meta user
network profile
| bio | website | |
|---|---|---|
| location | Madrid, Spain | |
| age | 49 | |
| visits | member for | 2 years, 6 months |
| seen | 5 hours ago | |
| stats | profile views | 146 |
|
May 4 |
comment |
Do these inequalities regarding the gamma function and factorials work? Is $x$ an integer? if not what does it mean $n \vert (x+1)$? |
|
Apr 2 |
comment |
Best Fake Proofs? (A M.SE April Fools Day collection) It seems that Euler himself was confused with this. See [here]( webspace.utexas.edu/aam829/1/m/Euler_files/EulerMonthly.pdf) |
|
Jan 4 |
comment |
Exact power of $p$ that divides the discriminant of an algebraic number field The statement will be easier to follow if you suffix the $\beta$'s. If I understand, there are $f_i$ elements in $B_i$, say $\beta_{ik}$ for $1\le k \le f_i$ and then the $n$ elements $\alpha_1,\dots,\alpha_n$ are $\alpha_{ij}\beta_{ik}$ for $1\le i\le r$, $1\le j\le e_i$ and $1\le k \le f_i$. In addition I think the $x \equiv \alpha_{ij} \pmod{Q_i^{e_i}}$ should read $x \equiv 1 \pmod{Q_i^{e_i}}$. If it is so, then the "large number $N$" means possibly: $N \ge e_i$ for every $i$ as you are then free to chose $\alpha_{ij}$ with the stronger constraint $\alpha_{ij} \equiv 0 \pmod{Q_h^N}$. |
|
Dec 1 |
comment |
Galois group and the Quaternion group I have corrected the right hand side of $(\theta^2-6-2\sqrt{3})^2$, sorry for that. I have also added some computations in the end to help you with the automorphisms. |
|
Nov 10 |
comment |
Early history of lower bounds on the prime counting function I'm sorry I've made an incorrect statement, I could not edit it so I have removed it. |
|
Nov 6 |
comment |
What would be complexity of computing $3^{n^n}$? If you are computing $3^{n^n}$ modulo an integer $M$ and you know it's totient function $\varphi(M)$ then you can compute $n^n$ in $O(\log n)$ multiplications $\pmod{ \varphi(M)}$ and then rise $3$ to the result in $O(\log M)$ multiplications mod $M$ finding the result in polynomial time. The problem is to find $\varphi(M)$. Some times it can be done in polynomial time (for example if $M$ is prime then $\phi(M)=M-1$), but in general if $M$ is composite finding $\varphi(M)$ is equivalent to finding the factorization of $M$. |
|
Nov 5 |
comment |
Proving a simple inequality I find $$f(3) = \frac{\tfrac{2}{3}\log 2}{\log 6} = 0.2579... > f(4) = \frac{\tfrac{1}{2}\log 3}{\log 12} = 0.2210...$$ Are you sure the inequality is correctly written? |
|
Nov 1 |
comment |
Half the rationals? A good candidate for your set $X$ with a very simple description is the set of reduced fractions $a/b$ with $$ a\cdot b \equiv 0 \pmod{3} $$ |
|
Nov 1 |
comment |
Half the rationals? I think there something missing in the right hand side of the inequality $\sum_{k=1}^{n-1} a_k\phi(k) < 1/2$, I suppose you mean $<\frac{1}{2}\sum_{k=1}^{n-1} \phi(k)$? On the other hand, even if it is probably true, I can't see how to prove that this construction works for all the intervals $Y$. |
|
Nov 1 |
comment |
Half the rationals? I think that the proportion of fractions with even numerator and denominator bounded by $N$ is aproximately one third, and the same for even denominator. |
|
Apr 19 |
comment |
Burgess on quadratic residues and non-residues No that's an error I correct it |
|
Apr 19 |
comment |
Burgess on quadratic residues and non-residues I have edited the answer to make it more clear. I hope this helps. |
|
Dec 20 |
comment |
Consequence of Bertrand's Postulate It proves that for every $N$ there exist a $k$ such that there are $N$ (or more) pairs of consecutive primes with common difference $k$. The problem if there are $N$ consecutive primes with common difference $k$ is AFAIK open, though is related to Green-Tao theorem. |
|
Dec 19 |
comment |
Consequence of Bertrand's Postulate user19012: It was my fault there was a flaw in the argument. I have corrected it. |
|
Sep 26 |
comment |
How many ways can I make six moves on a Rubik's cube? @Henry: You are right. actually the number I gave allows 18 legal moves while the OP ask for 12 (90 deg twists) so the number of different positions after at most 6 moves is a lot smaller: 1056772. Actually the method given by him (make six random moves until the puzzle is solved) fails for the 96696 positions after 5 moves if it does not take this into account. |
|
Sep 24 |
comment |
The number of ones in a binary representation of an integer @Matt: Why 16x increase? I'm not sure I agree, as you are substituing a few cheap operations with table lookup which needs slow memory access and messes the cache. |
|
Sep 12 |
comment |
Bound for divisor function $\mathrm{Li} x$ is the logarithmic integral $=\int_2^x \frac{ dt}{\log t}$ see here. |
|
May 5 |
comment |
Example of a rational function such that : $(f(x))^{3} + (g(x))^{3} + (h(x))^{3}=x$ @lhf: In the page linked by Gerry you have several solutions and links as: $(m^3-3^6n^9)^3 + (-m^3+3^5mn^6+3^6n^9)^3 + (3^3m^2n^3+3^5mn^6)^3$ $= m(3^2m^2n^2 +3^4mn^5+3^6n^8)^3$. |
|
Mar 13 |
comment |
The number of symmetric polynomials of n degree I don't understand what you mean by invariant, do you mean homogenous?, all the terms have the same total degree? take this four examples with $n=3$ and $k=2$, a) $x^3+y^3$, b) $x^2y + xy^2$, c) $x^3+y^3 + x + y$ d) $x^2y + xy^2+x+y$, which are you counting as degree 3? |
|
Mar 11 |
comment |
Find maximum divisors of a number in range PARI-GP is a great program to perform number theoretic calculations. The algorithm I've used is the same I have described, but PARI-GP knows how to handle big numbers and that makes everything easier. |
