Theorem: Vaughan's Identity

Let $\Lambda$ be von Mangoldt's function.
Let $\mu$ be the Möbius function.
Then for $y, z \ge 1$ and $n > z$:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y}} \map \mu d \map \ln {\frac n d} - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z}} \map \mu d \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z} } \map \mu d \map \Lambda c$
where $\divides$ denotes divisibility.

Proof:
By Sum Over Divisors of von Mangoldt is Logarithm:
:$\ds \ln n = \sum_{d \mathop \divides n} \map \Lambda d$
Hence:
{{begin-eqn}}
{{eqn | l = \map \Lambda n
      | r = \sum_{d \mathop \divides n} \map \mu d \map \ln {\frac n d}
      | c = Möbius Inversion Formula
}}
{{eqn | r = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \map \ln {\frac n d} + \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \map \ln {\frac n d}
      | c = 
}}
{{end-eqn}}
Taking the second summation in that last line:
{{begin-eqn}}
{{eqn | l = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \map \ln {\frac n d}
      | r = \sum_{\substack {d \mathop \divides n \\ d \mathop > y} } \map \mu d \sum_{c \mathop \divides n / d} \map \Lambda c
      | c = 
}}
{{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y} } \map \mu d \map \Lambda c
      | c = 
}}
{{eqn | r = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z} } \map \mu d \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \, \map \Lambda c
      | c = 
}}
{{end-eqn}}
Again, taking the second summation in that last line:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \map \Lambda c = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z} } \map \mu d \map \Lambda c$
Putting this together:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \map \ln {\frac n d} + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, c \mathop > z} } \map \mu d \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, c \mathop \le z} } \map \mu d \map \Lambda c$
It remains to be shown that:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = 0$
The summation is expressed as:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = \sum_{\substack {c \mathop \le z \\ c \mathop \divides n} } \map \Lambda c \sum_{d \mathop \divides \frac n c} \map \mu d$
Now we have $c \le z < n$, so:
:$\dfrac n c > 1$
Therefore, by Sum of Möbius Function over Divisors:
:for each $c$ the inner sum vanishes.
This shows that:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = 0$
as required.
{{qed}}
{{Namedfor|Robert Charles Vaughan|cat = Vaughan}}
Category:Analytic Number Theory
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