the Mobius function

Franklin T. Adams-Watters franktaw at
Thu Dec 16 03:24:24 CET 2004

In general, multiplicative functions are "really" only defined on the positive integers.  If you want to extend them to all integers while retaining the multiplicative property, first note that gcd(-1, -1) = 1, so you must have f(-1)^2 = f(1) = 1, so f(-1) = 1 or -1.  Then for general n, f(-n) = f(-1) * f(n); thus the extension is either symmetric or anti-symmetric.  In the case of an anti-symmetric extension, the only consistent value for f(0) is 0; for a symmetric extension, any value for f(0) will work.  Since the latter case provides no information about what value to use, I would generally say that f(0) = 0 is the correct choice.  (Of course, there are some exceptions; the constant function f(x) = 1 is multiplicative, and any value other than f(0) = 1 would be a very strange choice in this case.)

In the case of the Mobius function, there is an additional argument in favor of mu(0) = 0: mu(n) = 0 whenever m^2 | n for some m > 1, and of course every such m^2 | 0.

To summarize: I would have no problem with a program rejecting 0 as out of range for mu; but if it returns a value, it should be 0.

Emeric Deutsch <deutsch at> wrote:

>As far as I know, the Mobius function (mu) is defined only for 
>positive integers.
>Is there a natural or customary assignment for n=0 ?
>Maple yields mu(0)=-1. 

Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067

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