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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)) |