## Description

1. Consider the following non-linear equation:

f(x) := x

2 − 0.7x = 0 on [0.5, 1]

(a) Show that f(x) has exactly one root on [0.5,1] without solving the equation.

(b) Consider the bisection algorithm starting with the interval [0.5, 1], i.e. consider

[a1, b1] = [0.5, 1] and p1 = 0.75. Find the minimum number of iterations required

to approximate the solution with an absolute error of less than 10−5

.

2. Suppose f(x) is continuous on [a, b], and f(x) ∈ [a, b] for any x ∈ [a, b]. Show that f has

at least one fixed point on [a, b]. (Hint: use Intermediate Value Theorem.)

3. Given the following sequence {pn}∞

n=0:

(

pn+1 =

p

2

n+3

2pn

p0 is given.

(a) Calculate p1 and p2 with p0 = 3.

(b) Find all the possible limits of sequence {pn}∞

n=0 (for all possible p0).

(c) Show by definition that the given sequence is actually a sequence generated by Newton’s method to find a solution of the equation x

2 − 3 = 0.

4. Consider the following non-linear equation:

f(x) := x

2 − 3 = 0 on [0, 4]

(a) Write the formula for the next term in the sequence in terms of the two previous

terms for the secant method pn to solve the above equation, with the starting point

at p0 =

1

2

and p1 = 3, and calculate p2 and p3.

(b) Calculate p2 and p3 in the sequence of the method of false position pn to solve the

above equation, with the starting point at p0 =

1

2

and p1 = 3.

1

## 5. (Programming problem)

Solve the following equation by the following methods:

sin(x) − x = 0 on [0,

π

2

]

(a) Write a complete program to implement the secant method to solve the equation

within an accuracy of at least 10−5

(using |pn −pn−1|) with initial values p0 and p1 as

π

4

and 3π

8

respectively. Plot the graph of the sequence {pn} generated by the secant

method.

(b) Write a complete programme to implement the Newton’s Method to solve the equation within an accuracy of at least 10−5

(using |pn − pn−1|) with initial value p0 as

π

4

. Plot the graph of the sequence {pn} generated by Newton’s method.

2