Fixed Point Iteration - Divergence to Convergence

Question

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

hamam_Abdallah · Answer 1 · 2018-05-31T00:03:41.130

1

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 $$c=\frac {1}{1-g'(\alpha)} $$

edited May 31 '18 at 00:03

answered May 28 '18 at 23:50

hamam_Abdallah

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 – user565684 May 29 '18 at 02:14
@user565684 i just added the last line. – hamam_Abdallah May 30 '18 at 00:42
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. – user565684 May 30 '18 at 02:09
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. – user565684 May 30 '18 at 02:58
@user565684 Yes {}{}{}{}{{ – hamam_Abdallah May 31 '18 at 00:04
Thank you so much – user565684 May 31 '18 at 03:36

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