the Mobius function

Franklin T. Adams-Watters franktaw at netscape.net
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 duke.poly.edu> 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.

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