Proof: (related to "Möbius Inversion Formula") 

• By assumptions, $\alpha,\beta$ are arithmetic functions with $\beta(n)=\sum_{d\mid n}\alpha(d).$
• For any divisor $d\mid n$ we have, therefore, $$\beta\left(\frac nd\right)=\sum_{b\mid \frac nd}\alpha(b).$$
• Multiplying both sides of the equation by the Möbius function of $\mu(d)$ results in
  $$\mu(d)\beta\left(\frac nd\right)=\sum_{b\mid \frac nd}\mu(d)\alpha(b).$$
• The sum over all divisors $d\mid n$ results in
  $$\sum_{d\mid n}\mu(d)\beta\left(\frac nd\right)=\sum_{d\mid n}\sum_{b\mid \frac nd}\mu(d)\alpha(b)=\sum_{b\mid n}\sum_{d\mid \frac nb}\mu(d)\alpha(b).$$
• In the last step, we could exchange the indices $d$ and $b$ because as the index $b$ runs through all divisors $b\mid \frac nd$ so does $d$ run through all complementary divisors $d\mid \frac nb.$
• Now, the term $\alpha(b)$ does not depend on the index $d$ of the inner sum and we can use the distributivity law to get
  $$\sum_{b\mid n}\sum_{d\mid \frac nb}\mu(d)\alpha(b)=\sum_{b\mid n}\alpha(b)\sum_{d\mid \frac nb}\mu(d).$$
• The last sum can be significantly simplified using the sum of Möbius function over divisors and we get
  $$\sum_{b\mid n}\alpha(b)\sum_{d\mid \frac nb}\mu(d)=\sum_{b\mid n}\alpha(b)[b=n]=\alpha(n).$$
• We have shown
  $$\alpha(n)=\sum_{d\mid n}\mu(d)\beta\left(\frac nd\right).$$

| | | | created: 2019-04-06 22:57:06 | modified: 2019-04-06 23:03:00 | by: bookofproofs | references: [701], [1272]