Moebius Transform Sum: Understanding the mu(x) Function for Prime Numbers

[lokofer]

let be the sum (over all the divisors d of n):

$$f(n)= \sum_{d|n} \mu (n/d)g(d)$$ my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

$$f(p)= \mu (p)g(1) + \mu (1) g(p)$$ is that correct?...now the question is to know what's the value of mu(x) function for x=1 or p.

 

[shmoe]

let be the sum (over all the divisors d of n):

$$f(n)= \sum_{d|n} \mu (n/d)g(d)$$ my question is if n=prime then you have only 2 numbers 1 and p that are divisors so you get:

$$f(p)= \mu (p)g(1) + \mu (1) g(p)$$ is that correct?

Correct.

...now the question is to know what's the value of mu(x) function for x=1 or p.

Step #1 when trying to learn about mobius inversion and such:

Look at the definition of the mobius function.

Complete this step and $$\mu(1)$$ and $$\mu(p)$$ will be apparant.