Questions tagged [mobius-function]
Questions on the Möbius function μ(n), an arithmetic function used in number theory.
413
questions
1
vote
1
answer
126
views
Using the binomial formula in the form $(k - k)^n$
While proving a certain property of the number theoretic mobius function, namely that it is invertible in the monoid of multiplicative functions and its inverse is ...
- 583
0
votes
1
answer
35
views
lower bound of $\sum_{n=1}^x \frac{\mu(n)}{n}$
Denote by $\mu$ the Mobius function. Poussin showed that
$$
\sum_{n=1}^x \frac{\mu(n)}{n} = O(1/\log x),
$$
and there are further improvements since. I wonder what is known about lower bound of ...
- 175
2
votes
1
answer
46
views
Why $ \prod\limits_{n=1}^{\infty} \biggl (\phi(q^n)^{\mu(n)} \biggr)= 1-q $?
Playing with Euler $\phi $ function (not to be confused with the totient function, here another reference), I found this curious identity (I calculated it for various $q$ with Mathematica and it holds)...
- 913
1
vote
0
answers
40
views
Comparing two series expressions for $1/\zeta(s)$. What can be said about their complex roots?
The following two expressions involving the inverted Riemann $\zeta(s)$ functions are well known:
\begin{align}
\frac{1}{\zeta(s)} &= \sum_{n=1}^\infty \frac{\mu(n)}{n^s} \\
-\frac{\zeta'(s)}{\...
- 3,191
2
votes
1
answer
53
views
Möbius function of distributive lattice only takes values $\pm 1$ and $0$.
In this Wikipedia article, I found the statement
[...] shares some properties with distributive lattices: for example, its Möbius function takes on only values 0, 1, −1.
My question is: How it can ...
- 23.1k
0
votes
1
answer
32
views
Prove that $\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = 1$
In Wikipedia entry for Mertens function it says that
From [Lehman, R.S. (1960). "On Liouville's Function". Math. Comput. 14: 311–320.] we have that $$\sum_{d=1}^{n} M(\lfloor n/d \rfloor) = ...
- 1,190
2
votes
1
answer
47
views
Which is the error in this application of Möbius inversion formula?
In Wikipedia the following generalisation of the Möbius inversion formula is given (and proved):
Suppose $F(x)$ and $G(x)$ are complex-valued functions defined on the interval $[1, ∞)$ such that
$$G(...
- 1,190
0
votes
0
answers
113
views
Riemann Hypothesis follows from the statement $M\left(x\right)=o_x(x^{\frac{1}{2}+\varepsilon})$
Recall that the Mertens function is defined via:
$$M(n):=\sum_{n\ge x\ge 1} \mu(x)$$
Where $\mu$ is the Möbius function.
Littlewood proved that if $M\left(x\right)=o_x(x^{\frac{1}{2}+\varepsilon})$
...
- 119
2
votes
0
answers
57
views
What did I get wrong in this Mobius function question? [closed]
$f(n):=\sum\limits_{d\mid n}\mu(d)\cdot d^2,$ where $\mu(n)$ is the Möbius function. Compute $f(192).$
First, I found all of the divisors of 192 by trial division by primes in ascending order:
$D=\{...
- 637
0
votes
0
answers
56
views
Sequence notation?
Cor. Mobius Inversion Formula for multiplicative functions.
Let $f,F$ be multiplicative functions such that $F(n)=\sum\limits_{d\mid n}f(d)$. Then $f(n)=\sum\limits_{d\mid n}\mu(\frac nd)F(d)$.
Proof.
...
- 637
2
votes
1
answer
62
views
Why is Möbius function's co-domain $\mathbb{C}$?
I am new to the concept of partially ordered sets.
Here's my professor's definition of a Möbius function from her lecture notes:
The inverse of zeta function with relative to the convolution product ...
- 309
0
votes
0
answers
54
views
Generalization of Möbius inversion formula
A generalization of Möbius inversion formula guarantees that if we have $G(n)=\sum_{k=1}^{n} F\left(\frac{n}{k}\right)$, then $F(n)=\sum_{k=1}^{n} \mu(k) G\left(\frac{n}{k}\right)$.
If we have $\sqrt{...
- 1,190
2
votes
0
answers
78
views
Meaning of $M(n)=O\left(x^{\frac{1}{2}+\epsilon}\right)$
I am trying to fully understand the implications of $M(n)=O\left(n^{\frac{1}{2}+\epsilon}\right)$, where $M(n)$ is Mertens function, being equivalent to Riemann Hypothesis.
(i) Is the equivalence ...
- 1,190
2
votes
0
answers
67
views
Möbius inversion formula and $\sum_{k\leq n} \frac{\mu(k)}{k}$
I have tried to apply what is stated at the Generalizations of Möbius inversion formula section of Wikipedia to bound $$\sum_{k\leq n} \frac{\mu(k)}{k}$$ The application seems simple and ...
- 1,190
4
votes
0
answers
181
views
Bounding a partial sum with Möbius inversion formula
I am trying to bound the partial sum $$S(n)=\sum_{k=1}^{\sqrt{n}} \mu(k) \pi\left(\frac{n}{k}\right)$$
Where $\pi(x)$ is the prime counting function, and $\mu(x)$ is the Möbius function.
Empirical ...
- 1,190