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Inverting $\displaystyle\sum_{d|n} \mu(d) \lambda(d)=2^{\omega(n)}$ into $\displaystyle\sum_{d|n} \lambda(n/d) 2^{\omega(d)}=1$ ,where $n \geq1$, by using Mobius Inversion Formula.

I'm able to solve the latter without Inversion, and in problem too it's not necessary to use inversion, but I'm fascinated to know how to do that, because $2^{\omega(n)}$ comes to L.H.S side from R.H.S and becomes $2^{\omega(d)}$, also $\lambda(d)$ is converted into $\lambda(n/d)$ Please help. and if it's not possible then comment, I'll remove this problem from MSE.

Erick Wong
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mnulb
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1 Answers1

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(You should specify that $\lambda$ refers to the Liouville lambda function and not the Carmichael lambda function, which is also quite common.)

One way to do this is to use the total multiplicativity of $\lambda$ (and the convenient fact that $\lambda = 1/\lambda$) to write $\lambda(n/d)$ as both the product and the quotient of $\lambda(n)$ and $\lambda(d)$.

We rewrite the first sum as $$\lambda(n) \sum_{d\mid n} \mu(d) \lambda(n/d),$$ or in functional notation $\lambda \cdot (\mu \star \lambda)$. Here I'm using $\cdot$ to mean pointwise multiplication and $\star$ to mean Dirichlet convolution.

Since this is equal to $2^{\omega(n)}$ by the first equality, we multiply/divide both sides by $\lambda$ to get $\mu \star \lambda = \lambda \cdot 2^{\omega(n)}$, and Möbius inversion then gives $\lambda = 1 \star (\lambda \cdot 2^{\omega(n)})$. Writing this back in summation form gives $$\sum_{d\mid n} \lambda(d) 2^{\omega(d)} = \lambda(n),$$ and dividing/multiplying through by $\lambda(n)$ yields the desired equality.

Erick Wong
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  • It's a little more natural to view this in terms of the Abelian group of non-zero multiplicative functions under convolution. Letting $f$ denote the $2^\omega(n)$ function, the given equality is $(\mu\cdot\lambda)\star 1 = f$ and the desired equality is $\lambda \star f = 1$, so it amounts to showing that $\lambda$ is the convolution inverse of $\mu\cdot\lambda = \mu^2$. – Erick Wong Dec 30 '16 at 00:58
  • I can't seem to figure out the logic behind the Möbius inversion showing $$\sum_{d\mid n} \mu(d) \lambda(\frac{n}{d}) = \lambda(n) 2^{\omega(n)} \Rightarrow \lambda(n) = \sum_{d\mid n} \lambda(d) 2^{\omega(d)}$$. Could anyone elaborate on how you can do this? – GhostyOcean Nov 22 '21 at 21:38
  • @GhostyOcean The Möbius inversion formula says that in general if $F(n) = \sum_{d\mid n} \mu(d) G(n/d)$, then $G(n) = \sum_{d\mid n} F(d)$. This is directly applicable to the formula in your comment. – Erick Wong Nov 23 '21 at 02:58
  • My textbook says the converse is true for the relationship (that is, if $G = \sum_{d\mid n} F(d)$, then $F(n)=\sum_{d\mid n} \mu (d) G(\frac{n}{d}$). Is it the case that the relationship is an if and only if? – GhostyOcean Nov 23 '21 at 04:29
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    @GhostyOcean Yes it is, nearly by definition but not quite. Start from $F(n) = \sum_{d\mid n} \mu(d) G(n/d)$, then define $G'(n) := \sum_{d\mid n} F(d)$. Suppose $G' \ne G$, with $m$ being the smallest place where $G(m) \ne G'(m)$. By Möbius we have $F(m) = \sum_{d\mid m} \mu(d) G'(m/d)$ but we also have by assumption $F(m) = \sum_{d\mid m} \mu(d) G(m/d)$. But these two sums differ by exactly $G(m) - G'(m)$, a contradiction. – Erick Wong Nov 23 '21 at 06:38
  • @GhostyOcean You can also see this very easily, but more abstractly, if you already know that Dirichlet convolution is associative (which isn't too hard to prove). Then just convolve both sides by the all-ones function $\mathbb 1$, and cancel out the $\mathbb 1 * \mu$. – Erick Wong Nov 23 '21 at 06:55
  • Thank you for the detailed responses! This helps a lot, and it shows that I still have more work to do! – GhostyOcean Nov 23 '21 at 19:16