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Apr
6
revised Solving $z=w/2-\sin(tw)/(2t)$ for $w$
deleted 140 characters in body
Apr
4
accepted Solving $z=w/2-\sin(tw)/(2t)$ for $w$
Apr
4
reviewed Approve Solving $z=w/2-\sin(tw)/(2t)$ for $w$
Apr
4
asked Solving $z=w/2-\sin(tw)/(2t)$ for $w$
Apr
2
accepted Question on branches and $\iff$.
Mar
30
comment Question on branches and $\iff$.
Thank you. So if I restrict my attention to the principal branch (and I will also assume $f$ and $g$ are continuous) then $f+ig=0\iff e^{f(x)}\cos(g(x))+ie^{f(x)}\sin(g(x))=1$ ?
Mar
30
revised Question on branches and $\iff$.
added 68 characters in body
Mar
30
revised Question on branches and $\iff$.
added 8 characters in body
Mar
30
asked Question on branches and $\iff$.
Mar
30
revised $(\delta,\varepsilon)$ Proof of Limit
added 8 characters in body
Mar
30
revised $(\delta,\varepsilon)$ Proof of Limit
added 523 characters in body
Mar
30
revised $(\delta,\varepsilon)$ Proof of Limit
added 2 characters in body
Mar
30
answered $(\delta,\varepsilon)$ Proof of Limit
Mar
27
comment What does $O\left(\frac{1}{\log\log T}\right)$ mean?
Thanks. That's what I thought. So basically what the paper says is that as $T\to\infty$ the number of zeros outside the region is some constant multiple of $1/(\log\log T)$ ?
Mar
26
revised What does $O\left(\frac{1}{\log\log T}\right)$ mean?
added 16 characters in body
Mar
26
asked What does $O\left(\frac{1}{\log\log T}\right)$ mean?
Mar
25
comment Series with $e^{\frac{1}{n}}$
Shouldn't that be $+O(n^{-4})$ in your equality?
Mar
25
comment Series with $e^{\frac{1}{n}}$
You could first try expanding $e^{1/n}$ to some order ($O(1/n^k)$). Presumably the value of $k$ will be related to the existence of $1/(2n^2)$ in your summand. Then simplify if possible, and see what happens from there. Haven't tried it so can't be 100% sure, but that's what I'd do first.
Mar
24
comment Complex integration confusion
Try $z=e^{i\theta}$, where $\theta\in[0,2\pi]$.
Mar
21
comment Why Does $ \sum\limits_{k=0}^n \begin{pmatrix} n+1 \\ k+1 \end{pmatrix} p^{k+1} (1-p)^{n-k} $ sum to $ (1-(1-p)^{n+1}) $?
The binomial theorem states that: $$(a+b)^n=\sum_{k=0}^n{n\choose k}a^kb^{n-k}.$$