Mobius function summary
Properties : \ (\ sum_ {D |} n-\ MU (D) = [n-==. 1] \)
This number can be a combination of properties permit, if the image point is Pascal's Triangle.
Because the identity: \ (\ sum_ I = {0}} ^ {n-(-. 1) ^ {n-NC_} ^ {I} = 0 \) .
Mobius inversion:
in the form of a:
Known: \ (G (n-) = \ sum_ {D |} n-F (D) \) , there are: \ (F (n-) = \ sum_ {D | n- } \ MU (D) G (\ FRAC {n-} {D}) \) .
proven as follows:
\ [\ sum_ {D | n-} \ MU (D) G (\ FRAC {n-} {D}) = \ sum_ {d | n} \ mu
(d) \ sum_ {k | \ frac {n} {d}} f (k) \] exchange sequence summing with:
\ [\ sum_ {D |} n-\ MU ( d) \ sum_ {k | \ frac {n} {d}} f (k) = \ sum_ {k | n} f (k) \ sum_ {d | \ frac {n} {k}} \ mu (d ) = \ sum_ {k | n
} f (k) [\ frac {n} {k} = 1] = f (n) \] it is proved.
form II:
known: \ (G (n-) = \ sum_ {n-| D} F (D) \) , there are: \ (F (n-) = \ sum_ {n-| D} \ MU (\ FRAC {D} {n-}) G (D) \) .
prove as follows:
order \ (K = \ FRAC {D} {n-} \) , there are:
\ [\ sum_ {n-| D} \ MU (\ FRAC {D} {n-}) G (D) = \ sum_ { k = 1} ^ {\ infty
} \ mu (k) g (nk) = \ sum_ {k = 1} ^ {\ infty} \ mu (k) \ sum_ {nk | t} f (t) \] exchange summation order, are:
\[ \sum_{k=1}^{\infty}\mu(k)\sum_{nk|t}f(t)=\sum_{n|t}f(t)\sum_{k|\frac{t}{n}}\mu(k)=\sum_{n|t}f(t)[\frac{t}{n}=1]=f(n) \]
故得证.
: Common technique (application)
a property:
\ [\ lfloor \ FRAC {\ lfloor \ FRAC {X} {A} \ rfloor} {B} \ rfloor = \ lfloor \ FRAC {X} {ab &} \ rfloor \]
demonstrated as follows:
order \ (Y = \ lfloor \ FRAC {X} {A} \ rfloor, Z = \ lfloor \ FRAC {X} {ab &} \ rfloor \) , apparently: \ (X = ZAB + C, C \ ab & Leq \) .
equation both sides while divided by \ (A \) and the rounding has: \ (ZB + Y = \ lfloor \ FRAC {C} {A} \ rfloor \) .
simultaneously further divided by \ (B \) and the rounding: \ (\ lfloor \ FRAC {\ lfloor \ FRAC {X} {A} \ rfloor} {B} \ rfloor = Z + \ lfloor \ FRAC {\ lfloor \ FRAC {C} {A} \ rfloor} {B} \ rfloor \) .
after a relatively clearly.
Example One:
seeking: \ (\ sum_ {I =. 1} ^ {n-} \ sum_ {J =. 1} ^ {m} [GCD (I, J) =. 1] \) .
\ [\ Sum_ {I =. 1 } ^ {n} \ sum_ { j = 1} ^ {m} [gcd (i, j) = 1] = \ sum_ {i = 1} ^ {n} \ sum_ {j = 1} ^ {m} \ sum_ {d | gcd (i, j)} \ mu (d) = \ sum_ {i = 1} ^ {n} \ sum_ {j = 1} ^ {m} \ sum_ {d | i, d | j} \ mu (d) \]
explain: from the first step to the second step and more out of a style, is equivalent to a process of inclusion and exclusion, considering all the \ ((i, J) \) , for their \ (gcd \) to the inclusion-exclusion: first, plus all of the circumstances, then subtract \ (gcd \) to \ (2,3,5 ... \) multiple of, but it will reduce more, so add it back.
After further conversion to obtain:
\ [\ sum_ {D} \ MU (D) \ sum_ {I =. 1} ^ {n-} \ sum_ {D | I} \ sum_ {J =. 1} ^ {m} \ sum_ { d | j} 1 = \ sum_
{d} \ mu (d) \ lfloor \ frac {} {d} \ rfloor \ lfloor \ frac {m} {d} \ rfloor \] n the latter part of the direct divisible block, and then pretreated \ (\ mu \) prefix and on the line.
Example II:
seeking: \ (\ sum_ {I}. 1 = n-^ {} \ sum_ {J} = ^ {m}. 1 GCD (I, J) \)
The first comparison and the like, we have to change that:
\ [\ SUM {D} \ sum_ {I}. 1 = n-^ {} \ sum_ {J} = ^ {m}. 1 [GCD (I, J) = d] = \ sum {d } \ sum_ {i = 1} ^ {\ frac {n} {d}} \ sum_ {j = 1} ^ {\ frac {m} {d}} [gcd (i, j) = 1] \]
The gains Example:
\ [\ SUM {D} \ sum_. 1 {I} = {^ \ n-FRAC {} {D}} \ sum_. 1} = {J ^ {\ {m FRAC } {d}} [gcd ( i, j) = 1] = \ sum {d} \ sum_t \ mu (t) \ lfloor \ frac {n} {dt} \ rfloor \ lfloor \ frac {m} {dt} \ rfloor \]
order \ (T dt = \) , then the above equation can be turned into:
\ [= \ {SUM \ FRAC {T} {T}} \ sum_ {T} \ MU (\ FRAC {T} {D }) \ lfloor \ frac {n } {T} \ rfloor \ lfloor \ frac {m} {T} \ rfloor = \ sum_ {T} \ lfloor \ frac {n} {T} \ rfloor \ lfloor \ frac {m } {T} \ rfloor \ sum_ {t | T} \ mu (t) \ frac {T} {t} = \ sum_T \ lfloor \ frac {n} {T} \ rfloor \ lfloor \ frac {m} {T } \ rfloor \ varphi (T) \]
Finally, to say why Euler transformation functions:
Euler function has a very useful properties: \ (\ sum_ {D |} n-\ varphi (D) = n-\) , then the main proof from \ (f (n) = \ sum_ {d | n} \ varphi (d) \) this function is a multiplicative function starts. At the same time there is also a useful properties: \ (\ varphi (P ^ m) = ^ P ^ {m-MP. 1} \) .
With this bit inversion formula have:
\ [\ varphi (n-) = \ sum_ {D |} n-\ MU (D) \ {n-FRAC {D}} \]
for the two cases, the pre-divisible block + Euler processing functions on the line.
Relevant examples:
leaving the pit to be filled