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Quiz #2
CSC2800 (Spring Term 2007)
Full score: 10 % + 1% bonus
Name:_______________________________ (Student ID: ________________________)
Problem 1:
The values of f(x) at 5 different points are plotted in the above graph. The horizontal line in the graph represents
the x-axis.
(a) Suppose the False-Position method is used to approximate the zero of f(x), and x1 and x5 are selected as the
two initial points. Show the location of the estimated root after one iteration in the graph. You may need a
ruler in order to locate the estimated root.
[1%]
(b) Suppose f(x) is continuous and contains N roots between x1 and x5, and none of the roots is a multiple-root
(roots that overlap). What can you say about N besides the fact that N ≥ 1?
[1%]
N is an odd number.
Problem 2:
Suppose R is a positive number. When the sequence defined by the iterative formula xn1  xn (2  Rx n )
converges to a non-zero value, what is the value of xn?
[2%]
When the sequence converges, xn+1 = xn, thus
xn1  xn (2  Rx n )  xn  xn (2  Rx n )  1  (2  Rx n )  0  1  Rx n  Rx n  1  xn  1 / R
Problem 3:
Let f ( x )  x 3  R , where R is a positive number. Based on the Newton-Raphson method, derive the updating
formula for finding the zero of f(x).
f ( xn )
x  R 3x  x n  R 2 x n  R
 xn  n 2  n

2
2
f ' ( xn )
3x n
3x n
3x n
3
xn1  xn 
[2%]
3
3
3
Problem 4:
Each sequence of numbers in part (a) and (b) represents the errors, ei = xi+1 – xi (i = 1, 2, 3, …), of some converging
sequence xi. Estimate the convergent rate of the corresponding sequence.
(a) 1.05-1, 1.05-3, 1.05-5, 1.05-7, 1.05-9, …, 1.05-(2i-1), …
[1%]
log 1.05 ( 2i 1)
 (2i  1) log 1.05
2i  1
 lim
 lim
1

(
2
i

1
)
i  log 1.05
i   ( 2i  1) log 1.05
i  2i  1
Thus the sequence is said to converge linearly.
p  lim
(b) 10, 9, 8, 7, 5, 2, 1, 10-1, 10-2, 10-4, 10-8, 10-16, 10-32, 0
[1%]
log 1032  32 log 10

2
log 1016  16 log 10
Thus the sequence is said to converge quadratically.
p
Problem 5:
(a) State an advantage of using the Secant method over the Newton-Raphson method in roots finding. [1%]
With the Secant method, there is no need to derive f’(x) directly.
(b) State an advantage of using the Bisection method over the Newton-Raphson method in roots finding. [1%]
With the Bisection method, there is no need to derive f’(x) directly.
With the Bisection method, convergence is guaranteed.
Bonus:
e1sin x  e cos x
Find the zero of f ( x ) 
in [-2, 2] accurate to 8 decimal places. Assume π = 3.141592659.
x  2 x 2 e sin x
2
2
2
e1sin x  e cos x
f ( x)  0 
 0  e1sin x  e cos x  0  e1sin x  e cos x  1  sin x  cos 2 x  1  sin x  1  sin 2 x
2 sin x
x  2x e
2
 sin x  sin x  0  sin x(sin x  1)  0  sin x  0 or sin x  1  x  0 or x   / 2
2
Since f(0) does not exist, the only zero in [-2, 2] is –π / 2 or -1.57079633.