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seen Jun 10 at 12:27

I'm not who you think I am!


Jul
2
awarded  Curious
Jun
4
awarded  Nice Question
Apr
5
awarded  Nice Question
Feb
18
comment Find an isomorphism between two graphs
Yeah, seriously, dude, it's a cube. Just look at it. Label the vertices and you're done.
Feb
14
asked Variational characterization of gradient?
Nov
19
awarded  Critic
Nov
15
accepted Why does it appear that Willmore energy is always zero?
Nov
15
awarded  Commentator
Nov
15
comment Add vector on vector's tip and rotation
Ok, then compute the length $a=|u+v|$; the vector you want is then $av/|v|$.
Nov
15
awarded  Yearling
Nov
15
answered Cardioid: converting parametric form into polar coordinates
Nov
15
answered Add vector on vector's tip and rotation
Nov
15
comment Why does it appear that Willmore energy is always zero?
This was a good exercise -- thanks!
Nov
15
answered Why does it appear that Willmore energy is always zero?
Nov
9
comment Why does it appear that Willmore energy is always zero?
Hey Will - no book, but here's how I derived it. Stokes' says $\int_M d\alpha = \int_{\partial M} \alpha$ for $\alpha$ an $(n-1)$-form on an $n$-manifold. Then $\int_M df \wedge \star \alpha = \int_M d(f\star\alpha)-fd\star\alpha=\int_{\partial M} f\star\alpha - \int_M fd\star\alpha$ and the boundary integral in the final expression vanishes because there is no boundary. Letting $f=H$ and $\alpha=N^\flat$ we get the statement made in the original post. (Recall that $\nabla \cdot N = \star d\star N^\flat$ and $\nabla H = (dH)^\sharp$.)
Nov
9
accepted Fitting a quadratic polynomial of a special form
Nov
9
asked Why does it appear that Willmore energy is always zero?
Nov
20
comment Fitting a quadratic polynomial of a special form
Ok, but still there are always a small number of solutions. It's surprising to me that these solutions are so hard to characterize.
Nov
20
asked Fitting a quadratic polynomial of a special form
Oct
28
accepted Proof that vector area is a boundary integral?