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
xn−xn+1=xn−12(xn+axn)=12xn(x2n−a).
Secondly, we obtain that
x2n−a=14(xn−1+axn−1)2−a=x2n−14−a2+a24x2n−1=14(x2n−1−2a+a2x2n−1)=14(xn−1−axn−1)2≥0.
Hence, xn≥xn+1 and xn is bounded from below since x2n≥a for each n≥2.
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.
No comments:
Post a Comment