All Random Solved Random Open
3 solved out of 8 shown (show only solved or open)
Let $f$ be a Rademacher multiplicative function: a random $\{-1,0,1\}$-valued multiplicative function, where for each prime $p$ we independently choose $f(p)\in \{-1,1\}$ uniformly at random, and for square-free integers $n$ we extend $f(p_1\cdots p_r)=f(p_1)\cdots f(p_r)$ (and $f(n)=0$ if $n$ is not squarefree). Does there exist some constant $c>0$ such that, almost surely, \[\limsup_{N\to \infty}\frac{\sum_{m\leq N}f(m)}{\sqrt{N\log\log N}}=c?\]
Note that if we drop the multiplicative assumption, and simply assign $f(m)=\pm 1$ at random, then this statement is true (with $c=\sqrt{2}$), the law of the iterated logarithm.

Wintner [Wi44] proved that, almost surely, \[\sum_{m\leq N}f(m)\ll N^{1/2+o(1)},\] and Erdős improved the right-hand side to $N^{1/2}(\log N)^{O(1)}$. Lau, Tenenbaum, and Wu [LTW13] have shown that, almost surely, \[\sum_{m\leq N}f(m)\ll N^{1/2}(\log\log N)^{2+o(1)}.\] Harper [Ha13] has shown that the sum is almost surely not $O(N^{1/2}/(\log\log N)^{5/2+o(1)})$, and conjectured that in fact Erdős' conjecture is false, and almost surely \[\sum_{m\leq N}f(m) \ll N^{1/2}(\log\log N)^{1/4+o(1)}.\] This was proved by Caich [Ca23].

Additional thanks to: Mehtaab Sawhney
Is it true that all except $o(2^n)$ many polynomials of degree $n$ with $\pm 1$-valued coefficients have $(\frac{1}{2}+o(1))n$ many roots in $\{ z\in \mathbb{C} : \lvert z\rvert \leq 1\}$?

For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). If $S_n(t)$ is the number of roots of $\sum_{1\leq k\leq n}\epsilon_k(t)z^k$ in $\lvert z\rvert \leq1$ then is it true that, for almost all $t\in (0,1)$, \[\lim_{n\to \infty}\frac{S_n(t)}{n}=\frac{1}{2}?\]

Erdős and Offord [EO56] showed that the number of real roots of a random degree $n$ polynomial with $\pm 1$ coefficients is $(\frac{2}{\pi}+o(1))\log n$.

See also [521].

Additional thanks to: Zachary Chase
For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). Does there exist some constant $C>0$ such that, for almost all $t\in (0,1)$, \[\max_{\lvert z\rvert=1}\left\lvert \sum_{k\leq n}\epsilon_k(t)z^k\right\rvert=(C+o(1))\sqrt{n\log n}?\]
Salem and Zygmund [SZ54] proved that $\sqrt{n\log n}$ is the right order of magnitude, but not an asymptotic.
For any $t\in (0,1)$ let $t=\sum_{k=1}^\infty \epsilon_k(t)2^{-k}$ (where $\epsilon_k(t)\in \{0,1\}$). What is the correct order of magnitude (for almost all $t\in(0,1)$) for \[M_n(t)=\max_{x\in [0,1]}\left\lvert \sum_{k\leq n}\epsilon_k(t)x^k\right\rvert?\]
A problem of Salem and Zygmund [SZ54]. Chung showed that, for almost all $t$, there exist infinitely many $n$ such that \[M_n(t) \ll \left(\frac{n}{\log\log n}\right)^{1/2}.\] Erdős (unpublished) showed that for almost all $t$ and every $\epsilon>0$ we have $\lim_{n\to \infty}M_n(t)/n^{1/2-\epsilon}=\infty$.
Is it true that all except at most $o(2^n)$ many degree $n$ polynomials with $\pm 1$-valued coefficients $f(z)$ have $\lvert f(z)\rvert <1$ for some $\lvert z\rvert=1$? What is the behaviour of \[m(f)=\min_{\lvert z\rvert=1}\lvert f(z)\rvert?\]
Random polynomials with independently identically distributed coefficients are sometimes called Kac polynomials - this problem considers the case of Radamacher coefficients, i.e. independent uniform $\pm 1$ values. The first problem asks whether $m(f)<1$ almost surely. Littlewood [Li66] conjectured that the stronger $m(f)=o(1)$ holds almost surely.

The answer to both questions is yes: Littlewood's conjecture was solved by Kashin [Ka87], and Konyagin [Ko94] improved this to show that $m(f)\leq n^{-1/2+o(1)}$ almost surely. This is essentially best possible, since Konyagin and Schlag [KoSc99] proved that for any $\epsilon>0$ \[\limsup_{n\to \infty} \mathbb{P}(m(f) \leq \epsilon n^{-1/2})\ll \epsilon.\] Cook and Nguyen [CoNg21] have identified the limiting distribution, proving that for any $\epsilon>0$ \[\lim_{n\to \infty} \mathbb{P}(m(f) > \epsilon n^{-1/2}) = e^{-\epsilon \lambda}\] where $\lambda$ is an explicit constant.

Additional thanks to: Mehtaab Sawhney
Let $a_n\geq 0$ with $a_n\to 0$ and $\sum a_n=\infty$. Find a necessary and sufficient condition on the $a_n$ such that, if we choose (independently and uniformly) random arcs on the unit circle of length $a_n$, then all the circle is covered with probability $1$.
A problem of Dvoretzky [Dv56]. It is easy to see that (under the given conditions alone) almost all the circle is covered with probability $1$.

Kahane [Ka59] showed that $a_n=\frac{1+c}{n}$ with $c>0$ has this property, which Erd\H{s} (unpublished) improved to $a_n=\frac{1}{n}$. Erd\{o}s also showed that $a_n=\frac{1-c}{n}$ with $c>0$ does not have this property.

Solved by Shepp [Sh72], who showed that a necessary and sufficient condition is that \[\sum_n \frac{e^{a_1+\cdots+a_n}}{n^2}=\infty.\]

Let $a_n\in \mathbb{R}$ be such that $\sum_n \lvert a_n\rvert^2=\infty$ and $\lvert a_n\rvert=o(1/\sqrt{n})$. Is it true that, for almost all $\epsilon_n=\pm 1$, there exists some $z$ with $\lvert z\rvert=1$ (depending on the choice of signs) such that \[\sum_n \epsilon_n a_n z^n\] converges?
It is unclear to me whether Erdős also intended to assume that $\lvert a_{n+1}\rvert\leq \lvert a_n\rvert$.

It is 'well known' that, for almost all $\epsilon_n=\pm 1$, the series diverges for almost all $\lvert z\rvert=1$ (assuming only $\sum \lvert a_n\rvert^2=\infty$).

Dvoretzky and Erdős [DE59] showed that if $\lvert a_n\rvert >c/\sqrt{n}$ then, for almost all $\epsilon_n=\pm 1$, the series diverges for all $\lvert z\rvert=1$.

Let $d_k(n)$ be the expected distance from the origin after taking $n$ random steps from the origin in $\mathbb{Z}^k$ (conditional on no self intersections). Is it true that \[\lim_{n\to \infty}\frac{d_2(n)}{n^{1/2}}= \infty?\] Is it true that \[d_k(n)\ll n^{1/2}\] for $k\geq 3$?