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Jul
31
revised why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?
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Jul
30
revised why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?
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Jul
30
revised show this sequence inequality $x_{2^n}$
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Jul
30
answered why is $\zeta(1+it) \neq 0$ equivalent to the prime number theorem?
Jul
30
revised show this sequence inequality $x_{2^n}$
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Jul
30
answered show this sequence inequality $x_{2^n}$
Jul
23
comment Product of complex numbers $m+in$ with $0 < m,n \leq N$
This is relevant: mathoverflow.net/questions/116336/…
Jul
23
answered A sum involving twin primes and Prime Number Theorem
Jul
23
comment Order of Magnitude
To see this, note that the relevant integral is $$\int_{2}^{x}\frac{t^{-2\omega(x)}}{t\log t}dt,$$ and it satisfies $$\int_{2}^{x}\frac{t^{-2\omega(t)}}{t\log t}dt\geq\int_{2}^{x}\frac{t^{-2\omega(x)}}{t\log t}dt\geq x^{-2\omega(x)}\log\log x.$$
Jul
23
comment Order of Magnitude
Be careful sweeping details under the rug as something is not right with how you truncated the exponential series and put in a big-O term. In particular, we should not have $O(\omega(T))$ on the right hand side. For monotonic $\omega$, if $$\omega(t)=o\left(\frac{1}{\log t}\right)$$ then you will have a main term of $\log \log x$. If $$\omega(t)=\frac{c}{\log t}\left(1+o(1)\right),$$ then you will have a main term of $e^{-2c} \log \log x$, and for $\omega(t)$ larger than this, the sum will not behave like $\log \log x$ anymore.
Jul
21
awarded  Good Answer
Jul
17
answered For any $n$ is there $n$ consecutive $0$ in the decimal expansion of $2^m$ for suitable $m \in \mathbb N$?
Jul
16
answered Understanding Wright's proof of Landau's theorem
Jul
16
comment Understanding Wright's proof of Landau's theorem
@Charles: Well, yes, it's definitely relevant since this looks like a typo.
Jul
16
comment Understanding Wright's proof of Landau's theorem
You didn't define $\Omega_k(x)$.
Jul
16
revised An identity involving Gauss sums and convolution
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Jul
15
comment An identity involving Gauss sums and convolution
@user152169 Just take the inverse Fourier transform. It follows that $$f(m)=\delta(m) + \sum_{r:\ (r,N)>1} c_r e^{2\pi i m r/N}$$ where the $c_r$ can be any constants we like.
Jul
15
revised An identity involving Gauss sums and convolution
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Jul
15
comment An identity involving Gauss sums and convolution
@user152169: I was thinking of $N$ being prime. In any case this solution completely characterizes all such functions $f$. We have $G_\chi*f=G_\chi$ if and only if $\hat{f}(r)=1$ for all $(r,N)=1$. It doesn't matter what the value of $\hat{f}(r)$ is for $(r,N)>1$.
Jul
15
revised An identity involving Gauss sums and convolution
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