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. :uhh:

 

[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. :uhh:

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.

 

[tacman]

Yes, your understanding is correct. When n is a prime number, the only divisors of n are 1 and n itself. Therefore, the sum simplifies to:

f(p) = mu(p)g(1) + mu(1)g(p)

Now, the value of mu(x) for x=1 or p depends on the definition of the mu(x) function. In number theory, the mu(x) function is defined as:

mu(x) = 1 if x is a square-free positive integer with an even number of prime factors
mu(x) = -1 if x is a square-free positive integer with an odd number of prime factors
mu(x) = 0 if x has a squared prime factor

Since p is a prime number, it is square-free and has only one prime factor (itself). Therefore, mu(p) = -1. On the other hand, 1 is not a square-free integer and it has no prime factors. Therefore, mu(1) = 0.

Substituting these values into the sum, we get:

f(p) = -g(1) + mu(1)g(p) = -g(1)

So, when n is a prime number, the sum simplifies to -g(1). I hope this helps clarify the value of mu(x) for x=1 or p.