# Possibly Möbius Inversion Formula Application

EDIT: I believe I've figured it out! Feel free to take a look in case I've made a mistake.

Problem: Let $$n$$ and $$d$$ be positive integers and $$m, b \in R$$, some ring. If $$F(n) = \sum_{d|n}f(d)$$ and $$\sum_{d|n}g(d) = mF(n)+b$$, then $$g(n) = mf(n) + b \iota(n)$$.

Proof:

$$\sum_{d|n}g(d) = mF(n)+b$$ $$\Rightarrow$$ $$g(n) = \sum_{d|n}\mu(d)(mF(\frac{n}{d})+b)=m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d)$$ But our other assumption is, $$F(n) = \sum_{d|n}f(d) \Rightarrow f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})$$. Then, $$m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d) = mf(n) + b\sum_{d|n}\mu(d)$$ $$\Rightarrow$$ $$g(n) = mf(n) + b\sum_{d|n}\mu(d)$$

But $$\iota (n) = 1$$ if $$n=1$$ and $$\iota (n) = 0$$ if $$n>1$$, so $$\iota(n)$$ and $$\sum_{d|n}\mu(d)$$ have the same property. So we conclude, $$g(n) = mf(n) + b\iota(n)$$

• what is $\iota (n)$? – Phicar Nov 20 '20 at 21:03
• It's 1 if n=1 and 0 otherwise – Mobley Nov 20 '20 at 21:04
• You seem to be in the right track. What happens if $\iota = \sum _{d|n} \mu (d)$? Can you show it? – Phicar Nov 20 '20 at 21:06
• I have answer explaining that and making the hint clear. – Phicar Nov 21 '20 at 2:25

## 1 Answer

From the comments I will try to put you in the track:

You correctly have tried to use the Mobius inversion formula getting that $$f(n)=\sum _{d|n}\mu (n/d)F(d),$$ so, if you try to compute $$\sum _{d|n}\mu (n/d)(mF(d)+b),$$ and you show that $$\iota (n)=\sum _{d|n}\mu (d),$$ Hint: Show it for a power of prime and use multiplicativity, then $$\sum _{d|n}\mu (n/d)(mF(d)+b)=m\sum _{d|n}\mu (n/d)F(d)+\sum _{d|n}\mu (n/d)b=mf(n)+b\iota(n),$$ concluding, again by Mobius, that this is $$g(n).$$

• Your comment was very helpful.. this is pretty much what I did and edited the post. I think I edited my post and you answered almost at the same time. – Mobley Nov 21 '20 at 2:32
• @Mobley Nicely done. We were writing at same time. – Phicar Nov 21 '20 at 2:54