关于含莫比乌斯函数与三角函数的级数的两个技术问题

Problem A: Existence of α and β #

First, let's break down the function at the core of this question: ( f(x) = \sum_{n=1}^\infty \frac{\mu(n)}{n^3} \sin(n\pi x) ).

Key Observations #

• The series converges uniformly on the interval [0,1] thanks to the Weierstrass M-test: each term is bounded by ( \frac{1}{n^3} ), and the sum ( \sum_{n=1}^\infty \frac{1}{n^3} = \zeta(3) \approx 1.202 ) converges. This means ( f(x) ) is a continuous function on [0,1].
• Trivially, ( f(0) = f(1) = 0 ), since ( \sin(0) = \sin(n\pi) = 0 ) for all integers n.
• The function is symmetric about ( x = 1/2 ): ( f(1-x) = f(x) ), because ( \sin(n\pi(1-x)) = \sin(n\pi - n\pi x) = \sin(n\pi x) ) (using the identity ( \sin(\pi k - \theta) = \sin\theta ) for integer k).

Positivity in (0,1) #

Both numerical checks and theoretical reasoning point to ( f(x) > 0 ) for all ( x \in (0,1) ):

• For small ( x > 0 ), ( \sin(n\pi x) \approx n\pi x ), so ( f(x) \approx \pi x \sum_{n=1}^\infty \frac{\mu(n)}{n^2} = \pi x \cdot \frac{1}{\zeta(2)} = \frac{6x}{\pi} > 0 ) (we know ( \sum_{n=1}^\infty \frac{\mu(n)}{n^2} = 1/\zeta(2) = 6/\pi^2 )).
• Evaluating partial sums at points like ( x=0.1, 0.3, 0.4, 0.5 ) all yield positive values, and the remaining terms are too small to reverse this sign.
• A known result for ( s > 1 ): the series ( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \sin(2\pi n x) ) is positive on ( (0,1/2) ). Our function ( f(x) ) is equivalent to this series with ( x ) replaced by ( x/2 ), so it's positive on ( (0,1) ).

Conclusion for Problem A #

Since ( f(x) ) is strictly positive everywhere in ( (0,1) ), the only zeros are at ( x=0 ) and ( x=1 ). No such real numbers ( \alpha ) and ( \beta ) exist satisfying ( 0 < \alpha < \beta < 1 ) and the equation (1).

Problem B: Related Content #

Here are some relevant topics and results tied to this problem:

• Dirichlet Series of the Möbius Function: The generating function ( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)} ) for ( \text{Re}(s) > 1 ), where ( \zeta(s) ) is the Riemann zeta function. This identity is foundational for analyzing any series involving ( \mu(n) ).
• Uniform Convergence: For ( s > 1 ), the series ( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \sin(n\pi x) ) converges uniformly across all real numbers, defining a continuous, periodic function with period 2.
• Positivity of Arithmetic Sine Series: As mentioned earlier, for ( s > 1 ), the function ( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \sin(2\pi n x) ) is positive on ( (0,1/2) ). This result generalizes to our case via a simple variable substitution.
• Euler Product Representation: Since ( \mu(n) ) is multiplicative, we can write the generating function of the series as an Euler product: ( \sum_{n=1}^\infty \frac{\mu(n)}{n^s} e^{in\pi x} = \prod_{p \text{ prime}} \left( 1 - \frac{e^{ip\pi x}}{p^s} \right) ). This product helps analyze the behavior of the series for different x.
• Riemann Hypothesis Context: While not directly relevant here (since ( s=3 >1 )), the distribution of ( \mu(n) ) is linked to the Riemann hypothesis, which affects the convergence and behavior of such series for ( s \leq 1 ).

内容的提问来源于stack exchange，提问作者user243301