mm-1359 === Subject: Re: Is there anything special about this matrix? > Matrix(2,2,[[1-b1,b1],[-b2,1+b2]]) Both rows add to 1? So [[1],[1]] is an eigenvector with eigenvalue 1... === Subject: triangle number problem Which integers can be written as the quotient of two triangular numbers? --- J K Haugland http://www.neutreeko.com === Subject: Re: triangle number problem > Which integers can be written as the quotient of two triangular numbers? I think that T(m)/T(n) = k has infinitely many solutions if k isn't a square. If k is a square (greater than 1 ...), there are only finitely many solutions, often nothing. But T(8*T(2k))/T(2k) = (8k+2)^2, k >= 1 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: triangle number problem Ignacio Larrosa Ca.96estro a .8ecrit: >>Which integers can be written as the quotient of two triangular numbers? > I think that T(m)/T(n) = k has infinitely many solutions if k isn't a > square. > If k is a square (greater than 1 ...), there are only finitely many > solutions, often nothing. > But > T(8*T(2k))/T(2k) = (8k+2)^2, k >= 1 answer : most of them. an infinite number of solutions. Proof : a*(a-1)/2=k*b*(b-1)/2, equation (1). Let u=2*a-1 and v=2*b-1 a,b>1 integers <=> u,v>1 odd integers (1) <=> u^2-1=k*(v^2-1) that is u^2-k*v^2=1-k equation (2) This is a generalized Pell equation (look for Pell on the web) One of its fundamental solutions is u=1, v=1 so (2) has allways at least one odd solution. This one is not suitable for solving (1) and we have to find other solutions. Let X and Y be the fundamental solutions of the Pell equation (3) u^2-k*v^2=1. X and Y!=0 allways exist if k is not a square. So we have now to distinguish two cases : 1) k is not a square the (1,1) solution of eq (2) results in an infinite number of solutions : u_(n+1) = X*u_n + k*Y*v_n, v_(n+1) = X*v_n + Y*u_n starting from u_0 = v_0 = 1 because X and Y are allways of opposite parity, and if k even, X is odd, so every (u_n,v_n) is odd and an infinite number of solutions for eq (1). Simple example : k=195 (why not) X=14, Y=1 (u,v)=(1,1) (209,15) ... so the least solution a=105 b=8 5460=28*195 A more complicated example : k=61 (seems easier??? hummm) X=1766319049 Y=226153980 is the least solution of (3) !!! (u,v)=(1,1) (15561711829,1992473029) ... a=7780855915 b=996236515 30270859381104815655 (yes a triangular number !) is 61 times 496243596411554355 (yes another triangular number !) Searching other fundamental solutions of (2) less than X,Y is not so easy... so let's say this is the least solution we have found. 2) k=m^2 is a square, the only solution of (3) is X=1, Y=0. equation (2) is u^2-(m*v)^2=(u-m*v)(u+m*v)=1-m^2=-p*q u+m*v=p and u-m*v=-q give all solutions of (2) The number of solutions is finite, as finite ways of k-1=p*q. We need to keep only the (u,v) odd and >1 Some k give solutions, some don't k=36 (p,q)=(35,1) ... u=17 v=3 so a=9 b=2 k=100 (p,q)=(99,1) ... u=49 v=5 so a=25 b=3 k=196 (p,q)=(195,1) ... u=97 v=7 so a=49 b=4 Seems to be enhanced to define which squares are suitable... PS : Ignacio's formula gives not all suitable squares. 196=14^2 is not a (8k+2)^2 number and T(48)=1176=196*T(3) 100 is the first Ignacio's number T(24)=300=100*T(2) As 36 and 196 are (8k-2)^2, may be all solutions seems to be : all the (8k+/-2)^2 squares and all the non square numbers ? -- chephip at free dot fr === Subject: Re: triangle number problem > Ignacio Larrosa Ca.96estro a .8ecrit: >>Which integers can be written as the quotient of two triangular numbers? >> > I think that T(m)/T(n) = k has infinitely many solutions if k isn't a > square. > If k is a square (greater than 1 ...), there are only finitely many > solutions, often nothing. > But > T(8*T(2k))/T(2k) = (8k+2)^2, k >= 1 > PS : > Ignacio's formula gives not all suitable squares. > 196=14^2 is not a (8k+2)^2 number and T(48)=1176=196*T(3) > 100 is the first Ignacio's number T(24)=300=100*T(2) > As 36 and 196 are (8k-2)^2, may be all solutions seems to be : > all the (8k+/-2)^2 squares and all the non square numbers ? Yes, actually this is more general: T(8*T(k))/T(k) = (4k + 2)^2, k >= 1 -- Ignacio Larrosa Ca.96estro A Coru.96a (Espa.96a) ilarrosaQUITARMAYUSCULAS@mundo-r.com === Subject: Re: triangle number problem Ignacio Larrosa Ca.96estro a .8ecrit: >>Ignacio Larrosa Ca.96estro a .8ecrit: >>Which integers can be written as the quotient of two triangular numbers? >> [...] > Yes, actually this is more general: > T(8*T(k))/T(k) = (4k + 2)^2, k >= 1 For example : m=195=4k-1, T[675]=38025*T[3] m=980=4k, T[2400]=960400*T[2] Only 21 non 4k+2 squares out of the first 5000 squares are solutions. To compare with the 25% 4k+2 squares in this range. (if my program is good of course...) m=35,99,195,204,323,483,675,899,980,1155,1189,1443,1763,2115,2499,2716, 2915,3363,3843,4355,4899 The first 4k+1 which is not a square triangle : m=9701=4k+1, T[23762]=94109401*T[2] (square triangles are T[n]=m^2=m^2*T[1]= 1, 36, 1225, 41616, ...) Conclusion (up to now) : All the (4k+2)^2 plus some of the (4k)^2 and (4k+/-1)^2 and all the non square. A lot of (4k)^2 and (4k+/-1)^2 are not quotient of two triangles. -- chephip at free dot fr === Subject: Re: Diagonalizable matrices > Hi all, > Let k be an algebraically closed field and let n be a natural number. > Consider the space M of all square matrices with n lines, n rows and > entries in k. Let D be the subset of M of all diagonalizable matrices. > My question is: is D a dense subset of M with respect to the Zariski > topology? Of course, the answer is yes if k has characterist 0, because > D contains all matrices whose characteristic polynomial p has n distinct > roots and this is equivalente to (p,p') = 1. But if the characteristic > of k is greater than 0, this argument fails. > Jose Carlos Santos Yes it is dense. The reason is that any matrix with distinct eigenvalues is diagonalizable over an algebraically closed field. And that is Zariski open because the characteristic polynomial of the matrix has coefficients that are polynomials in the entries of the matrix and the discriminant of the characteristic polynomial is also a polynomial in the entries and the non-vanishing of this discriminant is therefore Zariski open. Since every non-empty open is dense in Zariski, the conclusion follows. This gives a very elegant proof of the Cayley-Hamilton theorem. === Subject: Re: Diagonalizable matrices >>Let k be an algebraically closed field and let n be a natural number. >>Consider the space M of all square matrices with n lines, n rows and >>entries in k. Let D be the subset of M of all diagonalizable matrices. >>My question is: is D a dense subset of M with respect to the Zariski >>topology? Of course, the answer is yes if k has characterist 0, because >>D contains all matrices whose characteristic polynomial p has n distinct >>roots and this is equivalente to (p,p') = 1. But if the characteristic >>of k is greater than 0, this argument fails. > Yes it is dense. The reason is that any matrix with distinct > eigenvalues is diagonalizable over an algebraically closed field. And > that is Zariski open because the characteristic polynomial of the > matrix has coefficients that are polynomials in the entries of the > matrix and the discriminant of the characteristic polynomial is also a > polynomial in the entries and the non-vanishing of this discriminant > is therefore Zariski open. Since every non-empty open is dense in > Zariski, the conclusion follows. This gives a very elegant proof of > the Cayley-Hamilton theorem. thought was: p has n distinct roots => (p,p') = 1 => res(p,p') <> 0 but the first implication fails in characteristic greater than 0. Jose Carlos Santos === Subject: Recurrence Hi Given the following recurrence: T(n, 1) = 2n - 1 T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) I want to prove that T(n,d) in O(n^d), i.e. exists c > 0, n_0 in |N forall n >= n_0: T(n, d) <= n Using induction over d yields the following induction step: d-1 -> d: T(n,d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) <=(IH) c * n^(d-1) + 2 * c * sum(i^(d-1), i=1..n-1) <= 2 * c * sum(i^(d-1), i=1..n) Thus we have to show that n^d / (2 * sum(i^(d-1), i=1..n)) >= 1 forall n >= n_0 Alternatively we could prove: lim(n->infinity) n^d / (2 * sum(i^(d-1), i=1..n)) >= 1 or lim(n->infinity) (2 * sum(i^(d-1), i=1..n)) / n^d = 0 Does not look too complicated but somehow I don't see the solution yet. Could you give me a hint? Thomas === Subject: Re: Recurrence > Given the following recurrence: > T(n, 1) = 2n - 1 > T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) > I want to prove that T(n,d) in O(n^d), i.e. > exists c > 0, n_0 in |N forall n >= n_0: T(n, d) <= n this should be: exists c > 0 exists n_0 in |N forall n >= n_0: T(n, d) <= n^d (1) Indeed the proof is extremely trivial since: T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) <= (2n - 1) * T(n, d-1) (2) And thus: T(n, d) = (2n - 1)^d in O(n^d) since we can choose c = 2^d The disadvantage of this simple approach is that one receives the impression that large constants are involved in O(n^d) which is only due to the very rough approximation in (2). So, I'm searching a c > 0 which does not depend on d (or if it has to depend on d then it should be at least significantly smaller than 2^d). Does anyone know a smarter approximation or approach leading to a smaller c > 0? BTW, if c > 0 does not depend on d we have proven: exists c > 0 exists n_0 in |N exists d_0 in |N forall n >= n_0 forall d >= d_0: T(n, d) <= n^d which is of course a stronger statement than (1). Thomas === Subject: Re: Recurrence >Given the following recurrence: >T(n, 1) = 2n - 1 >T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) >I want to prove that T(n,d) in O(n^d), i.e. >exists c > 0, n_0 in |N forall n >= n_0: T(n, d) <= n >Using induction over d yields the following induction step: >d-1 -> d: T(n,d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) > <=(IH) c * n^(d-1) + 2 * c * sum(i^(d-1), i=1..n-1) > <= 2 * c * sum(i^(d-1), i=1..n) >Thus we have to show that n^d / (2 * sum(i^(d-1), i=1..n)) >= 1 forall n >= n_0 >Alternatively we could prove: >lim(n->infinity) n^d / (2 * sum(i^(d-1), i=1..n)) >= 1 >lim(n->infinity) (2 * sum(i^(d-1), i=1..n)) / n^d = 0 >Does not look too complicated but somehow I don't see the solution >yet. Could you give me a hint? When you see sum in a recurrence, try to eliminate it. If we replace n by n+1, we have T(n+1, d) = T(n+1, d-1) + 2*sum(T(i,d-1), i=1..n) T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) Subtract to get T(n+1, d) - T(n, d) = T(n+1, d-1) - T(n, d-1) + 2 * T(n, d-1) = T(n+1, d-1) + T(n, d-1) Now prove that, for each d >= 1, there exists a polynomial f_d of degree d such that T(n, d) = f_d(n) (n >= 1) -- After California's recall election, wildfires Schwartz-en-ed the Bush-lands on its geographic right (when we wanted the forests to be Green). pmontgom@cwi.nl Home: San Rafael, California Microsoft Research and CWI === Subject: Re: Recurrence Hi Peter > When you see sum in a recurrence, try to eliminate it. > If we replace n by n+1, we have > T(n+1, d) = T(n+1, d-1) + 2*sum(T(i,d-1), i=1..n) > T(n, d) = T(n, d-1) + 2 * sum(T(i, d-1), i=1..n-1) > Subtract to get > T(n+1, d) - T(n, d) = T(n+1, d-1) - T(n, d-1) + 2 * T(n, d-1) > = T(n+1, d-1) + T(n, d-1) > Now prove that, for each d >= 1, there exists a polynomial > f_d of degree d such that > T(n, d) = f_d(n) (n >= 1) Please forgive my ignorance but how does T(n+1, d) - T(n, d) = T(n+1, d-1) + T(n, d-1) help me proving that T(n, d) = sum(a_i * n^i, i=0..d)? Thomas === Subject: Re: prime ideals of Z[x] > What are all the prime ideals of Z[x]? See my prior post -Bill Dubuque === Subject: Need help solving an axactly soluble ode may not be equiped to solve. I am not looking for people to furnish a solution, as I am hoping to do that myself, but I would be greatful for any hints or clues, things to look for, but please do not reply with a completed exact solution. the initial value problem is this y'=1-t4y, y(0)=1 the solution is given to us, but I want to come up with it y(t)=(4t-3+19exp(4t))/16 Any tips? I cannot find an obvious way to make it separable. My level of expertise is beginner so any tips rendered, i'd hope they'd be directed to a beginner. I have found much of the help given here i Josh === Subject: Re: Need help solving an axactly soluble ode > may not be equiped to solve. I am not looking for people to furnish a > solution, as I am hoping to do that myself, but I would be greatful > for any hints or clues, things to look for, but please do not reply > with a completed exact solution. > the initial value problem is this > y'=1-t4y, y(0)=1 > the solution is given to us, but I want to come up with it > y(t)=(4t-3+19exp(4t))/16 > Any tips? y' + t^4 y = 1 is linear. you need to find an 'integrating factor' m, such that m y' + m t^4 y = d [m t^4 y] / dt once you find such m, multiply both sides of your eqn and integrate. > I cannot find an obvious way to make it separable. My level of > expertise is beginner so any tips rendered, i'd hope they'd be > directed to a beginner. I have found much of the help given here i > Josh === Subject: Re: Need help solving an axactly soluble ode >may not be equiped to solve. I am not looking for people to furnish a >solution, as I am hoping to do that myself, but I would be greatful >for any hints or clues, things to look for, but please do not reply >with a completed exact solution. >the initial value problem is this >y'=1-t4y, y(0)=1 >the solution is given to us, but I want to come up with it >y(t)=(4t-3+19exp(4t))/16 It might help to write the problem correctly. That solution is a solution to y' = 1-t + 4y The standard formula for a linear constant-coefficient equation will handle this one. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Need help solving an axactly soluble ode >may not be equiped to solve. I am not looking for people to furnish a >solution, as I am hoping to do that myself, but I would be greatful >for any hints or clues, things to look for, but please do not reply >with a completed exact solution. >the initial value problem is this >y'=1-t4y, y(0)=1 >the solution is given to us, but I want to come up with it >y(t)=(4t-3+19exp(4t))/16 > It might help to write the problem correctly. That solution is a solution to > y' = 1-t + 4y > The standard formula for a linear constant-coefficient equation will handle > this one. > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia > Vancouver, BC, Canada V6T 1Z2 Yes it would. I ought to use the preview function and proofread a little more closely. josh === Subject: Re: Need help solving an axactly soluble ode > may not be equiped to solve. I am not looking for people to furnish a > solution, as I am hoping to do that myself, but I would be greatful > for any hints or clues, things to look for, but please do not reply > with a completed exact solution. > the initial value problem is this > y'=1-t4y, y(0)=1 > the solution is given to us, but I want to come up with it > y(t)=(4t-3+19exp(4t))/16 > Any tips? First, are you sure this is actually a solution? If y(t)=(4t-3+19exp(4t))/16 then y(0) = (4(0)-3+19exp(4*0))/16 = 1, which is good but y'(t) = (1/4)(t-19exp(4t)) while 1-t4y = 1-(1/4)(t)(4t-3+19exp(4t)) and these don't look very similar. Indeed, evaluating at t=0 one gets y'(0) = (1/4)(0-19exp(4*0))=-19/4 but 1-t*4*y evaluated at t=0 = 1 - 0*4*y(0) = 1. Ok, so the solution you have is *NOT* a solution to this differential equation. Assuming you still want a solution to y'=1-4ty, you might try writting this as y' + 4ty = 1 then multiply by exp(2t^2). This won't give you a closed form, but you can write the answer in terms of the error function. Best wishes, Mike === Subject: Re: Need help solving an axactly soluble ode > may not be equiped to solve. I am not looking for people to furnish a > solution, as I am hoping to do that myself, but I would be greatful > for any hints or clues, things to look for, but please do not reply > with a completed exact solution. > the initial value problem is this > y'=1-t4y, y(0)=1 > the solution is given to us, but I want to come up with it > y(t)=(4t-3+19exp(4t))/16 > Any tips? > First, are you sure this is actually a solution? If > y(t)=(4t-3+19exp(4t))/16 > then > y(0) = (4(0)-3+19exp(4*0))/16 = 1, > which is good but > y'(t) = (1/4)(t-19exp(4t)) > while > 1-t4y = 1-(1/4)(t)(4t-3+19exp(4t)) > and these don't look very similar. Indeed, evaluating at t=0 > one gets > y'(0) = (1/4)(0-19exp(4*0))=-19/4 > but > 1-t*4*y evaluated at t=0 = 1 - 0*4*y(0) = 1. > Ok, so the solution you have is *NOT* a solution to this > differential equation. Assuming you still want a solution > to y'=1-4ty, you might try writting this as > y' + 4ty = 1 > then multiply by exp(2t^2). This won't give you a closed > form, but you can write the answer in terms of the error > function. > Best wishes, > Mike Ah well wouldn;t you know it. friggin keyboard foiled me again. errata: the ode i entered earlier should read: y'=1-t+4y, with y(0)=1 and the previously given solution. I keyed it in wrong. this changes everything that people have posted. It is esactly soluble because I am told it is. The solution given is the correct solution. Sorry. Could you recheck? josh === Subject: Re: Need help solving an axactly soluble ode . . . > Ah well wouldn;t you know it. friggin keyboard foiled me again. > errata: the ode i entered earlier should read: > y'=1-t+4y, with y(0)=1 still linear. see my other post. > and the previously given solution. I keyed it in wrong. this changes > everything that people have posted. It is esactly soluble because I am > told it is. The solution given is the correct solution. Sorry. Could > you recheck? > josh === Subject: Re: Need help solving an axactly soluble ode It's not separable Look up linear first-order differential equations.. > may not be equiped to solve. I am not looking for people to furnish a > solution, as I am hoping to do that myself, but I would be greatful > for any hints or clues, things to look for, but please do not reply > with a completed exact solution. > the initial value problem is this > y'=1-t4y, y(0)=1 > the solution is given to us, but I want to come up with it > y(t)=(4t-3+19exp(4t))/16 > Any tips? > I cannot find an obvious way to make it separable. My level of > expertise is beginner so any tips rendered, i'd hope they'd be > directed to a beginner. I have found much of the help given here i > Josh === Subject: convergency (task) Hallo, I have the following task but I don't know where to start. :-( task: ### Examine the following continuation x0 = 1; x_(n+1) = 1+ 1/x_n. Does this continuation converge on a limiting value? Please specify this value, if it converges. Tip: Show that for a possible limiting value g the following is true: g^2 = 1+g and g > 1; g is from |R. g is the golden intersection. Show that |x_n-g| <= 1/g^(n+1). ### Karl. [P.S. I think this continuation has something to do with Fibonacci.] === Subject: Re: convergency (task) > Hallo, > I have the following task but I don't know where to start. :-( > task: > ### > Examine the following continuation x0 = 1; x_(n+1) = 1+ 1/x_n. > Does this continuation converge on a limiting value? Please > specify this value, if it converges. > Tip: Show that for a possible limiting value g the following is true: > g^2 = 1+g and g > 1; g is from |R. g is the golden intersection. > Show that |x_n-g| <= 1/g^(n+1). > ### > Karl. > [P.S. I think this continuation has something to do with Fibonacci.] If the sequence has a limiting value then it must be that x_(n+1) - x_n = (1+1/x_n) - x_n must converge to zero. This suggests that the limiting value, x, might be such that 1 + 1/x - x = 0. === Subject: Re: convergency (task) > Hallo, > I have the following task but I don't know where to start. :-( > task: > ### > Examine the following continuation x0 = 1; x_(n+1) = 1+ 1/x_n. > Does this continuation converge on a limiting value? Please > specify this value, if it converges. Look up Golden Ratio. Bob Kolker === Subject: Arrrgh... stuck on simple (I think) complex analysis proof Been staring at this for hours and can't seem to find a proof that seems needlessly complex or just plain doesn't work... can someone see a simple argument? I would be much appreciated - I don't even need this result for anything but somehow I just can't let it go. Let f be a continuous function on the unit circle T = {|z| = 1}. Show that f can be approximated uniformly on T by a sequence of polynomials in z if and only if f has an extension F that is continuous on the closed disk {|z| <= 1} and analytic on the interior {|z| < 1}. Hint. To approximate such an F, consider dilates F[subscripted r](z) = F(rz). I can't get either direction to work... if F is given then the hint seems to imply that you can obtain the polynomial sequence from the power series expansion of F, but how to show uniform convergence in a simple way? If the polynomial sequence is given then it's easy to use the Cauchy integral formula on the polynomials, take the limit, and get F analytic (limit of the integral = integral of the limit because of the uniform convergence, and analyticity follows from the continuity of the limit funciton f). But how to show that F is continuous at points on the unit circle? I found this in Complex Analysis by Gamelin, section V.4, problem 14. === Subject: Re: Arrrgh... stuck on simple (I think) complex analysis proof >Been staring at this for hours and can't seem to find a proof that >seems needlessly complex or just plain doesn't work... can someone see >a simple argument? I would be much appreciated - I don't even need >this result for anything but somehow I just can't let it go. >Let f be a continuous function on the unit circle T = {|z| = 1}. Show >that f can be approximated uniformly on T by a sequence of polynomials >in z if and only if f has an extension F that is continuous on the >closed disk {|z| <= 1} and analytic on the interior {|z| < 1}. Hint. >To approximate such an F, consider dilates F[subscripted r](z) = >F(rz). >I can't get either direction to work... if F is given then the hint >seems to imply that you can obtain the polynomial sequence from the >power series expansion of F, but how to show uniform convergence in a >simple way? The power series need not converge to F at points of the boundary. But that's not what the hint suggests - read the hint again... >If the polynomial sequence is given then it's easy to use the Cauchy >integral formula on the polynomials, take the limit, and get F >analytic (limit of the integral = integral of the limit because of the >uniform