Answer to Question #282845 in Real Analysis for abc

Question #282845

Check the convergence of the sequence defined by ๐‘ข๐‘›+1 = ๐‘Ž /(1+๐‘ข๐‘›) where ๐‘Ž > 0, ๐‘ข1 > 0.


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

Solution:


"u_{n+1}=1+1\/u_n"


mapย "u\\to 1+1\/u"ย can be extended to a Moebius transformation of the Riemann sphere

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

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

Its fixed points are:

"a=(1+\\sqrt 5)\/2,b=(1-\\sqrt 5)\/2"

obtained by solving the equation

"z^2-z-1=0"

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

"w=\\phi(z)=\\frac{z-a}{z-b}\\implies z=\\phi^{-1}(w)=\\frac{a-bw}{1-w}"

The fixed points now are w = 0 andย "w=\\infin"

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

"\\tilde{T}=\\phi \\circ T \\circ \\phi^{-1}"ย , then:

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


since

"\\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, whileย "\\infin"ย is repelling. This allows to conclude that in the original setting all initial pointsย "u_0\\neq b"

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


So, the sequence converges.


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