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 $g(n) = mf(n) + b \iota(n)$.


$$\sum_{d|n}g(d) = mF(n)+b$$ $$\Rightarrow$$ $$g(n) = \sum_{d|n}\mu(d)(mF(\frac{n}{d})+b)=m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d)$$ 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, $$g(n) = mf(n) + b\iota(n)$$

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

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 $g(n).$

  • $\begingroup$ 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. $\endgroup$ – Mobley Nov 21 '20 at 2:32
  • $\begingroup$ @Mobley Nicely done. We were writing at same time. $\endgroup$ – Phicar Nov 21 '20 at 2:54

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