convergence, and analyticity follows from the continuity of >the limit funciton f). But how to show that F is continuous at points >on the unit circle? Don't use the Cauchy formula. Use the Maximum Modulus Theorem, and look up uniform convergence and Cauchy sequence in a book on adbanced calculus. >I found this in Complex Analysis by Gamelin, section V.4, problem 14. David C. Ullrich === Subject: Re: Arrrgh... stuck on simple (I think) complex analysis proof If F is given consider the sequence of functions (defined on |z| = 1) F[r](z) = F(rz), where r is a monotone sequence of reals approaching 1, say r = 1/(1+n) with n a natural number. Since F is continuous F[r](z) uniformly approaches F(z) as r approaches 1. [This is the point I was missing before - it holds because the unit circle is a closed set.] Let E/2 be the worst case error value of |F(z) - F[r](z)|. Since rz lies within the radius of convergence of F, F converges uniformly there, and we can take a partial sum P[r](z) that is uniformly within E/2 of F[r](z). This partial sum is a polynomial in (z) as well as in (rz). For all z on the unit circle |F(z) - P[r](z)| <= |F(z) - F[r](z)| + |F[r](z) - P[r](z)| < E which goes to zero as r approaches 1. So the P[r](z) are the desired polynomials in z. [My only nagging doubt here is that this argument would seem to apply to an F defined on any circle, not just the unit circle.] If the sequence of polynomials is given they converge and therefore form a Cauchy sequence. The difference between any two of them is an analytic function and therefore reaches its maxium modulus on the boundary of the unit disk. So the polynomials form a uniform Cauchy sequence on the closed unit disk and therefore converge uniformly. [Interestingly none of the math books I had included Cauchy's criterion for uniform convergence, just Cauchy's criterion for convergence - too obvious an extension to state explicitly no doubt. But I did find explicit mention of it on the web.] -The polynomials converge uniformly on the closed unit disk and are continuous, therefore they converge to a continuous function there. -The polynomials converge uniformly on the open unit disk and are analytic, therefore they converge to an analytic function there. Ian >Been staring at this for hours and can't seem to find a proof that >seems needlessly complex or just plain doesn't work... can someone see >a simple argument? I would be much appreciated - I don't even need >this result for anything but somehow I just can't let it go. >Let f be a continuous function on the unit circle T = {|z| = 1}. Show >that f can be approximated uniformly on T by a sequence of polynomials >in z if and only if f has an extension F that is continuous on the >closed disk {|z| <= 1} and analytic on the interior {|z| < 1}. Hint. >To approximate such an F, consider dilates F[subscripted r](z) = >F(rz). >I can't get either direction to work... if F is given then the hint >seems to imply that you can obtain the polynomial sequence from the >power series expansion of F, but how to show uniform convergence in a >simple way? > The power series need not converge to F at points of the boundary. > But that's not what the hint suggests - read the hint again... >If the polynomial sequence is given then it's easy to use the Cauchy >integral formula on the polynomials, take the limit, and get F >analytic (limit of the integral = integral of the limit because of the >uniform convergence, and analyticity follows from the continuity of >the limit funciton f). But how to show that F is continuous at points >on the unit circle? > Don't use the Cauchy formula. Use the Maximum Modulus Theorem, > and look up uniform convergence and Cauchy sequence in a book > on adbanced calculus. >I found this in Complex Analysis by Gamelin, section V.4, problem 14. > David C. Ullrich === Subject: Re: Arrrgh... stuck on simple (I think) complex analysis proof >If F is given consider the sequence of functions (defined on |z| = 1) >F[r](z) = F(rz), where r is a monotone sequence of reals approaching 1, say >r = 1/(1+n) with n a natural number. Since F is continuous F[r](z) uniformly >approaches F(z) as r approaches 1. [This is the point I was missing before - >it holds because the unit circle is a closed set.] Or more precisely: This works because the closed _disk_ is a _compact_ set, which implies that f is uniformly continuous. >Let E/2 be the worst case >error value of |F(z) - F[r](z)|. Since rz lies within the radius of >convergence of F, F converges uniformly there, and we can take a partial sum >P[r](z) that is uniformly within E/2 of F[r](z). This partial sum is a >polynomial in (z) as well as in (rz). For all z on the unit circle |F(z) - >P[r](z)| <= |F(z) - F[r](z)| + |F[r](z) - P[r](z)| < E which goes to zero as >r approaches 1. So the P[r](z) are the desired polynomials in z. [My only >nagging doubt here is that this argument would seem to apply to an F defined >on any circle, not just the unit circle.] Well you can relax, it's true for any closed disk. (In fact it's true in much greater generality, although it's not so easy to prove; Runge's theorem, included in most complex books, gives a generalization, and Mergelyan's theorem gives the result under weaker hypotheses yet.) >If the sequence of polynomials is given they converge and therefore form a >Cauchy sequence. The difference between any two of them is an analytic >function and therefore reaches its maxium modulus on the boundary of the >unit disk. So the polynomials form a uniform Cauchy sequence on the closed >unit disk and therefore converge uniformly. [Interestingly none of the math >books I had included Cauchy's criterion for uniform convergence, just >Cauchy's criterion for convergence - too obvious an extension to state >explicitly no doubt. But I did find explicit mention of it on the web.] >-The polynomials converge uniformly on the closed unit disk and are >continuous, therefore they converge to a continuous function there. >-The polynomials converge uniformly on the open unit disk and are analytic, >therefore they converge to an analytic function there. >Ian >>Been staring at this for hours and can't seem to find a proof that >>seems needlessly complex or just plain doesn't work... can someone see >>a simple argument? I would be much appreciated - I don't even need >>this result for anything but somehow I just can't let it go. >> >>Let f be a continuous function on the unit circle T = {|z| = 1}. Show >>that f can be approximated uniformly on T by a sequence of polynomials >>in z if and only if f has an extension F that is continuous on the >>closed disk {|z| <= 1} and analytic on the interior {|z| < 1}. Hint. >>To approximate such an F, consider dilates F[subscripted r](z) = >>F(rz). >> >>I can't get either direction to work... if F is given then the hint >>seems to imply that you can obtain the polynomial sequence from the >>power series expansion of F, but how to show uniform convergence in a >>simple way? >> The power series need not converge to F at points of the boundary. >> But that's not what the hint suggests - read the hint again... >>If the polynomial sequence is given then it's easy to use the Cauchy >>integral formula on the polynomials, take the limit, and get F >>analytic (limit of the integral = integral of the limit because of the >>uniform convergence, and analyticity follows from the continuity of >>the limit funciton f). But how to show that F is continuous at points >>on the unit circle? >> Don't use the Cauchy formula. Use the Maximum Modulus Theorem, >> and look up uniform convergence and Cauchy sequence in a book >> on adbanced calculus. >>I found this in Complex Analysis by Gamelin, section V.4, problem 14. >> David C. Ullrich David C. Ullrich === Subject: Re: Mathematical puzzle? >The puzzle goes something like this: >sqrt(x+sqrt(x+ ....)) = 2 >Solve for x. How about: Sqrt[x+Sqrt[x+Sqrt[x+...]]] = x Solve for x... this time there's two solutions, 2 and... Or how about another variation on the problem: Sqrt[6+Sqrt[6+Sqrt[6+Sqrt[6+...]]]]] = ? Solve for ? adam === Subject: Re: Line Integrals - Parametrizing Line Segments ath: meganewsservers.com (x,y,z) coordinates? Your method seems be a little test/see-if-it-works type. >Hi. I have a question about parametrizing line segments. Here is how my book >explains it: >The directed line segment that begins at a = (a1,a2) adn ends at b = >(b1,b2) can be parametrized by setting > r(u) = (1 - u)a + ub u subset of [0,1] > you mean that u is within the interval [0,1], not a subset. > (a, b, r are vectors) >Can you please explain how they came up with this parametrization. They had >a sample problem involving vertices (1,0), (0,1), and (-1,0) but it still >did not make sense. Another problem involved vertices with (2,0), (0,2), and >(-2,0) and the explanation was not clear. Each involved three curves. > I think they are trying to get across the point that in order to > parametrize a curve (a line in space (a shoelace stretched between two > points)) you only need _one_ parameter. > That is, take your equation: > r = (1-u)a + ub > and plug in u = 0 > you get r = a (the begining point) > then take your equation and plug in u = 1 > you get r = b (the end point) > they could have equally as well parametrized the curve by: > r(u) = a*(1-u)^2 + (u^5)*b u within [0,1] > because at u=0, r=a and at u=1, r=b...but it's simpler to > see with the linear parameterization. > adam === Subject: Re: Line Integrals - Parametrizing Line Segments , > (x,y,z) coordinates? Your method seems be a little test/see-if-it-works > type. In any number of dimensions, let upper case letters represent points/vectors and lower case represent scalars, then: Given two distinct points/vectors, P(t) = A + t*(B - A) = (1-t)*A + t*B, with parameter t, represents a paramentrization of the line through points A and B. P(0) = A, P(1) = B, P(1/2) is midpoint of the segment from A to B for 0 < t < 1, P(t) is between A and B, for 1 < t, B is between A and P(t), and for t < 0, A is between P(t) and B A slightly altered form of the above replaces 1-t with s, so that Q(s,t) = s*A + t*B , where it is required that s + t = 1. These, s and t, are sometimes called homogenous coordinates or barycentric coordinates of the line. Barycentic coordinates can be expanded to higher dimensions. Given points A_0, ..., A_n in a space of dimension n, so that the n vectors A_2 - A_0, A_3 - A_0, ..., A_n - A_0 are linearly independent, then any point has unique representation P = t_0*A_0 + ...+ t_n*A_n, such that t_0 + ...+ t_n = 1. An alternate, more standard but equivalent representation is P = A_0 + t_1*(A_1 - A_0) + ... + t_n*(A_n - A_0)). Hope this is of some use. === Subject: Re: Line Integrals - Parametrizing Line Segments > (x,y,z) coordinates? Your method seems be a little test/see-if-it-works > type. Regardless of the number of dimensions, r(t) = [start] + t*[stop - start] for 0 <= t <=1 parameterizes the line segment from [start] to [stop]. If you prefer 0 <= t <= 5, r(t) = [start] + (t/5)*[stop - start]. If you prefer 0 <= t <= 1/2, r(t) = [start] + 2*t*[stop - start]. If you prefer 3 <= t <= 4, r(t) = [start] + (t -3)*[stop - start]. If you prefer 7 <= t <= 10, r(t) = [start] + ((t - 7)/3)*[stop - start]. -- Paul Sperry Columbia, SC (USA) === Subject: Re: Line Integrals - Parametrizing Line Segments ath: meganewsservers.com were different? Say, [0,2]? > <2683e$3fbef144$80dcc27d$7580@msgid.meganewsservers.com>, > Hi. I have a question about parametrizing line segments. Here is how my book > explains it: > The directed line segment that begins at a = (a1,a2) adn ends at b = > (b1,b2) can be parametrized by setting > r(u) = (1 - u)a + ub u subset of [0,1] > (a, b, r are vectors) > Can you please explain how they came up with this parametrization. They had > a sample problem involving vertices (1,0), (0,1), and (-1,0) but it still > did not make sense. Another problem involved vertices with (2,0), (0,2), and > (-2,0) and the explanation was not clear. Each involved three curves. > I will use upper case for points/vectors and lower case for scalars. > If you want a point on the segment between A and B, one can get it > by first going to A and then part of the way from A towards B along > (B-A), i.e. your position vector will be something like > R(u) = A + u(B - A), where 0 < u < 1. > Note that when u = 0, this gives A and when u = 1 this gives B. > Rearranging terms gives R(u) = (1-u) A + u B, as desired. === Subject: 0/0 Thread Posting Question The 0/0 thread was interesting to me, and I particularly enjoyed the post by Phil C. about 0 being a relational marker rather than a real number, for it showed an arguably coherent attempt to get at the underlying logic of the very nature of 0 and the resulting effect in math processes. I wanted to post a response saying no more than that, but I saw no link allowing replies to be posted. Would someone please explain why? Very Respectfully, Ray === Subject: Re: 0/0 Thread Posting Question > The 0/0 thread was interesting to me, and I particularly enjoyed the > post by Phil C. about 0 being a relational marker rather than a real > number, for it showed an arguably coherent attempt to get at the > underlying logic of the very nature of 0 and the resulting effect in > math processes. > I wanted to post a response saying no more than that, but I saw no > link allowing replies to be posted. Would someone please explain why? messages from years back (it has all the old DejaNews archives for a start), and it threads messages together simply because they have the same title, even if they are unrelated and posted years apart. The message you wanted to reply to was posted in 1999, and Google does not allow you to reply to such old messages. You should be able to reply to any of the recent 0/0 posts. === Subject: Re: 0/0 Thread Posting Question > messages from years back (it has all the old DejaNews archives for a > start), and it threads messages together simply because they have the > same title, even if they are unrelated and posted years apart. The > message you wanted to reply to was posted in 1999, and Google does not > allow you to reply to such old messages. You should be able to reply > to any of the recent 0/0 posts. THANK YOU! (and Mr. Feldman, too), I found the post again and noted the 1999 year date. This is not the first time that I've seen where I could not post a reply, and I very much appreciate the answer. Very Respectfully, Ray === Subject: Re: 0/0 Thread Posting Question > The 0/0 thread was interesting to me, and I particularly enjoyed the > post by Phil C. about 0 being a relational marker rather than a real > number, for it showed an arguably coherent attempt to get at the > underlying logic of the very nature of 0 and the resulting effect in > math processes. > > I wanted to post a response saying no more than that, but I saw no > link allowing replies to be posted. Would someone please explain why? > messages from years back (it has all the old DejaNews archives for a > start), and it threads messages together simply because they have the > same title, even if they are unrelated and posted years apart. The > message you wanted to reply to was posted in 1999, and Google does not > allow you to reply to such old messages. You should be able to reply > to any of the recent 0/0 posts. You can still reply to an old post via Google. Here's how: Read the old message. Open another window and read one of the newer messages. Click reply to this message. Go back to the first window. Copy the contents of the Your message to the clipboard (Ctrl/A, Ctrl/C in windows). Go to the other window (the one with the new message). Highlight the entire contents of the Your message window. Paste. Now you can reply to the old message. Disclaimer: JMHO Alan E. Feldman === Subject: Re: 0/0 Thread Posting Question > The message you wanted to reply to was posted in 1999, and Google > does not allow you to reply to such old messages. You should be > able to reply to any of the recent 0/0 posts. > You can still reply to an old post via Google. Yes, I think you are right. What I should have said is that they no longer provide links for such replies. > Here's how: > Read the old message. Open another window and read one of the newer > messages. Click reply to this message. Go back to the first window. > Copy the contents of the Your message to the clipboard (Ctrl/A, > Ctrl/C in windows). Go to the other window (the one with the new > message). Highlight the entire contents of the Your message window. > Paste. > Now you can reply to the old message. If you just copy the old message text into the reply form you will end up with an incorrect References line in the header - as far as newsreaders are concerned it will still be a reply to the recent message, not the old one. I think the correct way to do it is to copy the Message-ID from the URL of the message you want to reply to, and add it on to the end of this: Then load the resulting URL. This seems to work, in as much as you get a reply form with the correct message quoted in it, and previewing shows the correct References line. I haven't actually tried pressing the 'Post message' button, however. === Subject: Re: About big numbers... >What's the biggest number that can be made with 'three' nines? > Refer to the other post on Big numbers. I would go with the power tower answer. > 999 > Power tower means > 35 = 5^(5^5) > 56 = 6^(6^(6^(6^6))) > So 999 > = 9^9^9^9^9 ... ^ 9 > There will be 99 's in that list, and this number itself is HUGE I just realized that we could have [(9!)!]^[(9!)!]^[(9!)!] or as many levels of factorization as we desire! === Subject: Re: About big numbers... >What's the biggest number that can be made with 'three' nines? > Refer to the other post on Big numbers. I would go with the power tower answer. > 999 > Power tower means > 35 = 5^(5^5) > 56 = 6^(6^(6^(6^6))) > So 999 > = 9^9^9^9^9 ... ^ 9 > There will be 99 's in that list, and this number itself is HUGE Better is (9!)^(9!)^(9!), where (9!) = 9 factorial? === Subject: Re: Math factorization, other side > >> >> > Now I've given enough time for quite a few of you to think about the > factorization I talked so much about, and given you other information > to allow you to form your opinions based on the data available to you. > > My assumption is that many of you have some internal opinion about > your own thinking ability: both your basic intelligence level, your > ability to follow a rational argument, and your trust level for > experts. > > Now let's go from the other side by considering, yet again, a > >> factorization: >> > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > For some of you a quick way to serious mental confusion and > consternation is to simply try and answer the question of whether or > not f_1(x), f_2(x), and f_3(x) even exist. > >Yes, they exist. They are functions from the integers (for instance) >to the algebraic numbers. (Note, they are *not* functions from the >integers to the algebraic integers.) Moreover, many such functions >exist and will fit the bill. > > >>In particular, it's not hard to show that f_1(1), f_2(1), and f_3(1) >>will not be algebraic integers. >> >> >> >>Rick >> > > > Then show it. Go on, help Dik Winter out as he's drowning here. > > Show that with > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > Come on Rick Decker, I dare you. > > Hehe. I'm not going to respond until you double-dog dare me. > Besides, Dik has already produced the answer in this thread. > His divisibility argument can easily be generalized to > show that for any integral x, f_1(x) and f_2(x) will not be > algebraic integers (though I did erroneously claim that f_3(x) > also wasn't), and that 22 f_1(x) and 22 f_2(x) will be > algebraic integers. Look it over--it's simple long division. > Rick That's wrong. In any event, you still lose as you need 7 f_1(x) to be an algebraic integer as that's what Dik Winter and others have been debating me about for so long. Remember I gave a solution for the a's where a_1(x) = 7 f_1(x), and a_1(x) IS an algebraic integer. But you just said that 22 f_1(x) is an algebraic integer. Last I checked 22 and 7 were coprime in the ring of algebraic integers. Any smart-ass comments now Rick Decker? James Harris === Subject: Re: Math factorization, other side > Remember I gave a solution for the a's where a_1(x) = 7 f_1(x), and > a_1(x) IS an algebraic integer. Did you? Please repost your solution for the 'a's, as I can't seem to find them. Alternatively, you can post your solution for the 'f's, since it is trivial to recover the 'a's from them. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Math factorization, other side ... > Show that with > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > > Come on Rick Decker, I dare you. > > Hehe. I'm not going to respond until you double-dog dare me. > Besides, Dik has already produced the answer in this thread. Not for this case. He has tricked you as he has tricked Nora and me. by 49, so the assumption was that it came from that polynomial. It did not. The question is about arbitrary algebraic integer functions f1, f2 and f3 (and I think James needs them only for integer x). What it amounts to: is for every integer x the resulting value the product of three algebraic integers of the described form? I do not know the answer. (James does not either.) I would think that a proof of this (either way) is just as difficult as FLT. