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bio website mrcactu5.herokuapp.com/…
location New York, NY
age 30
visits member for 4 years
seen 5 hours ago

Data Scientist @ Explorer Media


Nov
7
comment ODE satisfied by $\,f(\xi) = \int_0^1 \frac{e^{-\xi x}}{\sqrt{1 - x^2}}dx$
@GitGud Look for some relationship between the first few derivatives: $f(\xi), f'(\xi), f''(\xi), \dots $
Nov
7
asked ODE satisfied by $\,f(\xi) = \int_0^1 \frac{e^{-\xi x}}{\sqrt{1 - x^2}}dx$
Nov
6
revised When is it easy to write down the Bhargava S-factorial?
added 32 characters in body
Nov
3
revised Does Hlawka Inequality follow from Triangle Inequality?
added 288 characters in body
Nov
3
asked Does Hlawka Inequality follow from Triangle Inequality?
Oct
31
answered Probability (X >Y) when X and Y have the same distribution?
Oct
29
asked What does “of independent interest” mean in math papers?
Oct
27
revised Using Dirichlet's hyperbola method and Dirichlet's formula
reading the photograph and substantial cleaning up of text
Oct
27
revised crazy number of divisors chart on Wikipedia - $\Omega(n)$
added 467 characters in body; edited title
Oct
27
revised crazy number of divisors chart on Wikipedia - $\Omega(n)$
added 467 characters in body; edited title
Oct
27
revised crazy number of divisors chart on Wikipedia - $\Omega(n)$
added 467 characters in body; edited title
Oct
27
asked crazy number of divisors chart on Wikipedia - $\Omega(n)$
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
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.