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location  New York, NY  
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Data Scientist @ Explorer Media. Statistics, Geometry and Physics.
4m

revised 
Brun's algorithm
added 540 characters in body 
12m

comment 
Brun's algorithm
@CatherinePfaff do any of these examples on these slides by Shweiger match your definition? 
15h

answered  Max flow, min cut on manifolds 
15h

answered  Brun's algorithm 
16h

comment 
Brun's algorithm
possibly, stuff by Valerie Berthé (pdf)? 
Jul 19 
asked  why is this result about Gaussian analytic functions equivalent to the Crofton formula 
Jul 17 
comment 
Hilbert Matrix and Approximation Theory
@DavidSpeyer Legendre polynomials also arise as the matrix elements of $SO(3)$ representations. I got interested because of Hilbert's use of Minkowski theorem and geometry of numbers, but never collected my thoughts on this. 
Jul 17 
comment 
Hilbert Matrix and Approximation Theory
@DavidSpeyer Do you think Chebyshev polynoimals are a better choice than Legendre polynomials that Hilbert picked? In Legendre basis this is a rational quadratic form so it can take arbitrarily small values as $n \to \infty$, I guess. 
Jul 8 
revised 
Degrees of maps from curves to $\mathbb P^1$
typo... cuves  i was rolling on the floor for 10 minutes 
Jul 6 
asked  1D TQFT in FreedHopkinsLurieTeleman 
Jul 6 
accepted  DijkgraafWitten TQFT vs. Representation Theory? 
Jul 4 
revised 
conjectures regarding a new Renyi information quantity
added 46 characters in body 
Jul 4 
revised 
conjectures regarding a new Renyi information quantity
added 218 characters in body 
Jul 2 
comment 
Angles and proportions occurring in Lsystem fractals
You may be intersted in this book Fractal Geometry: Mathematical Foundations and Applications by Kenneth Falconer. 
Jul 2 
awarded  Socratic 
Jul 2 
awarded  Inquisitive 
Jul 2 
awarded  Curious 
Jun 30 
comment 
Free Boson Correlator $ \langle X(z)X(w) \rangle = \ln z  w $
@MarcelBischoff $\langle x(z)x(w) \rangle =  \ln (zw)$ is not always positive right? In fact $$\langle x(z)x(z+1) \rangle =  \ln 1 = 0$$ And it is hard to define the norm $x(z) = \langle x(z)x(z) \rangle$ which would be divergent. So we can't put $x(z)$ into our Hilbert space. 
Jun 29 
revised 
Free Boson Correlator $ \langle X(z)X(w) \rangle = \ln z  w $
added 517 characters in body; edited title 
Jun 29 
comment 
Free Boson Correlator $ \langle X(z)X(w) \rangle = \ln z  w $
@AndréHenriques p22: the holomorphic part $x(z)$ is not a conformal field. Under a conformal map the metric transforms like $$ ds^2 \mapsto \frac{\partial f}{\partial z} \frac{\partial \overline{f}}{\partial \overline{z}} ds^2 $$ conformal "fields" $\phi(z, \overline{z})$ transform like differential forms, so that $ \phi(z, \overline{z}) dz^h d\overline{z}^{h'}$ is invariant: $$ \phi \mapsto \big(\frac{\partial f}{\partial z}\big)^h \big( \frac{\partial \overline{f}}{\partial \overline{z}}\big)^{h'} \phi $$ Maybe $x(z)$ can have a logarithmic singularity or something. 