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Non-uniqueness, polynomial factors It's easy to demonstrate that polynomials can be factored in an infinite number of ways--that is, there isn't a unique factorization. I've started showing that by giving you the other side of a factorization I've used for a while by showing (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where f_1(0) = f_2(0) = f_3(0) = 0. That expression represents one way to show a *family* of factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. Now I'll do some simple algebraic manipulations with that expression. Ok, let g_3(x) = f_3(x) + 3, so I have (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22 and now, I multiply both sides by 7, which gives (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), which gives (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) which should look familiar to some of you. Now then, what is required for the a's to be algebraic integer functions? James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. Once again you fail to restrain yourself in your definitions and then proceed to attack other mathematicians for restraining themselves when they do things. When proper legitimate mathematicians talk about unique factorization of polynomials over a field, they mean that for this unique factorization, they are restricting themselves to a _fixed_ unique factorization domain R and a _fixed_ polynomial ring R[x_i], where i ranges over some _fixed_ index set T. That ring is a unique factorization domain. When a mathematician wants to do something like adding a new variable to the variable set or by taking the ring modulo some funky set of relations, (s)he will explicitly declare that this is happening instead of automatically doing it all the time mentally and therefore claiming that the unique factorization theorem I mentioned isn't true. This is what you seem to do. Or do I misunderstand you, given the way you cite the algebraic closure of C (which makes this unique factorization theorem especially easy to prove for the one-variable polynomial ring over that field) which you probably don't understand how to prove yourself? ---- David === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. You have not shown that that is possible with algebraic integer functions. So we can only assume they are algebraic functions. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? 7 f1(x) and 7 f2(x) are algebraic integer functions. I see no other requirement at this moment. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Non-uniqueness, polynomial factors Adjunct Assistant Professor at the University of Montana. [.snip.] >You have not shown that that is possible with algebraic integer functions. >So we can only assume they are algebraic functions. I would suggest algebraic number-valued functions, since algebraic functions has a different meaning (functions obtained from polynomial functions by use of quotients, positive fractional exponents, addition, and multiplication...) It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Non-uniqueness, polynomial factors > [.snip.] >You have not shown that that is possible with algebraic integer functions. >So we can only assume they are algebraic functions. > I would suggest algebraic number-valued functions, since algebraic > functions has a different meaning (functions obtained from > polynomial functions by use of quotients, positive fractional > exponents, addition, and multiplication...) Wouldn't you also need functional inversion to handle something like the inverse function of x^5 + x (as a function from the reals to the reals)? Radicals don't get you that one. ---- David === Subject: Re: Non-uniqueness, polynomial factors Adjunct Assistant Professor at the University of Montana. >> [.snip.] >>You have not shown that that is possible with algebraic integer functions. >>So we can only assume they are algebraic functions. >> I would suggest algebraic number-valued functions, since algebraic >> functions has a different meaning (functions obtained from >> polynomial functions by use of quotients, positive fractional >> exponents, addition, and multiplication...) >Wouldn't you also need functional inversion to handle something like >the inverse function of x^5 + x (as a function from the reals to the >reals)? Radicals don't get you that one. Probably. The point being that algebraic function already has a different meaning, so what Dik is refering to is an algebraic number-valued function. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Non-uniqueness, polynomial factors >> >>You have not shown that that is possible with algebraic integer functions. >>So we can only assume they are algebraic functions. >> >> I would suggest algebraic number-valued functions, since algebraic >> functions has a different meaning (functions obtained from >> polynomial functions by use of quotients, positive fractional >> exponents, addition, and multiplication...) > >Wouldn't you also need functional inversion to handle something like >the inverse function of x^5 + x (as a function from the reals to the >reals)? Radicals don't get you that one. > Probably. The point being that algebraic function already has a > different meaning, so what Dik is refering to is an algebraic > number-valued function. Indeed. But when answering to James I get sloppy on occasion. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Non-uniqueness, polynomial factors > [.snip.] >>You have not shown that that is possible with algebraic integer functions. >>So we can only assume they are algebraic functions. > I would suggest algebraic number-valued functions, since algebraic > functions has a different meaning (functions obtained from > polynomial functions by use of quotients, positive fractional > exponents, addition, and multiplication...) Hmm. Breaking your vow, Arturo? In any case, it's nice to hear from you again under this tent. Rick === Subject: Re: Non-uniqueness, polynomial factors Adjunct Assistant Professor at the University of Montana. >> [.snip.] >You have not shown that that is possible with algebraic integer functions. >So we can only assume they are algebraic functions. >> I would suggest algebraic number-valued functions, since algebraic >> functions has a different meaning (functions obtained from >> polynomial functions by use of quotients, positive fractional >> exponents, addition, and multiplication...) >Hmm. Breaking your vow, Arturo? Not at all. This is what I said: I will not stop posting in your threads (no such thing, this is a public forum); but I will keep my word from the offer before. I will not follow up on anything else you post, not even by piggybacking. I may reply to others who ask specific mathematical questions. It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Non-uniqueness, polynomial factors >> [.snip.] > You have not shown that that is possible with algebraic integer > functions. So we can only assume they are algebraic functions. >> I would suggest algebraic number-valued functions, since algebraic >> functions has a different meaning (functions obtained from >> polynomial functions by use of quotients, positive fractional >> exponents, addition, and multiplication...) > Hmm. Breaking your vow, Arturo? In any case, it's nice to hear > from you again under this tent. No, this was an answer to a mathematical query of *another* poster. BTW, algebric functions can be obtained by other operations, like the inverse bijection of x^5+x > Rick === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? You mean, for example, a_1(x) = 0, a_2(x) = 0 {both for all x}, and a_3(x) = 60025*x^3 - 3675*x^2 - 72*x + 3, perhaps ? Andrzej > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factors >>(5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = >> 49(300125 x^3 - 18375 x^2 - 360 x + 22) >>which should look familiar to some of you. >>Now then, what is required for the a's to be algebraic integer functions? > You mean, for example, a_1(x) = 0, a_2(x) = 0 {both for all x}, > and > a_3(x) = 60025*x^3 - 3675*x^2 - 72*x + 3, > perhaps ? > Andrzej Heh. Actually, though, there might be some more general result hiding in all of this. Suppose that we have a cubic (for starters) polynomial p in Z[x] satisfying [some condition] and a factorization (t f_1(x) + s)(t f_2(x) + s)(t f_3(x) + t) = p(x) (where s, t, u in Z) valid for all integers x. Then [some conclusion to the effect that the only case in which the f's can be algebraic integer-valued functions is when the factorization is in some sense (like the above) trivial]. Just killing time before dinner, Rick === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken as *conclusive* proof that you cannot answer. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. Hey, it's not my fault you don't know enough mathematics to handle the question. I notice that the usual suspects are quiet. Where is Dik Winter, Nora Baron, or Rick Decker now? The question is a simple one, what is required for the a's to be algebraic integer functions? James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factors James Harris > C. Bond > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > > I've started showing that by giving you the other side of a > factorization I've used for a while by showing [same old ] > Now then, what is required for the a's to be algebraic integer functions? > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. What is an algebraic integer, Harris? === Subject: Re: Non-uniqueness, polynomial factors > > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > > Now I'll do some simple algebraic manipulations with that expression. > > Ok, let g_3(x) = f_3(x) + 3, so I have > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > and now, I multiply both sides by 7, which gives > > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > which should look familiar to some of you. > > Now then, what is required for the a's to be algebraic integer functions? > > James Harris > > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ > > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be > taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. > I notice that the usual suspects are quiet. > Where is Dik Winter, Nora Baron, or Rick Decker now? > The question is a simple one, what is required for the a's to be > algebraic integer functions? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ Since Harris does not offer any incentives for coming up with an answers to his questions, I suggest that they be ignored. I am certain that everyone can find better things to do than to pander to Harris' already oversized ego. === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > > Now I'll do some simple algebraic manipulations with that expression. > > Ok, let g_3(x) = f_3(x) + 3, so I have > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > and now, I multiply both sides by 7, which gives > > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > which should look familiar to some of you. > > Now then, what is required for the a's to be algebraic integer functions? > > James Harris > > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ > > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. > I notice that the usual suspects are quiet. > Where is Dik Winter, Nora Baron, or Rick Decker now? Sorry, but I am not continuously on-line reading news. > The question is a simple one, what is required for the a's to be > algebraic integer functions? a1, a2 and a3 are algebraic integer functions when 7 f1, 7 f2 and f3 are. I do not think there are other requirements. But as you have not shown yet that 7 f1, 7 f2 and f3 *are* algebraic integer functions, we are in limbo... -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Non-uniqueness, polynomial factors > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. It's your fault if *you* don't know enough mathematics to handle the question. Note that you still didn't answer. The ball's in *your* court. > I notice that the usual suspects are quiet. > Where is Dik Winter, Nora Baron, or Rick Decker now? > The question is a simple one, what is required for the a's to be > algebraic integer functions? Your failure to answer is conclusive proof that you cannot answer. -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: Non-uniqueness, polynomial factors > It's easy to demonstrate that polynomials can be factored in an > infinite number of ways--that is, there isn't a unique factorization. > > I've started showing that by giving you the other side of a > factorization I've used for a while by showing > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > where f_1(0) = f_2(0) = f_3(0) = 0. > > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > > Now I'll do some simple algebraic manipulations with that expression. > > Ok, let g_3(x) = f_3(x) + 3, so I have > > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > > and now, I multiply both sides by 7, which gives > > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > > which should look familiar to some of you. > > Now then, what is required for the a's to be algebraic integer functions? > > James Harris > > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ > You tell us, Mr. Magnificent Smart Ass. Your failure to respond will be taken > as *conclusive* proof that you cannot answer. > Hey, it's not my fault you don't know enough mathematics to handle the > question. I'm sure C. Bond can answer his own question, he's just testing you. You are not smart enough to realize that. David Moran > I notice that the usual suspects are quiet. > Where is Dik Winter, Nora Baron, or Rick Decker now? > The question is a simple one, what is required for the a's to be > algebraic integer functions? > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Non-uniqueness, polynomial factors Discussion, linux) > I'm sure C. Bond can answer his own question, he's just testing you. You are > not smart enough to realize that. Why would you suppose that? Do *you* know the answer? If not, why suppose that it's easy enough for C. Bond to know the answer? Maybe it's very difficult. Maybe it's ill-posed. I won't make any guesses about anyone's knowledge unless they indicate what they actually know. Let's leave the posing to James, shall we? -- Mathematicians are rather important in the infrastructures of many organizations that protect civilization. I've determined that they are a consistent security risk, and seem to have other agendas, other loyalties beyond loyalty to their respective nations. -- James Harris === Subject: Re: Non-uniqueness, polynomial factors > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > which should look familiar to some of you. > Now then, what is required for the a's to be algebraic integer functions? > James Harris A fool can ask questions that wise men cannot answer. It is not clear that the fool asking this question has any idea of the answer to his own question. If past history is anything to go by, he is asking, rather than stating, because he doesn't know the answer, and will steal the result from anyone foolish enough to give him the answer. === Subject: [JSH] Just Stupid Harrassment of Mathematicians: Re: Non-uniqueness, polynomial factors > Now then, what is required for the a's to be algebraic integer functions? Once again, the top number theorist in the universe realizes that all of his arguments are fatally flawed and now turns the table once again! What happened, again? Have you finally realized, after a year of chest pounding, that you have no clue of what you speak? You are such a nasty person and I would not like to know you personally as you have no class and do not know how to have a reasonable discussion on a topic. Month upon month of objections has once again brought you to a point of begging for others to show you a proof of something that you cannot possibly do yourself. You are so sad, but many of us love the ABSOLUTE NONSENSE that you continue to spew out! I think you are up to eight years now of wasting your life. Perhaps more alcohol is in order! Please take some algebra courses! === Subject: Re: [JSH] Just Stupid Harrassment of Mathematicians: Re: Non-uniqueness, polynomial factors > Now then, what is required for the a's to be algebraic integer functions? > Once again, the top number theorist in the universe realizes that all of his > arguments are fatally flawed and now turns the table once again! > What happened, again? Have you finally realized, after a year of chest > pounding, that you have no clue of what you speak? > You are such a nasty person and I would not like to know you personally as > you have no class and do not know how to have a reasonable discussion on a > topic. > Month upon month of objections has once again brought you to a point of > begging for others to show you a proof of something that you cannot possibly > do yourself. > You are so sad, but many of us love the ABSOLUTE NONSENSE that you > continue to spew out! I think you are up to eight years now of wasting your > life. Perhaps more alcohol is in order! > Please take some algebra courses! James claims that he uses PDE's, but he has shown repeatedly that he has a poor grasp on algebra and calculus, much less DE. He really needs to take some math classes and maybe he'd see we're not the liars he makes us out to be. David Moran === Subject: Re: Really big number > Here's another attempt at defining a really big number. > Let G be Graham's number and A: N^2 -> N be Ackermann's function. > Let f: N^2 -> N be defined as: > f(i, 0) = i > f(i, j) = f(A(i, i), j-1) if j>0 > My number is f(G, G). > This number is pretty big. Has anyone ever thought of a bigger number? I'm willing to bet your number is smaller than the 1000th busy beaver number === Subject: death by trig please help me with this: *show* me how to find *EXACT* value of x for: cot x = 3, pi< x < 3pi/2 thank you very, very much!!! === Subject: Re: death by trig > please help me with this: > *show* me how to find *EXACT* value of x for: > cot x = 3, pi< x < 3pi/2 > thank you very, very much!!! You can do arctangents in terms of complex logarithms. Is that what you want? -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: death by trig >please help me with this: >*show* me how to find *EXACT* value of x for: >cot x = 3, pi< x < 3pi/2 tan x = 1/3 The solution in (-pi/2, pi/2) is arctan(1/3), and tan has period pi, so x = pi + arctan(1/3). x is not a rational multiple of pi. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: death by trig > The solution in (-pi/2, pi/2) is arctan(1/3), and tan has period pi, so > x = pi + arctan(1/3). > x is not a rational multiple of pi. Maybe the teacher wanted to hear something like tetrahedral angle? -- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.de For our chemistry workgroup,remove math from the address === Subject: Re: death by trig You can work from the defn of cot and the trig identity (cosx)^2+(sinx)^2 = 1 so you have: (i) cot(x) = 3 by defn: (ii) cot(x) = cos(x)/sin(x) so (i) and (ii) imply (iii) cos(x) = 3 sin(x) Now using the trig identity (cosx)^2+(sinx)^2 = 1 for this problem, we find: (iv.a) (3sinx)^2+(sinx)^2 = 1 (iv.b) sin(x) = +/- sqrt(1/10) I don't see any easy way of finding the exact value for x other than simply computing its arcsin on a calculator, etc. The answer itself does not appear to indicate that it comes from a simple combination of trig functions on the standard unit circle. MB > please help me with this: > *show* me how to find *EXACT* value of x for: > cot x = 3, pi< x < 3pi/2 > thank you very, very much!!! === Subject: Re: death by trig > please help me with this: > *show* me how to find *EXACT* value of x for: > cot x = 3, pi< x < 3pi/2 > thank you very, very much!!! I would start this by taking the reciprocal of both sides tan x = 1/3 x=arctan(1/3) Since you need a 3rd quadrant angle, add pi, so x=arctan(1/3)+pi or approximately 3.463 David Moran === Subject: Salaries of Scientists Drop During the past two years, the salaries of most scientists have decreased significantly. See http://www.jupiterscientific.org/sciinfo/sciencesalaries.html Salary data for mathematicians is also at the site. -------------------------------------------------------------- Jupiter Scientific -- dedicated to the promotion of science and scientific education through books, the internet, and other means of communication http://www.jupiterscientific.org 777777777777777777777777777777777777777777777777777777777777777777777 === Subject: Re: Salaries of Scientists Drop > During the past two years, the salaries of > most scientists have decreased significantly. > See > http://www.jupiterscientific.org/sciinfo/sciencesalaries.html > Salary data for mathematicians is also at the site. High school salaries are low .9a an indication that state and local governments do not put sufficient money into public education. Is that a logical inference? Surely the test of whether sufficient money is put into public education (or anything else) is the _outcome_. -- G.C. === Subject: Re: Salaries of Scientists Drop > > During the past two years, the salaries of > most scientists have decreased significantly. > See > http://www.jupiterscientific.org/sciinfo/sciencesalaries.html > > Salary data for mathematicians is also at the site. > High school salaries are low ? an indication that state and local > governments do not put sufficient money into public education. like i replied in a previous follow-up that did not show up, high school salaries are not low due to lack of tax payer money being dumped into the sham that is public education. there is plenty of funding. what keeps the salaries low and the abuse of the profession high is the bureaucrats who control the sham that is public education. bureaucrats get the highest salaries and get to divert funds for their political/bureaucratic infrastructures. > Is that a logical inference? no, see above. > Surely the test of whether sufficient money is put into public education > (or anything else) is the _outcome_. neither the outcome nor salaries stand to be any better by dumping even more tax payer money into the sham that is public education. why? because the bureaucrats who control education will not have it any other way. ps posting from mathforum.org bites... === Subject: Re: Salaries of Scientists Drop Pay Scales for Educators Across All Academic Subjects Hey, when you include law schools and business schools, those academic salaries really look impressive! When they conclude: full professors are now paid a little better than industrial scientists it is a bit strange. Do yoiu think full professors in scientific subjects are paid better than industrial scientists in the same subject. