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5h
answered Why is $\lim\limits_{N\to\infty}x^{N+1}=0$, where $|x|<1$?
1d
comment Find the axis of rotation of a rotation matrix by $INSPECTION$ (NOT by solving $Kv=v$)
Here, "cyclically permutes..." just means $K\mathbf{e}_{1} = -\mathbf{e}_{2}$, $K(-\mathbf{e}_{2}) = \mathbf{e}_{3}$, and $K\mathbf{e}_{3} = \mathbf{e}_{1}$ (as indicated by the arrows). These equations can be verified by computing the products of $K$ with the appropriate column vectors, or by noting that the $i$th column of $K$ is $K\mathbf{e}_{i}$ (a general property of matrix multiplication).
1d
answered Find the axis of rotation of a rotation matrix by $INSPECTION$ (NOT by solving $Kv=v$)
1d
reviewed Close How to show that the norms are not equivalent on any infinitely dimensional closed subspace of $C[0,1]$
1d
reviewed Close Abstract Algebra: Prove that every field has only trivial ideals
2d
reviewed Close Hi I was thinking about a problem and have a question:
2d
reviewed Close Show $\lim\limits_{a \rightarrow + \infty} \int_0^{\infty} \frac{y}{1+y^2}e^{-ay} dy =0 $
2d
reviewed Close Performance Index for sports
2d
reviewed Close What is the shared secret? in Diffie-Hellman
2d
reviewed Close Squares modulo 2^n
2d
comment Manifolds and their dimension
An inner product on $\mathbf{R}^{3}$ is a (symmetric, positive-definite) real-valued function on $\mathbf{R}^{3} \times \mathbf{R}^{3}$; you probably want to identify an inner product with a matrix satisfying suitable properties, then determine the dimension of the set of all such matrices.
2d
comment Stokes' Theorem and Surface Independence Failure
The divergence really does vanish everywhere except the origin. If it didn't, this example wouldn't be interesting: There would be no reason to "expect" (i) $F$ to be the curl of another field $A$ or (ii) the flux of $F$ across a surface to depend only on the boundary, etc.
Apr
21
reviewed Close Cut Edges Question
Apr
21
reviewed Close max members in an infinite set
Apr
21
reviewed Looks Good What is a harmonic conjugate of $u=Arg(z)$?
Apr
21
reviewed Leave Closed Proof that a product of two quasi-compact spaces is quasi-compact without Axiom of Choice
Apr
21
reviewed Approve suggested edit on Solution to this problem?
Apr
21
comment Stokes' Theorem and Surface Independence Failure
You're very welcome. Incidentally, cutting in the middle is all right; check the respective directions of boundary traversal and corresponding surface normals. :)
Apr
21
comment Stokes' Theorem and Surface Independence Failure
It's possible I misunderstood your question. There exist vector fields $A_{1}$ and $A_{2}$ defined on, say, the complement of the positive $z$-axis and the complement of the negative $z$-axis, such that $\nabla \times A_{i} = F$. (This is not trivial, but follows because $\nabla \cdot F = 0$, and the indicated regions are contractible.) But since $F$ has non-zero flux over the unit sphere, there is no vector field $A$ on the complement of the origin such that $F = \nabla \times A$.
Apr
21
answered Stokes' Theorem and Surface Independence Failure