Largest squarefree divisor

It is known that $\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i$ for $x_i \in \mathbb{R}^+$ . So, to extract the squarefree divisors of $n$ only, we make use of the Möbius function and take the following sum:

$\sum_{d|n} \mu^2(d)d^k = n^k \sum_{d|n} \mu^2(d)\left( \frac{n}{d} \right)^{-k} = n^k (\mu^2 * \text{Id}_{-k})(n)$

Taking the $k$ th root and letting $k \to \infty$ :

$\max_{d|n} \mu^2(d)d = \lim_{k \to \infty} \left( n^k (\mu^2 * \text{Id}_{-k})(n) \right)^{1/k} = n \lim_{k \to \infty} (\mu^2 * \text{Id}_{-k})(n)^{1/k}$

Note:

$*$ is the Dirichlet convolution, a binary operator defined on functions $\mathbb{N} \longrightarrow \mathbb{C}$ such that $\displaystyle (f*g)(n) = \sum_{d|n} f(d)g(\frac{n}{d})$ .

$\text{Id}_{k}(n) = n^k$ for $n \in \mathbb{N}$ , $k \in \mathbb{C}$ .

#NumberTheory

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

Mind if you explain the first step? The

$\displaystyle \lim_{k \to \infty} \left( \sum_i x_i^k \right)^{1/k} = \max_i x_i$

Julian Poon - 5 years, 8 months ago

It just means $\displaystyle \lim_{k \to \infty} \left( \sum_{i=1}^n x_i^k \right)^{1/k} = \max_{i=1,2,\ldots ,n} x_i = \max \{ x_1,x_2, \ldots ,x_n \}$ .

Jake Lai - 5 years, 8 months ago

No, as to why it is so. Also mind if you also define what $\text{Id}$ means? Thanks

Julian Poon - 5 years, 8 months ago

@Julian Poon – This is a clean, nice proof. Also I've added clarifications to the note.

Jake Lai - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$