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# Partial sums

(Redirected from Summatory functions)

Given an arithmetic function ${\displaystyle f:\mathbb {N} \mapsto G}$ where G is any additive group, the partial sums of f, or summatory function of f, is the function

${\displaystyle F:\mathbb {N} \to G;~~n\mapsto \sum _{k=0}^{n}f(k)}$

The analog definition holds, with index k=0 replaced by k=1 (in all that follows) in case the function f is rather defined on positive integers N*.

One could denote the map ${\displaystyle f\mapsto F}$ by the symbol Σ, i.e., ${\displaystyle F:=\Sigma f}$. Then the inverse map ${\displaystyle \Delta :F\mapsto f}$ is that of the first differences (for indices > 0),

${\displaystyle \Delta F(k)=F(k)-F(k-1)=f(k)}$,

where the last equality also holds for k=0 when we agree that ${\displaystyle F(-1)}$ stands here for zero (i.e., ${\displaystyle \Delta F(0)=F(0)=f(0)}$).

(In other cases the convention ${\displaystyle \Delta 'F(k):=F(k+1)-F(k)=f(k+1)}$ may be preferable, but then the relation ${\displaystyle \Delta \circ \Sigma =\Sigma \circ \Delta =\operatorname {id} }$ does not hold any more. Yet another convention would be ${\displaystyle {\hat {F}}(n):=\Sigma 'f(n):=\sum _{k=0}^{n-1}f(k)}$ such that ${\displaystyle {\hat {F}}(n+1)-{\hat {F}}(n)=f(n)}$, i.e., ${\displaystyle \Delta '\circ \Sigma '=\operatorname {id} }$, but not ${\displaystyle \Sigma '\circ \Delta '=\operatorname {id} }$, (as is easily seen from the fact that ${\displaystyle \Sigma 'f(0)=0}$ for any f, which might have f(0) ≠ 0.)

### Examples

The Mertens function is the summatory Möbius function.

The identity map on the positive integers N* is the summatory function of the constant function ${\displaystyle f(n)=1}$ for all n > 0.

The square pyramidal numbers (A000330) are the partial sums of the squares.