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seen Jun 2 '10 at 7:34

Jul
25
awarded  Yearling
Jan
25
awarded  Good Answer
Feb
8
awarded  Yearling
May
18
answered Examples of asymptotic formulas with optimal error term
May
18
comment Examples of asymptotic formulas with optimal error term
I think you lifted your result about the number of prime factors of an integer from a certain paper on the arXiv. I would like to point out that sharper results (than included in that paper) follow easily from the Selberg-Delange method: indeed you can get an asymptotic expansion for the number of integers n <= x with Omega(n) = k (mod l). For l > 2 the asymptotic expansion takes the form x/l + c_1*x/(log x) + c_2*x/(log x)^2 + ... with constant c_i. When l = 2 all the c_i's vanish and the error term is expected to be O(x^{1/2+epsilon}), which is in fact equivalent to RH.
May
18
awarded  Nice Answer
May
16
comment Quick proofs of hard theorems
As Persi Diaconis puts it (when discussing Hardy-Ramanujan's proof): "Impressive as the argument is, to a probabilist, the project seems out of focus; they are proving the weak law of large numbers by using the local central limit theorem. If all that is wanted is their theorem, there are much easier arguments. With all their work, one could reach much stronger conclusions". See www-stat.stanford.edu/~cgates/PERSI/papers/Hardy.pdf for the rest of Persi's nice article.
May
16
answered Quick proofs of hard theorems
Apr
22
awarded  Nice Answer
Apr
14
comment What's an example of a transcendental power series?
Ah! I noticed only now that you also want elementary proofs. Well, I don't think the papers cited above use big machinery, but that's subjective of course.
Apr
14
answered What's an example of a transcendental power series?
Mar
26
answered Generalized binomial coefficients and Gaussian density
Mar
26
comment Generalized binomial coefficients and Gaussian density
[There is a small typo in the last line: multiply by exp(-log n * it)]
Mar
26
comment Generalized binomial coefficients and Gaussian density
Note that by Taylor's theorem n^{exp(it)-1} = exp(log n * it - (log n)*t^2/2 + O((log n)*t^3)) and 1/Gamma(exp(it)) = 1 + O(t). Therefore n^{exp(it)-1}/Gamma(exp(it)) = exp(log n * it - (log n) * t^2/2 + O((log n)*t^3)). Now what happens when you multiply the above formula by exp(-log n * t) and then substitute t/sqrt(log n) for t? You get e^{-t^2/2} in the limit!! :-)
Mar
23
comment Asymptotics of infinite Gauss sums
Perhaps you mean a = O(T^(1/2-epsilon))?
Mar
17
comment An elementary number theoretic infinite series
The main contribution to the sum will come from the integers n <= N, with (1/2)loglog N + O((loglog N)^(1/2 + epsilon)) prime factors. This is again a consequence of... the Selberg-Delange method.
Mar
17
awarded  Commentator
Mar
17
comment An elementary number theoretic infinite series
oups... i meant sum(z^w(n), n <= X) ~ C * X * (log X)^{z-1} in my comment above.
Mar
17
answered An elementary number theoretic infinite series
Mar
17
comment An elementary number theoretic infinite series
The correct guess is C*(log N)^(1/2). Look-up the Selberg-Delange method. (The point is that sum(z^w(n), n <= X) ~ C*(log X)^(z-1) and (1/2)^w(n) is essentially the same as 1/d(n), so by partial summation we get sum(1/kd(k),k <= X) ~ C*(log X)^(1/2-1+1))