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5h
answered Limit notation, tried factorising.
5h
comment Limit notation, tried factorising.
@rert588 : You don't always have to work to compute a limit! In this case plugging in is definitely the only sane way to proceed ; anything else would look crazy.
5h
comment This n can not be odd
Godement is a very good mathematician. Please verify that you copied his statement correctly.
5h
comment This n can not be odd
By the way, there is something wrong in the statement since 3) is true for all $n$ and the other statements do not involve $n$.
5h
comment This n can not be odd
@Walter : he means $m \equiv 12 \pmod{pq}$.
5h
comment This n can not be odd
What the hell? Where did you find this
19h
answered Is there a unique homomorphism of $\mathbf Z $ into $A$?
20h
revised Is there an alternative intuition for solving the probability of having one ace card in every bridge player's hand?
added 238 characters in body
20h
answered Is there an alternative intuition for solving the probability of having one ace card in every bridge player's hand?
21h
comment Orthonormal basis for the null space of almost-Householder matrix
@hayer : Um, I think you meant eigenvalue. If $Hv = 0 = 0v$, then $v$ is an eigenvector of $H$ with eigenvalue $0$.
21h
comment Orthonormal basis for the null space of almost-Householder matrix
@hayer : I am afraid there is not much more I can detail. Just try thinking about it for a little while, or maybe ask a more precise question?
21h
comment Orthonormal basis for the null space of almost-Householder matrix
@haver : $Hv = v - \langle v,v \rangle v = v-v = 0$.
21h
comment Orthonormal basis for the null space of almost-Householder matrix
@haver : Since $Hv = v - \langle v,v \rangle v = v - v = 0$, $v$ lies in the kernel and has eigenvalue $0$. Any vector $w$ orthogonal to $v$ satisfies $Hw = w$, so it is an eigenvector with eigenvalue $1$. Therefore if we take an orthonormal basis for the subspace of vectors orthogonal to $v$, say $\{w_1,\cdots,w_{n-1}\}$, then $\{w_1,\cdots,w_{n-1},v \}$ is an orthonormal basis for $V$ and over this basis, the matrix for $H$ looks like $$ \begin{bmatrix} 1 & 0 & & & \\ 0 & 1 & \ddots & & \\ & \ddots & \ddots & \ddots & \\ & & \ddots & 1 & 0 \\ & & & 0 & 0 \end{bmatrix} $$
1d
answered Orthonormal basis for the null space of almost-Householder matrix
2d
comment Is there something between summation and integration?
@Solomonoff'sSecret : The classical construction of this complete/translation invariant/containing all intervals measure on the reals is given by the completion of the Haar measure which would give you the Lebesgue measure. The problem with this construction on $\mathbb Q$ is that $\mathbb Q$, even though Hausdorff, is not locally compact. It's also for this reason that the construction works better when we "patch the holes" in $\mathbb Q$ to recover local compactness. It's not a proof that there is no such measure on $\mathbb Q$, but it is a reason to believe why it fails to exist.
2d
comment Prove sum of $\sin$ of angles is greater than $\sin$ of sum of angles
@Alex R. : I know, but it's pretty obvious that I used the "sledgehammer-no-thinking-required" method, just applying the usual tools of optimization. Your comment was already suggested in other answers, so I gave a different approach. It's not a bad thing to be able to compute things, otherwise why do we learn tools?
Jun
28
answered Prove sum of $\sin$ of angles is greater than $\sin$ of sum of angles
Jun
28
comment Prove sum of $\sin$ of angles is greater than $\sin$ of sum of angles
It cannot be "fairly obvious" if you can't prove it!
Jun
26
comment Irreducibility of the $k$-secant variety of the Veronese variety.
@Jan-MagnusØkland : If $E$ is a projective bundle $f : E \to X$ over the irreducible base $X$, pick a open subset $U \subseteq X$ such that $f^{-1}(U) \to X$ is isomorphic to the trivial bundle $U \times \mathbb P^n \to U$, so that $f^{-1}(U)$ is irreducible. Since $f^{-1}(U)$ is open (hence dense in $E$), $E$ is irreducible.
Jun
23
comment Why does $\cos(x) + \cos(y) - \cos(x + y) = 0$ look like an ellipse?
+1 just for the plots. :D