After a few dishes on the dog courage, learning the Mobius inversion - find themselves even dog food not. . .
Some define the basic function of number theory
\(id(n)=n\)
\(1(n)=1\)
\(\epsilon(n)=\left\{\begin{array}{rcl}1&n=1\\0&n\neq 1\end{array}\right.\)
These arithmetical functions are completely multiplicative function.
(Multiplicative function satisfies \ (F (A \ Times B) = F (A) \ Times F (B) \) , \ (GCD (A, B) =. 1 \) , and completely multiplicative function satisfies \ (F (a \ B Times) = F (a) \ F Times (B) \) , without satisfying (gcd = 1 \) \ conditions)
Dirichlet convolution
\((f*g)_{(n)}=\sum_{d|n}f(d)\times g(\frac{n}{d})\)
This thing is commutative, associative, distributive law.
- Commutative proof:
\((f*g)_{{n}}=\sum_{d|n}f(d)\times g(\frac{n}{d})=\sum_{d|n}f(\frac{n}{d})\times g(d)=(g*f)_{(n)}\)
- Associativity of proof:
\(((f*g)*h)_{(n)}=\sum_{d|n}h(\frac{n}{d})\sum_{k|d}f(k)\times g(\frac{d}{k})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{d|n}\sum_{k|d}f(k)\times g(\frac{d}{k})\times h(\frac{n}{d})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{k|n}\sum_{q|\frac{n}{k}}f(k)\times g(q)\times h(\frac{n}{kq})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{k|n}f(k)\sum_{q|\frac{n}{k}}g(q)\times h(\frac{n}{kq})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(f*(g*h))_{(n)}\)
- Distributive law of proof:
\((f*(g+h))_{(n)}=\sum_{d|n}f(d)\times(g(\frac{n}{d})+h(\frac{n}{d}))\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{d|n}f(d)\times g(\frac{n}{d})+f(d)\times h(\frac{n}{d})\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(f*g)_{(n)}+(f*g)_{(n)}\)
- \ (f * \ Epsilon = f \) , is very simple, not a card.
\ (\ varphi \) defined functions
\ (\ varphi (n) \ ) is defined as: \ (. 1-n-\) in all \ (n-\) number of prime numbers.
\ (\ varphi \) of the product and proof of
\ (\ varphi \) is a multiplicative function
We \ (1-a \ times b \) these numbers written as a rectangle:
\(\left(\begin{array}{cc}1&2&...&a\\a+1&a+2&...&2a\\.&.&...&.\\.&.&...&.\\.&.&...&.\\(b-1)a+1&(b-1)a+2...&...&ab\end{array}\right)\)
For each row:
- Since \ ((a, b) = 1 \) so that each row has \ (\ varphi (a) \ ) number and \ (A \) prime.
For each column:
- Since this \ (B \) number \ (\% b \) the remainder different from each other, so that each column will have \ (\ varphi (a) \ ) number and \ (B \) prime.
In summary: \ (\ varphi (A \ B Times) = \ varphi (A) \ Times \ varphi (B) \) , i.e., \ (\ varphi \) is a multiplicative function.
\ (\ varphi \) Some properties of functions
- If \ (\ varphi (n-) \) , \ (n-\) is a prime number, set \ (n-= P ^ C \) , then the \ (\ varphi (n) = p ^ cp ^ {c-1} \ ) . (Not a card)
- \(\varphi(n)=n\times\prod_{i=1}^k1-\frac{1}{p_i}\)
prove:
设\(n=\prod_{i=1}^{k}p_i^{c_i}\)。
\(\varphi(n)=\prod_{i=1}^k\varphi(p_i^{c_i})\)
\(\varphi(n)=\prod_{i=1}^kp_i^{c_i}\times(1-\frac{1}{p_i})\)
\(\varphi(n)=n\times\prod_{i=1}^k1-\frac{1}{p_i}\)
QED
- \(\varphi*1=id\)
prove:
Set \ (F (n-) = \ sum_ {D |} n-\ varphi (D) \) .
\(\because (n,m)=1\)时
\(f(n)\times f(m)=\sum_{i|n}\varphi(i)\times\sum_{j|m}\varphi(j)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{i|n}\sum_{j|m}\varphi(i)\times \varphi(j)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\sum_{ij|nm}\varphi(ij)\)
\(\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =f(n\times m)\)
\ (\ therefore f (n) \) is a multiplicative function.
