Re: Normal to a plan> You need 3d to get a normal to a plane. Any normal to a plane has to go at right angles to every line in the > plane, but in 2d, the plane and all its lines are the am trying to implement a collision detection algorithm > and cant figure out how Id be able to do this in 2D. What would plane, rather than a geometric plane, it is an entirely different problem.In one form the problem might be stated as: you are given two aircraft with given positions and constsant velocities in the plane and want to know whether the minimum distance between them gets smaller that some preset amount.If this is not a fair statement of your problem, can you give more details of what your problem statement actually posted a new version of his proof of a core error on October 12. He made a simple algebra mistakewhich he has since acknowledged. The interesting thing about this is that, when the algebramistake is corrected, it leads immediately to a proof thathim main conclusions are wrong. Here is an excerpt from his post of Oct. 12:>For me there have been two perspectives as I work to figure out how to>explain the definition problem in mathematics with LOTS of opposition,>and I wonder about mathematicians so dedicated to attacking an>argument that is clearly correct.I remind of that as I present what should finish their ability to>distract, as Ive seen a strange and dedicated effort to ignore the>actual math, and simply toss up just about anything rather than face>the truth.All variables are in the ring of algebraic integers unless otherwise>stated.Let 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 letR(m) = (m^3 f^4 - 3m^2 f^2 + 3m) x^3 - 3(-1+mf^2 )x u^2 + u^3 fso P(m) = f^2 R(m).Now considerP(m) = (a_1 x + uf)(a_2 x + uf)(a_3 x + uf)where the as are given by the following cubic:>[***] a^3 + 3(-1+mf^2)a^2 - f^2(m^3 f^4 - 3m^2 f^2 + 3m).> This is correct so far.>Then it must be true that the following factorization existsR(m) = (b_1 x + u)(b_2 x + u)(b_3 x + uf)> Note that what is implied here is that: b1 = a1 / f, b2 = a2 / f, and b3 = a3.>where the bs are given by the following cubic:b^3 + 3(-1+mf^2)b^2 - (m^3 f^4 - 3m^2 f^2 + 3m),> This is where he made the mistake. Note that a1/f = b1, or a1 = f*b1. Substituting this into the polynomial [***] above that the as satisfy and simplifying, one obtains[1] f*b1^3 + 3*(-1 + m*f^2)*b1^2 - (m^3*f^4 - 3*m^2*f^2 + 3*m). Clearly b2 is a root of the same polynomial. Harriss goal is to show that b1 and b2 are algebraicintegers for all integer values of m. Since b1 = a1/f and b2 = a2/f, this would imply that that a1 and a2 aredivisible by f in the algebraic integers. If his original equation for the bs were correct, thiswould follow immediately. It isnt correct. The correct expression is [1]. The key thing here is that the correct polynomial [1] which b1 and b2 satisfy is NON-MONIC. Notice that coefficient f in front of b1. If the polynomial [1] is irreducible and primitive,the ball game is over: because in that case, b1 andb2 CANNOT be algebraic integers. So, for what values of m and f is [1] irreducibleand primitive? Answer: MOST. For example, let f = 5 and m = 1.Then [1] becomes: 5*b1^3 + 72*b1^2 - 553 = 0.easily checked to be irreducible, primitive, and,most of all, NON-MONIC. Therefore b1 is not analgebraic integer. Therefore a1 is not divisibleby f = 5. It is tempting to gloat over this. Harriss ownblunder, corrected, leads to an *immediate* proof that his current argument is wrong. More importantly, itshows not only that the argument is wrong, but also that his *central claim* is wrong. This impliesthat his argument cannot be fixed. He has been pursuing this theme for many months now.He has submitted a paper on Advanced PolynomialFactorization, in which this is the central claim,to several journals, even apparently getting it published in a vanity journal put out by the MegaSociety. He has written to mathematicians and evenvisited one of his old math profs at Vanderbilt toexpound upon his proof. He has offered $100,000 to people who will helphim implement a computer program to verify hisproof, provided, of course, he wins the $800,000 Abel prize. His central claim here was an essential part of hisclaimed short proof of Fermats Last Theorem. It is tempting to gloat. This strutting, bragging,insulting, pontificating, mathematically incompetent dimwit has plagued sci.math for over 8 years with bogus proofs. I doubt this is the end of this. Logically it should be. Logic however is not Mr. Harrissstrong suit. Anyway, it looks certain that I am not going to get a share of that $100,000. Nora the odd and you could say esoteric error in> core mathematics with such a short, and rather simple argument, the> issue now is how long until mathematicians decide that theyd rather> have correct mathematics versus the *belief* that they had been> perfect in keeping error out of the collected body of work that is> called mathematics.Really, I can not understand your work, not even the core error problem.I apologize for replying in a public forum, but Ive chosen, or beenguided, to work under certain constraints. The danger of e-mail is thetemptation to say different things to different people. It would bebetter for me, professionally, to hide my support for you, but I refuseto do it.> My work is out there and rather easy to go over as can be seen at the> Hong Konk math site:See http://mathdb.math.cuhk.edu.hk/forum/e_show.php?msg=782And I send people there because their allowal of the use of LaTeX> makes for a *much* better presentation, and given the *social* issues> Im facing, I need all the help I can get.The social issues. Indeed. What help could I be to you with themathematical issues? Even herd mathematicians like Magidin are betterat math than I. The fact is, you dont *need* any help with themathematical issues.Im still working on your Plan. Ive very busy right now, but Imfinding ways to make the time for it. It will take several weeksat least. As I began to conceive/receive it, I realized that itwould do no good if you had a Bush-like outlook on life. You seemedannoyed by my detour into politics, but this is what I mean: youbeing concerned with *mathematical* issues is like Bush trying tobuild bigger bombs to fight terrorism. America is already powerfulenough, and doesnt need to become all-mighty so that all nations ofthe world cower at the thought of incurring its wrath. According toBush, terrorists and the people who cheer for them are insane and evil.According to Bush, a vast segment of the Islamic world is insane and evil.But George Bush is simpleminded, so perhaps we should forgive him hissimplistic viewpoint. Hatred for America is engendered by . . . Americasbetter teeth. Now heres what I want you to do . . . Only a suggestion,mind you: Im all for freedom and human rights and democracy (my ideasare also better than yours), so do whatever you want. What? What are youdoing? Now I have to beat the crap out of you! Oh, boo hoo, you wantyour natural resources back? Too late, I ate them. Ha ha. Hey! Donthit me there! Evil terrorist! I have a date with Brittany Spears tonight!Im not saying that youve installed any puppet dictatorships, James, butthere are aspects to Americas foreign policy arrogance that remind me ofyou. Sure, its a natural reaction to being better than everyone else.Better at building weapons, better at doing math, easy to think better inevery way. Easy to wonder, Im so great, but they all hate me . . . theymust be wrong. They must be evil.This is why I doubt that working out your Plan is even worth the effort.No one likes being told they have to change their behavior to achieve theirgoals. My guide (or muse, or the Holy Spirit, or whatever it is that seemsto be sending this Plan to me) argues that because you so fervently want toachieve your goal and because you realize that what youre doing isntworking, that you are receptive to new ideas. But I think your attitude ismore like, Let Ferry write his little plan, and if I dont like it, Illjust laugh at it. So, yes, Im resisting having to do this pointless work,but still the Holy Spirit (or guide, or muse, or whatever) is prodding me on.So please, just tell me, Oh, you want me to roll over and play Mr. NiceGuy! No dice! and then I can forget this whole hassle. At first I was sohonored that the Holy Spirit was choosing to work through me . . . oh, itsprobably a load of rubbish anyway. Me wanting to be important and imaginingthat I was getting divine inspiration. The whole thing just seems ridiculous.Never mind.> Some of you are now facing the reality of the human brain versus any> fantasy you might have had about being completely rational. Human> beings are NOT rational creatures but necessarily rely on social> forces to determine what they believe.You are creatures of society.You may have believed that your mathematical knowledge was based> completely on logic and rationality, but human beings dont work that> way; its built-in to your wiring NOT to work that way.Some of you must learn to be more than human.You know, I dont think genetics come into it. Im in no position toknow, but heres what the little voice in my head says: The chief difference between James Harriss and the Establishments mathematical systems lies not in the validity of each: rather it is simply that the world has chosen to accept the Establishment viewpoint. Reality is created by consciousness more than you know. James Harris is attempting to create a new Reality to displace the old, but his arguments simply *do not pertain* within the current Reality. The current Reality is §awed, of course, by Goedelian indeterminancies, and is ripe for replacement by something superior . . .Yeah. Sounds like a load of crap. Like Langans CTMU consciousnesscreates the world crap. Forget it.Mathematics is an absolute truth. Cbeck.The physical world is what it is. Check.I apologize for all my inconsistencies. I started writing this in oneframe of mind, and ended up in another. A little voice in my headtelling me how James Harris can conquer the (mathermatical) world?People are going to start telling *me* to take my meds . . .> You must learn to be truly rational, for the first times in your> lives.So its about time, as I wait, and wonder, how many of you can handle> the truth.And how many of you prefer the fantasy which was the world you> believed in, which actually never existed, except in your> imaginations; your wishes for a nicer world, where your wishes matter.> James HarrisMy imagination. Sigh. Im sure thats all it was. Sorry for wastingyour time.