(8 marks) Let the functions *g*_{1 }and *g*_{2 }be defined by

- Show that 1 is a fixed point of
*g*_{1 }and*g*_{2}, and that any fixed point of*g*_{1 }or*g*_{2 }is a root of the equation*x*^{3 }+ 4*x*^{2 }− 5 = 0. - Suppose that we take
*x*_{0 }sufficiently close to 1 (but not equal to 1) and would like to generate a sequence {*x*} by the fixed-point iteration_{n}*x*=_{n }*g*_{1}(*x*−1) (_{n}*n*≥ 1) or by the fixed-point iteration*x*=_{n }*g*_{2}(*x*−1) (_{n}≥ 1). If you wish to have faster convergence (to the point 1), which of the two fixed-point iterations would you choose? Justify your answer. (You need to make your choice by a theoretical result, not by comparing the first few iterates.)^{n }

- (6 marks) Let
*g*(*x*) = (2*x*^{2 }+ 3)^{1/4}.- Let
*p*_{0 }= 1. Find*p*_{10 }by the fixed-point iteration

- Let

*p _{n}*

(You should use MATLAB to do this. But just report *p*_{10 }with at least 10 digits. )

- Let = 1. Use Steffensen’s method to find and. (You should use MATLAB to do this. But just report and, with at least 10 digits. )
- Show that √3 is a fixed point of
*g*. Which of*p*_{10 }and gives a better approximation to this fixed point?

- (6 marks) Let
*f*(*x*) =*x*^{4 }+ 3*x*^{3 }− 4*x*^{2 }+ 3*x*− 2.

Find *f*(1*.*6) and *f*^{′}(1*.*6) by Horner’s method using a table form (with the help of a calculator).

- (5 marks) Let
*f*(*x*) =*x*^{3 }−*x*−1 and*x*_{0 }= 1*.*1*,x*_{1 }= 1*.*2*,x*_{2 }= 1*.*3 be approximations to a solution of*f*(*x*) = 0. Use Mu¨ller’s method to find the next approximation*x*_{3}. - (5 marks) Suppose that Newton’s method is applied to find the solution
*p*= 0 of the equation = 0. It is known that, starting with any*p*_{0 }*>*0, the sequence {*p*} produced by the Newton’s method is monotonically decreasing (i.e.,_{n}*p*_{0 }*> p*_{1 }*> p*_{2 }*>*··) and converges to 0.

Prove that {*p _{n}*} converges to 0 linearly with rate 2

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