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Salaries of Scientists Drop > When they conclude: > full professors are now paid a little better than industrial > scientists > it is a bit strange. Do yoiu think full professors in scientific > subjects are paid better than industrial scientists in the same > subject. It is possible that full professors are now paid better than industrial scientists. But the gap will shrink as more and more students realize it's not worth anymore to attend college, as there are no jobs for educated people in US. What will happen to talented individuals then? They'll become smart criminals as in Russia. === Subject: Re: Salaries of Scientists Drop > When they conclude: > full professors are now paid a little better than industrial > scientists > it is a bit strange. Do yoiu think full professors in scientific > subjects are paid better than industrial scientists in the same > subject. > It is possible that full professors are now paid better than industrial scientists. > But the gap will shrink as more and more students realize it's not worth anymore to > attend college, as there are no jobs for educated people in US. > What will happen to talented individuals then? They'll become smart criminals as in > Russia. When education becomes outlawed, only outlaws will be educated. Jon Miller === Subject: Re: Salaries of Scientists Drop >> What will happen to talented individuals then? They'll become smart >> criminals as in Russia. >When education becomes outlawed, only outlaws will be educated. It's *Doctor* Evil. I didn't spend six years in evil medical school to be called 'Mister', thank you very much. === Subject: Re: Salaries of Scientists Drop > When they conclude: > full professors are now paid a little better than industrial > scientists > it is a bit strange. Do yoiu think full professors in scientific > subjects are paid better than industrial scientists in the same > subject. > It is possible that full professors are now paid better than industrial > scientists. > But the gap will shrink as more and more students realize it's not worth > anymore to > attend college, as there are no jobs for educated people in US. > What will happen to talented individuals then? They'll become smart criminals > as in > Russia. Those who go to college merely to qualify for higher paying jobs are missing the point. === Subject: Re: Salaries of Scientists Drop >>When they conclude: >>full professors are now paid a little better than industrial >>scientists >>it is a bit strange. Do yoiu think full professors in scientific >>subjects are paid better than industrial scientists in the same >>subject. > It is possible that full professors are now paid better than industrial scientists. > But the gap will shrink as more and more students realize it's not worth anymore to > attend college, as there are no jobs for educated people in US. > What will happen to talented individuals then? They'll become smart criminals as in > Russia. Astute. As I've put it before the choice is becoming obvious: 1) A tax on wealth. 2) Attacks on wealth. === Subject: Re: Salaries of Scientists Drop > During the past two years, the salaries of > most scientists have decreased significantly. > See > http://www.jupiterscientific.org/sciinfo/sciencesalaries.html > Salary data for mathematicians is also at the site. > -------------------------------------------------------------- > Jupiter Scientific -- dedicated to the promotion of science > and scientific education through books, the internet, > and other means of communication > http://www.jupiterscientific.org > 777777777777777777777777777777777777777777777777777777777777777777777 That's good, right? It means more people are interested in science. -- Ron Hardin rhhardin@mindspring.com On the internet, nobody knows you're a jerk. === Subject: Parseval's Theorem to show that Zeta(2) is Pi^2 over 6 Hi everybody I'm having a hard time doing this homework problem. Here's the problem, straight out of Rudin: Let f(x) = x for 0 <= x , 2Pi. Apply Parseval's theorem to show that Zeta(2) is (Pi^2)/6. (of course, Rudin doesn't use Zeta, but it's much easier to type than the sum). Parseval's theorem says that the integral of the modulus of the function f from -Pi to Pi equals the sum of the moduli of the Fourier coefficients of f squared from -inf to inf. Here's the first problem. Rudin doesn't tell us how to deal with the fact that f is not defined on (-Pi, 0). I did a bunch of scratch work and figured out that it really doesn't matter, so long as you transform your function from f(x) = x to g(x) = x + Pi. (I wish Rudin would have developed the section on Fourier analysis a little more. This is in Principles of Mathematical Analysis, by the way) The real trouble is that I keep getting that the moduli of the Fourier coefficients are all 1/(n^2). This is trouble, since the integral of |f|^2 from 0 to 2Pi is 8 (Pi^3)/3. So I'm off by a factor of 8! I've done the integrals about 20 times each, and I've verified with Mathematica, so I know the integrals are right. Which leads me to suspect that I'm integrating over the wrong region. (I'm doing (0, 2Pi)) Can anybody give me a hint? Alex === Subject: Re: Parseval's Theorem to show that Zeta(2) is Pi^2 over 6 > Hi everybody > I'm having a hard time doing this homework problem. Here's the > problem, straight out of Rudin: > Let f(x) = x for 0 <= x , 2Pi. Apply Parseval's theorem to show that > Zeta(2) is (Pi^2)/6. (of course, Rudin doesn't use Zeta, but it's > much easier to type than the sum). > Parseval's theorem says that the integral of the modulus of the > function f from -Pi to Pi equals the sum of the moduli of the Fourier > coefficients of f squared from -inf to inf. No, Parseval says that (1/2Pi)*(integral of the modulus squared of the function f from -Pi to Pi) equals the sum of the moduli squared of the Fourier coefficients of f from -inf to inf. > Here's the first problem. Rudin doesn't tell us how to deal with the > fact that f is not defined on (-Pi, 0). I did a bunch of scratch work > and figured out that it really doesn't matter, so long as you > transform your function from f(x) = x to g(x) = x + Pi. (I wish Rudin > would have developed the section on Fourier analysis a little more. > This is in Principles of Mathematical Analysis, by the way) > The real trouble is that I keep getting that the moduli of the Fourier > coefficients are all 1/(n^2). No, you get that for the moduli *squared* (which is probably what you mean). More importantly, what about the 0th Fourier coefficient? === Subject: Re: Parseval's Theorem to show that Zeta(2) is Pi^2 over 6 > Hi everybody > > I'm having a hard time doing this homework problem. Here's the > problem, straight out of Rudin: > > Let f(x) = x for 0 <= x , 2Pi. Apply Parseval's theorem to show that > Zeta(2) is (Pi^2)/6. (of course, Rudin doesn't use Zeta, but it's > much easier to type than the sum). > > Parseval's theorem says that the integral of the modulus of the > function f from -Pi to Pi equals the sum of the moduli of the Fourier > coefficients of f squared from -inf to inf. > No, Parseval says that (1/2Pi)*(integral of the modulus squared of the > function f from -Pi to Pi) equals the sum of the moduli squared of the > Fourier coefficients of f from -inf to inf. Yes, you're right. > Here's the first problem. Rudin doesn't tell us how to deal with the > fact that f is not defined on (-Pi, 0). I did a bunch of scratch work > and figured out that it really doesn't matter, so long as you > transform your function from f(x) = x to g(x) = x + Pi. (I wish Rudin > would have developed the section on Fourier analysis a little more. > This is in Principles of Mathematical Analysis, by the way) > > The real trouble is that I keep getting that the moduli of the Fourier > coefficients are all 1/(n^2). > No, you get that for the moduli *squared* (which is probably what you > mean). More importantly, what about the 0th Fourier coefficient? You're right again. I've just been working the n'th coefficient, and I keep getting 1/(n^2), so I would assume that if n = 0, the coefficient is not defined. Am I missing something? Alex === Subject: Re: Parseval's Theorem to show that Zeta(2) is Pi^2 over 6 >> Hi everybody >> >> I'm having a hard time doing this homework problem. Here's the >> problem, straight out of Rudin: >> >> Let f(x) = x for 0 <= x , 2Pi. Apply Parseval's theorem to show that >> Zeta(2) is (Pi^2)/6. (of course, Rudin doesn't use Zeta, but it's >> much easier to type than the sum). >> >> Parseval's theorem says that the integral of the modulus of the >> function f from -Pi to Pi equals the sum of the moduli of the Fourier >> coefficients of f squared from -inf to inf. >> No, Parseval says that (1/2Pi)*(integral of the modulus squared of the >> function f from -Pi to Pi) equals the sum of the moduli squared of the >> Fourier coefficients of f from -inf to inf. >> >Yes, you're right. >> Here's the first problem. Rudin doesn't tell us how to deal with the >> fact that f is not defined on (-Pi, 0). I did a bunch of scratch work >> and figured out that it really doesn't matter, so long as you >> transform your function from f(x) = x to g(x) = x + Pi. (I wish Rudin >> would have developed the section on Fourier analysis a little more. >> This is in Principles of Mathematical Analysis, by the way) >> >> The real trouble is that I keep getting that the moduli of the Fourier >> coefficients are all 1/(n^2). >> No, you get that for the moduli *squared* (which is probably what you >> mean). More importantly, what about the 0th Fourier coefficient? >You're right again. I've just been working the n'th coefficient, >and I keep getting 1/(n^2), so I would assume that if n = 0, the >coefficient is not defined. Am I missing something? Yes. That coefficient cannot be undefined! The 1/n^2 doesn't work for n = 0. But somewhere in the derivation of the 1/b^2 there is a step which is not valid unless n is non-zero (in particular, there's a step where you divided by n, and you can't do that when n = 0.) So you have to calculate the 0-th coefficient by another method. >Alex David C. Ullrich === Subject: Re: Parseval's Theorem to show that Zeta(2) is Pi^2 over 6 >> >> More importantly, what about the 0th Fourier coefficient? > Yes. That coefficient cannot be undefined! The 1/n^2 doesn't > work for n = 0. But somewhere in the derivation of the 1/b^2 there > is a step which is not valid unless n is non-zero (in particular, > there's a step where you divided by n, and you can't do that > when n = 0.) So you have to calculate the 0-th coefficient by > another method. > David C. Ullrich Alex === Subject: Cubic Equation: Cardan's Method I have a question about using Cardan's Method to solve a cubic equation: Okay, so say we have Ax^3 + Bx^2 + Cx + D = 0. We make the substitution x = y-B/3 to eliminate the x^2 term. Then we get something of the form A'x^3 + C'x + D'=0. Divide everything by A' so that the coefficient of x^3 is 1. and we end up with an equation of the form t^3 + pt + q=0 Now, we set t = u+v to obtain u^3 + v^3 + (3uv + p)(u+v)+q = 0 and impose the condition uv = -p/3 to eliminate the term in the above equation. So the problem is reduced to solving the system u^3 + v^3 = -q uv = -p/3 We can find u and v easily, as u^3 and v^3 are the roots of the equation x^2 + qx - p^3/27=0 So after we find u^3 and v^3 we take their cube roots, add the result, and that gives us one root of the cubic equation. My question is how do you find the other roots? Obviously you can use synthetic division but this is not really feasible if the root just found is irrational or complex. There must be some way to extract them using the system of equations u^3 + v^3 = -q uv = -p/3 Secondly, it seems that a cubic has three real roots if and only if 27q^2+4p^3=0 (where p and q are as above). Does this mean that the probability of a cubic having three real roots is equal to 0, if the coefficients A, B, C, and D are chosen randomly from (0,1)? Nobody === Subject: Re: Cubic Equation: Cardan's Method > I have a question about using Cardan's Method to solve a cubic > equation: > Okay, so say we have Ax^3 + Bx^2 + Cx + D = 0. > We make the substitution x = y-B/3 to eliminate the x^2 term. Then we > get something of the form A'x^3 + C'x + D'=0. Divide everything by A' > so that the coefficient of x^3 is 1. and we end up with an equation > of the form > t^3 + pt + q=0 > Now, we set t = u+v to obtain > u^3 + v^3 + (3uv + p)(u+v)+q = 0 > and impose the condition uv = -p/3 to eliminate the term in the above > equation. So the problem is reduced to solving the system > u^3 + v^3 = -q > uv = -p/3 > We can find u and v easily, as u^3 and v^3 are the roots of the > equation > x^2 + qx - p^3/27=0 > So after we find u^3 and v^3 we take their cube roots, add the result, > and that gives us one root of the cubic equation. My question is how > do you find the other roots? Obviously you can use synthetic division > but this is not really feasible if the root just found is irrational > or complex. There must be some way to extract them using the system > of equations > u^3 + v^3 = -q > uv = -p/3 Once you find u and v, let w be either of the primitive cube roots of 1, w = (-1 +- sqrt(3))/2, so that w^2 + w + 1 = 0, then the three roots are r1 = u + v r2 = u*w + v*w^2 = u*w + v/w r3 = u*w^2 + v*w = u/w + v*w > Secondly, it seems that a cubic has three real roots if and only if > 27q^2+4p^3=0 (where p and q are as above). Does this mean that the > probability of a cubic having three real roots is equal to 0, if the > coefficients A, B, C, and D are chosen randomly from (0,1)? Actually when 27*q^2 + 4*p^3 < 0, u and v are complex conjugates, and it follows that u*w and v/w and also u/w and v*w are complex conjugates , so all three roots are sums of a pair of conjugates and so are real. > Nobody === Subject: Re: Cubic Equation: Cardan's Method Nobody > Okay, so say we have Ax^3 + Bx^2 + Cx + D = 0. [and get] > t^3 + pt + q=0 > Now, we set t = u+v to obtain > u^3 + v^3 + (3uv + p)(u+v)+q = 0 > and impose the condition uv = -p/3 to eliminate the term in the above > equation. So the problem is reduced to solving the system > u^3 + v^3 = -q > uv = -p/3 > We can find u and v easily, as u^3 and v^3 are the roots of the > equation > x^2 + qx - p^3/27=0 > So after we find u^3 and v^3 we take their cube roots, add the result, > and that gives us one root of the cubic equation. My question is how > do you find the other roots? You have three choices for each of the cube roots (say w, aw, and aaw, where a = (-1+sqrt(3))/2). That gives you nine possibilities for the original cubic, but only three choices will satisfy uv = -p/3. The solution in surds of the cubic must involve complex numbers even if all three roots are real. I read somewhere that Cardano, Tartaglia, and Ferrari had little in the way of algebraic notation, and used geometric constructions in their solutions of the cubic and quartic. I haven't seen these constructions, but they must have been pretty gross :) LH === Subject: Re: Cubic Equation: Cardan's Method > I read somewhere that Cardano, Tartaglia, and Ferrari had little in the way > of algebraic notation, and used geometric constructions in their solutions > of the cubic and quartic. I think that is incorrect, at least for Cardano. -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ === Subject: Re: Cubic Equation: Cardan's Method Larry Hammick Correction > a = (-1+sqrt(3))/2) should be > a = (-1+sqrt(-3))/2) of course. LH === Subject: Maths stuff You are sitting at home, playing with your little earth model you got for christmas, and see two towns on it. One lies at [80 00 00 N 007 45 00 E] and the other at [70 00 00 S 007 45 00 E]. Now you are becoming curious. You want two questions answered: 1. How far away are these two towns from each other? 2. If your car drives 200 km/h, how long will it take you to drive from one town to the other? === Subject: Re: Maths stuff > You are sitting at home, playing with your little earth model you got > for christmas, and see two towns on it. > One lies at [80 00 00 N 007 45 00 E] and the other at [70 00 00 S 007 > 45 00 E]. > Now you are becoming curious. You want two questions answered: > 1. How far away are these two towns from each other? > 2. If your car drives 200 km/h, how long will it take you to drive > from one town to the other? I assume that, given the lattidues and longitudes, you want to know the shortest distance between 2 towns on the surface of the Earth. If the approximation that the Earth is spherical is made, it is fairly easy to give a general answer to 1. Let r_A and r_B be the position vectors of towns A and B with respect to the centre of the Earth, and let R be the radius of the Earth. Then d, the distance between A and B is given by d = R*a, where a is the angle (in radians) between vectors r_A and r_B. This is just the definition of angle. If you want to use degrees for angle a, d = R*a*pi/180. Now, how is the angle a found? r_A dot r_B = R^2 * cos(a), so, a = arccos{(r_A dot r_B)/R^2} What about r_A dot r_B ? r_A dot r_B = x_A * x_B + y_A * y_B + z_A * z_B = R^2 * [sin(t_A)*cos(p_A)*sin(t_B)*cos(p_B) + sin(t_A)*sin(p_A)*sin(t_B)*sin(p_B) + cos(t_A)*cos(t_B)] Lattitude and longitude relate directly to spherical coordinates theta = t and phi = p. Longitude is p (take west to be negative) and t = 90 - lattitude (take south to be negative). I've chosen the convention for phi and theta usually used by physicists, which, I think, is opposite to the convention usually used by mathematicians. George === Subject: Re: Maths stuff > You are sitting at home, playing with your little earth model you got > for christmas, and see two towns on it. > One lies at [80 00 00 N 007 45 00 E] and the other at [70 00 00 S 007 > 45 00 E]. > Now you are becoming curious. You want two questions answered: Indeed I am. 80 deg N, 7 deg 45' E is North tip of Norwegian Sea 70 deg S, 7 deg 45' E is the South end of and entrance of the Ross Sea. > 1. How far away are these two towns from each other? There are no towns there. At best a research station at North tip of Norwegian Sea and at most a research vessel at the entrance of the Ross Sea. > 2. If your car drives 200 km/h, how long will it take you to drive > from one town to the other? Car? Surely you jest. For 200 km/hr you better charter a piper cub plane and be prepared for in flight refueling unless you intend the 200 km/hr to be average flight time including stop overs. That seems more practical. As were along the same longitude, the nautical distance to cover is 150 degrees along the great circle of 7 deg 45' E. Now you need to look up the radius of the Earth, pi and the oblateness of the Earth. Because of rotation, the N-S axial diameter is shorter than the equatorial diameter. Here now you have big problem, but much time during the thousands of miles of slow flight time, to read up and learn about arc lengths of ellipses and other oblate figures. They are no simple matter, being left to advance calculus courses. === Subject: Re: Maths stuff > You are sitting at home, playing with your little earth model you got > for christmas, and see two towns on it. > One lies at [80 00 00 N 007 45 00 E] and the other at [70 00 00 S 007 > 45 00 E]. > Now you are becoming curious. You want two questions answered: > 1. How far away are these two towns from each other? > 2. If your car drives 200 km/h, how long will it take you to drive > from one town to the other? And then you sigh, Father Christmas doesn't have this problem, because he always flies from the North Pole and he can tell how far his journey will be from the latitude of his destination. === Subject: how to use matlab to do multi-variable curve fitting? I want to do a curve fitting which has three variables... and in the following format: y=a+b*t1.^c+d*t2.^e+f*t3.^g, a, b, c, d, e, f, g are the values I need to determine... mymodel = fittype('a+b*t1.^c+d*t2.^e+f*t3.^g','independent','t1','independent','t2','i ndependent','t3'); fit1= fit(xx,xx1, xx2, yy,mymodel) where xx, xx1, xx2 are my input variables, and yy are the desired outputs.. but matlab does not recognize the above command and report error... I tried a number of times but fail to make it... Can anybody tell me how to do that? -Walala === Subject: Re: how to use matlab to do multi-variable curve fitting? Please see www.estlab.com > I want to do a curve fitting which has three variables... and in the > following format: > y=a+b*t1.^c+d*t2.^e+f*t3.^g, > a, b, c, d, e, f, g are the values I need to determine... > mymodel = fittype('a+b*t1.^c+d*t2.^e+f*t3.^g','independent','t1','independent','t2','i > ndependent','t3'); > fit1= fit(xx,xx1, xx2, yy,mymodel) > where xx, xx1, xx2 are my input variables, and yy are the desired outputs.. > but matlab does not recognize the above command and report error... > I tried a number of times but fail to make it... > Can anybody tell me how to do that? > -Walala === Subject: Re: Reason for usage of term two's complement? >I vaguely remember long, long, ago working on a computer where the >common way to write binary numbers was in octal, and there they >talked about 7's complement and 8's complement. I've worked on such computers, and I never heard the integer representation called anything but ones' complement. But I did hear a different, absolutely appalling, usage: octal K for 512. K, of course, is 1000 and not 1024. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Non self-identity at 10:02 PM, W. Dale Hall said: >objects to be equal to themselves. Despite my admittedly casual >efforts at understanding the arguments and assertions, I find the >topic to be hard to motivate, and harder to follow. I can't speak to motivation, but it is perfectly proper to construct a consistent Mathematical system and investigate what consequences its axioms have. The work may turn out to have no utility[1], it may turn out to be fruitful, or we may leave it for a later era to decide. Think of it as a game, with a warning that even if you contrive the rules to eliminate[2] the likelihood of practical applications, someone else may discover applications; even if you intend to apply it to practical problems, it may turn out to be useless. >I would appreciate being set straight), is that we >*typically* deal with entities that cannot be distinguished by a >mere listing of the elements. Actually, it turns out that if we are careful with our definitions that they can be distinguished by their elements. >After all, all finite- >dimensional manifolds can be viewed as constituting different >entities existing on a set of cardinality c. A manifold is a list of elements, one of which is an atlas, and a distinct manifold does have distinct elements. The ambiguity with manifolds is how we deal with equivalent atlases. Again, we have to be careful with our definitions, and there are several different ways to handle it. >Similarly, all Lie groups are among the entities existing on that same set. And again, A Lie group has different elements from the underlying topological space and from the underlying group. >Virtually all the mathematics I've seen involves entities that are >*more* than the underlying sets, No. You're just not used to precise definitions. >If all the tempest is about the claim that there exist things that >are not *just* sets, it seems a bit overblown. That depends upon what is is. The things that one deals with in most of Mathematics are just sets, but when you start talking about, e.g., large categories, then you need a bit more machinery. I see no need to abandon extensionality, but even if it is not necessary there is no reason to not investigate how much you can do without it. [1] In the Mathematical sense; utility to other Mathematicians, not to practical problems. [2] G. H. Hardy's work in Number Theory, -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: neighborhood -topology~~~ at 01:02 PM, Eli said: >The two definitions are NOT equivalent. Yes they are. >If you use the Lipschutz definition >of neighborhood, you have to use a corresponding definition of >continuous, which mentions open sets explicitly. No. In both cases, f is continuous a f iff for every neighborhoold N of f(x), f^-1(N) contains a neighborhood of x. