Fixed point obtaining g's

Question

I currently am working on learning more about fixed point method. Finding equations that satisfy the constraints of a g function can sometimes require a bit of engineering. I have come across one that many would consider simple. Yet, I have been stuck on it for some time now.

Here it is $ f(x) = x^2 - x - 2 = 0 $ on $[1.5,3]$.

I have tried many things; however, I have yet to successfully discover one that maps domain to range for both $g$ and $g^\prime$.

Would anyone be able to give me a guiding hand? Perhaps there are cool tricks.

Answer 1

When trial and error does not work, try a trick for $g(x)$ using Newton's method

$$x= g(x) = x - \dfrac{f(x)}{f'(x)} = x - \dfrac{x^2-x-2}{2 x-1}$$

This may not always work.

Answer 2

As per Renaissance reasoning put terms on both sides so that all coefficients are positive $$ x^2=2+x $$ and then solve for the higher-degree side to get the iteration $$ x_{n+1}=\sqrt{2+x_n}. $$ This is a standard exercise for a monotonously converging recursive sequence.

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As you have the interval, you can also shift the variable as $x=1.5+h$ or $x=3-h$. The resulting equation might be more intuitive to transform into a fixed-point iteration. $$ 0=\frac94+3h+h^2-\frac32-h-2=h^2+2h-\frac54. $$ As $h$ is now assumed to be small, the focus has to be on the linear and constant term. One could make the iteration $$h_{n+1}=\frac5{4(2+h_n)}$$ out of this.

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For the Newton approach it is sometimes interesting to balance the degrees of the terms, for instance by considering $\tilde f(x)=\frac{f(x)}x=x-1-\frac2x$ so that $$ x_{n+1}=x_n-\frac{x_n-1-\frac2{x_n}}{1+\frac2{x_n^2}} =x_n\frac{x_n^2+2-(x_n^2-x_n-2)}{x_n^2+2} =x_n\frac{x_n+4}{x_n^2+2} $$ The visible advantage here to the original Newton iteration is that the denominator has no singularities on the real line.