-- | Jim Ferry | Center for Simulation |+------------------------------------+ of Advanced Rockets || http://www.uiuc.edu/ph/www/jferry/ +------------------------+| jferry@[delete_this]uiuc.edu | Heres an important answer from me, and readers should read the lead> off paragraph from Nora Baron and my reply as I divide f^2 off so that I consider P(m)/f^2. The point is that then you have P(0)/f^2 = u^2 (3x + uf) as the constant term. Seems too simple readers? Well, if P(m) = g_1 g_2 g_3, and you have that at m=0, g_1 = uf, g_2 = uf, and g_3 = 3x + uf then what happens with P(m)/f^2? Lets say you have w_1 w_2 w_3 = f^2, so that you have g_1/w_1 = uf/w_1, g_2/w_2 = uf/w_2, and g_3/w_3 = (3x+uf)/w_3. Now then how many of you find yourselves incapable of figuring outwhat w_1, w_2, and w_3 are here? The answer is that w_1 = w_2 = f, while w_3 = 1, if f is coprime to 3. And thats given by, what? By looking at P(0)/f^2, of course. The math is basic.Whats extraordinary to me is how effective posters like Nora Baronhave been at questioning a very simple technique.The idea is that if you have terms or variables independent of onevariable, then of course its value doesnt matter, but it can be inthe way, when youre doing your analysis.So, since the variables value doesnt matter anyway, you can just setit to 0 to clear it out while you focus on terms or variablesindependent of it.Yet, this Nora Baron has questioned that technique!!!It seems that all it takes to shoot down a technique thats been inanalysis for, oh I dont know how long, is to claim that its aspecial Ok, I figured it out. If your logic were correct it would apply> to other functions. Yup. Let Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Say Q(m) is factored in the form Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Now note Q(0)/f^2 = x + f. That is, Q(0) = f * f * (x + f). Therefore when m = 0, we can say a1 = 0, a2 = 0,> and a3 = 1. Note that when m = 0, a1 and a2 are divisible by f. Yup. Now by your logic, for values of m other than 0,> a1 and a2 must be divisible by f and a3 is relatively> prime to f. Do you agree with this? This is a test of your> method. We need to know before we go to the next step. Nora B. The answer is that yes, but with the qualification that m not equal 1,> but I have a better explanation for why than the wacky reply when I> initially freaked out. Look again at your Q(m) = (a1*x + f)*(a2*x + f)*(a3*x + f). Here the constant terms for the factors are revealed to be f, f and x + f where it just so happens that for your a3 you have something> like--yup, I know some will probably not like this, but its the> reality--introducing h for the functions h(m), a3 = h_1(m) - mx h_2(m) + x + f where h_(1) has f^{2/3} as a factor and h_2(1)=1. And yes, the same thing can happen with my argument, but it requires mf^2 = 1 but both m and f are integers in that argument, so that condition> doesnt occur. You have suggested elsewhere that I need to solve> for the as with expression.> Lets do that. Recall that Q(m) = f^2*m*x^3 + f^2*(1 - m)*x + f^3. Let m = 1:> [1] Q(1) = f^2*x^3 + f^3 = f^2*(x^3 + f).> Assume this is factored in the form Q(1) = (a1*x + f)*(a2*x + f)*(a3*x + f). The roots of this polynomial are> [2] r1 = -f/a1, r2 = -f/a2, and r3 = -f/a3.> Of course from [1] I can compute the roots> explicitly. They are:> r1 = -f^{1/3} r2 = -f^{1/3}*(-1 + sqrt(3))/2 and r3 = -f^{1/3}*(-1 - sqrt(3))/2. and a3, since ai = -f/ri :> a1 = f^{2/3}, a2 = f^{2/3} * (-1 - sqrt(3))/2, and a3 = f^{2/3} * (-1 + sqrt(3))/2. Note that a1, a2, and a3 are all algebraic integers. Note that each of them has a factor of f^{2/3}. Note that none of them have f as a factor.> No surprises here. What has this got to do> with your wanting m*f^2 to equal 1 ? (If I didnt say they were before, well they are now.) For the more adventurous, check Q(m) with m NOT equal to 1, and yes,> you will find that *two* of the as have a factor that is f.> See my other post on this.> Nora B.> PS: Oh yes, you also asked if I would be convinced if> a computer verifies your proof. I would say *no*. I know exactly what is wrong with your> proof already. If someone said a computer verified> your proof, I would have to have a guarantee of at least> three things:> (1) That the computer program and the computing hardware> were infallible. (2) That the translation of your proof into the programs> language was done correctly. (3) That the person reporting the results could be trusted.> All of these three conditions will be harder to ascertain> than what I already know about your proof. I cannot see how> (1) can be guaranteed at all. As for (2), checking that would> probably considerably more tedious than what I have already done.> And for (3), the most likely person to report positive results> would be the least trustworthy. So no. But dont let me stop you. Go ahead and carry it out.> Maybe you will convince somebody else.Not to mention that unless someone had the code to look at against theoutput, it wouldnt tell you a thing.> It might be a fun as you gave me a scare. If any of you out there think you can find fault with my conclusion> here, please try. It is math after all. And its also a lot of fun.> James HarrisDavid w/f=ma?Cut sure out to be held to a standard of knowing that a slash / is a> symbol for a mathematical operationA slash [/] has several meanings Gene: Like using kilos; youve got toconsider the context(;^); does it mean mass or does it mean weight - andwhat is the difference? Dont all masses have weight that will vary withlocation? Isnt the ratio of an objects weight, divided by the accelerationat which it free fall a constant; at any and all particular locations; atany particular point in time? Isnt this constant a measure of the objectsmass and/or inertia? You can bet on it Gene!using ft as a symbol for> the multiplication of force by time--What do you think ft means Gene? Maybe I should write that out for you sothat you wont think it means do you think ft means Gene? Maybe I should write that out for you so>that you wont think it means mass.Whatever I think it means, it probably wont be the same thing that itis meant to mean in a Donald G. Shead message.You are an idiot, Donald.Gene Nygaardhttp://ourworld.compuserve.com/homepages/Gene_Nygaard/= === ==Subject: Re: How long must physics put up w/f=ma? The answer to the question How long must physics put up w/f=ma? is:> As long a F = ma describes reality And just what does that have to do with *w/f=ma*? You dont read very carefully, do you?>He read it as I meant it to be read Tom: As How long must physics put up_with_ f=ma? Sorry about the misunderstanding; my fault: It does look as ifphysics is putting up _w/f = ma_(:-!) Tom put up w/f=ma?In sci.math, Donald G. Shead:> The answer to the question How long must physics put up w/f=ma? is: As long a F = ma describes reality And just what does that have to do with *w/f=ma*? You dont read very carefully, do you?> He read it as I meant it to be read Tom: As> How long must physics put up _with_ f=ma?> Sorry about the misunderstanding; my fault:> It does look as if physics is putting up _w/f = ma_(:-!)The alternative ofm = f/a = w/gdoesnt appear to buy one much at all, being a simple algebraictransform off = maw = mgIts far simpler to conceive of a reference mass mthan to have to find a reference force or tension,although a reference tension can be done with some care.However thermometric calibration (necessary to generate theproper pressure/force) may itself require a known mass,in this case a known mass of liquid introduced into thethermometer bulb/pipette combination, which is manufacturedto a known volume of the bulb and a known thickness/radiusof the tube.Fortunately for you, that known mass can be specified asa volume at the freezing temperature of the liquid (eitheralcohol or mercury), with controlled pressure, usually ator near 1 atmosphere. Unfortunately, pure ethyl alcoholis hygroscopic, and cannot be fractionally distillated;mercury is poisonous, although AIUI its far worse asmercuric oxides than in its customary liquid form, as usedin older-model thermometers. (Newer models apparentlyare electronic, and are simply stuck in the ear.)Other methods include fabrication of the somewhat familiarbimetallic strip. However, I dont know offhand howthe bend thereof corresponds to the thicknesses of thetwo metals. (There may be a small corrective factorhere, if at all.)Still another method involves a variant of the bimetallicstrip: the thermocouple. Id have to look up exactlyhow to calibrate a thermocouple.One rather silly method might be to use the blackbodyradiation formula and heat everything to a knowntemperature, as measured by the glow of the apparatus.Once the temperature is known, of course, the force canbe generated in any of several interesting ways; the mostobvious being the introduction of a known amount of gas(known consistency, volume, temperature, and pressure)into a piston of known volume encased in a larger chamberof known consistency, temperature, and pressure. Asimple Hookes Law device (a spring scale) can be usedto monitor the resultant force, with appropriate correctionsfor the movement of the piston and the spring end.Once calibrated, the spring scale can be used to measureother forces, assuming someone doesnt bend the springout of shape. However, its far easier to just use aniridium alloy reference mass, and duplicate it, usingsimple pan scales.I frankly fail to see what problem youre solving, ifindeed youre solving one at all.I note that your notion of generating a known force witha given volume of water is not practical, as g varies by0.4% worldwide. In fact, g varies throughout the day,as both Sun and Moon pull at us; sensitive instrumentscan now measure this variation. (These pulls have longbeen known to create the tides.)[.sigsnip]-- #191, ewill3@earthlink.netIts still legal to go The answer to the question How long must physics put up w/f=ma? is:> As long a F = ma describes realityThe reality is that the total force F = the effective [net] force minus thefrictional - or other resisting force [f = F-uw]; which leaves F = wa/g Pollard-rhoIn Maple, the ïifactor command has the option of using thePollard rho method. The help window contains the comment:The pollard base method accepts an additional optional integer: ifactor(n,pollard,k), which increases the efficiency of the method when one of the factors is of the form k*m+1.Im wondering what the change in algorithm behind the increasein Pollard-rho> In Maple, the ïifactor command has the option of using the> Pollard rho method. The help window contains the comment:The pollard base method accepts an additional optional integer: > ifactor(n,pollard,k), which increases the efficiency of the method > when one of the factors is of the form k*m+1.Im wondering what the change in algorithm behind the increase> in efficiency is. The iterative function becomes x -> x^k+cThis forces differences in terms to have the form (y^k+c) - (x^k+c) = y^k-x^k.This form, being cyclotomic, has many factors, in particular those of the form m*d+1 where d|k, including d=k.Riesels Prime Numbers, and Computer MEthods for FactorisationandCrandall & Pomerances Prime Numbers, a Computational Perspectiveboth have descriptions of this technique.Phil-- Unpatched IE vulnerability: Security zone transferDescription: Automatically opening IE + Executing attachmentsPublished: March 22nd 2002Reference: infinity normSuppose a N by N matrix A is irreducible, symmetric and diagonal dominant(not strictly).Let || . || be any induced norm.It is known that if there exist T such that||Ax|| > T ||x||.Then|| inv(A) || < inv(T).For example when the norm is L2 norm, thenT = lambda_m, the smallest eigenvalue of A.Now my question is:What is the best T when || . || is an infinity norm?I just find a trvial value of T, i.e.T = lambda_m / sqrt(N).Remarks:When A = tridiag(-1,2,-1), thenT 16th 1984Im sure this bollocks means something to someone...Still, heres some more contrived crap. Menai Bridge is a village on the island of Anglesey, and shares its name with the suspension bridge to the mainland. Here I prove Menai Bridge is Hell.M=Roman for 1000I=Roman for 1.ENA obviously means Enable Negative Arithmetic, or subtract.1000-1=999.A bridge is something that takes you from one place to another. Thinking abstractly, it can take you from one state to another. One possible change of state is rotating pi radians, and this is indeed further confirmed as the intended translation by the design of the bridge which with its two legs looks just like a pi symbol.999 rotated pi-rad is 666 - the number of the Beast.Therefore Menai Bridge is Hell.Dave.