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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

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hint

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

It converges if

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

or

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

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

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

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  • $\begingroup$ 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

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