Sergey Finsky
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 Jan 6 accepted Cantor-Bendixson theorem proof Jan 5 asked Cantor-Bendixson theorem proof Dec 23 comment Number of representable as sum of 2 squares Yes, it'd be nice if somebody posted whole proof, because it's know quite straightforward. But thank you very much, I would never expect to see asymptotics for this problem, thought, there'll be only some bounds. Dec 23 accepted Number of representable as sum of 2 squares Dec 23 revised Number of representable as sum of 2 squares added 6 characters in body Dec 23 comment Number of representable as sum of 2 squares Sure, I meant $\Omega$. Thank you. Dec 23 asked Number of representable as sum of 2 squares Dec 23 comment Asymptotics with prime of form 4k+3 Yes, but in the paper we consider $L$-function, which is but it's definition a series (not finite), so we have to somehow jump from the infinite series to finite. (It's easy to do when the convergence in series is uniform, but it's obviously not the case) I mean, your previous comment is obviously true, but why $\sum_{p \equiv a \mod d \le n} \frac{1}{p} \sim \frac{\log(\log(n))}{\phi (d)}$. Do you see what I mean? Dec 23 comment Asymptotics with prime of form 4k+3 Thank you very much for your answer, but I have one question left: in papers it's proven that $\sum_{p \equiv a \pmod d} \frac{1}{p^{s}} \sim \dfrac1{\phi(d)}\sum_{p} \frac{1}{p^{s}}$ when $s \to +1$. But how does original problem follows from this? (It is not obvious) Dec 22 accepted Asymptotics with prime of form 4k+3 Dec 22 revised Asymptotics with prime of form 4k+3 there was a mistake in sum of prime fractions Dec 22 comment Asymptotics with prime of form 4k+3 Thank you, sure I did. =) Dec 22 asked Asymptotics with prime of form 4k+3 Dec 4 accepted Asymptotic expansion of the integral $\int_2^x \frac{e^t}{t} dt$ for $x \to \infty$ Dec 4 asked Asymptotic expansion of the integral $\int_2^x \frac{e^t}{t} dt$ for $x \to \infty$ Nov 23 comment Lebesgue measure is invariant under isometry Thank you guys very much. I just want to supply your beautiful answers with some links. proof that translation invariant measure is Lebesgue measure up to a constant factor Nov 22 asked Lebesgue measure is invariant under isometry Nov 18 awarded Supporter Nov 18 accepted Closed surjective map Nov 17 comment Power sums, fast algorithm Sorry, really, there is a misunderstanding in terminology.