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I use the Möbius inversion with polynomials as e.g. in the well-known inversion formula of the cyclotomic polynomials.
So I have $$p_{2n}(x)=\prod_{d|n}(2q_d(x))^{\mu(\frac{n}{d})}$$ Now I get the correct ( CAS tested ) formula
$$2q_n(x)=\prod_{d|n}p_{2d}(x)$$ But : if I separate the factor 2 into $$\prod_{d|n}2^{\mu(\frac{n}{d})}$$ I calculate $$2^{\sum_{d|n}\mu(\frac{n}{d})}$$ and this is $1$ for $n>1$ and so dissappears. So would result the wrong formula $$q_n(x)=\prod_{d|n}p_{2d}(x)$$ Where is the error in my reasoning ?

Wolfgang Tintemann
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1 Answers1

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You can't use Mobius inversion here because you need $$p_{2n}(x)=\prod_{d\mid n} q_d(x)^{\mu(n/d)}$$ for all $n$, but it is not true for $n=1$.

If you defined $r_n(x)=p_{2n}(x)$ if $n>1$ and $r_1(x)=p_{2}(x)/2=q_1(x)$ if $n=1$, then you'd have:

$$r_n(x)=\prod_{d\mid n} q_d(x)^{\mu(n/d)}$$ for all $n$. Then you'd get that:

$$q_d(x)=\prod_{d\mid n} r_n(x) = \frac 1 2 \prod_{d\mid n} p_{2d}(x)$$

The mistake is thinking that the statement of MI (additive form) is:

$$\left(g(n)=\sum_{d\mid n} f(d)\right)\iff\left(f(n)=\sum_{d\mid n} \mu\left(\frac n d\right)f(d)\right)$$

The correct statement is:

$$\left(\forall n:g(n)=\sum_{d\mid n} f(d)\right)\iff\left(\forall n:f(n)=\sum_{d\mid n} \mu\left(\frac n d\right)f(d)\right)$$

One of my favorite fake proofs makes this error.

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Thomas Andrews
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  • Thanks for that correct derivation of the result. But i am still confused why I cant take the product of the 2 factor out. Please correct the typo on the right side of your last two formulas : g instead of f as factor at the $\mu$-function. – Wolfgang Tintemann Jun 26 '14 at 18:47
  • You can take that product out, but that product is $2$ if $n=1$ and $1$ otherwise. So you don't get the result you think you do... – Thomas Andrews Jun 26 '14 at 19:34
  • You get $p_{2n}(x)=f(n)\cdot \prod q_d(x)^{\mu(n/d)}$, where $f(n) =2$ if $n=1$ and $f(n)=1$ otherwise. So you can apply Mobius inverson to $p_{2n}(x)/f(n)=r_n(x)$, but not to $p_{2n}(x)$. – Thomas Andrews Jun 26 '14 at 19:45
  • Thats clear now. Thank you. – Wolfgang Tintemann Jun 27 '14 at 13:47