Von Mangoldt Equivalence
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Theorem
For $n \in \N_{>0}$, let $\map \Lambda n$ be the von Mangoldt function.
Then:
- $\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$
is logically equivalent to the Prime Number Theorem.
Proof
Observe:
\(\ds \sum_{n \mathop = 1}^N \map \Lambda n\) | \(=\) | \(\ds \map \Lambda 1 + \map \Lambda 2 + \cdots + \map \Lambda n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 + \map \ln 2 + \map \ln 3 + \map \ln 2 + \map \ln 5 + 0 + \map \ln 7 + \map \ln 2 + \map \ln 3 + 0 + \cdots\) |
Notice this sum will have:
- as many $\map \ln 2$ terms as there are powers of $2$ less than or equal to $N$
- as many $\map \ln 3$ terms as there are powers of $3$ less than or equal to $N$
and in general, if $p$ is a prime less than $N$, $\map \ln p$ will occur in this sum $\floor {\map {\log_p} N}$ times.
But:
- $\map \ln p \floor {\map {\log_p} N} \sim \map \ln p \map {\log_p} N = \map \ln N$
so:
- $\ds \sum_{p \text{ prime} \mathop \le N} \map \ln p \floor {\map {\log_p} N} \sim \sum_{p \text{ prime} \mathop \le N} \map \ln N = \map \pi N \map \ln N$
Therefore:
- $\ds \sum_{n \mathop = 1}^N \map \Lambda n \sim \map \pi N \map \ln N$
and so if:
- $\ds \lim_{N \mathop \to \infty} \frac 1 N \sum_{n \mathop = 1}^N \map \Lambda n = 1$
then:
- $\ds \lim_{N \mathop \to \infty} \frac 1 N \map \pi N \map \ln N = 1$
and vice versa.
But this last equation is precisely the Prime Number Theorem.
Hence our statement regarding the von Mangoldt function is logically equivalent to the Prime Number Theorem.
$\blacksquare$
Source of Name
This entry was named for Hans Carl Friedrich von Mangoldt.