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arXiv 2107.00461math.NT
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关于Minkowski问号函数的导数

On the derivative of the Minkowski question-mark function

Dmitry Gayfulin

摘要:

Minkowski问号函数$?(x)$是定义在$[0,1]$区间上的连续严格增函数。众所周知,这个函数的导数,如果存在的话,只能取两个值:$0$和$+\infty$。还已知的是,在点$x=[0;a_1,a_2,\ldots,a_t,\ldots]$处的导数$?'(x)$的值与算术平均值$(a_1+a_2+\ldots+a_t)/t$的极限行为有关。特别是,N。Moshchevitin和A. Dusthova表明,如果$a_1+a_2+\ldots+a_t<κ_1 t$,其中$κ_1 = 2\log\bigl({\frac{1+\sqrt{5}}{2}}\bigr)/\log{2}= 1.3884\ldots$,那么$?'(x)=+\infty$。他们还证明了常数$κ_1$是不可改进的。我们考虑一个对偶问题:如果$?'(x)=0$,那么$a_1+a_2+\ldots+a_t-κ_1 t$的数量可以有多小?我们得到了这个量的不可改进的估计。

英文摘要:

The Minkowski question-mark function $?(x)$ is a continuous strictly increasing function defined on $[0,1]$ interval. It is well known fact that the derivative of this function, if exists, can take only two values: $0$ and $+\infty$. It is also known that the value of the derivative $?'(x)$ at the point $x=[0;a_1,a_2,\ldots,a_t,\ldots]$ is connected with the limit behavior of the arithmetic mean $(a_1+a_2+\ldots+a_t)/t$. Particularly, N. Moshchevitin and A. Dushistova showed that if $a_1+a_2+\ldots+a_t<κ_1 t$, where $κ_1 = 2\log\bigl({\frac{1+\sqrt{5}}{2}}\bigr)/\log{2}= 1.3884\ldots$, then $?'(x)=+\infty$. They also proved that the constant $κ_1$ is non-improvable. We consider a dual problem: how small can be the quantity $a_1+a_2+\ldots+a_t-κ_1 t$ if $?'(x)=0$? We obtain the non-improvable estimates of this quantity.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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