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| visits | member for | 2 years, 9 months |
| seen | May 16 at 1:55 | |
| stats | profile views | 222 |
Tyler Hilton
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1d |
awarded | Notable Question |
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Apr 13 |
awarded | Popular Question |
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Apr 9 |
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complex analysis poles and residues Okay, I am pretty confused about it. I will read it over and over. thanks for your help. |
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Apr 9 |
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complex analysis poles and residues Okay so you have that for all $n < D$ the denominator can not be zero. What if $n$ is a multiple of 2? wouldnt that make the denominator zero? |
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Apr 9 |
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complex analysis poles and residues Okay and last question. You say $z^k = 1$ iff $z = e^{2\pi i \frac{m}{k}}$ what is m in this case? |
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Apr 9 |
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complex analysis poles and residues Could you explain your last line. why is it analytic at $z = e^{2\pi i 1/d}$ |
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Apr 9 |
accepted | complex analysis poles and residues |
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Apr 9 |
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complex analysis poles and residues Is there a reason why the author does the "max" thing. ie what is the reasoning behind showing that $|F(z)| \leq A\frac{.}{.}$ |
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Apr 9 |
asked | complex analysis poles and residues |
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Apr 9 |
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using Dirichlet transforms to show infinity of primes The main paper is here, so maybe this can help you? math.uga.edu/~pollack/infprimes-final.pdf |
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Apr 9 |
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using Dirichlet transforms to show infinity of primes Could you share why the infinitude of primes follow? |
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Apr 8 |
asked | using Dirichlet transforms to show infinity of primes |
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Apr 5 |
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A proof that the Dirichlet and Möbius transforms are inverses of each other I am confused by your answer. It would be appreciated if you can expand on it if you have time. |
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Apr 5 |
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A proof that the Dirichlet and Möbius transforms are inverses of each other Your last paragraph says that the product can be written as a Dirichlet series. Is there a reason why? |
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Mar 27 |
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Jacobian and Newton's method Is this Numerical methods by burden? if so, it does contain a fair bit of errors. |
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Mar 25 |
accepted | Showing the sum over primes is equal to an integral |
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Mar 25 |
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Showing the sum over primes is equal to an integral I am very confused. Thats what I had though, no? I set f(n) = log(n) |
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Mar 25 |
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Showing the sum over primes is equal to an integral I believe I was thinking of $\log$. since the antiderivative of $1/u$ is $\log$ correct |
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Mar 25 |
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Showing the sum over primes is equal to an integral I see. What is the correct method I am looking for then? |
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Mar 25 |
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Showing the sum over primes is equal to an integral can i define it in terms of cases? log n if n is prime, 0 otherwise? |