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: neighborhood -topology~~~ at 08:29 PM, hot-girl said: >f^-1{b}={1,2} : not in Tx....thus not open {1} in Tx[1] 1 in {1} {1} C {1,2} So f is continuos at 1. [1] There's a problem with your notation; if you're defining the topology in terms of neighborhoods then you need to define the neighborhoods of each point; only an open set is a neighborhood of each of its points. So you should really have written, e.g., Tx_1, Ty_b. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: (curve fitting) which curve model is good for fitting this curve(image attached: 44KB) www.estlab.com please > Please take a look for me... which curve model can be used to fit this > curve? please see that attaced image... > Currently I am using this matlab code: > mymodel = fittype('a+b*cos(c*t+d)','independent','t'); > fit1= fit(xx,yy,mymodel) > but it does not give very good fitting... > can anybody give me some help ? > -Walalal === Subject: Re: Transfinite Aleph and Beth ordinal sequences >> The Aleph sequence is the next largest cardinal or initial >> ordinal. >> The Beth sequence is the next cardinal exponentiation of 2. >> aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| } >> aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal >> beth_0 = aleph_0; beth_tau+1 = 2^beth_tau >> beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal >> Do these sequence, especially the Beth sequence have fixed points? >for any f normal (see below), the sequence f(0), f(f(0), f(f(f(0),... >has a limit, which is a fixed point. This works for aleph and for >beth. >> Is Beth_(omega_0) a fixed point? >> Beth_(omega_0) = 2^Beth_(omega_0)? >No, as 2^A>A (from Cantor). Anyway, Beth (omega_0)= 2^(2^...)>> >omega_0, so Beth(Beth(omega0))>= Beth(omega_0+1)>Beth(omega0_)... omega_0 is a fixed point of Beth; Beth_(omega_0) = omega_0 Beth_(omega_0 + 1) = 2^Beth_(omega_0) = 2^omega_0 I was claiming Beth_(omega_0) = 2^(omega_0) thinking Beth is exponentation of 2 which it is, but only for successor ordinals, and at the same time claiming omega_0 is a fixed point of Beth ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences > >> The Aleph sequence is the next largest cardinal or initial > >> ordinal. > >> The Beth sequence is the next cardinal exponentiation of 2. > >> aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| } > >> aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal > >> beth_0 = aleph_0; beth_tau+1 = 2^beth_tau > >> beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal > >> Do these sequence, especially the Beth sequence have fixed points? > >for any f normal (see below), the sequence f(0), f(f(0), > f(f(f(0),... >has a limit, which is a fixed point. This works for > aleph and for >beth. > >> Is Beth_(omega_0) a fixed point? > >> Beth_(omega_0) = 2^Beth_(omega_0)? > >No, as 2^A>A (from Cantor). Anyway, Beth (omega_0)= 2^(2^...)>> > >omega_0, so Beth(Beth(omega0))>= Beth(omega_0+1)>Beth(omega0_)... > omega_0 is a fixed point of Beth; Beth_(omega_0) = omega_0 No, no: read again your definition : Beth_0 *is* already omega_0, and Beth is monotonous, so Beth_1>omega_0 (it is omega_1 if CH holds) and Beth(omega_0) is huge.... > Beth_(omega_0 + 1) = 2^Beth_(omega_0) = 2^omega_0 > I was claiming Beth_(omega_0) = 2^(omega_0) thinking Beth is > exponentation of 2 which it is, but only for successor ordinals, > and at the same time claiming omega_0 is a fixed point of Beth > ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences >> The Aleph sequence is the next largest cardinal or initial >> ordinal. >> The Beth sequence is the next cardinal exponentiation of 2. >> aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| } >> aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal >> beth_0 = aleph_0; beth_tau+1 = 2^beth_tau >> beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal > for any f normal (see below), the sequence f(0), f(f(0), > f(f(f(0),... >has a limit, which is a fixed point. > This works for aleph and for >beth. >> omega_0 is a fixed point of Beth; Beth_(omega_0) = omega_0 >No, no: read again your definition : Beth_0 *is* already omega_0, and >Beth is monotonous, so Beth_1>omega_0 (it is omega_1 if CH holds) >and Beth(omega_0) is huge.... The first fixed point of Beth is Beth(omega_0) Beth(Beth(omega_0)) = Beth(omega_0) In the event of CH, Beth(omega_0) = Aleph(omega_0) = omega_(omega_0) Beth_(omega_0) is much much bigger that the scope of most mathematics. However Aleph_(omega_0) shan't make that claim. I'd suspect some model has been or could be constructed with Aleph_(omega_0) = c or Aleph_(omega_0) = 2^c Riddle of the Day: are inaccessible ordinals outside the observable universe? ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences > >> The Aleph sequence is the next largest cardinal or initial > >> ordinal. > >> The Beth sequence is the next cardinal exponentiation of 2. > >> aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| > } >> aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal > >> beth_0 = aleph_0; beth_tau+1 = 2^beth_tau > >> beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal > > for any f normal (see below), the sequence f(0), f(f(0), > > f(f(f(0),... >has a limit, which is a fixed point. > > This works for aleph and for >beth. > >> omega_0 is a fixed point of Beth; Beth_(omega_0) = omega_0 > >No, no: read again your definition : Beth_0 *is* already omega_0, > and >Beth is monotonous, so Beth_1>omega_0 (it is omega_1 if CH > holds) >and Beth(omega_0) is huge.... > straight. The first fixed point of Beth is Beth(omega_0) > Beth(Beth(omega_0)) = Beth(omega_0) No , wrong again (see my previous posts). As Beth(alpha)=Beth(beta) => alpha=beta, and Beth (omega0)>omega0... > In the event of CH, > Beth(omega_0) = Aleph(omega_0) = omega_(omega_0) > Beth_(omega_0) is much much bigger that the scope of most mathematics. > However Aleph_(omega_0) shan't make that claim. > I'd suspect some model has been or could be constructed with > Aleph_(omega_0) = c This is impossible (a classical result of Hart.9ag), as c^omega0=c, and it is easy to prove that Aleph_(omega_0)^omega0 >Aleph_(omega_0) >or Aleph_(omega_0) = 2^c Same remark, same impossibility... > Riddle of the Day: > are inaccessible ordinals outside the observable universe? > ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences === Subject: Re: Transfinite Aleph and Beth ordinal sequences >> aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| } >> aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal >> beth_0 = aleph_0; beth_tau+1 = 2^beth_tau >> beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal > for any f normal (see below), the sequence f(0), f(f(0), > f(f(f(0),... >has a limit, which is a fixed point. > This works for aleph and for >beth. >> The first fixed point of Beth is Beth(omega_0) >> Beth(Beth(omega_0)) = Beth(omega_0) >No, wrong again (see my previous posts). >As Beth(alpha)=Beth(beta) => >alpha=beta, and Beth (omega0)>omega0... Beth_0 = omega_0 Beth_1 = c Beth_omega0 = sup { Beth_n | n < omega0 } Beth(0) = omega0 Beth^2(0) = Beth_omega0 [ = Aleph_omega0 = omega_omega0 with GCH ] that's whopping and it's only the very beginning, like in orbit easily circumscribing almost all mathematics. Beth^3(0) = Beth_(Beth_omega0) leaves the solar system Beth^4(0) much much bigger yet, leaves the galaxy Beth^5(0) leaves the large Virgo galactic cluster ... The first fixed point of Beth is Beth^omega0(0) = sup { Beth^n(0) | n < omega0 } = Union { Beth^n(0) | n < omega0 } which is given to us by the axiom of replacement in conjunction with omega0, from the axiom of infinity The next fixed point of Beth is Beth^omega0(Beth^omega0(0) + 1) ie Beth^omega0(ordinal successor(Beth^omega0(0))) Further Oswald Veblen's theorem asserts phi, the fixed points of Beth is continuous, that phi itself has even bigger fixed points, etc. ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences === > Subject: Re: Transfinite Aleph and Beth ordinal sequences > aleph_0 = omega_0; aleph_tau+1 = least { xi | aleph_tau < |xi| } > aleph_tau = sup { aleph_xi | xi < tau }, tau limit ordinal > beth_0 = aleph_0; beth_tau+1 = 2^beth_tau > beth_tau = sup { beth_xi | xi < tau }, tau limit ordinal >> for any f normal (see below), the sequence f(0), f(f(0), >> f(f(f(0),... >has a limit, which is a fixed point. >> This works for aleph and for >beth. > >> The first fixed point of Beth is Beth(omega_0) > >> Beth(Beth(omega_0)) = Beth(omega_0) > >No, wrong again (see my previous posts). > >As Beth(alpha)=Beth(beta) => > >alpha=beta, and Beth (omega0)>omega0... > Beth_0 = omega_0 > Beth_1 = c > Beth_omega0 = sup { Beth_n | n < omega0 } > Beth(0) = omega0 > Beth^2(0) = Beth_omega0 [ = Aleph_omega0 = omega_omega0 with GCH ] > that's whopping and it's only the very beginning, like in orbit > easily circumscribing almost all mathematics. > Beth^3(0) = Beth_(Beth_omega0) leaves the solar system > Beth^4(0) much much bigger yet, leaves the galaxy > Beth^5(0) leaves the large Virgo galactic cluster > ... > The first fixed point of Beth is > Beth^omega0(0) = sup { Beth^n(0) | n < omega0 } > = Union { Beth^n(0) | n < omega0 } > which is given to us by the axiom of replacement in > conjunction with omega0, from the axiom of infinity > The next fixed point of Beth is > Beth^omega0(Beth^omega0(0) + 1) > ie > Beth^omega0(ordinal successor(Beth^omega0(0))) > Further Oswald Veblen's theorem asserts phi, the fixed points of Beth > is continuous, that phi itself has even bigger fixed points, etc. All this looks ok now. For your next mind-boggling trip, may i suggest you have a look at inaccessible cardinals? > ---- === Subject: Re: Transfinite Aleph and Beth ordinal sequences > I'd suspect some model has been or could be constructed with > Aleph (omega 0) = c > This is impossible (a classical result of Hart.9ag), as c^omega0=c, Is that the correct spelling? I've also seen it as Hartogs, with an s and without an umlaut. Or are they two different people with similar names? Anyway, I think you're referring to a result of J. Konig (with umlaut of course). === Subject: Re: Line Integrals - Parametrizing Line Segments === >Subject: Re: Line Integrals - Parametrizing Line Segments >(x,y,z) coordinates? Your method seems be a little test/see-if-it-works >type. Yeah I have tips. Try to draw some different curves in 3-D, for example: The curve parametrzied by: x=t y=t z=t for t running from 0 to 1 starts at x=0,y=0,z=0 and makes a stright line to the endpoint x=1,y=1,z=1 In contrast the curve (also for t from 0 to 1): x=t y=t z=t^2 starts at x=0,y=0,z=0 and ends at x=1,y=1,z=1... but the shape of the curve is different, because the parametrization is different The first curve is nice and straight, whereas the second one is a little curvy. Or consider: x= t + (.1)Sin[10 t] y= t z= t^2 this curve has the same begining and endpoints as the first two, and it looks a lot like the second curve except that is has some small wiggles as it goes along in the x direction. Draw a bunch of curves like this... and if you have access to a computer fire up mathematica or some other such program and look at different 3d parametric plots. Here's a few more examples (for t running from 0 to 2 Pi): x=cos(t) y= sin(t) z= 0 parametrizes a circle in the x-y plane x=cos(t) y=sin(t) z=sin(t) parametrizes a ellipse whose projection on the x-y plane is a circle (the same circle as above). etc adam === Subject: Re: Dot product derivation >>As I see it, there are two 'separate' definition of dot product. >> >>One definition is that it is the multiplication of two vectors a~ >>and b~ where a~ = ax.i~ + ay.j~ + az.k~ and b~ = bx.i~ + by.j~ + bz.k~. >>This results in a~ dot b~ = ax.bx + ay.by + az.cz. >> >>The other definition is that it is the projection of one unit vector >>a~_ onto a second unit vector b~_ where a~_ = a~ / |a| and >>b~_ = b~ / |b|. In this case the definition is >>a~ dot b~ = |a|.|b|.cos(x) with x being the angle between the two >>vectors. This can also be written as cos(x) = a~_ dot b~_. >> ... >> Dave Langers gave one proof. Another approach would be to reduce the >> problem to 2D, by noting that the two vectors lie in a plane. You can >> choose an orthonormal basis for R^3 with two of the basis vectors >> lying in the plane. You will need to show that the inner product in >> R^3 is independent of the choice of orthonormal basis. >> John >How would this work? If we assume the definition > A.B=|A||B|cos(A,B) >then we can quickly deduce that this is independent of the cooridnate >system. And you can orient a cooridnate system so that the z >components of A and B are zero, but you then only get a two-terms >formula, ax bx + ay by, which does show what the OP wants but only >for that coordinate system in which az = bz = 0. But this still >doesn't show that the same quantitiy is equal to ax bx + ay by + az >bz in an arbitrary (Cartesian) coordinate system. IOW, while we know >the quantity AdotB remains fixed (as we rotate the coordinate >system), the values ax, bx, ..., all change and there is nothing here >to say that as those values change that the quantity ax bx + ay by + >az bz also doesn't change. >So I think you need more. >Disclaimer: JMHO >Alan E. Feldman No more than actually proving what I suggested. Define the dot product of two vectors a = (ax, ay, az) and b = (bx, by, bz) define ax*bx + ay*by + az*bz . It's easy to show that the value of this expression does not change if a and b are expressed in terms of another orthonormal basis. This follow easily from the facts that v.v = 1 if v has unit length, and v.w = 0 if v and w are orthogonal. Both of these results follow easily from the Pythagorean theorem. If we use an orthogonal basis having two vectors in the plane spanned by a and b, the dot product reduces to a 2D dot product, and the 2D proof that the dot product equals |A||B|cos(angle(A, B)) applies. John Mitchell === Subject: Re: Dot product derivation >> >>As I see it, there are two 'separate' definition of dot product. >> >>One definition is that it is the multiplication of two vectors a~ >>and b~ where a~ = ax.i~ + ay.j~ + az.k~ and b~ = bx.i~ + by.j~ + bz.k~. >>This results in a~ dot b~ = ax.bx + ay.by + az.cz. >> >>The other definition is that it is the projection of one unit vector >>a~_ onto a second unit vector b~_ where a~_ = a~ / |a| and >>b~_ = b~ / |b|. In this case the definition is >>a~ dot b~ = |a|.|b|.cos(x) with x being the angle between the two >>vectors. This can also be written as cos(x) = a~_ dot b~_. >> ... >> >> Dave Langers gave one proof. Another approach would be to reduce the >> problem to 2D, by noting that the two vectors lie in a plane. You can >> choose an orthonormal basis for R^3 with two of the basis vectors >> lying in the plane. You will need to show that the inner product in >> R^3 is independent of the choice of orthonormal basis. >> >> John >How would this work? If we assume the definition > A.B=|A||B|cos(A,B) >then we can quickly deduce that this is independent of the cooridnate >system. And you can orient a cooridnate system so that the z >components of A and B are zero, but you then only get a two-terms >formula, ax bx + ay by, which does show what the OP wants but only >for that coordinate system in which az = bz = 0. But this still >doesn't show that the same quantitiy is equal to ax bx + ay by + az >bz in an arbitrary (Cartesian) coordinate system. IOW, while we know >the quantity AdotB remains fixed (as we rotate the coordinate >system), the values ax, bx, ..., all change and there is nothing here >to say that as those values change that the quantity ax bx + ay by + >az bz also doesn't change. >So I think you need more. >Disclaimer: JMHO >Alan E. Feldman > No more than actually proving what I suggested. > Define the dot product of two vectors a = (ax, ay, az) and b = (bx, > by, bz) define ax*bx + ay*by + az*bz . It's easy to show that the > value of this expression does not change if a and b are expressed in > terms of another orthonormal basis. This follow easily from the facts > that v.v = 1 if v has unit length, and v.w = 0 if v and w are > orthogonal. Both of these results follow easily from the Pythagorean > theorem. If we use an orthogonal basis having two vectors in the plane > spanned by a and b, the dot product reduces to a 2D dot product, and > the 2D proof that the dot product equals |A||B|cos(angle(A, B)) > applies. > John Mitchell Yep, that works. However: You said, You will need to show that the inner product in R^3 is independent of the choice of orthonormal basis. You didn't specify which definition to start with. If you start with the ABcos definition, you immediately have that it is independent of choice of orthonormal basis. So by doing that one has already done what you suggested but that still doesn't show the equivalence of the two definitions in R^3. Disclaimer: JMHO Alan E. Feldman === Subject: Re: Dot product derivation >As I see it, there are two 'separate' definition of dot product. There are many. The two you give have conceptual problems. >One definition is that it is the multiplication of two vectors a~ >and b~ where a~ = ax.i~ + ay.j~ + az.k~ and b~ = bx.i~ + by.j~ + >bz.k~. That's only valid if your basis (i,j,k) is orthonormal. But that means that you must already have a metric in order to define the dot product, and if you already have a metric then you can define it without bothering with a basis. >The other definition is that it is the projection of one unit vector >a~_ onto a second unit vector b~_ where a~_ = a~ / |a| and b~_ = b~ >/ |b|. Again, that is only possible if you already have metric properties of your vectors. How do you define your progection? >I can't extend this to three dimensions Forget about defining it via coordinates or projects. Use the standard definition of a positive definite metric and take the subspace spanned by your two vectors. You can[1] then choose a basis in which two basis vectors are in that subspace and easily get your result. [1] Assuming that your two vectors are independent. But if they aren't then the result is trivial. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: Differentiate an inverse function (F^-1) with respect to the function F >>How does one differentiate the inverse of a function (say F^-1) with >>respect to the function itself? That is, what is (d/dF)(F^-1)? Does this >>even make sense? >Suppose F(G(x)) = x, then F'(G(x))G'(x) = 1; thus, G'(x) = 1/F'(G(x)). >What you are looking for sounds like G'(x)/F'(x) which would be > G'(x)/F'(x) = 1/(F'(G(x)) F'(x)) >Rob Johnson >take out the trash before replying >>It seems to me he's asking for G'(x) = d/dx G(x), since x = F in your >>notation. >Writing F(G(x)) = x means that G = F^{-1}, at least locally. It was not >intended to identify F(x) and x. >If G(x) = F^{-1}(x), then G'(x) = 1/F'(G(x)); that is, the reciprocal of >the derivative of F, evaluated at G(x). He asked not for (d/dx)G(x), >which would be G'(x), but (d/dF)G(x), which is G'(x)/F'(x). > ... I would write y = G(x), x = F(y), and assume the O.P. was asking for dy/dx, which can be expressed in the usual way as 1/F'(G(x)), as you indicated. Personally, I find your use of x as the argument to F a little confusing. In any case, the O.P. can clarify what he was asking if the answers given don't satisfy him. John === Subject: Re: Differentiate an inverse function (F^-1) with respect to the function F >> >>How does one differentiate the inverse of a function (say F^-1) with >>respect to the function itself? That is, what is (d/dF)(F^-1)? Does this >>even make sense? > >Suppose F(G(x)) = x, then F'(G(x))G'(x) = 1; thus, G'(x) = 1/F'(G(x)). >What you are looking for sounds like G'(x)/F'(x) which would be > > G'(x)/F'(x) = 1/(F'(G(x)) F'(x)) > >Rob Johnson >take out the trash before replying >> >>It seems to me he's asking for G'(x) = d/dx G(x), since x = F in your >>notation. >Writing F(G(x)) = x means that G = F^{-1}, at least locally. It was not >intended to identify F(x) and x. >If G(x) = F^{-1}(x), then G'(x) = 1/F'(G(x)); that is, the reciprocal of >the derivative of F, evaluated at G(x). He asked not for (d/dx)G(x), >which would be G'(x), but (d/dF)G(x), which is G'(x)/F'(x). > ... > I would write y = G(x), x = F(y), and assume the O.P. was asking for > dy/dx, which can be expressed in the usual way as 1/F'(G(x)), as you > indicated. Personally, I find your use of x as the argument to F a > little confusing. In any case, the O.P. can clarify what he was asking > if the answers given don't satisfy him. You guys are differentiating the inverse of F. That's not what the OP is asking; he's asking, what is the derivative of the inverse function, AS A FUNCTION OF F. As phrased, that question doesn't even require F to be differentiable. The only reasonable interpretation of this question is in the sense of the calculus of variations, but no one else seems to be trying to answer it that way. As a start: if we were asking this question in L(X,X), where X is a Banach space, then we answer it as follows: if we let U be the subset of invertible continuous linear maps from X to X, then U is open, and g(L) = L^(-1) maps U into L(X,X). As such it is Fr'echet-differentiable, since (L+h)^(-1) = L^(-1) - L^-1 h L^-1 + L^-1 h L^-1 h L^-1 - ... and therefore (L+h)^(-1) = L^(-1) - L^-1 h L^-1 + o(h), for ||h|| sufficiently small. Therefore Dg(L)(h) = -L^-1 h L^-1. Now consider homeomorphisms F from R^n to R^n. (Or even R to R, to start.) The question is, what space should we regard these as lying in? There's no natural way to regard the continuous functions from R to R as a Banach space; a locally convex space, sure, which can be quasi-normed as a Fr'echet space. I'm not very clear about differentiation in Fr'echet spaces. I know there are theories about differentiation in LCS's, and I'm vaguely aware that there are issues with things like the inverse function theorem, and even the chain rule, but I haven't looked at those