不动点法解数列的原理是什么?

请问这个方法的原理是什么?为什么可以这么做?
2024-12-02 19:14:27
推荐回答(3个)
回答1:

不动点法一般来说依据压缩映射原理,但也要看具体问题。

回答2:

递推式:
a(n+1)=(A*an+B)/(C*an+D)
(n∈N*,A,B,C,D为常数,C不为0,AD-BC不为0,a1与a2不等)

其特征方程为x=(A*x+B)/(C*x+D)

特征方程的根称为该数列的不动点

这类递推式可转化为等差数列或等比数列

1)若x=(A*x+B)/(C*x+B)有两个不等的根α、β,则有:

(a(n+1)-α)/(a(n+1)-β)=k*((an-α)/(an-β))

其中k=(A-α*C)/(A-β*C)

x=(A*x+B)/(C*x+D)

C*x^2+(D-A)*x-B=0

α不等于β
(D-A)^2+4*B*C不等于0

C*α^2+(D-A)*α-B=0
C*α^2-A*α=B-α*D

a(n+1)-α=(A*an+B-C*α*an-α*D)/(C*an+D)=(A*an-C*α*an+C*α^2-A*α)/(C*an+D)=(A-C*α)*(an-α)/(C*an+D)

a(n+1)-β=(A*an+B-C*β*an-β*D)/(C*an+D)=(A*an-C*β*an+C*β^2-A*β)/(C*an+D)=(A-C*β)*(an-β)/(C*an+D)

(a(n+1)-α)/(a(n+1)-β)=(A-α*C)/(A-β*C)*((an-α)/(an-β))


(an-α)/(an-β)=((A-α*C)/(A-β*C))^(n-1)*((a1-α)/(a1-β))


an=(β*(((A-α*C)/(A-β*C))^(n-1))*((a1-α)/(a1-β))-α)/(((((A-α*C)/(A-β*C))^(n-1))*((a1-α)/(a1-β))-1)

2)若x=(A*x+B)/(C*x+B)有重根α,则有

1/(a(n+1)-α)=1/(an-α)+k

其中k=(2*C)/(A+D)

x=(A*x+B)/(C*x+D)

C*x^2+(D-A)*x-B=0

C*α^2+(D-A)*α-B=0

α=(A-D)/(2*C)

a(n+1)-α=(A-C*α)*(an-α)/(C*an+D)

1/(a(n+1)-α)=((C*an+D)/(A-C*α))*(1/(an-α))

=1/(an-α)+(C*an+D-A+((A-D)/(2*C))*C)/((A-(A-D)/(2*C)*C)*(an-(A-D)/(2*C)))=1/(an-α)+(C*an+C*(D-A)/(2*C))/(((A+D)/2)*(an+(D-A)/(2*C)))
=1/(an-α)+(2*C)/(A+D)


1/(an-α)=(2*C*(n-1))/(A+D)+1/(a1-α)

an=1/((2*C*(n-1))/(A+D)+1/(a1-α))+α

以上即是证明过程……
至于更进一步的原理,涉及到高等数学的知识,就不仔细解释了……

回答3:

a(n+1)=(A*an+B)/(C*an+D)
(n∈N*,A,B,C,D为常数,C不为0,AD-BC不为0,a1与a2不等)
其特征方程为x=(A*x+B)/(C*x+D)
特征方程的根称为该数列的不动点
这类递推式可转化为等差数列或等比数列
1)若x=(A*x+B)/(C*x+B)有两个不等的根α、β,则有:
(a(n+1)-α)/(a(n+1)-β)=k*((an-α)/(an-β))
其中k=(A-α*C)/(A-β*C)
x=(A*x+B)/(C*x+D)
C*x^2+(D-A)*x-B=0
α不等于β
(D-A)^2+4*B*C不等于0
C*α^2+(D-A)*α-B=0
C*α^2-A*α=B-α*D
a(n+1)-α=(A*an+B-C*α*an-α*D)/(C*an+D)=(A*an-C*α*an+C*α^2-A*α)/(C*an+D)=(A-C*α)*(an-α)/(C*an+D)
a(n+1)-β=(A*an+B-C*β*an-β*D)/(C*an+D)=(A*an-C*β*an+C*β^2-A*β)/(C*an+D)=(A-C*β)*(an-β)/(C*an+D)
(a(n+1)-α)/(a(n+1)-β)=(A-α*C)/(A-β*C)*((an-α)/(an-β))

(an-α)/(an-β)=((A-α*C)/(A-β*C))^(n-1)*((a1-α)/(a1-β))

an=(β*(((A-α*C)/(A-β*C))^(n-1))*((a1-α)/(a1-β))-α)/(((((A-α*C)/(A-β*C))^(n-1))*((a1-α)/(a1-β))-1)
2)若x=(A*x+B)/(C*x+B)有重根α,则有
1/(a(n+1)-α)=1/(an-α)+k
其中k=(2*C)/(A+D)
x=(A*x+B)/(C*x+D)
C*x^2+(D-A)*x-B=0
C*α^2+(D-A)*α-B=0
α=(A-D)/(2*C)
a(n+1)-α=(A-C*α)*(an-α)/(C*an+D)
1/(a(n+1)-α)=((C*an+D)/(A-C*α))*(1/(an-α))
=1/(an-α)+(C*an+D-A+((A-D)/(2*C))*C)/((A-(A-D)/(2*C)*C)*(an-(A-D)/(2*C)))=1/(an-α)+(C*an+C*(D-A)/(2*C))/(((A+D)/2)*(an+(D-A)/(2*C)))
=1/(an-α)+(2*C)/(A+D)

1/(an-α)=(2*C*(n-1))/(A+D)+1/(a1-α)
an=1/((2*C*(n-1))/(A+D)+1/(a1-α))+α
以上即是证明过程……
至于更进一步的原理,涉及到高等数学的知识,就不仔细解释了……