bio  website  math.ucla.edu/~tao 

location  Los Angeles  
age  39  
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Professor of Mathematics at UCLA
1d

awarded  Nice Answer 
1d

comment 
How to prove this determinant is positive?
Seems like $e^{A_i}$, and hence $e^{A_1} \dots e^{A_n}$, lies in the split orthogonal group. Not sure how to prevent such an element from having a negative eigenvalue pair $\lambda,1/\lambda$ though. 
Apr 30 
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Is the set $ AA+A $ always at least as large as $ A+A $?
Ah, you're right. The argument I had in mind showed that the natural probability measure on $AA$ had small Fourier coefficients when $A \gg F^{1/2}$, but this is only enough to get $A+AA$ somewhat larger than $A$, and not positive density in $F$, unless $A$ is significantly closer to $F$ (I can believe $F^{2/3}$ is the natural limit of the method). 
Apr 30 
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Distribution of the sequence $\bigl (\frac{\phi(n)}{n}\bigr )_{n=1}^{\infty}$ in $[0,1]$
Bernoulli convolutions are reasonably explicit (their characteristic function is a Riesz product, so their distribution is the inverse Fourier transform of a Riesz product), although the formula might not be terribly tractable in practice. 
Apr 29 
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Distribution of the sequence $\bigl (\frac{\phi(n)}{n}\bigr )_{n=1}^{\infty}$ in $[0,1]$
If one were interested in the limiting distribution of $\frac{\phi(p1)}{p1}$ instead of $\frac{\phi(n)}{n}$, the only difference would be that the $I_p$ now have expectation $1/(p1)$ rather than $1/p$. 
Apr 29 
comment 
Distribution of the sequence $\bigl (\frac{\phi(n)}{n}\bigr )_{n=1}^{\infty}$ in $[0,1]$
Starting from $\frac{\phi(n)}{n} = \exp( \sum_p \log(1\frac{1}{p}) 1_{pn} )$, one can show (using the absolute convergence of $\sum_p \log(1\frac{1}{p}) / p$) that the limiting distribution $\mu$ of $\frac{\phi(n)}{n}$ is the distribution of $\exp( \sum_p \log(1\frac{1}{p}) I_p )$, where the $I_p$ are independent Bernoulli variables with expectation $1/p$. In other words, $\mu$ is the exponential of a certain Bernoulli convolution, which one would then typically expect to be rather singular in nature (as indicated by the references given in other answers here). 
Apr 29 
answered  Is the set $ AA+A $ always at least as large as $ A+A $? 
Apr 29 
comment 
Is the set $ AA+A $ always at least as large as $ A+A $?
@OliverRocheNewton By taking tensor powers one can get $AA+A/A+A$ arbitrarily small in the product ring $({\bf Z}/p{\bf Z})^n$, though this is cheating of course as this ring has some zerodivisors. It could well be that there is an ErdosSzemeredi type bound $AA+A \gg \min(A^2,F)$ whenever $A$ generates a finite field $F$, which would imply this weaker conjecture. (I think, by the way, that Fourier analysis gives this bound in the case $A \gg F^{1/2}$, which rules out modifying the example given in my answer.) 
Apr 28 
awarded  Good Answer 
Apr 28 
revised 
Is the set $ AA+A $ always at least as large as $ A+A $?
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Apr 28 
comment 
Is the set $ AA+A $ always at least as large as $ A+A $?
Fair enough; I was implicitly thinking asymptotically in my answer, and have now edited it to clarify. But perhaps $\{1,0,1\}$ is the only counterexample (this is about as close as one can get to being both closed under multiplication and addition, for a finite subset of the reals). 
Apr 28 
revised 
Is the set $ AA+A $ always at least as large as $ A+A $?
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Apr 28 
awarded  Nice Answer 
Apr 28 
answered  Is the set $ AA+A $ always at least as large as $ A+A $? 
Apr 26 
awarded  Enlightened 
Apr 26 
awarded  Nice Answer 
Apr 26 
revised 
Are reduced residue systems relative primorials an active area of research? If not, why not?
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Apr 26 
answered  Are reduced residue systems relative primorials an active area of research? If not, why not? 
Apr 23 
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Asymptotic limit of truncated Legendre sieve
@Lucia You are in good company: en.wikipedia.org/wiki/Legendre%27s_constant Indeed, the numerical discrepancy that caused Legendre to be off by about 8% may in fact be related to the one in this question. 
Apr 23 
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Asymptotic limit of truncated Legendre sieve
Formally, one has $M(z) =  \sum_{n>z} \mu(n)/n$, so the sum $\sum_{k=1}^\infty M(k) \log(1+\frac{1}{k})$ appears to telescope formally to $\sum_n \frac{\mu(n) \log n}{n} = 1$. One can presumably justify these formal computations by zeta regularization. 