Fixed Point Iteration - Divergence to Convergence

Please refer to the question in the image.

My attempts are as follows:

1) Sub in g(x)=x for a final result of h(x)=x, then x=alpha obtains the result, but not sure if there is more to it.

2) I am not sure as by (1) my result is simply h(alpha)=alpha, which means c=all real numbers (so I must be going in the wrong direction already)

3) I know for quadratic convergence h'(alpha)=0 and h''(alpha) does not equal zero, but I cannot get this due to (2)

Thank you for your help



$$h'(x)=1+(g'(x)-1)c $$

It converges if

$$-1 <h'(\alpha)<1$$


$$-1 <1+(g'(\alpha)-1)c <1$$ which gives

$$\frac{-2}{g'(\alpha)-1}<c <0$$

For the quadratic convergence, we need $$h'(\alpha)=0$$ which means $$c=\frac {1}{1-g'(\alpha)} $$

  • $\begingroup$ Is there more that I should realize about g'(alpha) other than it is >1 in order to obtain the values of c for which we get convergence? As for now I have (-2/(>1-1)))<c<0 $\endgroup$ – user565684 May 29 '18 at 2:14
  • $\begingroup$ @user565684 i just added the last line. $\endgroup$ – hamam_Abdallah May 30 '18 at 0:42
  • $\begingroup$ Thank you. One last question regarding quadratic convergence and the value of c. Is it possible here to have h'(alpha)=0 and at the same time h''(alpha)=not zero for quadratic convergence. I'm not seeing it. $\endgroup$ – user565684 May 30 '18 at 2:09
  • $\begingroup$ My thoughts are to take h'(alpha)=0 and solve for c where c=(-1)/(g'(alpha)-1), then solve h"(alpha)=cg"(alpha) with this c value to obtain h"(alpha)=-(g"(alpha))/(g'(alpha)-1)=not zero....I hope. $\endgroup$ – user565684 May 30 '18 at 2:58
  • $\begingroup$ @user565684 Yes {}{}{}{}{{ $\endgroup$ – hamam_Abdallah May 31 '18 at 0:04

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.