> 1984Sir:While it is obviously impossible to question the quality of yourmathematical skills, I would like to point out one single §aw in yourreasoning. Admittedly, Menai Bridge does yield the Dreaded Number of theBeast, when subjected to your brilliant mathematical operations.Nevertheless, this horrific Number is not the number of hell. It is thenumber of the Beast, which I understand to represent Satan.Your faithful servant,I remain,-Baruch> Im sure this bollocks means something to someone... Still, heres some more contrived crap. Menai Bridge is a village on> the island of Anglesey, and shares its name with the suspension bridge> to the mainland. Here I prove Menai Bridge is Hell. M=Roman for 1000> I=Roman for 1. ENA obviously means Enable Negative Arithmetic, or subtract.> 1000-1=999. A bridge is something that takes you from one place to another.> Thinking abstractly, it can take you from one state to another. One> possible change of state is rotating pi radians, and this is indeed> further confirmed as the intended translation by the design of the> bridge which with its two legs looks just like a pi symbol. 999 rotated pi-rad is 666 - the number of the Beast. Therefore Menai f(x)=1+x+3/2*x^2+8/3*x^3+125/24*x^4+54/5*x^5+16807/720*x^6+... coefficient> of x^n being (n+1)^(n-1)/n!. Let y=ln(1+x) so that f(y)=g(x). How does one> prove that (1+x)^g(x)=g(x)? Any ideas?See answer in sci.math.research under thread Hyperexponetial.-- Ioannishttp://users.forthnet.gr/ath/jgal/_____________________ ______________________Eventually, _everything_ is f(x)=1+x+3/2*x^2+8/3*x^3+125/24*x^4+54/5*x^5+16807/720*x^6+... coefficientof x^n being (n+1)^(n-1)/n!. Let y=ln(1+x) so that f(y)=g(x). How does oneprove that (1+x)^g(x)=g(x)? Any quadratic system:18=a(5)^2+b(5)+c69=a(12)^2+b(12)+c74=a(14)^2+b(14)+c Im not sure how to go about solving this. In class, weve only solvedsystems like this when on of the equations had x=0, so it rather easy tosolve for c. Can anybody point me in the Quadratic System Helpfrom slick_shoes@punkass.com:>18=a(5)^2+b(5)+c>69=a(12)^2+b(12)+c> 74=a(14)^2+b(14)+c Im not sure how to go about solving this. In class, weve only solved>systems like this when on of the equations had x=0, so it rather easy to>solve for c. Can anybody point me in the right direction on this studying some of this now, only having done so a little bit many yearsago. What Math course does your presented problem occur in? Anyway, without using matrices for your system, your variables are now a, b, and c. Finish multiplying to obtain the coefficients. You have threevariables to find, and three equations. You could solve for a,b,c bysuccessive substitutions and resubstitutions; or you could try adding orsubtracting multiples of an equation to another equation for elimination. Matrix operations can be easier if you understand following quadratic system:18=a(5)^2+b(5)+c> 69=a(12)^2+b(12)+c> 74=a(14)^2+b(14)+c> >What is this ... a(5)^2+b(5)+c ??a times the square of 5 plus b times 5 plus c ??If so, it is a system of linear equations for a,b,c, not you know of any examples of proofs which take infinitely> much paper to write? Theres no problem using software to prove> things involving _huge_ axioms of infinity... (because actually the> axioms are _finite_, not huge - theyre talking about huge sets.)If all we know is that a given predicate is recursive and it is trueof all values in our universal set, then to prove (all X)P(X) we cantalk about proving P(X) for all values of X if infinite proofs areallowed.But that only proves that (all X)P(X) is true, not that its provable. Or rather, (all X)|-P(X). So Im not so sure you dont need newrules of inference to get anything new from infinite proofs. It wouldbe nice if you didnt. [Note that under w-consistency, this imples~|-~(all X)P(X).]Charlie VolkstorfCambridge, MAhttp://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ Re: Question on Hilbert & Godel> ??? Do you know of any examples of proofs which take infinitely> much paper to write? Theres no problem using software to prove> things involving _huge_ axioms of infinity... (because actually the> axioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs. For example F.P.Ramsey noted that while the multiplicative axiom is not provable in the system of PM, it would be provable if we replaced the notion of propositional function in PM with a notion of function in extension simpliciter and allowed infinitely long proofs. Brouwer also thought that the finiteness of written proofs is simply a defect of written proofs, not a fact that shows that proofs (in intuition) must be finite. Georg Kreisel and some other platonistically oriented logicians have also toyed with the idea that humans might have the power to grasp infinite proofs (which obviously cant be written down).-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Tractatus Godel??? Do you know of any examples of proofs which take infinitely> much paper to write? Theres no problem using software to prove> things involving _huge_ axioms of infinity... (because actually the> axioms are _finite_, not huge - theyre talking about huge sets.)I cant see David Ullrichs op, so Im piggy backing. I didnt meaninvolving _huge_ axioms of infinity I meant involving axioms of hugeinfinities (i.e. much bigger than omega which ïthe usual axiom ofinfinity posits).-- know of any examples of proofs which take infinitely> much paper to write? Theres no problem using software to prove> things involving _huge_ axioms of infinity... (because actually the> axioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs.If a proof can be of infinite length (it is a mapping from N) thenwill your set of all possible proofs continue to be recursivelyenumerable? Otherwise, you are talking about a whole new game fromcomputers, logic, etc.Not that I think that the set of proofs must be r.e. In my ownaxiomitization of computer science, the universal set need not be r.e.[which is simply axiom TRUE(x)] - I call it unimplementableprogramming languages. Computer scientists assume that the universalset of values that a program can input or output is r.e., but we canconsider systems of real numbers.This is more general than Turing Machines (programs) or course, andthis assumption of the universal set being r.e. is left unstated insuch theorems as the fact that a recursive set must be r.e. That istrue only given that assumption [axiom TRUE(x)].Also note that this greatly simplifies formalizing Peanos Axioms(Peano Arithmetic) from 5 complex wffs to the single wff TRUE(x). Actually, TRUE(x) is one of only 3 axioms needed to formalize theTheory of Computation, as I describe below.1. TRUE(x)2. YIT(I,J,K)3. -~YES(x,x)Charlie VolkstorfCambridge, MAhttp://www.mathpreprints.com/math/Preprint/CharlieVolkstorf/ greatly simplifies formalizing Peanos Axioms>(Peano Arithmetic) from 5 complex wffs to the single wff TRUE(x). >Actually, TRUE(x) is one of only 3 axioms needed to formalize the>Theory of Computation, as I describe below.1. TRUE(x)>2. YIT(I,J,K)>3. -~YES(x,x)Say what? How about giving an actual first-order axiomatization of yourTheory. What you appear to have here are 3 constants (I, J, K), 2functions (~, YES), and 3 predicates (TRUE, YIT, -), all unrealated. Sothere must be a lot of background machinery in addition to your only 3axioms.-- ---------------------------| BBB b barbara minus knox at iname stop com| B B aa rrr b || BBB a a r bbb | | B B a a r b b | | Question on Hilbert & GodelX-DMCA-Notifications: http://www.giganews.com/info/dmca.html ??? Do you know of any examples of proofs which take infinitely much paper to write? Theres no problem using software to prove things involving _huge_ axioms of infinity... (because actually the axioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs. For >example F.P.Ramsey noted that while the multiplicative axiom is not >provable in the system of PM, it would be provable if we replaced the >notion of propositional function in PM with a notion of function in >extension simpliciter and allowed infinitely long proofs. Brouwer also >thought that the finiteness of written proofs is simply a defect of >written proofs, not a fact that shows that proofs (in intuition) must be >finite. Georg Kreisel and some other platonistically oriented logicians >have also toyed with the idea that humans might have the power to grasp >infinite proofs (which obviously cant be written down).I was aware that people did talk about infinitely long proofssometimes.My impression was that the things they _proved_ about theseinfinitely long proofs were proved using proofs of finite length,so those infinitely long proofs still have no bearing on thepossibility of checking things by machine, any more thantheorems about infinite sets cannot be checked by machine.The idea that people have toyed with the idea that humanscould grasp proofs that cannot be written down on finitelymuch paper is new to me. But surely toyed with is theright word here? I mean if a proof cannot be written downthen its hard to see how one person can communicatea precise specification of the proof to another Hilbert & Godel>??? Do you know of any examples of proofs which take infinitely>much paper to write? Theres no problem using software to prove>things involving _huge_ axioms of infinity... (because actually the>axioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs. For example F.P.Ramsey noted that while the multiplicative axiom is not provable in the system of PM, it would be provable if we replaced the notion of propositional function in PM with a notion of function in extension simpliciter and allowed infinitely long proofs. Brouwer also thought that the finiteness of written proofs is simply a defect of written proofs, not a fact that shows that proofs (in intuition) must be finite. Georg Kreisel and some other platonistically oriented logicians have also toyed with the idea that humans might have the power to grasp infinite proofs (which obviously cant be written down).