Question #280114

Check the convergence of the sequence defined by 𝑢𝑛+1 = 𝑎/ 1+𝑢𝑛 where 𝑎 > 0, 𝑢1 > 0.


1
Expert's answer
2022-01-03T09:07:11-0500

un+1=1+1uu_{n+1}=1+\frac{1}{u}


map u1+1uu\to 1+\frac{1}{u} can be extended to a Moebius transformation of the Riemann sphere

C{}:C\cup \{\infin\}:

zz+1z,T(0)=,T()=1z\to \frac{z+1}{z},T(0)=\infin,T(\infin)=1

Its fixed points are:

a=(1+5)2,b=(15)2a=\frac{(1+\sqrt 5)}{2},b=\frac{(1-\sqrt 5)}{2}

obtained by solving the equation

z2z1=0z^2-z-1=0

We now introduce a new complex coordinate w on C, related to z via

w=ϕ(z)=zazb    z=ϕ1(w)=abw1ww=\phi(z)=\frac{z-a}{z-b}\implies z=\phi^{-1}(w)=\frac{a-bw}{1-w}

The fixed points now are w = 0 and w=w=\infin

in terms of the new coordinate w the transformation T appears as

T~=ϕTϕ1\tilde{T}=\phi \circ T \circ \phi^{-1} , then:

T~:wbaw,T~(0)=0,T~()=\tilde{T}: w\to \frac{b}{a}w,\tilde{T}(0)=0,\tilde{T}(\infin)=\infin


since

ba=352=0.382\frac{b}{a}=\frac{3-\sqrt 5}{2}=-0.382

we can infer that the fixed point 0 is attracting with basin of attraction all of C, while \infin is repelling. This allows to conclude that in the original setting all initial points u0bu_0\neq b

lead to limnun=a\displaystyle \lim_{n\to \infin} u_n=a


So, the sequence converges.


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