Taken seniors zsy of PPT. Although there is no link to the original, but in order to show respect for seniors fruits of labor, where a paste links . zsy seniors usually is generous, lively and interesting classes, and each talked about their own knowledge very seriously. Recommended visit his blog to learn from.
We all know binomial generating function:
\ [F (X) = (. 1 + X) n-^ = \ sum_ K = {0}} ^ {n-\ dbinom {{K} n-X} ^ K \]
When we bring \ (x = -1 \) , the equation would give this:
\ [F (-1) = (. 1 -. 1) n-^ = \ sum_ K = {0}} ^ {n-\ dbinom {n} {k} - (
1) ^ k \] when \ (n = 0 \) , the left part there is no meaning. However, the right calculated exactly \ (\ Binom {0} = {0}. 1 \) . So we got a identity:
\ [\ sum_ {K = 0} ^ {n-} \ dbinom {n-} {K} (-. 1) ^ {K} = [n-= 0] = \ Epsilon (n-+. 1) \]
Suppose we know \ (F (n-) = \ sum_ {K = 0} ^ {n-} \ Binom {n-} {K} G (K) \) , then we How \ (f (k) \) represented \ (g (n) \) it?
We first piece together in the form of a binomial:
\ [G (n-) = \ sum_ {m} = 0 {n-^} [= n-m] G (m) = \ sum_ 0} = {m} ^ {n- \ epsilon (nm + 1) g
(m) \] then substituted into the above that Identities:
\ [G (n-) = \ sum_ {m = 0} ^ {n-} \ sum_ {K = 0} ^ {nm} (-1) ^ {k} \
dbinom {n} {m} \ dbinom {nm} {k} g (m) \] the meaning of the combination, it is easy to obtain:
\ [G (n-) = \ {m sum_ = 0} ^ {n} \
sum_ {k = 0} ^ {nm} (- 1) ^ {k} \ dbinom {n} {k} \ dbinom {nk} {m} g (m) \] Next, No need to exchange sum. To facilitate understanding, so that here I \ (of S_ {MK} = (-1) ^ {K} \ {n-dbinom} {K} \ {dbinom NK} {m} G (m) \) , then the formula becomes the original become \ (\ sum_ 0} = {m} ^ {n-\ sum_ K = {0}} ^ {nm of S_ {MK} \) .
\ [\ Begin {bmatrix} S_ {0,0} & S_ {0,1} & \ cdots & S_ {0, n-1} & S_ {0, n} \\ S_ {1,0} & S_ { 1,1} & \ cdots & S_ { 1, n-1} \\ \ vdots & \ cdots & \ vdots \\ S_ {n -1, 0} & S_ {n-1, 1} \\ S_ {n , 0} \ end {bmatrix}
\] clearly, a method of seeking the previous row by row, from left to right sum. We also order a one, summed from top to bottom. therefore:
\ [G (n) = \ sum_ {k = 0} ^ {n} \ sum_ {m = 0} ^ {nk} (- 1) ^ {k} \ dbinom {n} {k} \ dbinom {nk} {m} g (m) \
] made and \ (m \) independent coefficients:
\ [G (n-) = \ sum_ {K = 0} ^ {n-} (-. 1) ^ K \ dbinom {n-} { k} \ sum_ {m = 0
} ^ {nk} \ dbinom {nk} {m} g (m) \] notes \ (\ sum_ {m = 0 } ^ {nk} \ dbinom {nk} {m} g (m) \) is \ (F (NK) \) . Thus:
\ [G (n-) = \ sum_ K = {0}} ^ {n-(-. 1) ^ K \ {n-dbinom}} F {K (NK) \]
This is the inverse of the binomial. Through it, we can use the binomial sum launch of a bad calculation answer.
Ultimately, inversion or exclusion capacity and can not get away.