> I was aware that people did talk about infinitely long proofs> sometimes.My impression was that the things they _proved_ about these> infinitely long proofs were proved using proofs of finite length,> so those infinitely long proofs still have no bearing on the> possibility of checking things by machine, any more than> theorems about infinite sets cannot be checked by machine.This is what happens normally in proof theory when you (for various techincal reasons) allow for infinitely long proofs. Obviously if youre going to communicate them somehow using normally accepted means, you have to be able to squeeze the information into a finite space by means of some mathematical or formal device.> The idea that people have toyed with the idea that humans> could grasp proofs that cannot be written down on finitely> much paper is new to me. But surely toyed with is the> right word here? I mean if a proof cannot be written down> then its hard to see how one person can communicate> a precise specification of the proof to another person...Indeed. But communicability was never the aim of Brouwers intuitionistic mathematics, since according to him any communication necessarily distorts the intuitive constructions, especially formal sorts of comunication.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig Wittgenstein, Hilbert & GodelX-DMCA-Notifications: http://www.giganews.com/info/dmca.html> >??? Do you know of any examples of proofs which take infinitelymuch paper to write? Theres no problem using software to provethings involving _huge_ axioms of infinity... (because actually theaxioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs. For >example F.P.Ramsey noted that while the multiplicative axiom is not >provable in the system of PM, it would be provable if we replaced the >notion of propositional function in PM with a notion of function in >extension simpliciter and allowed infinitely long proofs. Brouwer also >thought that the finiteness of written proofs is simply a defect of >written proofs, not a fact that shows that proofs (in intuition) must be >finite. Georg Kreisel and some other platonistically oriented logicians >have also toyed with the idea that humans might have the power to grasp >infinite proofs (which obviously cant be written down). I was aware that people did talk about infinitely long proofs sometimes. My impression was that the things they _proved_ about these infinitely long proofs were proved using proofs of finite length, so those infinitely long proofs still have no bearing on the possibility of checking things by machine, any more than theorems about infinite sets cannot be checked by machine.This is what happens normally in proof theory when you (for various >techincal reasons) allow for infinitely long proofs. Obviously if youre >going to communicate them somehow using normally accepted means, you >have to be able to squeeze the information into a finite space by means >of some mathematical or formal device.> The idea that people have toyed with the idea that humans could grasp proofs that cannot be written down on finitely much paper is new to me. But surely toyed with is the right word here? I mean if a proof cannot be written down then its hard to see how one person can communicate a precise specification of the proof to another person...Indeed. But communicability was never the aim of Brouwers >intuitionistic mathematics, since according to him any communication >necessarily distorts the intuitive constructions, especially formal >sorts of comunication.Huh. I imagine youre serious, and I imagine youre right about thefacts of the case... The idea that we shouldnt worry about being able to explain our proofs to others strikes me as extremely wacky;much wackier than anything else Ive ever heard aboutintuitionism, for example.I mean, not to mention names, but do I get credit forproving Fermats Last Theorem if I publish a paperannouncing that I see a proof but I just cant explainit without using an _infinitely_ wide margin? Hilbert & Godel??? Do you know of any examples of proofs which take infinitely>much paper to write? Theres no problem using software to prove>things involving _huge_ axioms of infinity... (because actually the>axioms are _finite_, not huge - theyre talking about huge sets.)Some logicians have seriously considered infinitely long proofs. For example F.P.Ramsey noted that while the multiplicative axiom is not provable in the system of PM, it would be provable if we replaced the notion of propositional function in PM with a notion of function in extension simpliciter and allowed infinitely long proofs. Brouwer also thought that the finiteness of written proofs is simply a defect of written proofs, not a fact that shows that proofs (in intuition) must be finite. Georg Kreisel and some other platonistically oriented logicians have also toyed with the idea that humans might have the power to grasp infinite proofs (which obviously cant be written down).>I was aware that people did talk about infinitely long proofs>sometimes.My impression was that the things they _proved_ about these>infinitely long proofs were proved using proofs of finite length,>so those infinitely long proofs still have no bearing on the>possibility of checking things by machine, any more than>theorems about infinite sets cannot be checked by machine.This is what happens normally in proof theory when you (for various techincal reasons) allow for infinitely long proofs. Obviously if youre going to communicate them somehow using normally accepted means, you have to be able to squeeze the information into a finite space by means of some mathematical or formal device.>The idea that people have toyed with the idea that humans>could grasp proofs that cannot be written down on finitely>much paper is new to me. But surely toyed with is the>right word here? I mean if a proof cannot be written down>then its hard to see how one person can communicate>a precise specification of the proof to another person...Indeed. But communicability was never the aim of Brouwers intuitionistic mathematics, since according to him any communication necessarily distorts the intuitive constructions, especially formal sorts of comunication.> Huh. I imagine youre serious, and I imagine youre right about the> facts of the case... The idea that we shouldnt worry about being > able to explain our proofs to others strikes me as extremely wacky;> much wackier than anything else Ive ever heard about> intuitionism, for example.Well, Brouwer *was* much wackier than necessary for an intuitionist :)> I mean, not to mention names, but do I get credit for> proving Fermats Last Theorem if I publish a paper> announcing that I see a proof but I just cant explain> it without using an _infinitely_ wide margin? Weird.The situation is not quite this absurd. The idea is that humans have some sort of transcendental intuitions, which cant necessarily be effectively communicated at all, for example intuition about time, on which the natural numbers and the induction axiom are based. When one has ïintuited an infinite proof, which cant be communicated to someone lacking this sort of intuition, one can nevertheless use language to sort of nudge other people with necessary intuitions into ïintuiting the proof construction. So, the example would be something like this: you have effected a construction in your intuition which proves Fermats Last Theorem, and then - to convince others that you actually have done this - you give all sorts of hints and pointers, which allow others equipped with the same sort of intuitions you have to come to effect the same construction. No finite formalism necessarily captures your contstruction. This is somewhat a wacky notion, I grant you.-- Aatu Koskensilta (aatu.koskensilta@xortec.fi)Wovon man nicht sprechen kann, daruber muss man schweigen - Ludwig was never the aim of Brouwers>intuitionistic mathematics, since according to him any communication>necessarily distorts the intuitive constructions, especially formal>sorts of comunication. Huh. I imagine youre serious, and I imagine youre right about the> facts of the case... The idea that we shouldnt worry about being> able to explain our proofs to others strikes me as extremely wacky;> much wackier than anything else Ive ever heard about> intuitionism, for example. I mean, not to mention names, but do I get credit for> proving Fermats Last Theorem if I publish a paper> announcing that I see a proof but I just cant explain> it without using an _infinitely_ wide margin? Weird.Suppose you have a finite proof, and dont publish it.Would you get credit? No. But what would that say aboutthe validity of your proof?Likewise with infinite proofs; there could be a valid proofthat is infinite; the fact that nobody will ever get credit for it,has nothing to do with validity or non-validity.However, this point of view is quite Platonistic, isnt it?Hardly something one would expect from an on Hilbert & Godel sha1:dPX69REJtZ2A8ZVJwtR8npIhIGY=>Software is fine if youre a strict finitist. Some mathematicians likea set theory with _huge_ axioms of infinity in which to do theircategory theory. ??? Do you know of any examples of proofs which take infinitely> much paper to write? Theres no problem using software to prove> things involving _huge_ axioms of infinity... (because actually the> axioms are _finite_, not huge - theyre talking about huge sets.)Yes, the proofs are finite, but think of the diagrams! (He did saycategory theory after all.)-- Evariste Galois was clearly a passionate man. He tried to kill theking of France for instance... Remember that the French Revoloution*did* happen, so he wasnt really out of his times. -- JSH on Galois(1811 - 1832) prefiguring the French GodelX-DMCA-Notifications: http://www.giganews.com/info/dmca.htmlSoftware is fine if youre a strict finitist. Some mathematicians like>a set theory with _huge_ axioms of infinity in which to do their>category theory. ??? Do you know of any examples of proofs which take infinitely much paper to write? Theres no problem using software to prove things involving _huge_ axioms of infinity... (because actually the axioms are _finite_, not huge - theyre talking about huge sets.)Yes, the proofs are finite, but think of the diagrams! (He did say>category theory after all.)Hmm. Good Hilbert & Godel> What did Hilbert ask and claim concerning Foundations of Mathematics> (sets, predicate calculus), metamathematics, Logic, Incompleteness,> etc? Just a side note: Bourbaki was cited as the best example so far of mathematics> organized into a coherent framework. According to Andre Weil,> Perhaps the most important contribution of Bourbaki was to> carry out a famous proposal made by the great German mathe-> matician David Hilbert in 1900 that mathematics be placed on> a more secure foundation. He noted: Hilbert just said so,> and Bourbaki did it And just how did Bourbaki do as Weil claims? But didnt Hilbert actually claim that a lot more than that is> possible? Didnt he ask for (and claim that it must exist) a decision> procedure to determine if an arbitrary predicate calculus wff is valid More than that, an arbitrary wff of mathematics.Where did he ever claim that there was a decision procedure?Wir Mussen wisen, wir werden wissen. Speech in Konigsberg in 1930,now on his tomb in Gottingen.Wir werden wissen is We will know that might express determinationor prediction. (I have trouble with will in English, never SpacesNNTP-Posting-User: ,;upNQ$t/)!L]^R9Po%swv4 is S contractible to every point in S?