questions in thirty-plus years. Whatever, we need to look at homeomorphisms F which lie in the INTERIOR of the set of homeomorphisms. Surely there are such (e.g. surely the identity, if perturbed by a function which is small enough, will still be a homeomorphism). That may not be as easy as it sounds, since the usual Fr'echet quasinorm on C(R) allows ||h|| to be small even if h is large near infinity. But it's a step which has to be taken before we can do anything else. Query: does the set of homeomorphisms have nonempty interior in C(R) with the quasinorm ||f|| = sum_{n=1}^infty 2^(-n) * max{|f(x)|:|x|<=n}/(1+max{|f(x)|:|x|<=n}) Then we have to consider (F+h)^-1 - F^-1 as a function of h. To do that, F will almost certainly have to be of a certain amount of smoothness. Can we channel the discussion along these lines? Surely the question has been asked before, and answered. (Probably in the negative, but that's worth hearing about too.) There's some real functional analysis here, not freshman calculus. Of course, it's possible this is NOT what the OP means, that he's a college freshman, and he has abused notation and terminology to unpleasant limits. That wouldn't be unusual in sci.math. But the question is well-posed mathematically, and why not give him the benefit of the doubt? --Ron Bruck === Subject: Re: Differentiate an inverse function (F^-1) with respect to the function F >>Suppose F(G(x)) = x, then F'(G(x))G'(x) = 1; thus, G'(x) = 1/F'(G(x)). >>What you are looking for sounds like G'(x)/F'(x) which would be >> >> G'(x)/F'(x) = 1/(F'(G(x)) F'(x)) If F(G(x)) = x, there is no need for F(x), or F'(x) to be defined for any x for which G(x) or G'(x) is defined. For example, Let G(x) = sqrt(-x), for x < 0, and F(y) = y^2 for y > 0. === Subject: Re: Differentiate an inverse function (F^-1) with respect to the function F >>>Suppose F(G(x)) = x, then F'(G(x))G'(x) = 1; thus, G'(x) = 1/F'(G(x)). >>>What you are looking for sounds like G'(x)/F'(x) which would be >>> >>> G'(x)/F'(x) = 1/(F'(G(x)) F'(x)) >If F(G(x)) = x, there is no need for F(x), or F'(x) to be defined >for any x for which G(x) or G'(x) is defined. >For example, Let G(x) = sqrt(-x), for x < 0, >and F(y) = y^2 for y > 0. I think you meant F(y) = -y^2 for y > 0. That makes F(G(x)) = x, for x < 0. It is true that the domains of F and F^{-1} do not need to intersect, but in that case, (d/dF)F^{-1} has an empty domain. However, if the domains of F and F^{-1} do intersect, and on that intersection, F and F^{-1} are both differentiable, then on that intersection, (d/dF)F^{-1} exists and equals 1/(F'(F^{-1}(x)) F'(x)). Rob Johnson take out the trash before replying === Subject: Re: Basic Characteristic functions > Why is the above wrong? > g(hf(x)) = g(f(h^-1 x), isn't that right? since hf(x) = f(h^-1 x) by > definition > now h^-1 x is just an element u in X since G acts on X, so we have > g(f(u)) = f(g^-1 u) = f(g^-1 h^-1 x).............what step is wrong? And > why is The wrong step is the first equality. It is not true that g(hf(x)) = g(f(h^-1 x)). Yes, I know that hf(x) = f(h^-1 x). The problem is that g is *not* a function; it's an element of G. Jose Carlos Santos === Subject: Action device working model, Part II This is what has happened until now... > Hello everybody, after reading the ongoing controversy about the so > called action device, I decided to give it a try in all fairness. > After all, everyone is saying it should be so easy to build. So I got > myself some wood, steel angles, springs and bolts and built it. And > guess what, it really works! > I climbed on the roof (I didn't want the thing sticking to the > ceiling and then having to do a lot of explaining), I stretched my > arms, and let go. Wow, it imediately started moving in an orthogonal > angle to the plain, as had been described. The stupid thing is I must > have been holding it upside down or something, because instead of > going off into space, it shot straight downwards and landed on the > driveway. At that moment my mom's truck drove over it and reduced it > to a miserable heap of splinters... So I'm going to have to wait to > get new materials and give it a new try. I'll give you the news as > soon as I get to it. PART II Hi again everybody, I'm back in business. Shortly after that stupid accident, I got fresh materials and fresh insights. I built a new action device with a few changes however. First the angle of the device is a lot sharper, about 45 degrees, this should provide for greater push, then I painted marks on the thing to make sure which part is up and which one is down (figuring that out wasn't easy though). I must tell you of something I had already observed the first time around which I had kept to myself because it was an unconfirmed observation: When I tried to install the springs, for some strange reason, they resisted being stretched! I felt there where undiscovered forces in action even before completion of the device! After climbing to the roof I gave the thing a nice shove, just to make sure it landed on the lawn if something went wrong this time. Again, it worked ! the thing started moving up and away from me. Then fate struck again, I haven't figured out what exactly happened, my guess is that point d detacthed itself and continued the trip without the rest of the device, or something. I'll tell you when I find out. However there is something else I wan't to share with you, I carefully observed the trajectory of the device from the time I released it to the time it landed and found it was a parable. I'm going to try and figure the implicatins this has, but I already consider it a very interesting and revolutionary fact in itself. Guenther. === Subject: Re: Action device working model, Part II > However there is something else I wan't to share with you, I carefully > observed the trajectory of the device from the time I released it to > the time it landed and found it was a parable. I'm going to try and > figure the implicatins this has, but I already consider it a very > interesting and revolutionary fact in itself. Hi Guenther, I am very interested in your story, because I have had similar results with the Action Device that I have built. I'm not going to go into detail, because my patent applications are still pending. However, I sincerly doubt that the trajectory you observed was a parable, or a usually short fictitious story that illustrates a moral attitude or a religious principle [Merriam-Webster]. More likely, it was a parabola. See for details. I'm quite convinced that my Action Device works, because, although it didn't fly off into space immediately, I left it by the curb overnight in the hopes that it was just getting warmed up, and sure enough, the next morning it was gone! I'm left to conclude that it actually did fly off with uniform acceleration into space. Good luck, -Mike === Subject: Fuffy > fuffy > fuffy Fuffy has refined its tactics. First, all that Fuffy did was to keep the name of the most current poster in a thread from appearing on Google. Fuffy has now upped the ante by fuffying *his own* last post. By doing this, not only does Fuffy keep Google from displaying the name of the last real poster, but Fuffy also makes it much more difficult to see what the most recent posting was. Who is Fuffy to impose its likes and dislikes on the readership of these newsgroups, by keeping others from reading postings it doesn't like? Only a Mathie would stoop to something like this. The failure of anyone on sci.math to express disapproval of Fuffy is an index of how readily Mathies accomodate censorship. What is to be done about Fuffy? Duh-h-h. Don't ask me. I'm a Mathie. --John The mathematicians play their game very very well. I'd suggest that you never underestimate their cleverness, or their deviousness, or even their potential for wickedness. If you don't play their game their way, they will do everything they can to psychologically destroy you. === Subject: Re: Fuffy > fuffy > > fuffy > Fuffy has refined its tactics. First, all that Fuffy did was > to keep the name of the most current poster in a thread > from appearing on Google. Free clue for the clueless: If you want to see the threads have lots of sub-threads, and the default order makes it difficult to find recent comments in more than one. > What is to be done about Fuffy? > Duh-h-h. Don't ask me. Well, I suspect virtually every Google user but you already knew about sort by date and how to find recent posters. > I'm a Mathie. === Subject: Re: Fuffy > fuffy > > fuffy > > Fuffy has refined its tactics. First, all that Fuffy did was > to keep the name of the most current poster in a thread > from appearing on Google. > Free clue for the clueless: If you want to see the > threads have lots of sub-threads, and the default > order makes it difficult to find recent comments > in more than one. > What is to be done about Fuffy? > > Duh-h-h. Don't ask me. > Well, I suspect virtually every Google user but you > already knew about sort by date and how to find > recent posters. > I'm a Mathie. > People want to know who the last poster in a thread is, without having to do a Google Advanced Search. For example, if Google shows Justin Van Twinkie as the last poster, I won't bother to read the last posting because I know that van Twinkie has nothing interesting to say. However, if Rupert McCallum O'Thuggee has Fuffied the thread, I'll have to do a Google Advanced Search, in order to grok the van Twinkie presence. Hmm. Rupert McCallum O'Thuggee, Justin van Twinkie and Randy Poe. Do I detect a m.8enage .88 trois? --John Uncle Al is our reigning senior curmudgeon. === Subject: Re: Fuffy > fuffy > > fuffy > Fuffy has refined its tactics. First, all that Fuffy did was > to keep the name of the most current poster in a thread > from appearing on Google. Fuffy has now upped the ante by > fuffying *his own* last post. By doing this, not only does > Fuffy keep Google from displaying the name of the last real > poster, but Fuffy also makes it much more difficult to see > what the most recent posting was. > Who is Fuffy to impose its likes and dislikes on the > readership of these newsgroups, by keeping others from > reading postings it doesn't like? Only a Mathie would > stoop to something like this. > The failure of anyone on sci.math to express disapproval > of Fuffy is an index of how readily Mathies accomodate > censorship. > What is to be done about Fuffy? > Duh-h-h. Don't ask me. I'm a Mathie. > --John > The mathematicians play their game very very well. I'd suggest that > you never underestimate their cleverness, or their deviousness, or > even their potential for wickedness. If you don't play their game > their way, they will do everything they can to psychologically > destroy you. fuffy === Subject: Re: Fuffy > fuffy > > fuffy > > Fuffy has refined its tactics. First, all that Fuffy did was > to keep the name of the most current poster in a thread > from appearing on Google. Fuffy has now upped the ante by > fuffying *his own* last post. By doing this, not only does > Fuffy keep Google from displaying the name of the last real > poster, but Fuffy also makes it much more difficult to see > what the most recent posting was. > > Who is Fuffy to impose its likes and dislikes on the > readership of these newsgroups, by keeping others from > reading postings it doesn't like? Only a Mathie would > stoop to something like this. > > The failure of anyone on sci.math to express disapproval > of Fuffy is an index of how readily Mathies accomodate > censorship. > > What is to be done about Fuffy? > > Duh-h-h. Don't ask me. I'm a Mathie. > > --John > The mathematicians play their game very very well. I'd suggest that > you never underestimate their cleverness, or their deviousness, or > even their potential for wickedness. If you don't play their game > their way, they will do everything they can to psychologically > destroy you. > fuffy A footless finagler named Fuffy, Whose math was sterile and stuffy Censored for tuppence Until the come-uppance Of Rupert Mccallum O'Thuggee === Subject: Re: Fuffy > fuffy > > fuffy > > Fuffy has refined its tactics. First, all that Fuffy did was > to keep the name of the most current poster in a thread > from appearing on Google. Fuffy has now upped the ante by > fuffying *his own* last post. By doing this, not only does > Fuffy keep Google from displaying the name of the last real > poster, but Fuffy also makes it much more difficult to see > what the most recent posting was. > > Who is Fuffy to impose its likes and dislikes on the > readership of these newsgroups, by keeping others from > reading postings it doesn't like? Only a Mathie would > stoop to something like this. > > The failure of anyone on sci.math to express disapproval > of Fuffy is an index of how readily Mathies accomodate > censorship. > > What is to be done about Fuffy? > > Duh-h-h. Don't ask me. I'm a Mathie. > > --John > The mathematicians play their game very very well. I'd suggest that > you never underestimate their cleverness, or their deviousness, or > even their potential for wickedness. If you don't play their game > their way, they will do everything they can to psychologically > destroy you. > > fuffy > A footless finagler named Fuffy, > Whose math was sterile and stuffy > Censored for tuppence > Until the come-uppance > Of Rupert Mccallum O'Thuggee fuffy === Subject: Re: Mathematical pr0n > Maybe some scientist (of MATHEMATICS) can explain why mathematicians > are spending their time developing mathematical ways to generate pr0n? > http://perpetualocean.com/amgallery8.html > What's wrong with human pr0n? Why are mathematicians taking the first > step on the slippery slope towards ROBOT PR0N??? The phrase machine screw comes to mind. -- http://hertzlinger.blogspot.com === Subject: Re: Infinite groups G with Aut(G) finite >> Is there some characterization of infinite groups G with the >> automorphism group Aut(G) finite ? > Dunno, but as G/Z(G) embeds in Aut(G) then they would have to > have centre of finite index. If |G:Z(G)| is finite then > Aut(G) is finite iff Out(G) is. > How about this. Consider a semidirect product of Z^2 by C_3 > (cyclic of order 3) with C_3 acting with matrix > (0 1 //-1 -1). An outer automorphism of G must act on Z^2 > and G/Z^2 = C_3. Let's consider one acting trivially on C_3. > Then its action on Z^2 commutes with that of C_3 on Z_3 > which means that it must act like a power of the above matrix. > So there is a finite index subgroup of Out(G) acting > trivially on Z^2 and G/Z^2. Are all such automorphisms of G inner. > I think up to finite index they are....can't be bothered > with details right now :-) Ignore all that --- it was total bollocks :-) Here's an example of an abelian group with only the two obvious automorphisms. Let G = {a/b in Q: a, b in Z, b squarefree}. And there are uncountably many more examples where that came from :-) -- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html Needless to say, I had the last laugh. Alan Partridge, _Bouncing Back_ (14 times) === Subject: Square sum puzzle This I found as a price puzzle (so I don't give a solution) in a German science mag: Sort the numbers 1 to 16 in a row such that each adjacent pair sums to a square. My hypothesis: For n instead of 16, this is always possible except of a finite number of cases. (The netting increases.) -- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.de For our chemistry workgroup,remove math from the address === Subject: Re: Square sum puzzle >This I found as a price puzzle (so I don't give a >solution) in a German science mag: >Sort the numbers 1 to 16 in a row such that >each adjacent pair sums to a square. Solved. >My hypothesis: For n instead of 16, this is >always possible except of a finite number of >cases. (The netting increases.) Among the first n integers are there more than two integers j such that j+k is a square for exactly one integer k? === Subject: Re: Square sum puzzle >>This I found as a price puzzle (so I don't give a >>solution) in a German science mag: >>Sort the numbers 1 to 16 in a row such that >>each adjacent pair sums to a square. >Solved. >>My hypothesis: For n instead of 16, this is >>always possible except of a finite number of >>cases. (The netting increases.) >Among the first n integers are there more than two integers j such >that j+k is a square for exactly one integer k? Adding k to j for k from 1 to j-1 gives you consecutive integers from j+1 to 2j-1. There will be at least two squares among these if sqrt(2j-1)-sqrt(j+1)>2, which is true for j > 27. Let P(j) be the second positive integer k <> j such that j+k is a square. Here's a table of P(j) for j <= 27: j P(j) 1, 8 2, 14 3, 6 4, 12 5, 11 6, 10 7, 9 8, 17 9, 16 10, 15 11, 14 12, 13 13, 12 14, 11 15, 10 16, 20 17, 19 18, 31 19, 17 20, 16 21, 15 22, 14 23, 13 24, 12 25, 24 26, 23 27, 22 Thus n is not squaresumsortable if there are more than two integers j <= n with P(j) > n. So there are no squaresumsortable integers from 3 to 14, and 18 is not squaresumsortable either. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada V6T 1Z2 === Subject: Re: Square sum puzzle > Among the first n integers are there more than two integers j such > that j+k is a square for exactly one integer k? My arguing is: Let n=O(m^2). Pick a k from 1 to n, also O(m^2). You have k+1,k+2,...k+n with O(m^2) possibilities but m^2,m^2+2m+1,... are all square so I expect O(m) usable branches for each k. But note that this won't guarantee a connected path. -- Hauke Reddmann <:-EX8 Private email:fc3a501@math.uni-hamburg.de For our chemistry workgroup,remove math from the address === Subject: Re: Square sum puzzle > Sort the numbers 1 to 16 in a row such that > each adjacent pair sums to a square. > My hypothesis: For n instead of 16, this is > always possible except of a finite number of > cases. (The netting increases.) Except for a finite number of cases? Hmm... Let's call m squaresumsortable, if the condition holds: Numbers 1 to m can be sorted in a row such that each adjacent pair sums to a square. Enumerating these m as a(1), a(2), ... in ascending order makes a nice OEIS entry, doesn't it? [Are there any sequences in the OEIS like that one, i.e. with a(K+i) = a(K)+i for some K and all i?] Rainer Rosenthal r.rosenthal@web.de === Subject: Category Theory Question. Is the following true:- * A Natural Transformation is a Functor in * the Category of categories with the Morphisms being functors. If false, where and how badly. titled natural such-and-suches in standard math, that later turned out to be categorically Natural Transformations. He gave as example the natural mapping of vector space V to its double dual V**. What other ones are there? ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- relation between A and B == function from A to P(B) == multi-valued partial function from A to B ---------------------------------------------------------------------------- -- === Subject: Re: Category Theory Question. |Is the following true:- | |* A Natural Transformation is a Functor in |* the Category of categories with the Morphisms being functors. | |If false, where and how badly. hmm. it doesn't parse very well, not even well enough to say that it's really false in any obvious sense. it doesn't take that much to fix it so that it parses and is true, but there might be more than one way to do this, and it's not clear which of these ways might be the one that you're really trying to get at. here's one very simple way: for any pair of categories x and y, there's a category (called the category of functors from x to y) where: 1. the objects are the functors from x to y. 2. the morphisms between the objects are the natural transformations between those functors. or were you aiming for something trickier than that? (i was going to try to say a bit more here about why your original formulation didn't parse, but i didn't manage to think of an un-confusing way to explain it yet.) -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Category Theory Question. > |Is the following true:- > |* A Natural Transformation is a Functor in > |* the Category of categories with the Morphisms being functors. > |If false, where and how badly. > hmm. it doesn't parse very well, not even well enough to say that > it's really false in any obvious sense. it doesn't take that much > to fix it so that it parses and is true, but there might be more than > one way to do this, and it's not clear which of these ways might be > the one that you're really trying to get at. > here's one very simple way: for any pair of categories x and y, > there's a category (called the category of functors from x to y) > where: > 1. the objects are the functors from x to y. > 2. the morphisms between the objects are the natural transformations > between those functors. > or were you aiming for something trickier than that? > (i was going to try to say a bit more here about why your original > formulation didn't parse, but i didn't manage to think of an > un-confusing way to explain it yet.) Some other natural transformations: The equivalence between a boolean algebra and the algebra of clopens of its Stone space and similarly for a Stone space and the Stone space of its algebra of clopens. All sorts of things in algebraic topology, such as the homomorphism in homology induced by a continuous map of spaces, the Hurewicz homomorphism from the fundamental group of a space to the first homology group (and more generally, the homomorphism from a group to its commutator quotient). On the category of rings, the map from a ring to its quotient mod the Jacobson radical. Here are a few things that might have looked natural, but aren't. The inclusion of the center of a group into the group; the inclusion of a module into its injective envelope; the inclusion of a field in its algebraic closure. === Subject: Re: Category Theory Question. > |Is the following true:- > |* A Natural Transformation is a Functor in > |* the Category of categories with the Morphisms being functors. > |If false, where and how badly. > hmm. it doesn't parse very well > (...) Are you thinking of a 2-category, such as in e.g. Hovey, Model Categories? I haven't really read his section in too much detail but I think that's a similar thing. === Subject: Re: Category Theory Question. |> |Is the following true:- |> | |> |* A Natural Transformation is a Functor in |> |* the Category of categories with the Morphisms being functors. |> | |> |If false, where and how badly. |> hmm. it doesn't parse very well |> (...) | |Are you thinking of a 2-category, such as in e.g. Hovey, Model