5,391 reputation
926
bio website math.uwaterloo.ca/~j2rosen
location Waterloo, Ontario
age 29
visits member for 2 years, 6 months
seen 5 hours ago

I am a postdoctoral fellow in the department of Pure Mathematics at the University of Waterloo


Oct
10
revised 100 sequential parking spaces
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Oct
10
revised 100 sequential parking spaces
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Oct
10
answered 100 sequential parking spaces
Oct
2
reviewed Approve suggested edit on Compactness of the identity operator
Sep
20
comment Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
Very interesting! I am surprised to see this integral expressed in terms of the "single" polylogarithm.
Sep
17
awarded  Nice Answer
Sep
17
revised Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
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Sep
16
revised Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
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Sep
16
revised Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
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Sep
16
revised Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
added 2 characters in body
Sep
16
revised Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
added 2 characters in body
Sep
16
answered Closed form for ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
Sep
11
awarded  Nice Answer
Jul
27
answered The ring $R$ is a field iff $R$ has a finitely generated divisible module
Jul
26
comment Find the probability that a matrix is diagonalizable
In the computation for part (b), a diagonalizable matrix can have 0 as an eigenvalue, so the number of central diagonal matrices is $p$, and the number of diagonal increasing non-central matrices is $p(p-1)/2$.
Jul
26
revised Find the probability that a matrix is diagonalizable
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Jul
26
answered Find the probability that a matrix is diagonalizable
Jul
22
comment Is every affine scheme the complement of the closed point $x$ of the spectrum of a local ring $A$?
For $R$ algebraic over its prime field, it's not true in general that every subring of $R$ is a field (e.g. $R=\mathbb{Q}$, $A=\mathbb{Z}_{(p)}$). The argument still works for $R=\mathbb{F}_p$, though.
Jul
22
comment Is every affine scheme the complement of the closed point $x$ of the spectrum of a local ring $A$?
(1) A finite field cannot be the ring of fractions of a DVR. (2) There's a classification of those topological spaces homeomorphic to the spectrum of a ring, and I think it implies there always exists local $A$ with Spec$(A)\backslash\{m\}$ homeomorphic to Spec$(R)$ (en.wikipedia.org/wiki/Spectral_space)
Jul
2
awarded  Curious