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| bio | website | |
|---|---|---|
| location | Madrid, Spain | |
| age | 49 | |
| visits | member for | 2 years, 6 months |
| seen | 22 hours ago | |
| stats | profile views | 146 |
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May 4 |
comment |
Do these inequalities regarding the gamma function and factorials work? Is $x$ an integer? if not what does it mean $n \vert (x+1)$? |
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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) |
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Mar 9 |
awarded | Guru |
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Jan 19 |
answered | $n +1$th Fibonacci number modulo $n$ |
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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 $\alpha_{ij} \equiv 0 \pmod{Q_h^N}$. |
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Dec 1 |
revised |
Galois group and the Quaternion group added 1 characters in body |
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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. |
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Dec 1 |
revised |
Galois group and the Quaternion group Added explanation |
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Nov 30 |
answered | Galois group and the Quaternion group |
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Nov 30 |
answered | If coefficients are algebraic then the roots are algebraic |
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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. |
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Nov 8 |
awarded | Yearling |
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Nov 7 |
answered | Cover a disk with thin rectangles |
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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 $M$. |
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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? |
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Nov 3 |
revised |
Half the rationals? Added a generalization |
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Nov 3 |
answered | Half the rationals? |
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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 $$ a\cdot b \equiv 0 \pmod{3} $$ |
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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 $Y$. |
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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. |