(Ed Hook already answered this.)> If a topological space S is strongly contractible to some point> p in S, is S strongly contractible to every point in S?I dont think so. Consider the comb space in R^2: X = [0,1] x {0} union {0} x [0,1] union {1/n} x [0,1], n >= 1.Then X is strongly contractible to the origin (collapse everythingdown to [0,1]x{0}, and then left to (0,0)), but not to the point(0,1): during the course of any possible homotopy, the points (1/n,1)on the teeth of the comb would have to go down to the x-axis and sowould eventually get far away from the point (0,1), so there is nocontinuous homotopy fixing (0,1). (See Sieradskis book _An__Introduction to Topology and Homotopy_, examples 6-7 on page 308 andexample 4 on page 318.)> Whats an example of a space that is> contractible but not strongly contractible?Have you looked in the book _Counterexamples in Topology_? Theremight be answer there.> S is contractible to a when theres some continuous h:Sx[0,1] -> S with> for all x in S, h(x,0) = x, h(x,1) = a,> and strongly contractible when in addition> for all t, h(a,t) = a-- J. H. PalmieriDept of Mathematics, Box 354350 mailto:palmieri@math.washington.eduUniversity of Washington http://www.math.washington.edu/~palmieri/Seattle, WA === 98195-4350Subject: Re: Contractible Spaces Re: Contractible SpacesHi Ed> If a topological space S is contractible to some point p in S,> is S contractible to every point in S?> >Yes. Suppose S is contractible to a particular point >s0 in S, with F: S x [0,1] --> S giving a contraction >and s1 is any other point in S. Define G: S x [0,1] --> S by > F(s,2t) if t in [0,1/2] > G(s,t) = . > F(s1,2-2t) if t in [1/2,1] >Then G gives a contraction of S to s1.G(s,1/2) = F(s,1) = s0G(s,1/2) = F(s1,1) = s0G(s,1) = F(s1,0) = s1Ok, F(s1,2-2t) is a path from s0 to s1.Heres another way to visualize the homotopy.Let h be a homotopic contraction of S to a.As S is contractible, its path connected.Let p be a path from a to b. Define homotopy H H(x,r) = h(x,2r) when r in [0,1/2] H(x,r) = p(2r-1) when r in [1/2,1]H contracts S to b. If a topological space S is strongly contractible to some point p in S, is S strongly contractible to every point in S? > Probably not in general, although I dont know offhand > of any relevant example. There are theorems that say > things likeIf S is homogeneous and strongly contracts to a with homopty H(x,t), then S strongly contracts to b.Proof: Let h be auto-homeomorphism of S with h(a) = b. Then h(H(h^-1(x),t)strongly contacts S to b.Tho homogeneous is sufficient, its not necessary asa kite space, ie disk with line segement tail, shows. >If A is a closed subset of X and both A, X are ANRs[*], >then A is a deformation retract of X if and only if >A is a strong deformation retract of X. >Your question corresponds to the case A = {point} -- but its >not clear (to me, anyhow) just how to apply that theorem ... >Theres probably also an obstruction theory that applies >to your question ... >[*] ANR = absolute neighborhood retract -- I believe >the above theorem actually applies to spaces that are >ANRs for the class of compact metric spaces -- X is such >a thing if any imbedding of X as a closed subspace of a >compact metric space is a retract of some open neighborhood >of the image ...So a single point is an ANR. If S is contractible ANR, then S iscontractible to any point, hence strongly contactible to any point.> Whats an example of a space that is> contractible but not strongly contractible? S is contractible to a when theres some continuous h:Sx[0,1] -> S with> for all x in S, h(x,0) = x, h(x,1) = a,> and strongly contractible when in addition> for all t, h(a,t) = interested to understand when to use differentials equations ? Have you some popularized examples to explain me...For what are interested to understand when to use differentials equations ?> Have you some popularized examples to explain me...For what are they useful ?>1/ to compare tangents between themselves> A.>One problem that can be solved by solving an ODE is the falling parachutist problem:If a parachutist falls from 10,000 meters with initial velocity 0, then two forces act on him:- gravity- air dragIn the period before he opens his parachute, say the wind drag force is: k*v Newtons upwards, v being velocity in meters/second.Say gravity is downwards and equal to 1000 Newtons.So the total force is 1000 - k*v (Newtons downward).Assuming g = 10 m/sec/sec, the parachutist has a mass of 100 kg (parachute included).So the acceleration in the downwards direction is: 10 - (k/100)*v [ m/sec/sec ] .Since acceleration = dv/dt, then dv/dt = 10 - (k/100)*v ( m/s/s)Say k/100 = 1/5 = 0.2 . Then,dv/dt = 10 - v/5 , which is an ODE.Also, v at the time we can call 0 (start) is zero.How long will it take from start until the parachutist is at a height of 5000 meters?1 minute? 30 seconds? By solving the ODE, you get am a total newbie... I am interested to understand when to use differentials equations ?> Have you some popularized examples to explain me... For what are they useful ?> 1/ to compare tangents between themselves ?Differential equations turn up when we know some relationshipbetween a quantity and its derivative(s). They turn up almosteverywhere in science. Chemistry, biology, economics, sociology,etc.... and specially in physics.Heres two examples from physics:Maybe you know that acceleration is the time-derivative of thevelocity, and that velocity is the time-derivative of the position.Suppose for example that you have a spring, and that you havenoticed (with a dynamometer) that the force the spring is exertingis proportional to the elongation.Since you know Newtons law F = m*a,and since you know that F = -k*x,and that the a in Newtons law is the acceleration, i.o.w. that a = d^2x/dt^2,being the second time-derivative of the elongation x, you havefound that the elongation x of the spring satisfies the followingdifferential equation: -k * x = m * d^2x/dt^2.or written in a perhaps more familiar form x + k/m x = 0where x is the second time-derivative of x.This can be easily solved and it gives the elongation of thespring as a function of the time: x(t) = A*sin( Sqrt(k/m)*t + B )where A and B are some numbers. Knowing the initial conditionsof the problem can help find the values of A and B.In fact, for an even simpler example, suppose that there is noforce working on an object. Then Newtons first law reduces to a = 0,which amounts to x = 0which is a very simple differential equation all by itself.It is easily solved: x(t) = x0 + v*twhich says that the object travels in a straight line with aconstant velocity. If the number v is 0, then the object doesnot move at all and it is sitting still at location x0. Again, thenumbers x0 and v are again numbers that can be found ifyou know the initital algorithmHello!is anyone aware of a good free implementation for the graph isomorphismproblem for graphs of bounded genus and degree? Currently, a graphisomorphism algorithm for planar FormulaStirlings Formula_. His proof hinges on showing that phi(t) = sin(pi*t)/pi,where phi(t) is in product form. That is, phi(t) = t * Product(k = 1, 2, 3, ...; (1 - t*t/(k*k))).The last step in his proof of this stymies me. Feller first shows that: phi(2x) = 2phi(x) * phi(x + 1/2) / phi(1/2)In other words, he proves the double-angle formula for phi. So far, sogood.He then sets f(x) = log(pi * phi(t) / sin(pi*t)),asserts that f(2x) = (f(x) + f(x + 1/2)) / 2,and loses me. Can anyone explain? He says that it follows from thedouble-angle formula for sin(pi*x)/pi and the above formula for phi(2x).(I know there are other proofs that phi(t) = sin(pi*t)/pi. My questionis about Fellers.)not keep track of my steps. Now I cant find it again. Does anyonehave a URL? Please note, Im not talking about Fellers book, just my employers.James M. Stern stern@itgssi.com (213) 270-7955ITG Software Solutions, Inc.Culver City, CA phi(x + 1/2) / phi(1/2)In other words, he proves the double-angle formula for phi. So far, so> good.He then sets f(x) = log(pi * phi(t) / sin(pi*t)),asserts that f(2x) = (f(x) + f(x + 1/2)) / 2,and loses me. Can anyone explain? He says that it follows from the> double-angle formula for sin(pi*x)/pi and the above formula for phi(2x).... Presumably your definition of f should have t instead of x. I cant get Fellers formula either, but instead come up withf(2x) = f(x) + f(x + 1/2) - f(1/2).Would the rest of his proof follow from that, or is there a more seriousdifficulty? Ken Formula...> not keep track of my steps. Now I cant find it again. Does anyone> have a URL? Please note, Im not talking about Fellers book, just the1. M. I. Aissen, Some remarks on Stirling formula, A. M. M. 61 (1954) , 687-691. 2. H. Robbins, A remark on Stirlings formula, A. M. M. 62 (1955), 26-29 3. W. Feller, A direct proof of stirlings formula, A. M. M 74 (1967), 1223-1225. 4. R. A. Khan, A probabilistic proof of Stirlings formula, A. M. M. 81 (1974), 366-369. 5. C. S. Wong, A note on the central limit theorem, A. M. M. 84 (1997), 472. 6. C. R. Blyth and P. K. Pathak, A note on eazy proofs of Stirlings formula, A. M. M. 93 (1986), 376-379. 7. P. Diaconis and D. Freedman, An elementary proof of Stirlings formula, A. M. M. 93(1986), 123-125. 8. J. M. Patin, A very short proof of Stirlings formula, A. M. M. 96 (1989), 41-42. 