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一个数列方面比较有用的知识,发上来分享一下
By a generating function we mean a function f with formula f(x) in closed form,which, when expanded gives the terms of the sequence,so that for sufficiently small values of x we have
f(x)=A0+A1*x+A2*x^2+....An*x^n
By determining such functions we can not only obtain the terms of the sequence,but by giving x suitable values we can deduce the sums of some intereting serise. As an example,
f(x)=5/(1-2x)=5(1+2x+2^2*x^2+...2^n*x^n+...) so that An=5*2^n,which is a geometric sequence . The series expansion for f() is valid for |x|<1/2 and putting x=1/3 we get
5(1+2/3+4/9+8/27+....)=15
Let the sequence(An)be defined by A0=a A1=b A(n+2)=c*A(n+1)+dAn,where a,b,c,d are constants, then the generating function for the sequence (An) is given by
f(x)=[(a+(b-ca)x]/(1-cx-dx^2)
Example:
Find the n-th term of the sequence defined by A0=1,A1=11,A(n+2)=5A(n+1)-6An
Here a=1,b=11 c=5 d=-6 b-ac=6
So f(x)=(1+6x)/(1-5x+6x^2)=9/(1-3x) -8/(1-2x)
=9(1+3x+...3^n*x^n)-8(1+2x+...2^n*x^n+..)
=9*3^n-8^n
Find the sum of the infinite series whose n-th term is n^2/2^n
Let An=n^2
1/(1-x)^3=1+3x+6x^2+10x^3+....(n+1)*n*x^(n-1)/2
(1+x)/(1-x)^3=1+3x+6x^2+...(n+1)*n*x^(n-1)/2 +x+3x^2+....(n-1)*n*x^(n-1)/2
=1+4x+9x^2+...n^2*x^(n-1)
So f(x)=x(1+x)/(1-x)^3=x+4x^2+9x^3+...n^2*x^n,which is the generating function for the sequnce (An).
Let Bn=n^2/2^n
Sum of Bn,Tn=1/2 +4/2^2 +9/2^3+....n^2/2^n
Putting x=1/2 in th funtion f(x) ,the sum of Bn is (1+1/2)/2(1-1/2)^3=6
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