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Or adding the additive function additive function
Set \ (f, g \) is the additive function, \ (H = F + G \)
For the \ (\ forall n, m \ in Z _ +, gcd (n, m) = 1 \)
\(h(nm)=(f+g)(nm)=f(nm)+g(nm)=f(n)+f(m)+g(n)+g(m)=(\ f(n)+g(n)\ )+(\ f(m)+g(m)\ )=h(n)+h(m)\)
And readily available card, if and only if both of the fully additive, additive must also be fully additive
Multiplying the multiplicative function or multiplicative function
Set \ (\ boldsymbol f, \ boldsymbol g \) is a multiplicative function, \ (H = \ boldsymbol F \ CDOT \ boldsymbol G \)
For the \ (\ forall n, m \ in Z _ +, gcd (n, m) = 1 \)
\(h(nm)=(\boldsymbol f\cdot \boldsymbol g)(nm)=\boldsymbol f(nm)\cdot \boldsymbol g(nm)=\boldsymbol f(n)\cdot \boldsymbol f(m)\cdot \boldsymbol g(n)\cdot \boldsymbol g(m)=(\ \boldsymbol f(n)\cdot \boldsymbol g(n)\ )\cdot (\ \boldsymbol f(m)\cdot \boldsymbol g(m)\ )=h(n)\cdot h(m)\)
And readily available card, if and only if both are fully integrability, the multiplication must have completely integrability
Dealey Craig convolution product of a function or multiplicative function
Set \ (\ boldsymbol f, \ boldsymbol g \) is a multiplicative function, and \ (\ boldsymbol h = \ boldsymbol f * \ boldsymbol g \)
For the \ (\ forall n, m \ in Z _ +, gcd (n, m) = 1 \)
\(\displaystyle \boldsymbol h(nm)=(\boldsymbol f*\boldsymbol g)(nm)=\sum_{d\mid nm}\boldsymbol f(d)\boldsymbol g({nm\over d})\)
Considering the \ (n, m \) coprime, so there is no overlap prime factor
故 \(\displaystyle \boldsymbol h(nm)=\sum_{d\mid nm}\boldsymbol f(d)\boldsymbol g({nm\over d})=\sum_{d_1\mid n\bigwedge d_2\mid m}\boldsymbol f(d_1)\boldsymbol f(d_2)\boldsymbol g({n\over d_1})\boldsymbol g({m\over d_2})=(\sum_{d_1\mid n}\boldsymbol f(d_1)\boldsymbol g({n\over d_1})\ )\cdot (\sum_{d_2\mid n}\boldsymbol f(d_2)\boldsymbol g({m\over d_2})\ )=\boldsymbol h(n)\cdot \boldsymbol h(m)\)
Because of \ (d_1, d_2 \) are \ (n, m \) factor, both when it is not \ (1 \) when the contribution to Deeley Cray convolution are independent, the use of the multiplication principle can be distinguished; when both exist \ (1 \) when the function is \ (1 \) , the contribution of Dealey Cray convolution or independent