A very strange statement from Wolfram Alpha....

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Discussion Overview

The discussion revolves around the convergence of the Möbius function and its series representation. Participants explore the properties of the discrete Möbius function, denoted as μ(n), and its non-discrete counterpart, μ(x), particularly in the context of convergence tests and numerical computations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Experimental/applied

Main Points Raised

  • One participant defines the Möbius function μ(n) and discusses its series representation μ(x) = ∑ μ(n) x^n, asserting that it converges for 0 ≤ x < 1.
  • Another participant agrees on the convergence but notes that Mathematica fails to determine it for their specific sum involving μ(k) / 2^k.
  • A different participant presents an explicit expression for the series involving nested sums, indicating that the computational complexity exceeds the capabilities of Wolfram Alpha.
  • One participant shares an experiment using an old Pentium PC to compute values of μ(x) numerically, reporting good convergence in the range -0.99 < x < 0.99, but raises questions about behavior in the ranges -1 < x < -0.99 and 0.99 < x < 1.
  • This participant also mentions the oscillatory nature of the Marten's function M(n) and speculates that μ(x) may cross zero infinitely in the range 0.99 < x < 1.

Areas of Agreement / Disagreement

Participants generally agree on the convergence of the series for 0 ≤ x < 1, but there are differing opinions on the behavior of the series in other ranges and the computational challenges involved. The discussion remains unresolved regarding the specific behavior of μ(x) in those ranges.

Contextual Notes

Participants express uncertainty about the convergence tests and the implications of numerical results, particularly in ranges outside of -0.99 < x < 0.99. There are also limitations regarding the computational tools used and the complexity of the expressions involved.

chisigma
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The so called 'Moebious Function' is a discrete function defined as...

$$\mu(n)= \begin{cases} - 1 &\text{if n has an odd number of prime distinct fators} \\ 0 &\text{if n has one or more prime fractors with exponent greater than one}\\ 1 &\text{if n has an even number of prime distinct factors}\end{cases}\ (1)$$

Actually I'm spending some of my time around the [non discrete] function... $$\mu(x)= \sum_{n=1}^{\infty} \mu (n)\ x^{n}\ (2)$$

In any case is for $0 \le x < 1$ ...$$|\sum_{n=1}^{N} \mu (n)\ x^{n}| \le \sum_{n=1}^{N} x^{n}\ (3)$$

... I'm sure that for $0 \le x < 1$ the series (2) converges. It seems however not to be so obvious for 'Monster Wolfram'...

sum mu(n) x^n from 1 to infinity - Wolfram|Alpha

... according to that the convergence test fails because 'the ratio test in inconclusive'... I would be very happy if somebody clarifies mi ideas... Kind regards $\chi$ $\sigma$
 
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Well, the sum seems convergent to me as well . I ran it on Mathemaica on my PC and it still can not determine convergence .

I tried the following $$\sum_{k\geq 1} \frac{\mu(k)}{2^k } $$ still no response !
 
Today I have found the following 'explicit expression'...

$$\sum_{n=1}^{\infty} \mu(n)\ x^{n} = x - \sum_{a=2}^{\infty} x^{a} + \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b} - \sum_{c=2}^{\infty} \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b\ c} + \sum_{d=2}^{\infty} \sum_{c=2}^{\infty} \sum_{b=2}^{\infty} \sum_{a=2}^{\infty} x^ {a\ b\ c\ d} - ...\ (1)$$

No surprise about the fact that the amount of computation required overflows also the capability of 'Monster Wolfram' (Sweating) ...

Kind regards

$\chi$ $\sigma$
 
Today I decided to try a 'nice 'experiment'... I recovered from the cellar a very 'artifact', one of them first Pentium PC [improved after it was discovered that first examples failed in doing multiplications (Sadface)...] that I used about twenty year ago and conserve as 'souvenir' and used it to numerically compute some values of the function...

$$\mu (x) = \sum_{n=1}^{\infty} \mu(n)\ x^{n}\ (1)$$

The sum was interrupted after 4000 iteration and the range was $-.99 < x < .99$. The result is reported in the diagram...

http://www.123homepage.it/u/i72335019._szw380h285_.jpg.jfif

Clearly in this range we have 'good convergence' but an obvious question is: what does it happen in the range $-1 < x < -.99$ and $.99 < x < 1$?... The answer is not so easy and You can realize that considering that for x=1 we have...

$$\mu(1) = \lim_{n \rightarrow \infty} M (n)\ (2)$$

... where...

$$M(n) = \sum_{k=1}^{n} \mu(k)\ (3)$$

... is the so called 'Marten's function' , the behaviour of which for 'large' values of n is highly 'oscillatory' so that may be that the $\mu(x)$ in the range $.99 < x < 1$ crosses the zero infinite times (Sweating)... in any case I intend to do more 'investigations'...

Kind regards

$\chi$ $\sigma$
 

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