求多項式係數

1.Fn(x)= [Fn-1(x)]^2-4[Fn-1(x)]+4
Fn-1(x)= an-1+bn-1x+cn-1x^2+….
[Fn-1(x)]^2= (an-1)^2+2 (an-1)( bn-1)x+[2(an-1)( cn-1)+( bn-1)^2]x^2+……
4[Fn-1(x)]= 4an-1+4bn-1x+4cn-1x^2+….
2.故an=(an-1)^2-4an-1+4,
bn=2 (an-1)( bn-1)-4 bn-1,
cn=[2(an-1)( cn-1)+( bn-1)^2]-4 cn-1
3. (1)a1=4,a2=4,a3=4,….. ,an=4
(2) bn=2 (an-1)( bn-1)-4 bn-1=4 bn-1,=> bn= b1*4^(n-1)=-(4^n)
(3) cn=[2(an-1)( cn-1)+( bn-1)^2]-4 cn-1
=4( cn-1)+4^(2n-2)…=>cn=4^(n-1)*c1+4^(n-2)*16+….+ 4*16^(n-2)+ 16^(n-1)
cn=4^(n-1)+ 4^(n-2)+…+4^(2n-3)+ 4^(2n-2)= 4^(n-1) [(4^n)-1]/3

2010-02-25 21:34:48 補充：
訂正
cn=4^(n-1)+ 4^(n)+…+4^(2n-3)+ 4^(2n-2)= 4^(n-1) [(4^n)-1]/3

Let Fn(x)=Fn(x)=(((x-2)^2-2)^2-...-2)^2

Fn(x)=(Fn-1(x)-2)^2=[Fn-1(x)]^2-4Fn-1(x)+4

So an=(an-1)^2-4(an-1)+4
bn=2(bn-1)an-4bn-1
cn=2(cn-1)(an-1)+(bn-1)^2-4(cn-1)

Since a1=4,a2=4,...an=4
bn=8(bn-1)-4bn-1=4bn-1=4^2bn-2=...=4^(n-1)b1=-4^n
cn=8(cn-1)+4^(2n-2)-4(cn-1)=4(cn-1)+4^(2n-2)=4^(n-1)+4^(2n-2)

Example when n=2: (x^2-4x+2)(x^2-4x+2)=x^4-4x^3+2x^2-4x^3+16x^2-8x+2x^2-8x+4=x^4-8x^3+20x^2-16x+4

a2=4,b2=-16,c2=20