Answer to Question #282845 in Real Analysis for abc

Question #282845

Check the convergence of the sequence defined by 𝑒𝑛+1 = π‘Ž /(1+𝑒𝑛) where π‘Ž > 0, 𝑒1 > 0.


1
Expert's answer
2021-12-27T16:22:37-0500

Solution:


un+1=1+1/unu_{n+1}=1+1/u_n


map uβ†’1+1/uu\to 1+1/u can be extended to a Moebius transformation of the Riemann sphere

Cβˆͺ{∞}:C\cup \{\infin\}:

zβ†’z+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=(1βˆ’5)/2a=(1+\sqrt 5)/2,b=(1-\sqrt 5)/2

obtained by solving the equation

z2βˆ’zβˆ’1=0z^2-z-1=0

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

w=Ο•(z)=zβˆ’azβˆ’bβ€…β€ŠβŸΉβ€…β€Šz=Ο•βˆ’1(w)=aβˆ’bw1βˆ’ww=\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~=Ο•βˆ˜Tβˆ˜Ο•βˆ’1\tilde{T}=\phi \circ T \circ \phi^{-1} , then:

T~:wβ†’baw,T~(0)=0,T~(∞)=∞\tilde{T}: w\to \frac{b}{a}w,\tilde{T}(0)=0,\tilde{T}(\infin)=\infin


since

ba=3βˆ’52=βˆ’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 βˆž\infin is repelling. This allows to conclude that in the original setting all initial points u0β‰ bu_0\neq b

lead to lim⁑nβ†’βˆžun=a\displaystyle \lim_{n\to \infin} u_n=a


So, the sequence converges.


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