# Iteration for solving x=g(x).

$g(x) = \frac{x^2}{3}$

$P=3$

$p_0 = 3.5$

1)Graph $g(x)$, the line $y=x$, and the fixed point P (DONE)

2)Using the given starting value $p_0$, compute $p_1$ and $p_2$ (THE ANSWER MIGHT BE $p_1 = 4.083333$ , $p_2 = 5.537869$)

Determine geometrically if fixed point iteration converges (ANSWER: DIVERGES)

We might decide this using the analogical graphing structure:

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It's not at all clear what your question is. –  Thomas Andrews Feb 5 '13 at 18:18
By the way, I got $p_2=5.557870...$. Looks like you copied a digit wrong. –  Thomas Andrews Feb 5 '13 at 18:25
I need to know how to get those answers . It has to deal with g'(P) = 2. From figure 2.5 it's monotone divergent as 1< g'(P). But I also need to find $p_1$, $p_2$ It is probably related to the formula $|P-p_n| \le K^n |P-p_0|$. –  John Lennon Feb 5 '13 at 18:29
the answer says, $p_2 = 5.537869$ However, they are not always exact, so yours is probably correct. –  John Lennon Feb 5 '13 at 18:31
how did you get 5.557870...? –  John Lennon Feb 5 '13 at 19:58
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Writing $g(x)=x\cdot \frac{x}{3}$, we see that if $x>3$ then $g(x)>x$ and if $0\leq x<3$, then $g(x)<x$. So what we see is that, unless $p_0=3$, $g(p_0),g(g(p_0)),...$ is going to get steadily further away from $3$.
I am also wondering how you get $p_1$, $p_2$. Thank you. –  John Lennon Feb 5 '13 at 18:35
If $p_0$ is a bit to the right of the fixed point $p=3$, we have $p_1 = g(p_0) > p_0$, so the iteration takes you farther away from the fixed point, and it diverges. Similarly if $p_0$ was a bit to the left of the fixed point $p=3$, you'd have $p_1 = g(p_0) < p_0$. In that case it would converge to the other other fixed point $0$.