9. G. Marsaglia and J. C. W. Marsaglia, A new derivation of Stirling Approximation to n!, A. M. M. 97 (1990), numbers are three dimensional in the traditional RxRxR sense. This comes simply as a result of general summation, graphical interpretation involves a tatrahedral set of poles coming out of the origin. I agree that the four-signed domain is probably more new and interesting. There is a natural product and so the ability to perform the mandelbrot mapping on it exists. I am not aware of what this products equivalent is for the cartesian space. It may be new, and since it maps to the complex numbers for three-signed we could consider it to be of value. The trouble that I am having is in the transformation back to cartesian dimensions. This involves decomposing a tetrahedron into its three-dimensional components. It is just triginometry but I am struggling with it. Perhaps you will be able to do it. I would put the # pole right on the x-axis. Then the - pole in the x-y plane, then the last two wherever they fall. There seem to be some arbitrary choices in this transform that didnt exist back in three-signed. Anyhow, I think we should look at it graphically as a tetrahedral since that is the symmetrical mapping of four poles in RxRxR. There are a few maps that could be gotten working without a transform and just analyzing for example the minus-pound (-,#) plane. There are six of these graphs. I am coding up general sign math now. Once I have that I should be able to graph these and higher signs too. I have done the mandelbrot mapping for Y-space(three-signed) and it looks a whole lot like the standard complex one, as it should since the math works out to be equivalent, though I did not know it until after I did the graph. I have been more focused on physics thoughts and trying to stretch this math towards a source of stability. I am very happy to talk to someone who appreciates and understands this construct. Im still not sure that someone else hasnt done this math already. There is a guy really close to it with terplex numbers but I dont think its the same. He seems to be using reals as their basis which confuses me.Ive done some work on the four-signed math this weekend. It looks like > the key to making it work is carefully defining a reduced form. Under > certain definitions of reduced from, n-signed math exists in R^(n-1), > Under other definitions, n-signed math is isomorphic to C. I need to > finish some more details, then Ill let you know what I come up with. I > think that thinking in terms of n-tuples will make things easier when > dealing with general n-signed math. Otherwise you may need to switch to > subscripted signs.> Using my notation of (-,+,*,#), there are two possible reductions for (a,b,c,d).The tetrahedral interpretation uses: (a,b,c,d) = (a-m,b-m,c-m,d-m) where m = min(a,b,c,d).The planar interpretation uses: (a,b,c,d) = (a-m,b-n,c-m,d-n) wherem = min(a,c), n = min(b,c)Both reductions are consistent with multiplication and addition, however they behave quite differently.The planar is isomorphic to the complex plane by setting a=i,b=-1,c=-i,d=1 and preservers multiplication and addition.The tetrahedral maps to R^3, with no obvious interpretation of multiplication (to me), and addition corresponds to vector addition of (i,j,k) points. I didnt work out the mapping in detail since you posted it elsewhere.Personally, I think the tetrahedral reduction is more interesting, though I dont know what applications there may be.-- Will Twentymanemail: wtwentyman at copper dot for the tetrahedral angle.Its on http://mathforum.org/library/drmath/view/55023.html which Ifound simply by searching for tetrahedron angle on corners as: 180 - 2*arctan(sqrt(1/2)).which comes out to: 109.47122... degrees.This is the same value that I get plugging: 180 - arccos( 1/3 ) into a calculator.> space. Any point in RxRxR can be uniquely defined in four-signed math.> Since I dont have the math I cant say to have proven this. I can see> it though. Making this assumption and putting the #-pole in the i> direction (as in i,j,k) we get the following partial transform for a> four-signed value x:> a = n(x) - ( 1/3 )( m(x) + p(x) + s(x) )> where a is the i component of the three dimensional vector ai + bj +> ck.> n(x) is the # (number) component, m the - (minus), p the + (plus), and> s the * (star).> Now putting the minus pole in he ij plane and going in a left-handed> direction we see that the one third component yields an angle of:> pi - arccos( 1 / 3 )> where arccos is the inverse cosine.> This should be the angle from the center to any two corners of the> tetrahedron.> Please note that I am not proving this. Id like to find that angle in> a book somewhere or sharpen up my triginometry skills so that I could> verify this.> If this is true then there should be no problem with a clean RxRxR> transformation for four-signed math. The same concept should work> upward beyond Re: Fundamental Reason for High Achievements of Jews can be seen from a few excerpts from his posts below, Bob Kolker should remove the logs from his eyes. Bull. I have identified the enemy. The enemy is Islam. The dreadful > happenings of 9/11 prove it. The way to defend against the evil your > enemies do is to kill your enemies. The principle is clear. Help and protect your friends. Kill your > enemies. It is as simple as that. And we are ALL in this fight. If the > U.S. government. would give me a nuke, I would gladly blow it and myself > up at at the next Haj in Meccah. I would love taking a hundred thousand > of the bastards with me.Bob, if you are proud of your disgusting thoughts[1], thats yourbloody business. But, please, take that out of sci.math and letus concentrate on important issues, like when JSH will get that Abelprize and who he shares it with.-- Jesse HughesIts easy folks. Just talk about my approach to your favoritemathematician. If they cant be interested in it, theyvedemonstrated a lack of mathematical skill. -- James === of Jews>Subject: Re: Fundamental Reason for High Achievements of Jews>Message-id: Fundamental Reason for High Achievements of JewsMessage-id: Most historians believe that Jews avoid pork,> :> : because the ancient Jews associated pigs with leprosy,> :> : and pigs and people with leprosy were unclean.> :> > :> Name one historian who believes that, and give a citation to the> :> place where he says it.As I recall, this was in Tacitus Histories >which was written in the first century A.D.> > If thats the best you can do, I think that we can safely ignore>yourtheory.Its soooo easy Richard. Someone in your campshould have the information at his fingertips.GOOGLE: tacitus histories ~13,600 hits tacitus histories jews ~3,950 hits tacitus histories jews leprosy ~154 hits This is all irrelevant. Quoting from Tom Potter: 2. This is the computer age. Almost everyone has access to the first hand historical accounts, and can do wild card searches on the source material. It is STUPID to provide detailed cites, as these focus ONLY on the POINTS trying to be emphasized by the writer. It also STUPID to use second hand, accounts which have a racial, religious, national, or personal spin on them, rather than using the FIRST HAND historical accounts. Anything written by Tacitus would NOT be a FIRST HAND account, so only a STUPID person would allow himself to be brainwashed by Tacitus racial, religious, national, personal spin on history. It is interesting to see that Mensanator, like many people, has been brainwashed to think that Tacitus, who was one of the most unbiased, rational, and correct historians, put a racial, religious spin on history. Liar. YOU are the one who says that second hand accounts put a spin on history. Or are you going to deny that that was your quote? I wouldnt try it, I located it via a Google search, others can too. This attitude obviously has its roots in conditioning, as the works of Tacitus are far more rational and correct, than the bible, the Greek and Roman mythologies, etc. But they are not a FIRST HAND account. So your brainwashed opinion of their validity is irrelevant. Or are you now admitting that the quotation about second hand accounts was stupid. You cant have it both ways, so which is it? In other words, people who have been brainwashed to a point of view, have a great difficulty in accepting data that con§icts with their conditioning. In other words, Tom Potter is too stupid to realize that hes just been hoisted by his own petard.>It is interesting to see that Mensanator >does not comprehend that Herodutus, Tacitus, Josephus, >and the ancient Greek and Roman Historians, I comprehend perfectly well that Herodutus, et al, when writingabout the eras in which they lived, are giving FIRST HANDaccounts. >are the closest FIRST HAND accounts of ancient history,Everything is ancient history to _us_. But some things were ancient history to the historians. For you to fail to make thedistinction between what was contemporary versus historicto Tacitus is a fallacy on your part. >and that works of modern historians use these works >as their starting reference points.Now begins the hand-waving part of the argument. This has nobearing on the point under discussion.If you want to know the facts about ancient history, When they arent FIRST HAND accounts, there are no factsabout ancient history, only opinions.>you start with the works of the ancient historians, >so you can detect the spin put on >history by modern historians who impose their >present day agendas on the works.More hand-waving. The fact that modern historians put their ownspin on history has no bearing on whether ancient historians puttheir own spin on what was ancient history to them.writers from different religions and nations, >put a personal spin on the history, >and a lot of information is lost >as it is played through fallible, tuned filters.Hand-waving repeated.No doubt, some original historians put a personal >spin on their works, Aha!>but this does not seem to >be the case with Tacitus, Herodutus, Thucydides, Xenophon,>and Polybius, Seem? And what do you base this on Mr. Potter? Do you haveFIRST HAND accounts that predate Tacitus? Unless you have FIRST HAND accounts of the Jews having leprosy, then itseems to be just the opposite; Tacitus claim _was_ apersonal spin on history.>although one can detect a little Delphi Oracle >spin in Herodutus works.QED-->Tom has come in!!!!> I must reiterate: I dont recognize that your work is indeed correct. (Maybemy IQ is not high enough...) My position is that its not absurd to think itcould be correct rather than to assume, arrogantly, that if I dont understandit, it must be wrong. Im sickened by this carnival of dogs driven wild by rawmeat, and Im outraged that James is being treated like raw meat. James Harrisis a child of God. We are all children of God. Where did it all go so wrong?I am supporting James because if he is correct (and Im entirely unable to assigna number to the probability of that), it will have a profound impact on society.For the better, despite short-term chaos.Hmmm...you seem to have backtracked Jim Ferry, and your post can be> interpreted to mean that you are just humoring me.Backtracked? After the initial jolt I went a little overboard, calling you themessiah, etc. Ive certainly backed off that stance. Please consider howdizzying it is to be shaken out of the conventional mindset.> Well then, how about the core error problem?Cant you understand how independent terms are independent?I just dont get it! How many times do I have to tell you! Your mathematicalwritings make no sense to me! Are you going to be calling me a retard next?Oh, excuuuuse me for not being as smart as you, James! What happens to me inthe New World Order? Do you send me to the slaughterhouse with all the othermeat animals?Sure, Ive got some trinkets of intellect: M.I.T. degree, some Putnam medals,Ph.D. from Brown, joined Mega once, mathematician, rocket scientist, etc. Butthats a hell of long way from the serious hardware that youre interested in:Abel Prize, Fields Medal, Nobel Prize, Clay Prize, etc. And even the peoplewho have such hardware arent necessarily able to follow in your bold footsteps.So why are you on *my* case?I never claimed to understand your work. Maybe I had a §eeting §ash ofintuition that it was correct, but that was all.I was offering a different kind of help. Something to complement and completethe intense laser of your rational thought. Something to help bring your theoriesinto the human realm. You dont believe in the efficacy of such things, I guess,or you just want to sear your way through with that laser of yours, so JSH your ship has come in!!!!