mm-2699 === Subject: obtaining eigenvectors what is the best way of obtaining the eigenvectors of a matrix given you know its eigenvalues? performing gauusian elimination on the system (A - lambdaI) [where 'A' is the original matrix, 'lambda' a particular eigenvalue, and 'I' the identity matrix], and solving the system by back substitution seems a poor method. is there a better iterative method? === Subject: f(x,y,t) = f0*n*sin(k1*x + k2*y +w*t) Please, help me out with this problem, I really need to know if my solution is or isnÍt correct or if there is a simple way of getting the correct result. I have to find the wave length of the function f(x,y,t) = f0*n*sin(k1*x + k2*y +w*t) where n is a generic unitary vector. I considered the function at t=0 and I found 2 consecutive wave fronts: k1*x + k2*y = 0 k1*x + k2*y = pi i.e. y = - k1/k2 * x y = - k1/k2 * x +pi/k2 Finally I calculate the distance between these two lines D = (pi/k1)*(pi/k2) / sqrt[(pi/k1)^2 + (pi/k2)^2] I think this should be the wave length of the function. Please see if my solution is correct, and if itÍs wrong please explain to me how to find the wave length of this function. Also if there is a better way of finding the wave length of a generic wave I would really appreciate if you could explain it to me. === Subject: Really need help-proof hi If net force on the body is zero and also sum of torques is zero, then object doesn't rotate and wherever we put an axis, the sum of torques is always zero. I know it's true and it does make sense, but can you show me mathematical proof (or vector proof ) that no matter where the axis is, the sum of torques is always zero? In the following example no matter where we put an axis the sum of torques is zero(I already solve it, so no need for help there) _______|C|______ |A| |B| Board( weight of the board is negligible ) length is 4 meters and lays on top of boxes A and B. Third box C is on top of board and has mass 16 kilos. Length from A to C is 0.5 m and from A to B is 4 meters Please try keep it ( proof ) as simple as possible thank you very much === Subject: Re: Really need help-proof hi in a general the total torque cam be calculated in the following way: if for example we have 2 forces M = F1 x r1 + F2 x r2 where x is the vecor product (M, F1, F2, r2, r2 are vectors) Now the torque relative to a given axis z is M_z = |M|*cos(a) (M_z is no longer a vector but is the projection of your torque on the z axis; |M| is the length of the vector M) a is the angle between the vector M and the your axis z. Now it is trivially seen that if the total torques M is zero it is zero also the quantity |M|*cos(a) = M_z for all angles a The proof is that if you have a zero vector its projection is allways going to be zero no metter what axis you project it on. I hope I understood what your question was and I hope this answer can help. It's always tricky to prove what looks obvious. === Subject: Re: Newton's Method Using MATLAB/MATHEMATICA Hi. Not sure, but if you are just wanting to solve the equation using a program such as Mathematica still, than perhaps like this... equ={ x1^2+x2^2-1==0, 5*x1^2-x2-2==0 }; NSolve[equ,{x1,x2}] {{x1->-0.732260195229901, x2->0.6810249675906652}, {x1->0.47306976914813215, x2->-0.8810249675906656}, {x1->-0.47306976914813376, x2->-0.8810249675906672}, {x1->0.7322601952299002, x2->0.6810249675906642}} Simplify[Solve[equ,{x1,x2}]] {{ x2->(1/10)*(-1-Sqrt[61]), x1->(-(1/5))*Sqrt[(1/2)*(19-Sqrt[61])]}, {x2->(1/10)*(-1-Sqrt[61]), x1->(1/5)*Sqrt[(1/2)*(19-Sqrt[61])]}, {x2->(1/10)*(-1+Sqrt[61]), x1->(-(1/5))*Sqrt[(1/2)*(19+Sqrt[61])]}, {x2->(1/10)*(-1+Sqrt[61]), x1->(1/5)*Sqrt[(1/2)*(19+Sqrt[61])] }} HTH. :>) -- Dana DeLouis > for reference, the prob is this: > Solve: > f(x1, x2) = x1^2 + x2^2 - 1 =0 > f(x1, x2) = 5*x1^2 - x2 - 2 = 0 === The subject heading says it all. Your help has been, well, extremely helpful. You are the MAN! === Subject: Re: Twin Primes > I'm trying to find the proof for the following statement. > The integers n, n+2, form a pair of twin primes if and only if > 4[(n-1)!