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May
10
comment Which operators commute with integration?
I may have mistagged this as operator theory, feel free to edit if so.
May
10
asked Which operators commute with integration?
May
10
asked When is $\sum_{n,m=-\infty}^\infty \frac{1}{(n\omega_1+m \omega_2)^\alpha}\in \mathbb{R}$?
May
7
answered Integral $\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha$
May
2
accepted PID question in Ireland and Rosen
May
2
comment PID question in Ireland and Rosen
Thanks, I obviously just pretended to myself to learn the definition of an ID.
May
2
revised PID question in Ireland and Rosen
added 1 character in body
May
2
asked PID question in Ireland and Rosen
Apr
30
accepted Intuition behind normal subgroups
Apr
30
comment Intuition behind normal subgroups
@TobiasKildetoft Are there any other reasons or is that pretty much it?
Apr
30
revised Intuition behind normal subgroups
deleted 31 characters in body
Apr
30
asked Intuition behind normal subgroups
Apr
28
awarded  Popular Question
Apr
25
reviewed Reject suggested edit on Variance of importance sampling estimator
Apr
25
reviewed Approve suggested edit on Green's Theorem and Divergence (2D)
Apr
23
reviewed Approve suggested edit on Prove something is divisible by a prime
Apr
23
comment Be $G=\{ e,g_1,g_2,\ldots, g_n \}$, $|G|=n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.
@marcelolpjunior You mean you proved it for $n$ odd, then discarded the case of $n$ even because $n$ is never even? If so, that's correct.
Apr
23
revised Different methods of calculating $\zeta(s)$'s Laurent series.
edited title
Apr
23
comment Different methods of calculating $\zeta(s)$'s Laurent series.
@robjohn Would it be too radical a change to change the question to 'ways to calculate the Laurent series of zeta', or is it better to delete and reask?
Apr
23
answered Be $G=\{ e,g_1,g_2,\ldots, g_n \}$, $|G|=n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.