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Ben Green
Ben Green
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Timothy,

This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here:

www.math.uchicago.edu/~schlag/papers/POLTRAN.pdf

I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have).

I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors.

I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large).

EDIT: Thinking about it some more, it's not obvious to me even how one would show that $S(1/x) \neq 0$, though maybe this does follow from some kind of lower bound for linear forms in $\log p$. My point is that if there is deviation from the behaviour for a random sequence I would expect that one would see it near $\alpha = 0$ (and near other rationals with small denominator).

Timothy,

This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here:

www.math.uchicago.edu/~schlag/papers/POLTRAN.pdf

I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have).

I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors.

I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large).

Timothy,

This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here:

www.math.uchicago.edu/~schlag/papers/POLTRAN.pdf

I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have).

I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors.

I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large).

EDIT: Thinking about it some more, it's not obvious to me even how one would show that $S(1/x) \neq 0$, though maybe this does follow from some kind of lower bound for linear forms in $\log p$. My point is that if there is deviation from the behaviour for a random sequence I would expect that one would see it near $\alpha = 0$ (and near other rationals with small denominator).

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Ben Green
Ben Green
  • 4.8k
  • 2
  • 37
  • 35

Timothy,

This is likely to be a pretty difficult question I think. For a random sequence of $\pm 1$s in place of the von Mangoldt function $\Lambda(n)$ the answer is a little surprising: the infimum is basically $1/\sqrt{x}$, a result of Konyagin and Schlag. This is available here:

www.math.uchicago.edu/~schlag/papers/POLTRAN.pdf

I say surprising because most people, if they were given 10 seconds to guess the answer, would probably go for $\sqrt{x}$ (I certainly would have).

I'm not sure there's any real reason to suppose that the answer for the deterministic function $\Lambda(n)$ will be much different, except perhaps in logarithmic factors.

I think you have precisely no chance of saying anything useful about the $\alpha_x$, but maybe someone will prove me wrong! I would be surprised if they were not close to equidistributed, though there may be some repulsion effects away from rationals with small denominator (where $S(\alpha)$ will be large).