# Find an equation that using Fixed Point Iteration converges to -1.02

Let $f(x) = e^{x-2} + x^3 - x$, then using fixed point iteration, find all of the roots. I've already found two equations which converge to 0.163822 and ~0.788941.

The equation which converges toward 0.163822 is: $e^{x-2} + x^3$ with a guess of $0.1$

The equation which converges toward ~0.788941 is: $\frac{-e^{x-2} + x}{x}^{\frac{-1}{2}}$ with a guess of $1.5$.

I can't seem to find an equation which converges to the last root of approximately -1.02.

• Try another initial guess. Then, the iteration might converge to that root. – Peter Oct 17 '16 at 16:47
• The third possible function is $\ln(x-x^3)+2$ – Peter Oct 17 '16 at 16:50
• @Peter, I've tried that function, but it does not converge to -1. Is there a particular guess you had in mind? I've tried guesses of -1, 0, 0.1, and 1 and they all seem to go to infinity – user3370201 Oct 17 '16 at 17:07
• The fixpoint-iteration might fail for this root. I did not find a function and a guess doing the job yet. – Peter Oct 17 '16 at 17:45
• By the way, instead of $0$, the third root is $0.1638$. – Peter Oct 17 '16 at 17:50

Try $f(x)=\sqrt[3]{x-e^{x-2}}$, starting from a negative value. The rationale is that the cubic root has a slope less than $1$ for values less than $-1$.
$$-2 \\ -1.26375541197 \\ -1.09195226149 \\ -1.0438385854 \\ -1.02961125511 \\ -1.02533462938 \\ -1.02404271663 \\ -1.02365186038 \\ -1.0235335567 \\ -1.02349774382 \\ \vdots$$
$$1 \\ 0.858222649309 \\ 0.813807987157 \\ 0.798134216479 \\ 0.792376247446 \\ 0.790229882858 \\ 0.789425446205 \\ 0.789123338776 \\ 0.789009795416 \\ 0.788967109326 \\ \vdots$$