5,351 reputation
926
bio website math.uwaterloo.ca/~j2rosen
location Waterloo, Ontario
age 29
visits member for 2 years, 5 months
seen 29 mins ago

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


2d
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 ${\large\int}_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx$
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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
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
May
14
reviewed Approve suggested edit on The determinant of a special matrix
May
9
reviewed Approve suggested edit on Density on the square, expected value
May
3
reviewed Reject suggested edit on Prove that $\operatorname{rank}(A) + \operatorname{rank}(B) \ge \operatorname{rank}(A + B)$
Apr
23
revised Challenge: Demonstrate a Contradiction in Leibniz' differential notation
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