\(f(p^c)=\sum_{d|p^c}\varphi(d)=\varphi(1)+\varphi(p)+\varphi(p^2)+...+\varphi(p^{c-1})=p^c\)
设\(n=\prod_{i=1}^{k}p_i^{c_i}\)。
\(\therefore f(n)=\prod_{i=1}^kp_i^{c_i}=n\)
That \ (F = \ varphi ID = *. 1 \) .
QED
\ (\ varphi \) functions method for finding
Now that we have come before the \ (\ varphi \) the general term formula, so now think about \ (\ varphi \) should be how to find it.
I heard this guy called the Euler function, that - of course with a linear screen to find it.
While we can wire sieve to obtain prime numbers, the \ (\ varphi \) also gave updated.
Code:
void sieve(){
varphi[1]=1;
for(int i=2;i<=n;i++){
if (!vis[i]) p[++p[0]]=i,varphi[i]=i-1;
for (int j=1;j<=p[0]&&i*p[j]<=n;j++){
vis[i*p[j]]=1;
if (i%p[j]==0){
varphi[i*p[j]]=varphi[i]*p[j];
break;
}
varphi[i*p[j]]=varphi[i]*(p[j]-1);
}
}
}\ (\ mu \) defined functions
设\(n=\prod_{i=1}^{k}p_i^{c_i}\)。
\(\mu(n)=\left\{\begin{array}{rcl}1&n=1\\(-1)^k&\forall c_i,c_i\le1\\0&\exists c_i>1\end{array}\right.\)
\ (\ mu \) Some properties of functions
- \ (\ mu \) have multiplicative
prove:
Young \ (\ mu (a) = 0 \) some \ (\ Mu (B) = 0 \) , Provisions \ (\ mu (a \ times b) = \ mu (a) \ times \ mu (b) \ ) .
\ (\ because (a, b ) = 1 \ therefore a \) is the quality factor, the \ (B \) prime factors different from each other.
\(\therefore \mu(a\times b)=\mu(a)\times\mu(b),(a,b)=1\)
QED
- \ (\ M * 1 = \ epsilon \)
prove:
Can be converted to the original proposition: \ (\ sum_ {D |} n-\ MU (D) = [D =. 1] \)
Set \ (n-\) have \ (K \) a prime factor.
\(\mu*1=\sum_{d|n}\mu(d)=\sum_{i=0}^k(-1)^i\times C_k^i\)
Expand the \ (\ SUM \) , give:
\ (= U *. 1-C_k C_k ^ ^ 0 ^ C_k. 1 + 2 -... (+/-) C_m ^ m = (1-1 of) ^ m \) , intermediate with the binomial theorem.
\(\therefore \mu*1=\epsilon\)
QED
- \(id*\mu=\varphi\)
prove:
While the volume on both sides \ (1 \) .
\(\ \ \ \ \ \ id*\mu=\varphi\)
\(id*\mu*1=\varphi*1\)
\ (\ \ \ \ \ \ \ \ M * 1 = \ epsilon \)
QED
Mobius inversion
若\(g(n)=\sum_{d|n}f(d)\)则\(f(n)=\sum_{d|n}\mu(d)\times g(\frac{n}{d})\)
prove:
\(f(n)=\sum_{d|n}\mu(d)\times g(\frac{n}{d})\)
\(\ \ \ \ \ \ \ \ \ =\sum_{d|n}\mu(d)\sum_{i|\frac{n}{d}}f(i)\)
\(\ \ \ \ \ \ \ \ \ =\sum_{i|n}f(i)\sum_{d|\frac{n}{i}}\mu(d)\)
\(\ \ \ \ \ \ \ \ \ =\sum_{i|n}f(i)\times (\mu*1)_{[\frac{d}{i}]}\)
\(\ \ \ \ \ \ \ \ \ =\sum_{i|n}f(i)\times\epsilon(\frac{n}{i})\)
\(\ \ \ \ \ \ \ \ \ =f(n)\)
QED
In fact, there are other forms:
\(g(n)=∑_{d|n}f(n)\)，\(f(n)=∑_{d|n}μ(d)g(\frac{n}{d})\)
Prove similar, also requested the reader to deduce.