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Mar
23
comment a good introduction to Laplace Beltrami operator over differential manifolds?
Or, maybe, $X$ is a map $X: \mathbb{R}^2 \to\mathbb{R}^3$ and $\Delta X$ means "take the Euclidean Laplacian of each component of $X$ ", and the equation says you obtain the components of a vector field on the image of $X$ which is proportional to the normal vector field?
Mar
23
comment a good introduction to Laplace Beltrami operator over differential manifolds?
Yan, I can't make sense of your equation as it is written. The $\Delta$ operator acts on functions (or differential forms), not on manifolds. Even assuming the LHS was $\Delta_X$ in the sense of "the Laplace-Beltrami operator on the manifold $X$", the RHS is not a differential operator but a vector field on $X$. So, please, clarify.
Mar
22
comment a good introduction to Laplace Beltrami operator over differential manifolds?
Could you try to clarify the notation? For example, what does $\Delta X$ mean for $X$ a smooth surface in $\mathbb{R}^n$?
May
18
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Jan
18
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Jan
18
comment About 0.999… = 1
(Thank you for your valid answer. I accepted the other one because of the comment thread following it)
Jan
18
accepted About 0.999… = 1
Jan
18
comment About 0.999… = 1
Ok, thank you for the explanation.
Jan
18
comment About 0.999… = 1
We're interested in something like $(1-r)(1+r+r^2+\dots r^N)=1-r^N$. Which property (of ordered fields?) is being transferred? (it would be an occasion for me to understand what this transfer principle is about, in an elementary instance like this)
Jan
18
comment About 0.999… = 1
But why the formula $\sum_{i=1}^N r^i=(1-r^N)/(1-r)$, for $r$ real, is still valid when $N$ is infinite? (btw, yes I think "infinite" is the technical term used in nostandard analysis)
Jan
18
asked About 0.999… = 1
Mar
16
awarded  Scholar
Mar
16
accepted Complex elliptic curve for the “conjugate” lattice
Mar
16
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Mar
16
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Mar
16
revised Complex elliptic curve for the “conjugate” lattice
edited body
Mar
16
asked Complex elliptic curve for the “conjugate” lattice