Categories? I |haven't really read his section in too much detail but I think that's a |similar thing. i'm not sure what you mean. it's true though that bill taylor's question does vaguely suggest that he might be more or less unconsciously trying to re-invent 2-categories or something like that. -- [e-mail address jdolan@math.ucr.edu] === Subject: Re: Category Theory Question. Discussion, linux) > |Is the following true:- > |* A Natural Transformation is a Functor in > |* the Category of categories with the Morphisms being functors. > |If false, where and how badly. > hmm. it doesn't parse very well, not even well enough to say that > it's really false in any obvious sense. [...] > (i was going to try to say a bit more here about why your original > formulation didn't parse, but i didn't manage to think of an > un-confusing way to explain it yet.) How about this? If a natural transformation is a functor in the category of small categories, then it must map small categories to small categories, and functors to functors, subject to the usual conditions. Now, pick your favorite natural transformation. Here's a simple one. Let 1: Set -> Set be the functor that sends a set X to the singleton set {0}, which is terminal in Set. This functor sends a map X -> Y to the only place it can, namely, the identity {0} -> {0}. Let Id: Set -> Set be the identity functor. There is an obvious natural transformation Id -> 1. Its component at a set X is the unique morphism X -> {0}. Now: If transformations really are functors, what is this transformation supposed to do to the category Set? What should it do to the category of finite groups? To what categories should these be mapped and why? This was fairly longwinded and explicit, but I hope it works to indicate why transformations aren't functors like Bill asked. At least, I can't see any sensible way to interpret them as such. Maybe I'm dull, though. -- Run mathematicians, RUN!!! I'm coming for you. It may take a few months, but I'll get [computer verification of my proof] and then your lives will be ended as you previously knew it. -- JSH meets PVS === Subject: Re: Goldbach > Crandall & Pomerance, Prime Numbers, Section 1.2.3, let R(n) be > the number of pairs (p, q) of primes such that n = p + q, and give > a heuristic argument that R(n) is asymptotic to C n / (log n)^2, > where C is an explicitly given positive constant. Isn't this the topic of the Goldbach Comet ? It's well worth googling for the Golbach comet, it's a very beautiful thing, and eerily reminiscent of the rings of Saturn, in comet-tail form. ---------------------------------------------------------------------------- -- Bill Taylor W.Taylor@math.canterbury.ac.nz ---------------------------------------------------------------------------- -- Open the internet connection, please, HAL. I'm sorry Bill; I'm afraid I can't do that. ---------------------------------------------------------------------------- -- === Subject: Re: Proof of identity related to n! > How to prove this identity ? : > > n! = n^n - binomial(n,1)*(n-1)^n + binomial(n,2)*(n-2)^n > -binomial(n,3)*(n-3)^n + ... > Count injections from {1, ..., n} to itself via > the inclusion/exclusion principle. How to prove the more general identity that for every real a : n! = a^n - binomial(n,1)*(a-1)^n + binomial(n,2)*(a-2)^n - binomial(n,3)*(a-3)^n + ... ? === Subject: Re: Proof of identity related to n! >> How to prove this identity ? : >> >> n! = n^n - binomial(n,1)*(n-1)^n + binomial(n,2)*(n-2)^n >> -binomial(n,3)*(n-3)^n + ... >> Count injections from {1, ..., n} to itself via >> the inclusion/exclusion principle. >How to prove the more general identity that for every real a : >n! = a^n - binomial(n,1)*(a-1)^n + binomial(n,2)*(a-2)^n - >binomial(n,3)*(a-3)^n + ... ? Consider a(n,m) defined by --- n-k m b(n,m) = > C(n,k)(-1) k [1] --- k Multiply b(n,m) by x^m/m! and sum over m to get --- x^m > b(n,m) --- --- m! m --- --- n-k (kx)^m = > > C(n,k)(-1) ------ --- --- m! m k --- n-k kx = > C(n,k)(-1) e --- k x n = (e - 1) [2] Note that the power series for e^x-1 = x + ... Thus, the power series for (e^x - 1)^n = x^n + ... Compare the coefficients of x to get that for m < n, b(n,m) = 0 [3a] b(n,n) = n! [3b] Using [1] and [3], we get --- k n > C(n,k)(-1) (a-k) --- k --- --- k n-j j n-j = > > C(n,k)(-1) C(n,j)(-1) a k --- --- k j --- j j = > b(n,n-j) C(n,j)(-1) a --- j = n! [4] Rob Johnson take out the trash before replying === Subject: Re: Proof of identity related to n! > > How to prove this identity ? : > > n! = n^n - binomial(n,1)*(n-1)^n + binomial(n,2)*(n-2)^n > -binomial(n,3)*(n-3)^n + ... > > Count injections from {1, ..., n} to itself via > the inclusion/exclusion principle. > How to prove the more general identity that for every real a : > n! = a^n - binomial(n,1)*(a-1)^n + binomial(n,2)*(a-2)^n - > binomial(n,3)*(a-3)^n + ... ? Rewrite the above sum as: sum{k=0 to n} sum{j=0 to n} binomial(n,k) binomial(n,j) k^j a^(n-j) (-1)^(j+k) So, taking the k-sum in isolation: sum{k=0 to n} binomial(n,k) k^j (-1)^(j+k) = n! S(j,n) (-1)^(j+n), where S(j,n) is a Stirling number of the second kind. S(j,n) can be defined recursively as: S(1,1) = 1, S(1,n) = 0 for n > 1, S(j+1,n) = n * S(j,n) + S(j,n-1). And the k-sum identity can be proved using induction and the recursion definition. So, your sum is: n! sum{j=0 to n} S(j,n) (-1)^(j+n) a^(n-j) binomial(n,j) But... which can be seen inductively. And binomial(n,j) = 0 for j > n, which can be seen from the binomial coefficients' recursive formula as well. So, the only nonzero term is when j = n. And, from the recursive definitions of S(j,n) and binomial(n,j), we see that S(n,n) = binomial(n,n) = 1. So, we get the sum is n!. Leroy Quet === Subject: Re: Proof of identity related to n! > > How to prove this identity ? : > > n! = n^n - binomial(n,1)*(n-1)^n + binomial(n,2)*(n-2)^n > -binomial(n,3)*(n-3)^n + ... > > Count injections from {1, ..., n} to itself via > the inclusion/exclusion principle. > How to prove the more general identity that for every real a : > n! = a^n - binomial(n,1)*(a-1)^n + binomial(n,2)*(a-2)^n - > binomial(n,3)*(a-3)^n + ... ? That's absurd. When a = 0, every term is 0 except the last, while when a = 1, it is the binomial expansion of (1-1)^n. What it is is (a-1)^n. === Subject: Re: Proof of identity related to n! > > How to prove this identity ? : > > n! = n^n - binomial(n,1)*(n-1)^n + binomial(n,2)*(n-2)^n > -binomial(n,3)*(n-3)^n + ... > > Count injections from {1, ..., n} to itself via > the inclusion/exclusion principle. > > How to prove the more general identity that for every real a : > > n! = a^n - binomial(n,1)*(a-1)^n + binomial(n,2)*(a-2)^n - > binomial(n,3)*(a-3)^n + ... ? > > That's absurd. When a = 0, every term is 0 except the last, while > when a = 1, it is the binomial expansion of (1-1)^n. What it is is > (a-1)^n. ??? === Subject: Re: Degrees of trigonometric values over Q > Let a_n = cos(2*PI/n) > b_n = sin(2*PI/n) > c_n = tan(2*PI/n) > > since a_n,b_n,c_n lie in the cyclotomic field of order n over Q their > a_n does lie in the n-th cyclotomic field, but b_n and c_n needn't. > They certainly lie in the (4n)-th cyclotomic field > degrees over Q are divisors of phi(n) but are there formulas for the > degrees of a_n,b_n,c_n over Q ? > For n >= 3. a_n has degree phi(n)/2. > Note that b_n = cos(2pi(1/4 - 1/n)) and so has degree phi(m/2) > where m is the denominator of (1/4 - 1/n) in lowest terms > (unless this is 1 or 2). > I'll leave c_n as an exercise. :-) Easy exercise ? The following recent sequence from the OEIS contains the word hard in the description : Here is Sequence A089929 (this will take a moment): ID Number: A089929 URL: http://www.research.att.com/projects/OEIS?Anum=A089929 Sequence: 2,1,4,2,6,2,6,4,10,2,12,6,8,4,16,6,18,4,12,10,22,4,20,12,18, 6,28,8,30,8,20,16,24,6,36,18,24 Name: Algebraic degree of Cot[Pi/n]. Comments: Also the algebraic degrees of the surface area and volume of an n-prism with unit edge lengths. Links: Eric Weisstein's World of Mathematics, Prism Eric Weisstein's World of Mathematics, Regular Polygon See also: Adjacent sequences: A089926 A089927 A089928 this_sequence A089930 A089931 A089932 Sequence in context: A003958 A082729 A076686 this_sequence A065423 A008733 A004795 Keywords: nonn,hard,more,new,nice Offset: 3 === Subject: Re: Mathematical puzzle? To be honest I already had a discussion in a local user group thread, there I started out with my solution as an answer of somewhom else. Maybe I was a little too sure that I was right, and the discussion came close to flame. I don't want to see this repeated. Altough your reasoning was different than in the other group, I think noone has exposed the point yet I discovered. So if you are interested read on, and comment on it please: I start off with Nicolas statement: >However, one must note that sqrt(x+sqrt(x+ ....)) is not a valid >mathematical expression. Yes I also think so. > At least, we clearly know what such things mean, even if they're not > defined. As you say, using dots is like using notes to show what you > want to do without bothering to clearly define them. But in an > official proof, I suppose this isn't valid. But I may be an integrist > and the only one to think that way. Yes, I agree, and I'll go further as to say: Sometimes it is neccesary to specify in full what ... means, or else the problem is undefined (meaningless, ambiguous, ....). In our case one of the possible interpretations already have been given by Achava. But as I must say it is not the most intuitive to me. And it does something fundamental different, than the original setting suggests: He calculates from inside-out, whereas the problem suggests calculating from the outside-in. For me the ... normally mean on an informal basis repeat the step an infinite number of times or when an end is given like a ... b do something from a to b. To come out with my definition as how to interpret the ellipsis ... in this case: Take the equation: sqrt(x+2)=2 substitute 2 from the equation into itself: sqrt(x+ sqrt(x+2) ) = 2 Thats the same equation. continue sqrt(x+sqrt(x+2))=2 still the same. sqrt(x+sqrt(x+....)) still the same, provided that ... means: Repeat the substitution step an arbitrary number of times but don't try to go to the limit (inf), as this would render the expression uncalculatable and thus meaningless. The first square root only can be calculated when all other subexpressions or arguments have been evaluated. In a stack oriented computer implementation this would mean you have specified an recursion without a termination condition. As everyone knows this results in a stack fault of the machine. So in this puzzle the ... is the points misleading us to believing anything. I would say this puzzle is not a mathematical one, but more one of the puzzles that require you to think around the corner (don't know if know this idiom). Roland === Subject: Re: Mathematical puzzle? Oh so sorry, > To be honest I already had a discussion in a local user group thread, > there I started out with my solution as an answer OF somewhom else. But correct is: To be honest I already had a discussion in a local user group thread, there I started out with my solution as an answer TO somewhom else. The first might sound like I did not tell them my name. Of course I did. === Subject: Re: Mathematical puzzle? > Please don't blame me if of the following. > I have my opinions about it, but would like to hear yours beforehand, so > that I receive an unbiased answer. > I have seen this puzzle on a site in the internet. > http://mlug.ca/documents/column2/column_1strauss.shtml > The puzzle goes something like this: > sqrt(x+sqrt(x+ ....)) = 2 > Solve for x. Let L=sqrt(x+sqrt(x+ ....)). Then L = 2, and sqrt(x+L)=2 Hence sqrt(x+2)=2 ---> x+2 = 4 ---> x = 2 === Subject: Re: Mathematical puzzle? > Jonathan Miller grava .88 la saucisse et au marteau: > Why not? How about sigma(0. .oo) or integral() or any of several other > things? > At least, we clearly know what such things mean, even if they're not > defined. I don't agree with this statement, but it is not difficult to come up with a definition for sqrt(x + sqrt(x + sqrt(x + ... Let t_1 = sqrt(x) t_2 = sqrt(x + sqrt(x)) ... t_n = sqrt(x + sqrt(x + ... + sqrt(x))))) where there are n square root signs. Now we can take a limit as n -> 00. Then Mr Le Roux would have to add to his determination of the limit a proof that the limit exists. Noticing that (t_n)^2 - x = t_(n-1), we could also have recursively defined t_n as sqrt(x + t_(n-1)) > What is FWIW ? I haven't actually seen this before, but my guess is that it stands for For What It's Worth. achava === Subject: Re: Mathematical puzzle? === >Subject: Mathematical puzzle? >Please don't blame me if of the following. >I have my opinions about it, but would like to hear yours beforehand, so >that I receive an unbiased answer. >I have seen this puzzle on a site in the internet. >http://mlug.ca/documents/column2/column_1strauss.shtml >The puzzle goes something like this: >sqrt(x+sqrt(x+ ....)) = 2 >Solve for x. Something else interesting to consider is the function: (1/2)(1 + Sqrt[1 + 4x]) for positive x it looks to be very similar to Sqrt[x+Sqrt[x+Sqrt[x+Sqrt[x+... adam === Subject: Re: Please solve this Diophantine Problem. > I have an unsolved problem. It's about a Diophantine Equation. > The problem is ... Find, with proof, all positive integers a and b such that > a^4+(a^2-1)^2=b^2. . > I try to solve with this : > Let a, b be any real number > so: 2a^4 - 2a^2 +1 - b^2 = 0 > Let a^2 = x then 2x^2 - 2x +1 - b^2 = 0 and x >=0 > but the product of root is (1 - b^2)/2 >= 0 when -1 <= b <= 1 > then : b = 1 and a = 1 > But. Why (2, 5) is also a solution of this problem. ? > Anyone can help me. Please !!! (NOTE: In the following, unless otherwise specified, values that pop up out of nowhere are integers which the preceding equations imply must exist - It's tedious to keep having to mention this explicitly.) As GCD(a^2, a^2 - 1) = 1 there must be coprime integers m, n of opposite parity (and WLG both non-negative) such that one of the following cases holds: Case 1 a odd : a^2 - 1, a^2 = 2.m.n, m^2 - n^2 ------ As a is odd, the 2nd equation mod 4 implies m odd, n even, so that for some M, N coprime with M + N odd: a, m, n = M^2 + N^2, M^2 - N^2, 2.M.N In the 1st equation this gives: (M^2 + N^2)^2 - 1 = 4.M.N.(M^2 - N^2) (M^2 - N^2 - 2.M.N)^2 = 1 (M - N)^2 - 2.N^2 = +/- 1 of which (since 0 <= M - N, N) every integer solution is: (M - N) + N.sqrt(2) = (1 + sqrt(2))^k k = 0, 1, 2, .. Case 2 a even: a^2 - 1, a^2 = m^2 - n^2, 2.m.n ------ This is trickier, unless I've missed something obvious. As the first mod 4 implies m even, n odd, the second gives, with N odd: a, m, n = 2.M.N, 2.M^2, N^2, which back in the first gives: (2.M^2 - N^2 + 1)^2/4 + (2.M^2 - N^2 - 1)^2/4 = N^4 Adding the bracketed terms, we have with (u, v) = 1 and u + v odd: 2.M^2 - N^2, N^2 = u^2 + 2.u.v - v^2, u^2 + v^2 Adding these gives M^2 = u.(u + v), so that: u, u + v = U^2, V^2 But subtracting the two bracketed terms gives: 2.u^2 - (u + v)^2 = +/- 1 and hence: 2.U^4 - V^4 = +/- 1. where V is odd and we have the +ve or -ve option according as U is odd or even resp. As this has solutions U, V = 0, 1 and 1, 1 (which satisfy the remaining condition N^2 = U^4 + (U^2 - V^2)^2 incidently), neither sign can be excluded by congruence considerations alone. But presumably it's pretty easy to prove by descent that there are no other integer solutions. I've done the +ve option, which is sketched below, and I'll leave the -ve option as an exercise. (Reassigning letters ...) We have to show that x^4 + y^2 = 2.z^4 has only the integer solutions |x|, |y| = 1, 1. of y and m: x^2 + y, x^2 - y, z^2 = 2.(m^2 - n^2), 4.m.n, m^2 + n^2 Adding the first two gives x^2 = (m + n)^2 - 2.n^2, so that: m + n, n = P^2 + 2.Q^2, 2.P.Q and the third implies: m, n = R^2 - S^2, 2.R.S Then [ n = ] P.Q = R.S implies P, Q, R, S = p.q, r.s, p.r, q.s, with (p, s) = (q, r) = 1. In [ m + n = ] P^2 + 2.Q^2 = R^2 + 2.R.S - S^2 this gives finally: (q^2 - r^2).p^2 + 2.q.r.p.s - (2.q^2 + r^2).s^2 = 0 In the latter p = 0 implies s = 0 and/or q = r = 0, and s = 0 implies |q| = |r|, so that |q| = |r| = 1. If p.s != 0 then a necessary condition for p/s to be rational is that 2.q^4 = r^4 + t^2, which is of the same form as the original but with smaller integers. --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Please solve this Diophantine Problem. Uppps - sent my original posting to a wrong thread. sorry. -------------------- > I have an unsolved problem. It's about a Diophantine Equation. > The problem is ... Find, with proof, all positive integers a and b such that > a^4+(a^2-1)^2=b^2. . > I try to solve with this : > Let a, b be any real number > so: 2a^4 - 2a^2 +1 - b^2 = 0 > Let a^2 = x then 2x^2 - 2x +1 - b^2 = 0 and x >=0 > but the product of root is (1 - b^2)/2 >= 0 when -1 <= b <= 1 > then : b = 1 and a = 1 > But. Why (2, 5) is also a solution of this problem. ? > Anyone can help me. Please !!! a^4-1 + (aî-1)î = bî-1 (aî-1) ( (aî+1) + (aî-1)) = bî-1 (aî-1) 2aî = (bî-1) (aî-1) 4aî = 2(bî-1) (a-1)(a+1) 4 aî = 2(b-1)(b+1) frome here there are some solutions obvious: a=(1,-1,0) , b=(1,-1) (aî-1) 4aî = 2(bî-1) 4aîî - 4aî + 1 = 2(bî-1) + 1 (2aî-1)î = ( 2bî-1) (a,b)=(2,5) is a possible solution, but I wouldn't have seen it and also don't get further from here. Next would be a consideration of moduli, for instance 4, or 16, or remainders of 2bî-1 modulus any odd square-number ... Gottfried Helms -------------------------------------------------- When will a number of the form (2bî-1) equal a square of such a form.? Factorizing all occurences and see, whether we find a rule, which allows conclusion ---------------------------------------------------------------------------- --------------------------------------- ! p(n) = q(n) + 1 ! q(n+1) = q(n) (factorized) b ! 2bî-1 !=(2bî-1)î ! x * q(n) = (2bî-1)î-1 ---!------------------------------------------------------------------------ --------------------------------------- 1 ! 1 ! 1 = 0*6*4 + 1 ! 0 * 1*2 2 ! 7 ! 49 = 2*6*4 + 1 ! q1= 4*2* 2*3 4 * 1*(3!)î / (1!)î *3 3 ! 17 2*8+1 1*2 *8 +1 ! 289 = 12*6*4 + 1 ! * 6/1 = q1 * 3*4 / 1*2 4 * 2*(4!)î / (2!)î *4 4 ! 31 4*8-1 2*2 *8 -1 ! 961 = 40*6*4 + 1 ! *10/3 = q2 * 4*5 / 2*3 4 * 3*(5!)î / (3!)î *5 5 ! 49 6*8+1 2*3 *8 +1 ! ... = 100*6*4 + 1 ! *15/6 = q3 * 5*6 / 3*4 4 * 4*(6!)î / (4!)î *6 6 ! 71 9*8-1 3*3 *8 -1 ! = 210*6*4 + 1 ! *21/10 = q4 * 6*7 / 4*5 7 ! 97 12*8+1 3*4 *8 +1 ! = 392*6*4 + 1 ! *28/15 = q5 * 7*8 / 5*6 8 !127 16*8-1 4*4 *8 -1 ! = 672*6*4 + 1 ! *36/21 = q6 * 8*9 / 6*7 .......... ! ! ........... ---------------------------------------------------------------------------- --------------------------------------- When will a number of the form (2bî-1) equal a square of such a form? Additional example 49 .... is (2bî-1) for b=5 ( column 2, line 5) ... is (2aî-1)î for a=2 ( column 5, line 2) required according to column(6). Try an analytic study of the structures of the terms... to see, whether it is possible, that the two types can match. Gottfried Helms === Subject: Steady states of an ODE Hi all Apologies if this question is trivial too you, but please bear with this biologist! I am interested in finding the steady states of dN/dt = ((aN / (N+n)) - b)N - kN(N+n) with a,b,k and n constant. Clearly there is a trivial solution at N=0, but how do I find the other? All I seem to get is massively long algebra that seemingly leads nowhere! James Orange County === Subject: Re: Steady states of an ODE > Hi all > Apologies if this question is trivial too you, but please bear with > this biologist! > I am interested in finding the steady states of dN/dt = ((aN / (N+n)) > - b)N - kN(N+n) > with a,b,k and n constant. Clearly there is a trivial solution at N=0, > but how do I find the other? > All I seem to get is massively long algebra that seemingly leads > nowhere! > James > Orange County You can rearrange the equation as follows, where d(N+n) = dN, and A, B, C are functions of a, b, k: (N+n).d(N+n)/dt = A.(N+n)^2 + B.(N+n) + C If A = 0 Taking y := N+n for convenience, so the LHS is (d(y^2)/dt)/2, we have three cases: 1. A = B = 0: -------------- This implies: y^2 = 2.C.t + const 2. A = 0 (B != 0) ------------------ The equation is now of the form: y.dy/dt = B.y + C. y.d(y + C/B)/dt = B.