Luboš Motl
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 Feb 4 comment For a sequence, why must $\lim _{n→∞} {||x_n||} = ∞$, $\lim _{n→∞} {||x_n||} = 0$, or there exists a convergent subsequence with a nonzero limit? No, you are not right. If a sequence is unbounded, it does not imply that the limit is $\infty$. $0,1,0,2,0,3,0,4,0,5,\dots$ is unbounded but it has no limit because some of the entries are stuck at small values while others get arbitrarily large. Feb 4 comment Eigenvector with matrix amlost full with zeros This generalized Jordan form may be found by special Mathematica functions, see e.g. reference.wolfram.com/language/ref/JordanDecomposition.html Feb 4 comment Eigenvector with matrix amlost full with zeros Let me add that for such defective matrices, one may construct "something" instead of the list of eigenvectors, namely the Jordan normal form. In this 2 times 2 case, one has vectors $u$ and $v$ such that $Au=\lambda u$ is an eigenvector, but $(A-\lambda)v=u$. So the action of $A-\lambda$ on the "not quite eigenvector", the other vector, isn't quite zero, but is equal to the (previous) vector, in this case the eigenvector $u$. So one gets chains etc. Feb 4 comment Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ Dear @ArchisWelankar, all the terms in the sum are clearly positive numbers (square of some other positive numbers) so their sum has to be positive, too. Feb 4 answered Compute $(\sin4^\circ)^2 +(\sin8^\circ)^2+(\sin12^\circ)^2+\cdots+(\sin176^\circ)^2$ Jan 30 comment Are certain equations for orthogonal trajectories of a curve incomplete? $y=kx$ fails to include the special case formally "$y=\infty\cdot x$" which is the vertical line. Similarly, $x=ky$ fails to include the special case "$x=\infty\cdot y$", the horizontal line. But those are measure-zero i.e. "infinitely unlikely" values which allows you to say that it basically works. Geometrically, on a generic smooth curve like the circle, the tangents have these precise slopes (exactly vertical or exactly horizontal) at 2+2 places only (top, bottom, left, right) and indeed, you may describe the nearby points and calculate the tangent as a limit whenever the curve is smooth. Jan 30 revised How do you go about solving this recurrence? added 52 characters in body Jan 30 answered Difference between continuous and uncountable set. Jan 28 comment Solve $z^6=(z-1)^6$. Dear @user301068 - the case $k=0$ doesn't produce any solutions because for that value, the equation reduces to $z/(z-1)=1$ which is equivalent to $z=z-1$ and that has no solutions (because it's equivalent to $0=1$), except for (formally) $z=\infty$. So only the values $k=1,2,3,4,5$ produce five solutions. That shouldn't be surprising because your original equation is a 5th order algebraic equation - the term $z^6$ cancels when you expand the polynomials. Jan 28 comment Solve $z^6=(z-1)^6$. It's very hard to understand what you're missing. You present the calculation, divided to really small, correct, elementary pieces, but you still don't seem to be satisfied. What "doesn't work"? Or can't you solve the equation $z/(z-1) = K$ for a constant $K$? This is a linear equation $z=Kz-K$ i.e. after you multiply it by $z-1$ which is solved by $z=-K/(1-K)$. Jan 26 answered Convergence test of $S=\frac{1}{\ln 2} \sum_{k=1}^\infty \ln (1+\frac{1}{k(k+2)}) \ln k$ Jan 25 answered How can I prove that for a Killing vector $\nabla^a \nabla_a \xi^\mu = -R^b_a \xi^a$? Jan 25 comment How can I prove that for a Killing vector $\nabla^a \nabla_a \xi^\mu = -R^b_a \xi^a$? The equation you "got to" can't be right because some termshave the free $\mu$ index but the last term doesn't. Jan 25 answered How to prove $\sin3θ=3\sinθ-4\sin^3θ$ Jan 21 comment Finding Lie algebra isomorphisms This successful localization of "very different matrices" with the same commutator algebras is equivalent to finding "very different representations" of the Lie algebra (and therefore the Lie group, too). The Dirac matrices are matrices expressing the generators of the Lie algebra with respect to "spinor representations". The dimension of the spinor representation is the power of two. Instead of just two nonzero entries, $J_{ij}$ in the spinor representation have one nonzero entry in each row (and in each column). Jan 21 comment Finding Lie algebra isomorphisms Hi @user302234 - yes, I meant exactly that matrix, $1$ at $ij$ and $-1$ at $ji$, otherwise zeroes. The funny thing is that these matrices have the commutators $[J_{ij},J_{kl}] = J_{il}\delta_{jk}-...-...+...$, four similar terms in total, and one may find completely different matrices of different size that have exactly the same commutators. For example, one may find them in terms of the "Dirac gamma matrices" and their products. The Dirac matrices are $2^N\times 2^N$ matrices of various sorts, tensor products of many copies of Pauli matrices and the identity. They anticommute with each other. Jan 19 answered Finding Lie algebra isomorphisms Jan 19 comment Finding Lie algebra isomorphisms There are many ways to prove it, explicitly or by clever arguments, but I want to stress that these isomorphisms are absolutely fundamental, especially in physics. Among other consequences, they're the reason why spin-1/2 particles such as the electron (with the spin up or down) may exist in the 3D space. Jan 18 answered Limit as $x$ goes to infinity Jan 15 comment Normal Distribution sample mean and population mean? Sorry, the last step is $4472/5000=0.894$ standard deviations.