> >And I noticed that you said that Jim Ferry is on the team! How can I get>on the team if you wont help me when I am struggling? By the way, is>anybody else on the team? I think that all the team-members should have>well-defined roles for the up-coming battle.Sarcasm. Sigh.> So far the team as I call it are people who recognize that my workis indeed correct, and so far seem to only be very high IQ people.I must reiterate: I dont recognize that your work is indeed correct. (Maybe> my IQ is not high enough...) My position is that its not absurd to think it> could be correct rather than to assume, arrogantly, that if I dont understand> it, it must be wrong. Im sickened by this carnival of dogs driven wild by raw> meat, and Im outraged that James is being treated like raw meat. James Harris> is a child of God. We are all children of God. Where did it all go so wrong?I am supporting James because if he is correct (and Im entirely unable to assign> a number to the probability of that), it will have a profound impact on society.> For the better, despite short-term chaos.Hmmm...you seem to have backtracked Jim Ferry, and your post can beinterpreted to mean that you are just humoring me.Well then, how about the core error problem?Cant you understand how independent terms are independent?James revisitedstart at the upper right corner of table. If the the element of table beinglooked at is bigger than the key being searched, move left. if element =key, you have found it. Otherwise move down. Continue until you reachbottom left corner. if after this search you have not found it, it is not inthe table. With a table of m rows and n columns, maximum path length you cantrace is m + n, giving a O(m+n) wostcase algorithm> BASIS> -----> A two dimensional table have integers distributed this way (in this> examle, only odd numbers): 3,11,13,21,27,...> 7,13,17,29,35,...> 9,15,19,33,41,... Assume that this table is huge, so a fast algorithm is needed -> especially since many tables of this form must be searched many times. So the values always increase when going from left to right, top to> bottom, no matter where you start. That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1] PROBLEM> -------> Given such a table, t, can you locate a given integer N in polynomial> time - or - can you determine that N is not in the table in polynomial> time (or better, as in constant :-)? If yes, what does that algorithm look like? STATUS> ------> I tried the following approach: start in the middle, t[i_max / 2,> j_max /2] and see if its less than N. If so, the top/left quadrant of> the table can be eliminated because all values in that quadrant must> be smaller than N. Apply algorithm recursively to the remaining> quadrants. Of course, this is not a fast algorithm at all...> Any insight would be highly revisited> BASIS> -----> A two dimensional table have integers distributed this way (in this> examle, only odd numbers): 3,11,13,21,27,...> 7,13,17,29,35,...> 9,15,19,33,41,... Assume that this table is huge, so a fast algorithm is needed -> especially since many tables of this form must be searched many times. So the values always increase when going from left to right, top to> bottom, no matter where you start. That is, t[i,j] < t[i+1,j] and t[i,j] < t[i,j+1] PROBLEM> -------> Given such a table, t, can you locate a given integer N in polynomial> time - or - can you determine that N is not in the table in polynomial> time (or better, as in constant :-)? If yes, what does that algorithm look like? STATUS> ------> I tried the following approach: start in the middle, t[i_max / 2,> j_max /2] and see if its less than N. If so, the top/left quadrant of> the table can be eliminated because all values in that quadrant must> be smaller than N. Apply algorithm recursively to the remaining> quadrants. Of course, this is not a fast algorithm at all...> Any insight would be highly appreciated.Your quartering method is not as slow as you seem to think.For reasons similar to the analysis of a binary search of linearly> ordered data, this is about the best you can expect, and is about> O(log(max(R,C))), which is better than polynomial.If you work out the details, his method is O(m(lg m) + n(lg n)) time [assuming n rows, m columns, n>m, the latter in accord with hanthemans problem statement in the ] and is not as good as O(m+n) from following an of f(x)=2abs(x) ------ 1+x^2How would I start this the graph of f(x)=2abs(x)> ------> 1+x^2 How would I start this question?This function is even, i.e. f(-x) = f(x). Then sketch the graph of f(x) =2x/(1 + x^2) for x >= 0. and re§ect G(2,4) be the Grassmanian of 2-dimensional subspaces in R^4.I map G(2,4) -> RP^2 as follows. Given a plane g, choose anorthogonal basis e1, e2. Identifying R^4 with the quaternions inthe obvious way, form the quaternion u = e1 * e2^(-1). Thenu is a square root of -1, well defined up to a sign. The squareroots of -1 are naturally identified with the unit ball S^2 sittingin R^3 sitting in R^4 via (a,b,c) |-> (0,a,b,c), so u gives awell-defined element of S^2/(plus or minus 1) = RP^2.Question 1: Is there a standard name for this map?Question 2: Declare two elements of G(2,4) to be equivalent ifthey sit in the same fiber of the above map. That is, two planes are equivalent if they are both fixed by the same 90 degreerotation of R^4. Is there a standard name for this equivalencerelation?Steven E. all,Please help me with the following problem:Y=A*X*B, where X is the input, Y is the output, A, X, B, Y are NxN matrices;is there a way to absorb A, B into one big matrix C, which leads toY=C*X, which is a more common form for linear limit of measurable functions measurable?> >The question might seem stupid, Subject line certainly sounded like, lets call it an easy> so confused - is he having another one of those> senior moments or what?> >but Im not talking about REAL-VALUED>functions.Let (X,T_X) and (Y,T_Y) be topological spaces, and let B_X and B_Y>denote the Borel fields on them. (= generated by the open sets)It is not reasonable that the pointwise limit of a sequence of>measurable functions f_n : (X,B_X) --> (Y,B_Y) is also measurable. (By>measurable I of course mean that the inverse image of a set in B_Y is>in B_X.) This is because sequences dont mix well with topological>spaces.Im no doubt being stupid, but it seems to me its easy to show.> So whats wrong with this:Say f_n -> f pointwise. Assume that O is an open set in Y. Now> (*) f(x) is in O if and only if (ii) there exists N such that > f_n(x) is in O for all n > N...oops, thats not true. (i) implies (ii) but not conversely, because> f_n(x) could wander off to the boundary of O. I guess this> explains the relevance of the condition below - under that> condition (i) _is_ equivalent to (ii).Right. In general, if U is open thenf^{-1}(U) subset bigcup_{N=1}^infty bigcap_{n=N}^infty f_n^{-1}(U) subset f^{-1}(bar U),because f(x) in U ==> exists N (n ge N implies f_n(x) in U); and,if f_n(x) in U for all n, then lim_n f_n(x) in bar U. For U = cupG_m = cup overline{G_m}, therefore f^{-1}(U) = bigcup_m f^{-1}(G_m) subset bigcup_m bigcup_N bigcap_{nge N} f_n^{-1}(G_m) subset bigcup_m bigcup_N bigcap_{n ge N} f_n^{-1}(overline{G_m}) subset bigcup_m f^{-1}(overline{G_m}) = f^{-1}(U).I like Art§Dodgrs suggestion that the Borel field B_Y should begenerated by the Re: When is the limit of measurable functions measurable?X-DMCA-Notifications: http://www.giganews.com/info/dmca.html>The question might seem stupid, Subject line certainly sounded like, lets call it an easyso confused - is he having another one of thosesenior moments or what?>but Im not talking about REAL-VALUED>functions.Let (X,T_X) and (Y,T_Y) be topological spaces, and let B_X and B_Y>denote the Borel fields on them. (= generated by the open sets)It is not reasonable that the pointwise limit of a sequence of>measurable functions f_n : (X,B_X) --> (Y,B_Y) is also measurable. (By>measurable I of course mean that the inverse image of a set in B_Y is>in B_X.) This is because sequences dont mix well with topological>spaces.Im no doubt being stupid, but it seems to me its easy to show.So whats wrong with this:Say f_n -> f pointwise. Assume that O is an open set in Y. Now(*) f(x) is in O if and only if (ii) there exists N such that f_n(x) is in O for all n > N...oops, thats not true. (i) implies (ii) but not conversely, becausef_n(x) could wander off to the boundary of O. I guess thisexplains the relevance of the condition below - under thatcondition (i) _is_ equivalent to (ii).So its not such a stupid question, and I dont know exactlywhat the answer is. Whats a simple example showing itdoesnt always work?>But there ARE circumstances under which this is true. I remember a>theorem proved by Calderon in one or another class he was teaching>where he proved it was sufficient that T_Y have the property that every>open set U can be written as a countable union of open sets G_n with>the property that the closure of G_n is a subset of G_{n+1}. This>property is satisfied, e.g., by regular T1 second-countable spaces.Does anybody know the most general circumstances under which this>result is true? I doubt that second-countability is necessary; its>TOO global a condition. Probably first countability isnt enough.Alternatively, can anybody characterize the property of open sets U>being writable as such countable unions?Always wondering,>Ron Color TheoremGiven what you know about graphs and vertex coloring, which do youthink would be the easiest task?.1. Prove that a 5-chroma planar graph cannot exist.2. Prove that all planar graphs are 4-colorable.3. Prove that a 5-chroma true: A*X*B=X => A=I and B=I?What else results about A and B can I obtain from A*X*B=X?