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Number Theory A very strange statement from Wolfram Alpha...

chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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 \(\displaystyle \sum_{k\geq 1} \frac{\mu(k)}{2^k } \) still no response !
 

chisigma

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
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...



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

$\chi$ $\sigma$