二项式反演

也许更好的阅读体验

表达

若有(f(n)=sum_{i=0}^ninom{n}{i}g(i))
则有(g(n)=sum_{i=0}^n(-1)^{n-i}inom{n}{i}f(i))

---

证明

(egin{aligned}g(n)&=sum_{i=0}^n(-1)^{n-i}inom{n}{i}f(i)\ &=sum_{i=0}^n(-1)^{n-i}inom{n}{i}sum_{j=0}^iinom{i}{j}g(j)\ &=sum_{i=0}^nsum_{j=0}^i(-1)^{n-i}inom{n}{i}inom{i}{j}g(j)\ &=sum_{j=0}^nsum_{i=j}^n(-1)^{n-i}inom{n}{i}inom{i}{j}g(j)\ end{aligned})
在组合数中，有这么一个式子
(egin{aligned}inom{i}{j}inom{j}{k} &= frac{i!}{j!(i - j)!}frac{j!}{k!(j - k)!} \ &=dfrac {i!}{k!left( i-k ight) !}dfrac {left( i-k ight) !}{left( i-j ight) !left( j-k ight) !}\ &=dfrac {i!}{k!left( i-k ight) !}dfrac {left( i-k ight) !}{left( left( i-k ight) -left( j-k ight) ight) !left( j-k ight) !}\ &=egin{pmatrix} i \ k end{pmatrix}egin{pmatrix} i-k \ j-k end{pmatrix}end{aligned})
即
(egin{aligned}inom{i}{j}inom{j}{k}=egin{pmatrix} i \ k end{pmatrix}egin{pmatrix} i-k \ j-k end{pmatrix}end{aligned})
所以
(egin{aligned}原式&=sum_{j=0}^nsum_{i=j}^n(-1)^{n-i}inom{n}{j}inom{n-j}{i-j}g(j)\ &=sum_{j=0}^nsum ^{n-j}_{i=0}left( -1 ight) ^{n-i-j}egin{pmatrix} n \ j end{pmatrix}egin{pmatrix} n-j \ i end{pmatrix}g(j)\ &=sum_{j=0}^nleft(1-1 ight)^{n-j}egin{pmatrix} n \ j end{pmatrix}gleft(j ight) end{aligned})
此时要求(n != j)，此时(原式=0)
当(n=j)时，(原式=g(n))

到此，原式得证

如有哪里讲得不是很明白或是有错误，欢迎指正
如您喜欢的话不妨点个赞收藏一下吧