+1] = -n (mod n(n+2)). > I know that you can get this from Wilsons theorem but I'm at a loss as > to how one would go about it. > So if anyone has a link to this proof or is willing to jot it down > here I would appreciate it. Well ... here are some musings that you can probably turn into an honest-to-goodness proof: First, suppose that n, n+2 are a pair of twin primes. By Wilson's Theorem for n, (n-1)! + 1 = kn for some positive integer k. Then n + 4[(n-1)! + 1] = n + 4kn = (4k + 1)n and (since (n,n+2) = 1) the result will follow if you can show that 4k + 1 is a multiple of n+2. But (n+1)! + 1 = n(n+1)(n-1)! + 1 = n(n+1)(kn - 1) + 1 = kn^3 + (k-1)n^2 - n + 1 = k(-2)^3 + (k-1)(-2)^2 - (-2) + 1 (mod n+2) = -8k + 4(k-1) + 2 + 1 (mod n+2) = -(4k + 1) (mod n+2). By assumption, n+2 is *also* a prime, so you can apply Wilson's Theorem again to conclude that 4k + 1 = 0 (mod n+2)., finishing this half of the proof. To go the other direction, suppose that the given congruence holds. Then (firstly) n must be a prime, since otherwise (n-1)! + 1 = 1 (mod n) and the assumed congruence would imply that 4 = 0 (mod n) (which you can easily rule out). But. once you know that n is prime, you get that (n-1)! + 1 = kn for some k from Wilson's Theorem and then that n + 4[(n-1)! + 1] = (4k + 1)n exactly as above. Again, since (n,n+2) = 1, the assumed congruence implies that 4k + 1 is a multiple of n+2. But also, exactly as above, (n+1)! + 1 = -(4k + 1) (mod n+2). Hence Wilson's Theorem holds for n+2, which implies that n+2 is a prime. So n, n+2 are a pair of twin primes. [] === Subject: Re: Twin Primes >> I'm trying to find the proof for the following statement. >> The integers n, n+2, form a pair of twin primes if and only if >> 4[(n-1)!+1] = -n (mod n(n+2)). >> I know that you can get this from Wilsons theorem but I'm at a loss as >> to how one would go about it. >> So if anyone has a link to this proof or is willing to jot it down >> here I would appreciate it. > Well ... here are some musings that you can probably turn into > an honest-to-goodness proof: > First, suppose that n, n+2 are a pair of twin primes. By Wilson's > Theorem for n, (n-1)! + 1 = kn for some positive integer k. Then > n + 4[(n-1)! + 1] = n + 4kn = (4k + 1)n > and (since (n,n+2) = 1) the result will follow if you can show that > 4k + 1 is a multiple of n+2. But > (n+1)! + 1 = n(n+1)(n-1)! + 1 = n(n+1)(kn - 1) + 1 > = kn^3 + (k-1)n^2 - n + 1 > = k(-2)^3 + (k-1)(-2)^2 - (-2) + 1 (mod n+2) > = -8k + 4(k-1) + 2 + 1 (mod n+2) > = -(4k + 1) (mod n+2). > By assumption, n+2 is *also* a prime, so you can apply Wilson's > Theorem again to conclude that 4k + 1 = 0 (mod n+2)., finishing > this half of the proof. > To go the other direction, suppose that the given congruence holds. > Then (firstly) n must be a prime, since otherwise (n-1)! + 1 = 1 >(mod n) > and the assumed congruence would imply that 4 = 0 (mod n) (which > you can easily rule out). But. once you know that n is prime, you >get > that (n-1)! + 1 = kn for some k from Wilson's Theorem and then that > n + 4[(n-1)! + 1] = (4k + 1)n > exactly as above. Again, since (n,n+2) = 1, the assumed congruence > implies that 4k + 1 is a multiple of n+2. But also, exactly as >above, > (n+1)! + 1 = -(4k + 1) (mod n+2). Hence Wilson's Theorem holds for > n+2, which implies that n+2 is a prime. So n, n+2 are a pair of >twin > primes. [] in but it sure looks better. === Subject: Re: Twin Primes I'm trying to find the proof for the following statement. >> The integers n, n+2, form a pair of twin primes if and only if >> 4[(n-1)!+1] = -n (mod n(n+2)). >> I know that you can get this from Wilsons theorem but I'm at a loss as >> to how one would go about it. >> So if anyone has a link to this proof or is willing to jot it down >> here I would appreciate it. > Well ... here are some musings that you can probably turn into > an honest-to-goodness proof: > First, suppose that n, n+2 are a pair of twin primes. By Wilson's > Theorem for n, (n-1)! + 1 = kn for some positive integer k. Then > n + 4[(n-1)! + 1] = n + 4kn = (4k + 1)n > and (since (n,n+2) = 1) the result will follow if you can show that > 4k + 1 is a multiple of n+2. But > (n+1)! + 1 = n(n+1)(n-1)! + 1 = n(n+1)(kn - 1) + 1 > = kn^3 + (k-1)n^2 - n + 1 > = k(-2)^3 + (k-1)(-2)^2 - (-2) + 1 (mod n+2) > = -8k + 4(k-1) + 2 + 1 (mod n+2) > = -(4k + 1) (mod n+2). > By assumption, n+2 is *also* a prime, so you can apply Wilson's > Theorem again to conclude that 4k + 1 = 0 (mod n+2)., finishing > this half of the proof. > To go the other direction, suppose that the given congruence holds. > Then (firstly) n must be a prime, since otherwise (n-1)! + 1 = 1 >(mod n) > and the assumed congruence would imply that 4 = 0 (mod n) (which > you can easily rule out). But. once you know that n is prime, you >get > that (n-1)! + 1 = kn for some k from Wilson's Theorem and then that > n + 4[(n-1)! + 1] = (4k + 1)n > exactly as above. Again, since (n,n+2) = 1, the assumed congruence > implies that 4k + 1 is a multiple of n+2. But also, exactly as >above, > (n+1)! + 1 = -(4k + 1) (mod n+2). Hence Wilson's Theorem holds for > n+2, which implies that n+2 is a prime. So n, n+2 are a pair of >twin > primes. [] > in but it sure looks better. I was feeling a little guilty, so ... if you do go this direction, note that there's a point I glided over in describing the argument: what's true is that (n-1)! = 0 (mod n) if n is composite _and_ n > 4. So you need to be a bit careful when proving the <== direction ... === Subject: Re: Calculus/Trig I am not very good at the calculus, but if I read the problem correctly we are looking for some a such that -2(a^2 + a + 6)sin(2x) + a - 2 =/= 0 for any x on R. Or am I misunderstanding something? === Subject: Re: Sequence and Series >hi guys, >suppose I have this sequence( NOT series) >(nî-1)n! / n(n+1)! , n=1,2..... (n^2-1)*n! = (n-1)*(n+1)*n! = (n-1)*(n+1)! Each term in your series reduces to (n-1)/n >can I use ratio test to solve it? >for this series, determine if it converges or diverges. >how do you solve it? >infinity >Sum of (2 + sin n)/ ?(n-1) >n=2 <> === Subject: Solution of equation involving Logical and Arithmetic operator. Hi all, Is there any process by which equations involving logical and arithmetic operators can be solved. For example: if x and y are two integers, then x (xor) 3 + y (and) 4 = (not) 5 x (nand) 5 + y (and) 2 = 6 Something sort of this.. Just asking. Infinity === Subject: Re: Solution of equation involving Logical and Arithmetic operator. > Is there any process by which equations involving logical and > arithmetic operators can be solved. > For example: > if x and y are two integers, then Define xor or nand and not for _all_ integers positive and negative, _rigorously_. > x (xor) 3 + y (and) 4 = (not) 5 You'll find not especially knoty. Don't tell me, -5 = 1 + not 5. Prove it by your definitions. > x (nand) 5 + y (and) 2 = 6 === Subject: Re: Optimization > The line y = mx+b intersects the parabola y = x^2 at point A in the > second quadrant and point B in the first quadrant. Find the point P on > the arc AOB, where O is the origin, that maximizes the area of the > triangle PAB. Clearly the gradient of the parabola at P will be m (do you see why?). Do you know about differentiation yet? Mike. === Subject: Re: Optimization > The line y = mx+b intersects the parabola y = x^2 at point A in the > second quadrant and point B in the first quadrant. Find the point P on > the arc AOB, where O is the origin, that maximizes the area of the > triangle PAB. No. -- Darrell === Subject: Re: Optimization > The line y = mx+b intersects the parabola y = x^2 at point A in the > second quadrant and point B in the first quadrant. Find the point P on > the arc AOB, where O is the origin, that maximizes the area of the > triangle PAB. Ok, first calculate the ordinates of A and B and then the center and radius of AOB. Then figure out the maximum of area is when P is half way from A to B along AOB. So what's the coordinates of P? === Subject: Re: Optimization > The line y = mx+b intersects the parabola y = x^2 at point A in the > second quadrant and point B in the first quadrant. Find the point P on > the arc AOB, where O is the origin, that maximizes the area of the > triangle PAB. Would it help to note that the area of a triangle with vertices at (a,a^2), (b,b^2, and (c,c^2) is proportional to the determinant of the following matrix? M = [[ 1 a a^2 ] [ 1 b b^2 ] [ 1 c c^2 ]]