Fixed-point Theorem Problem

Question

Show that the sequence generated by the formula xn+1 =(2/3)*xn +a/(3xn^2) (n = 0, 1, 2, . . .) converges to a^1/3 for any x0 > 0. [Hint: first, use the fixed-point theorem to show that xn → a^1/3 whenever x0 > a^1/3, then show that if 0 < x0 < a^1/3, then x1 > a^1/3.]

Answer 1

The given sequence $x_(n+1)$ $=$ $(2/3)$$x_n$ $+$ $a/3$($x_n$)² As you have told to use fixed-point theorem

So consider the given sequence to be a function "$\phi$" Gives as $\phi(x)$ $=$ $2/3$$x$ $+$ $a/3x²$

By definition of fix point , if $x$ is a fixed point then $\phi(x)$$=$$x$

Thus , $2/3$$x$ $+$ $a/3x²$ $=$ $x$

$a/3x²$$=$ $x/3$

$x³$$=$$a$

$x$$=$ a^1/3

So a^1/3 is fixed point

Hence by the theorem $x_n$ converges to a^1/3