(y + C/B) Letting z := y + C/B, this becomes: (1 - C/(B.z)) . dz/dt = 1 and integrating this WRT t gives: z - C/B.log(z) = t + const 3. A != 0 ---------- In this case it's easiest to treat y as the independent variable and solve the following by partial fractions: dt/dy = y / (A.y^2 + B.y + C) If A.y^2 + B.y + C = 0 has two equal roots, say f, then you can express the equation in the form: A.dt/dy = 1/(y - f) + f/(y - f)^2 which can be integrated term by term to give: A.t = log(y - f) - f/(y - f) + const Otherwise, if the distinct roots are f, g, it can be expressed in the form: (f - g).A.dt/dy = f/(x - f) - g/(x - g) which again you can integrate term by term to give: (f - g).A.t = f.log(x - f) - g.log(x - g) + const --------------------------------------------------------------------------- John R Ramsden (jr@adslate.com) --------------------------------------------------------------------------- Eternity is a long time, especially towards the end. Woody Allen === Subject: Re: Steady states of an ODE > Hi all > Apologies if this question is trivial too you, but please bear with > this biologist! > I am interested in finding the steady states of dN/dt = ((aN / (N+n)) > - b)N - kN(N+n) > with a,b,k and n constant. Clearly there is a trivial solution at N=0, > but how do I find the other? > All I seem to get is massively long algebra that seemingly leads > nowhere! N' = N(aN/(N+n) - b) - kN(N+n) N' = 0 => aN^2/(N+n) - bN - kN(N+n) = 0 multiply by (N+n): aN^2 - bN(N+n) - kN(N^2 + 2nN + n^2) = 0 => N = 0 or aN - b(N+n) - k(N^2 + 2nN + n^2) = 0 => kN^2 + (2nk + b - a)N + bn + kn^2 = 0 and hence N = -(2nk + b - a)/2k +/- sqrt( (2nk + b - a) - 4kbn - 4k^2 n^2 ) / 2k = -(2nk + b - a)/2k +/- sqrt( 4ank + b^2 - a^2 ) / 2k -- P.A.C. Smith The vast majority of Iraqis want to live in a peaceful, free world. And we will find these people and we will bring them to justice. === Subject: Re: graphs of 0/1-polytopes again >Let G=(V,E) be an arbitrary finite graph. Now is it possible to >construct a 0/1-polytope (that is a polytope where all vertices have >coordinates in {0, 1}) P with vertex set U and an injection f:V-->U >such that: The restriction of graph(P) on f(V) gives a graph >isomorphic to G. >In other words: Can we construct a 0/1-polytope P such that there is >a subset of its vertices identified with V such that two vertices >are adjacent on P if and only if they share an edge in G? >Some extremal cases: >2)G is the complete graph K_n: On the n-simplex, all vertices are >adjacent. It can be realized in (n+1)-space using the canonical unit >vectors (0,...,0,1,0,...,0). >What about the graphs in between? Any ideas? Extend them to or embed them in a complete graph? ---- === Subject: Re: graphs of 0/1-polytopes again > >Let G=(V,E) be an arbitrary finite graph. Now is it possible to > >construct a 0/1-polytope (that is a polytope where all vertices have > >coordinates in {0, 1}) P with vertex set U and an injection f:V-->U > >such that: The restriction of graph(P) on f(V) gives a graph > >isomorphic to G. > >In other words: Can we construct a 0/1-polytope P such that there is > >a subset of its vertices identified with V such that two vertices > >are adjacent on P if and only if they share an edge in G? > >Some extremal cases: > >2)G is the complete graph K_n: On the n-simplex, all vertices are > >adjacent. It can be realized in (n+1)-space using the canonical unit > >vectors (0,...,0,1,0,...,0). > >What about the graphs in between? Any ideas? > Extend them to or embed them in a complete graph? No. If we restrict to node set V of a complete graph to a subset U, the graph on U will also be complete. -- Phyics is much too hard for physicists. reverse my forename for mail! === Subject: Re: graphs of 0/1-polytopes again === Subject: Re: graphs of 0/1-polytopes again >Let G=(V,E) be an arbitrary finite graph. Now is it possible to >construct a 0/1-polytope (that is a polytope where all vertices >have coordinates in {0, 1}) P with vertex set U and an injection >f:V-->U such that: The restriction of graph(P) on f(V) gives a >graph isomorphic to G. >In other words: Can we construct a 0/1-polytope P such that there >is a subset of its vertices identified with V such that two >vertices are adjacent on P if and only if they share an edge in G? >Some extremal cases: >2)G is the complete graph K_n: On the n-simplex, all vertices are >adjacent. It can be realized in (n+1)-space using the canonical >unit vectors (0,...,0,1,0,...,0). >What about the graphs in between? Any ideas? >> Extend them to or embed them in a complete graph? >No. If we restrict to node set V of a complete graph >to a subset U, the graph on U will also be complete. Add edges to G until a complete graph K. Embed K into above construction Keep nodes of K, remove edges from K as needed to obtain G. ---- === Subject: Re: graphs of 0/1-polytopes again > Add edges to G until a complete graph K. > Embed K into above construction > Keep nodes of K, remove edges from K as needed to obtain G. That's trivial. It gives us a proof that every graph is a subgraph of the graph of some 0/1-polytope. Really not too surprising. It is the first part Add edges to G until a complete graph K. which shall not be allowed. My basic idea was to show that every graph IS the graph of some 0/1-polytope - since this is obviously wrong because of connectivity reasons (see also the thread properties of graphs of 0/1-polytopes) the next natural question to ask is: Can we embed every G without adding or removing edges into the graph of some 0/1-polytope? Motivation: There is a result of Biilera and Sarangajaran who proved that every 0/1-polytope is combinatorially equivalent to a face of the asymmetric traveling salesman polytope ATSP(n) for some n (convex hull of all incidence vectors of the hamiltonian tours on the complete digraph K_n). Knowing that every G occurs as a part of the graph of some 0/1-polytope, we can conclude that G is a part of some ATSP(n)-graph. This shows that the graph of ATSP(n) is extremely complicated, restriction to a subset of its nodes can give ANYTHING. Remark: the same also applies in the symmetric case STSP(n) where the underlying complete graph is undirected. -- Phyics is much too hard for physicists. reverse my forename for mail! === Subject: Re: trig help No but textbooks and instructors are great help. Why do students refuse to go to the instructor for help when they're paying big bucks to have hjim/her there for this purpose??? === Subject: Conversion of angles I have a software package that aligns 3D data sets and I am wanting to test how accurately it achieves this. To do this I have two data sets that differ in orientation by a known amount of yaw, pitch and roll (these are defined from a fixed axis). When the algorithm has finished alignment it reports the yaw pitch and roll needed to align the data. However, as I know the correct rotations required I can find the errors in the three rotations. However, I wish to compare the accuracy of these results using different data sets, and instead of comparing the separate yaws pitchs and rolls, I wish to convert the errors in these rotations into a single rotation vector. For comparison purposes I am only interested in the magnetite or the rotation vector. My question is how do I convert from the three rotations to the required rotational geometry is quite complicated! Any help would be greatly appreciated Dave B === Subject: Log equation I 'm trying to find a way to solve the equation y = (1-x)/ Log(x) How can I find x when y is known ? Is there a numerical method ? with series development? -- Bernard Bour.8ee bernard@bouree.net === Subject: Re: Log equation > I 'm trying to find a way to solve the equation > y = (1-x)/ Log(x) > How can I find x when y is known ? > Is there a numerical method ? Newton-Raphson, perhaps? Carlos -- === Subject: Re: Log equation Please note, Bernard, that posting the same question _separately_ to different groups is a bad idea. Crossposting is better if you think that you really must post the same message in more than one group. Here's a copy of the answer which I recently posted to your question in sci.math.num-analysis . David ------------------------------------ > I 'm trying to find a way to solve the equation > y = (1-x)/ Log(x) > How can I find x when y is known ? That equation can be solved precisely in closed form using the Lambert W function. (Briefly, it is the inverse of z*e^z. For more information about the Lambert W function, see for example .) The solutions are given by x = y*LambertW(1/y * e^(1/y)) BTW, the Lambert W function is implemented, for example, in some well known computer algebra systems. (Note: In Mathematica, it is called ProductLog.) > Is there a numerical method ? > with series development? It should be reasonably easy to get a series development from the solution I gave you. Look at the first or second paper cited at the link above for appropriate series for Lambert W. David Cantrell === Subject: Factorizations: Connecting the arguments Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 Now consider (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where f_1(0) = f_2(0) = f_3(0) = 0. That expression represents one way to show a *family* of factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. Now I'll do some simple algebraic manipulations with that expression. Ok, let g_3(x) = f_3(x) + 3, so I have (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = 300125 x^3 - 18375 x^2 - 360 x + 22 and now, I multiply both sides by 7, which gives (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), which gives (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = 49(300125 x^3 - 18375 x^2 - 360 x + 22) Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, where x is in the ring of algebraic integers. Some smart regrouping of terms gives P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 where you should note that using v = -1 + 49x, gives P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 and now I let the a's be roots of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Notice that if I let x=0, I have P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) as the cubic defining the a's at x=0 is a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and a_2(0) to equal 0, which leaves a_3(0) with a value of 3. P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7) P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22). Now P(x) has a factor of 49 as P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x) which means that (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) has a factor of 49. However, the constant term of P(x)/49 is 22, which is verified by again setting x=0, which gives P(0)/49 = 22. But for two of the factors of P(x), the constant terms is 7, which is NOT a factor of 22. Therefore, *none* of the constant terms of P(x)/49 as they multiply to give 22 can have 7 as a factor. Given that the constant terms are independent of x's value, it must be the case that dividing P(x) by 49 divides the two constant terms equal to 7, by 7. But to divide 7 from those constant terms requires dividing through two of the factors, so (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 from reverse use of the distributive property, which gives constant terms that don't have 7 as a factor, as required. Which is (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 with which I started. James Harris My math discoveries, found for profit http://mathforprofit.blogspot.com/ === Subject: Re: Factorizations: Connecting the arguments > Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 > Now consider > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. [ Some boring stuff snipped here ] > (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > with which I started. > James Harris Which merely proves that James Steven Harris goes around in circles. === Subject: Re: Factorizations: Connecting the arguments [snip a boring argument that get us to a point we have been many times before and that nobody disputes] > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) > has a factor of 49. James then make a correct argument for the behaviour at x=0 > However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. James then incorrectly tries to extent this result to values of x not equal to 0, The factors are not polynomial factors so this does not work. Just because we divide the first two factors by 7 at x=0 does not mean we have to divide the first two factors by 7 at x not equal to 0. > But to divide 7 from those constant terms requires dividing through > two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. > Which is > (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > with which I started. Note f_1(x) and f_2(x) are not algebraic integers for all values of x. -William Hughes === Subject: Re: Factorizations: Connecting the arguments > Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 > Now consider > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. Note that you do *not* claim here that the f's are algebraic integer functions. If you with they are, please show such. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, > where x is in the ring of algebraic integers. > Some smart regrouping of terms gives > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 v is irrelevant at this point. > and now I let the a's be roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Can you show that the a's that you derived with multiplying the f's by 7, 7 and 1 *must* be roots of this cubic? Let's try another regrouping: P(x) = 7^2(2401 x^3 - 147 x^2 + 2x)(5^3) - 3(-1 + 44 x)(5)(7^2) + 7^3 and I let the a's be roots of: a^3 + 3(-1 + 44x)a^2 - 49(2401 x^3 - 147 x^2 + 2x). > Notice that if I let x=0, I have > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. My cubic defining the a's has exactly the same properties. > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22). > Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x) > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) > has a factor of 49. > However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. Yup, right upto this point. > But to divide 7 from those constant terms requires dividing through > two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 And this is a wrong conclusion. Again, the same wrong conclusion. Dividing off by three functions w1(x), w2(x) and w3(x) such that w1(0) = w2(0) = 7 and w3(0) = 1 gives exactly the same. And as long as w1(x), w2(x) and w3(x) multiply together to get 49, this is also a valid division. Note that the constant term of (5 a1(x) + 7) / w1(x) = (5 a1(0) + 7) / w1(0) = 7/7 = 1 (using the definition *you* gave for constant term). So each triple of functions w with the above definitions satisfy your requirement. And I may also note that this is an infinite family of triples. > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. Reverse use of the distributive property *is not valid*. E.g. in a ring, whenever (a + b)/c is in that ring, (a + b)/c = a/c + b/c is *only* true when all three are in that ring. So while it is true for (5 a1(x) + 7)/w1(x) for the w's I have previously shown, it is *not* true for those w's for (5 a3(x) + 22)/w3(x) Because neither 5 a3(x)/w3(x), nor 22/w3(x) are algebraic integers. But their *sum* is (with the w's I provided before). Note: w3(x) = gcd(5 a3(x) + 22, 7) which is not necessarily 1... Nor are w1(x) and w2(x) constant. > Which is > (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > with which I started. Yup, but you have not yet shown that f1 and f2 are algebraic integer functions. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ === Subject: Re: Factorizations: Connecting the arguments > Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 > Now consider > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, > where x is in the ring of algebraic integers. > Some smart regrouping of terms gives > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > and now I let the a's be roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). Since you say above that a_1(x) = 7*f_1(x), we must have that 7*f_1(x) is a root of the above polynomial also. This implies 7^3*[f_1(x)]^3 + 3*(-1 + 49*x)*7^2*[f_1(x)]^2 -49*(2401*x^3 - 147*x^2 + 3*x) = 0, 7*[f_1(x)]^3 + 3*(-1 + 49*x)*[f_1(x)]^2 - (2401*x^3 - 147*x^2 + 3*x) = 0. Now specializing to x = 1 and letting w = f_1(1) gives 7*w^3 + 144*w^2 - 2257 = 0. The polynomial in w on the left has integer coefficients, and is primitive, irreducible, and *non-monic*. Therefore its roots cannot be algebraic integers. Therefore f_1(1) cannot be an algebraic integer. Therefore f_1(x) cannot be an algebraic integer function. This contradicts your implicit assumption at the beginning. Why do you keep doing this ??? > Notice that if I let x=0, I have > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22). > Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x) > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) > has a factor of 49. > However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. > But for two of the factors of P(x), the constant terms is 7, which is > NOT a factor of 22. Therefore, *none* of the constant terms of > P(x)/49 as they multiply to give 22 can have 7 as a factor. > Given that the constant terms are independent of x's value, it must be > the case that dividing P(x) by 49 divides the two constant terms equal > to 7, by 7. There is an unspoken assumption here: that the product of the constant terms is the constant term of the product. No problem with that; it is a true fact! Moreover the product of the constant terms, after division by 49, is an integer: 22. Another true fact. However it is NOT true that each of the constant terms, after the division has occurred, is necessarily an algebraic integer. Two of them are (those corresponding to the 7's, but the third is not (the one corresponding to 22). Note well, *this is where you are making your mistake*. You think that whatever you divide 22 by, the result must be an algebraic integer. Not true! There is no reason to assume this! > But to divide 7 from those constant terms requires dividing through > two of the factors, so > (5 a_1(x)/7 + 1)(5 a_2(x)/7 + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > from reverse use of the distributive property, which gives constant > terms that don't have 7 as a factor, as required. What you have overlooked here is that there is *another* way to divide through by 49, and it works perfectly well with respect to the constant terms. There are algebraic integers r, s, and t such that 49 = r*s*t, and a_1(x)/r, 7/r, a_2(x)/s, 7/s, and (5*f_3(x) + 22)/t are all algebraic integers. Here r, s, and t are, like a_1(x), etc., dependent on x: one could write r = r(x), s = s(x), and t = t(x). Also note that the constant terms multiply out exactly as they should: (7/r)*(7/s)*(22/t) = (49/(r*s*t))*22 = 1*22 = 22. And again: 22/t is ***NOT*** an algebraic integer; moreover, contrary to what you are assuming, *** there is no reason it needs to be ***. > Which is > (5 f_1(x) + 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > > 300125 x^3 - 18375 x^2 - 360 x + 22 > with which I started. Yes, this is what you started with. You have successfully gone in a circle. The problem is, what you started with has no solutions in algebraic integer functions, if in addition you require, *** as you have said above *** that 7*f_1(x) is a root of a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x) = 0. To recap yet again: your claims here lead to a contradiction; they must therefore be false. Moreover, I have specified exactly where you are making an error: essentially in assuming that 22/t must be an algebraic integer. Your claims are false and I have pinpointed your error. Time to concede. Nora B. > James Harris > My math discoveries, found for profit > http://mathforprofit.blogspot.com/ === Subject: Re: Factorizations: Connecting the arguments > Let S(x) = 300125 x^3 - 18375 x^2 - 360 x + 22 > Now consider > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > where f_1(0) = f_2(0) = f_3(0) = 0. > That expression represents one way to show a *family* of > factorizations of the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. > Now I'll do some simple algebraic manipulations with that expression. > Ok, let g_3(x) = f_3(x) + 3, so I have > (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 g_3(x) + 7) = > 300125 x^3 - 18375 x^2 - 360 x + 22 > and now, I multiply both sides by 7, which gives > (5(7) f_1(x)+ 7)(5(7) f_2(x) + 7)(5 g_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > and now let a_1(x) = 7 f_1(x), a_2(x) = 7 f_2(x), and a_3(x) = g_3(x), > which gives > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 a_3(x) + 7) = > 49(300125 x^3 - 18375 x^2 - 360 x + 22) > Now let P(x) = 49 S(x) = 14706125 x^3 - 900375 x^2 - 17640 x + 1078, > where x is in the ring of algebraic integers. > Some smart regrouping of terms gives > P(x)= 7^2(2401 x^3 - 147 x^2 + 3x) (5^3) - 3(-1 + 49 x )(5)(7^2) + 7^3 > where you should note that using v = -1 + 49x, gives > P(x) = (v^3+1)(5^3) - 3v(5)(7^2) + 7^3 > and now I let the a's be roots of > a^3 + 3(-1 + 49x)a^2 - 49(2401 x^3 - 147 x^2 + 3x). > Notice that if I let x=0, I have > P(0) = (5(0) + 7)(5(0) + 7)(5(3) + 7) = 7(7)(22) > as the cubic defining the a's at x=0 is > a^3 - 3a^2, which has roots, 0, 0 and 3, and I've picked a_1(0) and > a_2(0) to equal 0, which leaves a_3(0) with a value of 3. > P(x) = (5 a_1(x) + 7)(5 a_2(x) + 7)(5 f_3(x) + 5(3) + 7) > P(x) = (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22). > Now P(x) has a factor of 49 as > P(x)/49 = 300125 x^3 - 18375 x^2 - 360 x + 22 = S(x) > which means that > (5 a_1(x)+ 7)(5 a_2(x) + 7)(5 f_3(x) + 22) > has a factor of 49. > However, the constant term of P(x)/49 is 22, which is verified by > again setting x=0, which gives P(0)/49 = 22. P(0) evaluates the polynomial at x=0. P(0)/49 = 22 P(1) evaluates the polynomial at x=1. P(1)/49 = 281412 P(2) evaluates the polynomial at x=2. P(2)/49 = 2326802 -- There are two things you must never attempt to prove: the unprovable -- and the obvious. -- Democracy: The triumph of popularity over principle. -- http://www.crbond.com === Subject: Re: differentiable manyfold at 10:40 PM, tern said: >Is it possible to define a differentiable structure on a set saying >that some real function is differentiable? Yes, if[1] it is 1-1 and the range is connected. You can generalize that approach, but it is more cumbersome than the standard approach and I don't see why you would want to bother. [1] That's actually stroner than necessary. -- Shmuel (Seymour J.) Metz, SysProg and JOAT Unsolicited bulk E-mail will be subject to legal action. I reserve the right to publicly post or ridicule any abusive E-mail. Reply to domain Patriot dot net user shmuel+news to contact me. Do === Subject: Re: differentiable manyfold excuse me , what is [1] ? May you explain with more words? === Subject: JSH: 22 versus 7 I just gave a post which *can* clear up the main issues if you follow it all the way through, but it's long, the issues are advanced, and I don't think many of you care what the mathematical truth is. The primary point that settles things is that 22 is coprime to 7 in the ring of algebraic integers. Posters using various primitive arguments, and mostly just *claiming* one position that they cannot prove, are stuck with their position forcing them to need 22 if you look one way, and 7 if you believe them, and look another. But you see 22 is not 7. By showing (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where f_1(0) = f_2(0) = f_3(0) = 0, I'm pushing the underlying polynomial to the fore, and for *that* polynomial the constant term is 22. Make no mistake, what I do is use special tools of analysis on the polynomial 300125 x^3 - 18375 x^2 - 360 x + 22. It doesn't take much to see that if the f's are linear functions making the factorization into polynomial factors that multiplying f_1(x) and f_2(x) by 22 will give you algebraic integer functions. But I've also used a more complicated solution for the f's where multiplying them times 7 gives algebraic integer functions. What gives? That's what I'm not going to explain in detail to you, except to say that the simplest explanation is that the definition for the ring of algebraic integers is screwed up enough to let you do some wacky stuff. Now then, consider the *demonstrated* apparent contradiction with (5 f_1(x)+ 1)(5 f_2(x) + 1)(5 f_3(x) + 22) = 300125 x^3 - 18375 x^2 - 360 x + 22 where f_1(0) = f_2(0) = f_3(0) = 0, as you look for 22 times f_1(x) and f_2(x) in one case, while in another you look at 7 times them with the guarantee that you get algebraic integer functions. If you have any sense at all, you might be now wondering, what does 7 have to do with anything? James Harris