Friday, 19 August 2016

limits - Showing the sequence converges to the square root




For any a>0, I have to show the sequence xn+1 = 12(xn+ axn)



converges to the square root of a for any x1>0.



If I assume the limit exists ( denoted by x) then,



x = 12(x+ ax) can be solved to x2=a



How could I show that it does exist?


Answer




As mentioned in the comments, we need to show that the sequence is monotonic and bounded.



First, we observe that
xnxn+1=xn12(xn+axn)=12xn(x2na).
Secondly, we obtain that
x2na=14(xn1+axn1)2a=x2n14a2+a24x2n1=14(x2n12a+a2x2n1)=14(xn1axn1)20.
Hence, xnxn+1 and xn is bounded from below since x2na for each n2.



Monotonic and bounded sequence converges. Denote the limit of the sequence x=lim. Then we have that
x=\frac12\Bigl(x+\frac ax\Bigr)\quad\iff\quad x=\sqrt a.


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