I'm learning fixed point iteration (first and second form).
My teacher said there are two forms:
g(x) = x - f(x)
g(x) = x - h(x)f(x)
where h(x) is an aux function.
suppose f(x) = x^2 + 2x -3.
Let f(x) = 0.
x^2 + 2x -3 = 0
x^2 = - 2x + 3
x = 3/x -2
So we let g(x) = 3/x -2, for a fixed point, and root, right?
To solve, we iterate and see convergence:
xk+1 = 3/xk -2
and pick an initial xk.
My teacher said that in second form:
from f(x) = x^2 + 2x -3:
g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)
with the aux function h(x) = -1 / (x^2 - 5)
Where did he get h(x) = -1 / (x^2 - 5) from? And where is
g(x) = (x + (x^2 + 2x -3)) / (x^2 - 5)
from?
thanks!
This is most probably not what they said, and it seems that typos due to carelessness are killing you here.
Actually, to solve $f(x)=0$ with $f(x)=x^2 + 2x -3$, they most probably suggested to solve $g(x)=x$ with $$g(x) = x + \frac{x^2 + 2x -3}{x^2 - 5}$$ that is, a function $g$ exactly of the general form $g(x)=x-f(x)h(x)$, for the auxiliary function $$h(x) = -\frac1{x^2 - 5}$$ They chose $h$ arbitrarily.
Supplementary material: This choice of auxiliary function $h$ is not especially suited to this precise function $f$, if you ask me. A better choice would be to get rid of the $x^2$ term in $f(x)$, using $$g_c(x)=x-\frac{x^2 + 2x -3}{x+c}$$ for some fixed positive $c$, that is, $g_c(x)=x-f(x)h_c(x)$ with $$h_c(x)=\frac1{x+c}$$ Now, a simple computation shows that $$g_c(x)=\frac{(c-2)x+3}{x+c}$$ hence the value $c=2$ is interesting, leading to the iteration of the function $$g_2(x)=\frac3{x+2}$$ Note finally that for every initial condition $x_0\geqslant0$, the sequence defined recursively by $x_{n+1}=g_2(x_n)$ converges to the positive root $x^*$ of $f$ at a geometric rate $\frac13$ (and that $x^*=1$, but this is irrelevant to the construction and to the remark).