table of Contents
The origin of Mobius function
For equation \ (\ displaystyle \ boldsymbol F ( n) = \ sum_ {d \ mid n} \ boldsymbol f (d) \)
If we are known \ (\ boldsymbol f \) can be recursive, quickly solved \ (\ boldsymbol F \)
But through this formula, if known \ (\ boldsymbol F \) How Anti solved \ (\ boldsymbol f \) it?
很显然 \(\displaystyle \boldsymbol F(n)=\sum_{d\mid n}\boldsymbol f(d)=\sum_{d\mid n}\boldsymbol f(d)\boldsymbol I({n\over d})=(\boldsymbol f*\boldsymbol I)(n)\)
The final step uses a definition of the convolution of Dealey Clay
Therefore, it is clear that the \ (\ boldsymbol F = \ boldsymbol f * \ boldsymbol I \)
Of course, we may assume that \ (\ boldsymbol I \) is inverse to \ (\ boldsymbol \ mu \)
则两边卷上\ (\ boldsymbol \ it \)得\ (\ boldsymbol F * \ boldsymbol \ mu = \ boldsymbol f * \ boldsymbol I * \ boldsymbol \ mu = \ boldsymbol f * \ boldsymbol \ epsilon = \ boldsymbol f \ )
In this way, we can get \ (\ boldsymbol f \) , which is the origin of Möbius function
Mobius function value
Obviously, according to the definition Mobius function: \ (\ boldsymbol \ * MU \ boldsymbol the I = \ boldsymbol \ varepsilon \)
First, \ (\ boldsymbol \ MU (. 1) =. 1, \ boldsymbol \ MU \) is a multiplicative function
Secondly, if \ (n \ neq 1 \) there
\(\displaystyle 0=\boldsymbol \varepsilon(n)=\sum_{d\mid n}\boldsymbol \mu(d)\boldsymbol I({n\over d})=\sum_{d\mid n}\boldsymbol \mu(d)\)
Since the Möbius function is multiplicative function, we need only consider the \ (\ boldsymbol \ mu (p ^ k) \) to the:
当 \(k=1\) 时 \(\displaystyle 0=\sum_{d\mid p}\boldsymbol \mu(d)=\boldsymbol \mu(p)+\boldsymbol \mu(1)\)
Tokuita \ (\ boldsymbol \ mu (p ) = - \ boldsymbol \ mu (1) = - 1 \)
当\ (a \ neq 1 \)时\ (\ displaystyle 0 = \ frac {d \ mid p ^ k} \ boldsymbol \ He (d) = \ frac {c = 0} ^ a \ boldsymbol \ it (p ^ c) = \ frac {c = 2} ^ a \ boldsymbol \ it (p ^ c) + \ boldsymbol \ he (p) + \ boldsymbol \ mu (1) = \ frac {c = 2} ^ a \ boldsymbol \ he (p ^ c) \)
Since this type of arbitrary \ (k> 1 \) have been set up, so we can quickly get \ (\ boldsymbol \ mu (p ^ c) = 0, c> 1 \)
Thus, we can simply summarized as \ (\ boldsymbol \ mu (p ^ k) = - [k = 1] \)
Or we write: \ (\ boldsymbol \ MU (P ^ K) = [P ^ 2 \ P ^ the nmid K] \ CDOT (-1) ^ {[P \ K ^ P MID]} \)
Therefore, the \ (n-\) we have \ (\ displaystyle \ boldsymbol \ mu (n) = \ prod_ {i = 1} ^ m [p_i ^ 2 \ nmid n] \ cdot (-1) ^ {[p_i \ mid n]} \)