(this is not a HW daughter has this equation and we dont know how to solve, approach such problems that provides you with aconceptual understanding of how natural numbers, integers, andfractions connect.Theorems are what math is made of. Definitions set the stage for thetheorems. Therefore, after a definition, Ill show you a theorem - aformula (and its proof) - and how to use it on anything like 1/6 - 2:Any natural number or integer x can be defined as a fraction x/1.Using this definition to see everything in terms of fractions, we canquickly and easily get a handle on such situations as 1/6 - 2 via abasic theorem that will last you or your daughter now matter how faryou or she would want to go in math. This is because this followingbasic theorem always works when we are to add (or subtract) fractionsno matter what numbers are used, even real or complex numbers, whichusually dont have least common denominators. I for one like the ideaof having at least one formula that always works in all contexts,rather than always having to use different formulas for differentcontexts.Theorem. a/b + c/d = (ad + bc)/bd. On paper, see this as: a c --- + --- = b d ad + bc--------- bd Proof. (a/b + c/d) = (a/b + c/d)(1) = (a/b + c/d)(bd/bd) = (abd/b +cbd/d)(1/bd) = (ad + bc)(1/bd) = (ad + bc)/(bd). QEDNotice that all we are doing is dividing the sum of the products ofthe different numerators and denominators by the product of thedenominators.For subtraction, just replace the plus sign with a minus sign.For 1/6 - 2, just apply the above definition on 2, replacing it with2/1, and then replace the letters of the alphabet with the numbers.You dont have to write all this out as follows. I just write it outto explicitly show you how to substitute into the formula:1/6 - 2 = 1/6 - 2/1 = (1*1 - 6*2) / (6*1) = (1 - 12) / 6 = -11 / 6.Note: -11 / 6 is also equal to -(11 / 6), which you can if you wantchange into -(1 + 5/6).Try out this formula with the definition above (turning an integer xinto a fraction x/1), even with really big numbers using a calculator.For practicing on the big numbers, write out the formula, then writeit out with the plugged in numbers, and then use the calculator. Withpractice, you wont have to write out so much each time. But alwayswrite out the formula first each time, since thats the real math, thegeneral truth. Neat, huh?Paul-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Number combinations> Can you give me a formula for calculating the no of possible> combinations possible from a quantity of options. To clarify my> requirement here are the the possibilies for three options1> 2> 3> 12> 13> 23> 123Therefore the answer is 7It looks like what you are asking for is the number of non-emptysubsets of an n-element set. The number of subsets of an n-element setis 2^n, which includes the empty set as a subset. Just subtract 1 toget the number of non-empty subsets, 2^n - 1.Your set has 3 elements, so 2^3 - 1 = 7, fitting your requirement.Paul-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Math Education Sequence from 7th to 12thAs a student, I had both algebras before the geometry. I loved it. We wereable to finish the entire geometry book.Later, the school in which I taught had the geometry between the algebras.Perhaps, because the students werebeing exposed to geometry a year earlier and didnt have the added algebracontent, I was never able to cover thematerial at the depth that I had wished.When I asked the Department head about it, she said that they did this wayto exposed the kids to more geometrybefore the SATs and also so that the seniors will have the Algebra II justbefore they headed to college.This explanation never satisfied me.Some math books are getting away from this sequencing and introducting thetopics when it seems most appropriate.Thus geometry is integrated in with the algebra, not taught separate fromit. I think this helps with relevance.However, I must confessed I retired and I have not seen any examples of whatI speak.> What will be the ideal sequence of Math education from 7th to 12th grade?? Is it PreAlgebra, Algebra, Geometry, Algebra II, Trigonometry and then> Calculus??>-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Math Education Sequence from 7th to 12th> When I asked the Department head about it, she said that they did this way> to exposed the kids to more geometry> before the SATs and also so that the seniors will have the Algebra II just> before they headed to college.> This explanation never satisfied me.> Yet another example of having to teach to the test.Here in Colorado, the schools are graded by the CSAP, which tests kids on materials that in many cases they havent even had yet. (I dont remember if this was the actual example, but it was something like testing 8th grade students on geometry)-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: Math Education Sequence from 7th to 12thOne institution that I attended used the sequence, algebra 1, geometry, algebra2; the next institution I attended used the sequence, elementary algebra,intermediate algebra, geometry. These both were good sequences. About theonly thing relevant is that the beginning algebra course came before thegeometry course to enable students to use good algebra skill to study and provethings in geometry. I found the two algebras to be easier than the geometry atboth institutions. These three courses were dedicated to their titles, andonly used a small amount of topic integration.G C-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: algebra bookI would highly recommend both Algebra the easy way and ForgottonAlgebraAlgebra the Easy Way is a very fun and interesting book, as a matterof fact it is writen in the format of a short fantasy novel, thecharacters derive algebra through the book to solve problems thatarize in their land of carmiro( i probably miss-spelled that) it alsohas matricies & determinants, complex numbers, proofs by mathmaticalinduction, and a lot of other neat highschool material.Forgotton Algebra is designed to be read by someone who had done thematerial before but long since fogoten it, however it doesn not havematricies and complex numbers-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: math and musicMusic is the pleasure the soul gains from counting without realizing thatit is counting.Liebnitz, I think.-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: curriculumDo you have a curriculum for Special Ed children k-5?-- submissions: post to k12.ed.math or e-mail to k12math@k12groups.orgprivate e-mail to the k12.ed.math moderator: kem-moderator@k12groups.orgnewsgroup website: http://www.thinkspot.net/k12math/newsgroup charter: by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, sci.math.symbolic, jerry:> u have a 4x4 patern and u have 1-16 # that have to = the same # > for magic square, Loh-shu, or D.9frers magic square.-- #191, ewill3@earthlink.netIts still legal to go 10 11 1213 14 15 16> You need to ask your question in sha1:qQ1+rASHWf3FRCBDULHlQqcBRaE=> u have a 4x4 patern and u 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 You need to ask your question in English.Prhps h wnts mgc 13 14 15 16 > You need to ask your question in English.He needs to ask a question. I am guessing that he wants to make a4x4 magic square.-- Julian V. NobleProfessor Emeritus of ^^^^^^^^^^^^^^^^^^http://galileo.phys.virginia.edu/~jvn/ God is not willing to do everything and thereby take away our free will and that share of glory that rightfully belongs to us. -- parabola>Im 62. When I shoot a basketball, my release point is 7 high. I>want to practice my three point shot, but I dont have a basketball>goal. Instead, I want to place a mark on a wall, which represents the>top of my shots arc.>I read somewhere that an initial trajectory of 45 degrees will yield>the best results. I dont know why that would be. In the absence of air resistance, for > a given initial speed a 45 degree initial angle would go the farthest > distance before coming down to the level of the release point, but > thats hardly relevant. I agree this is more of a physics issue. I dont know that 45 degreeswould be the best initial angle. With a parabola, there is symmetryin the arc. What I am really concerned about is the optimum angle ofapproach to the rim. I would think it best that the ball arrive withthe minimum of velocity (and a little backspin), so that shots that donot fall cleanly through, might rebound off the rim and still fall inthe hole. A more perpindicular approach would allow more shots to fallcleanly through the rim, but the increase in velocity would likelyresult in a miss for any shot that strikes the rim. One might alsoconsider the longer the path to the hole, the more pronouncd anyimperfection in the shot would arc in college is 199.....call it>20. A goal would be 10 high.No, I might call it 20 and 10: is for inches, and ï for feet.> >How far from the wall should I stand, and how high up is the target>mark such that if my shot strikes this mark at the top of my shooting>parabola, the ball would fall into the goal from 20.Why would you want to do it that way?> 1) its hard to judge whether your shot is at the top of its arc> when it hits the wall> 2) you wont have this mark to shoot at in real life.Better to paint a basketball net on the wall at the proper height> and practise shooting from the proper distance.Robert Israel israel@math.ubc.ca> Department of Mathematics http://www.math.ubc.ca/~israel > University of question.> My math professor discovered the §aw in Intels Pentium chip. (I like> to believe he was checking my homework at the time).> Where did I attend college?http://en.wikipedia.org/wiki/Pentium_FDIV_bugProfessor Thomas R. Nicely of Lynchburg College, Virginia.Bug reported found a very interesting site on mathematics. It will be usefulfor high school students. Youll find all theoretical and practicalstuff, and problems solution variants there.Check it deviationsI am studying competitive balance in a sports league. I have a book whichsuggests comparing the standard deviation of winning percentages of all teamsagainst the standard deviation of a hypothetical league, one which assumesthat all teams are of equal strength. For the hypothetical league, the SDwould be .500 / sqrt(n) where n is the number of games played by each team.The amount of competitive balance would be equal to the difference between thevariance in the hypothetical league and the actual league.My question is this: can I use an F-test to see if the difference between theleagues is significant? The reason I ask is that the books authors say theyused a t-test, but that seems needlessly complicated to me since the means areequal, and the amount of variance is the only thing being studied. And if Ican use the f-test, what is the degrees of freedom, the number of games playedby each team, or the number of teams?-- -------------------------------------------------------------- | best, | Youre doing a lot of choppin || ed | but no chips are §yin. | -------------------------------------------------------------- MathStatica or something else?I have seen MathStatica in action on their web page and it seems ratherimpressive - I was wondering which other tools handles symbolic math andstatistics on the same level?Links to comparisons between the tools would expressed above are my own.P.P.S. Remove dashes in mail else?I have seen MathStatica in action on their web page and it seems rather> impressive - I was wondering which other tools handles symbolic math and> statistics on the same level?Links in advance> /TorbenYou could try Maple and the free program Maxima, a descendant Macsyma,but I dont know how they compare to MathStatica.I am interested in Mathstatica and have bought the book thataccompanies it, but I am not going to try it seriously until it iscompatible with Mathematica 5 (the most recent release of thatprogram). Just reading the book recently helped me answer acolleagues question.