Question #280115

Check the convergence of the sequence defined by š‘¢š‘›+1 = (1 + 1 š‘¢š‘› ) , š‘¢1 > 0. Note that this is the sequence associated with the continued fraction expansion of the Golden ratio. 


Expert's answer

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


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

C∪{āˆž}:C\cup \{\infin\}:

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

Its fixed points are:

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

obtained by solving the equation

z2āˆ’zāˆ’1=0z^2-z-1=0

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

w=Ļ•(z)=zāˆ’azāˆ’bā€…ā€ŠāŸ¹ā€…ā€Šz=Ļ•āˆ’1(w)=aāˆ’bw1āˆ’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=āˆžw=\infin

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

T~=Ļ•āˆ˜Tāˆ˜Ļ•āˆ’1\tilde{T}=\phi \circ T \circ \phi^{-1} , then:

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


since

ba=3āˆ’52=āˆ’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 āˆž\infin is repelling. This allows to conclude that in the original setting all initial points u0≠bu_0\neq b

lead to lim⁔nā†’āˆžun=a\displaystyle \lim_{n\to \infin} u_n=a


So, the sequence converges.


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