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bio website mrcactu5.herokuapp.com/…
location New York, NY
age 29
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Data Scientist @ Explorer Media


Oct
17
revised Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
added 2 characters in body
Oct
15
revised Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
forgot an important sum sign
Oct
15
revised Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
connection to Virsasoro
Oct
15
revised Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
connection to Virsasoro
Oct
15
answered Nice derivation of $\sum_{n=1}^\infty \frac{1}{n} \left( \frac{q^{2n}}{1-q^n}+\frac{\bar q^{2n}}{1-\bar q^n}\right)=-\sum_{m=2}^\infty \ln |1-q^m|^2$
Oct
14
comment Correctly transforming ODEs
@resu I was going to suggest $y+1 \mapsto y$ for the first step. Where did these equations come from? What text?
Oct
13
comment What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$?
line 3 is very convenient that $x \mapsto x^k$ is the same as rearranging the terms
Oct
13
comment What is the value of $\sum_{m=1}^{19} \frac{1}{\zeta^{3m}+\zeta^{2m}+\zeta^{m}+1}$ with $\zeta=e^{2\pi i/19}$?
this is good you can use polynmial identity summing over the roots: $$ \frac{d}{dz} \big[\log p(z)\big]= \frac{p'(z)}{p(z)} = \sum \frac{1}{z-a} $$ where $p(z) = z^N-1$.
Oct
13
comment Show Laplace operator is rotationally invariant
@EmutheEmu By chain rule, your equation $u = x \cos \theta + y \sin \theta$ becomes $$\frac{\partial}{\partial u} = \frac{\partial}{\partial x} \cos \theta + \frac{\partial}{\partial y} \sin \theta $$ and a similar formula for second derivatives. Then you can plug in.
Oct
13
revised Show Laplace operator is rotationally invariant
added 643 characters in body
Oct
12
answered Show Laplace operator is rotationally invariant
Oct
12
revised Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$
added 226 characters in body
Oct
12
revised Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$
added 226 characters in body
Oct
12
revised Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$
added 226 characters in body
Oct
12
answered Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$
Oct
6
comment How to prove $\frac{(a_1 a_2\cdots a_n)^2-1}{8}\equiv\sum_{i=1}^n\frac{a^2_i -1}{8}\pmod 8$
Can you explain between step 2 and 3 - turning the product $\prod$ into sum $\sum$?
Oct
4
revised How prove this $\int_{a}^{b}f(x)dx=\frac{1}{2}(b-a)[f(a)+f(b)]-\frac{1}{12}(b-a)^3f''(\xi)$
added 242 characters in body
Oct
4
revised How prove this $\int_{a}^{b}f(x)dx=\frac{1}{2}(b-a)[f(a)+f(b)]-\frac{1}{12}(b-a)^3f''(\xi)$
added 988 characters in body
Sep
30
awarded  Explainer
Sep
29
answered Generalizations of $\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}$ and $\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}$