2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Try another initial guess. Then, the iteration might converge to that root. $\endgroup$ – Peter Oct 17 '16 at 16:47
  • $\begingroup$ The third possible function is $\ln(x-x^3)+2$ $\endgroup$ – Peter Oct 17 '16 at 16:50
  • $\begingroup$ @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 $\endgroup$ – user3370201 Oct 17 '16 at 17:07
  • $\begingroup$ The fixpoint-iteration might fail for this root. I did not find a function and a guess doing the job yet. $\endgroup$ – Peter Oct 17 '16 at 17:45
  • $\begingroup$ By the way, instead of $0$, the third root is $0.1638$. $\endgroup$ – Peter Oct 17 '16 at 17:50
1
$\begingroup$

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 $-1$.

enter image description here

$$ -2 \\ -1.26375541197 \\ -1.09195226149 \\ -1.0438385854 \\ -1.02961125511 \\ -1.02533462938 \\ -1.02404271663 \\ -1.02365186038 \\ -1.0235335567 \\ -1.02349774382 \\ \vdots$$

You can also get convergence to a positive root.

$$ 1 \\ 0.858222649309 \\ 0.813807987157 \\ 0.798134216479 \\ 0.792376247446 \\ 0.790229882858 \\ 0.789425446205 \\ 0.789123338776 \\ 0.789009795416 \\ 0.788967109326 \\ \vdots$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.