Solution:
u n + 1 = 1 + 1 / u n u_{n+1}=1+1/u_n u n + 1 β = 1 + 1/ u n β
map u β 1 + 1 / u u\to 1+1/u u β 1 + 1/ u can be extended to a Moebius transformation of the Riemann sphere
C βͺ { β } : C\cup \{\infin\}: C βͺ { β } :
z β z + 1 z , T ( 0 ) = β , T ( β ) = 1 z\to \frac{z+1}{z},T(0)=\infin,T(\infin)=1 z β z z + 1 β , T ( 0 ) = β , T ( β ) = 1
Its fixed points are:
a = ( 1 + 5 ) / 2 , b = ( 1 β 5 ) / 2 a=(1+\sqrt 5)/2,b=(1-\sqrt 5)/2 a = ( 1 + 5 β ) /2 , b = ( 1 β 5 β ) /2
obtained by solving the equation
z 2 β z β 1 = 0 z^2-z-1=0 z 2 β z β 1 = 0
We now introduce a new complex coordinate w on C, related to z via
w = Ο ( z ) = z β a z β b β
ββΉ β
βz = Ο β 1 ( w ) = a β b w 1 β w w=\phi(z)=\frac{z-a}{z-b}\implies z=\phi^{-1}(w)=\frac{a-bw}{1-w} w = Ο ( z ) = z β b z β a β βΉ z = Ο β 1 ( w ) = 1 β w a β b w β
The fixed points now are w = 0 and w = β w=\infin w = β
in terms of the new coordinate w the transformation T appears as
T ~ = Ο β T β Ο β 1 \tilde{T}=\phi \circ T \circ \phi^{-1} T ~ = Ο β T β Ο β 1 , then:
T ~ : w β b a w , T ~ ( 0 ) = 0 , T ~ ( β ) = β \tilde{T}: w\to \frac{b}{a}w,\tilde{T}(0)=0,\tilde{T}(\infin)=\infin T ~ : w β a b β w , T ~ ( 0 ) = 0 , T ~ ( β ) = β
since
b a = 3 β 5 2 = β 0.382 \frac{b}{a}=\frac{3-\sqrt 5}{2}=-0.382 a b β = 2 3 β 5 β β = β 0.382
we can infer that the fixed point 0 is attracting with basin of attraction all of C, while β \infin β is repelling. This allows to conclude that in the original setting all initial points u 0 β b u_0\neq b u 0 β ξ = b
lead to lim β‘ n β β u n = a \displaystyle \lim_{n\to \infin} u_n=a n β β lim β u n β = a
So, the sequence converges.
Comments
Leave a comment