COS2633 EXAM MEMO
MERGED.
university
of south africa
COS2633 NUMERICAL METHODS 1 October/November 2014
SOLUTIONS
QUESTION 1
(1.1)
(a) (3)Write the Newton formula (recurrence relation) to approximate3
√ 5.
3
√ 5 is a solution to the equation x 3 = 5, or equivalently, x3 − 5 = 0. So ...
(1.1)
√3
(a) Write the Newton formula (recurrence relation) to approximate 5. (3)
√3 3 3
5 is a solution to the equation x = 5, or equivalently, x − 5 = 0. So if we apply the Newton’s
method to the function f defined by f (x) = x 3 − 5, provided x 0 is given, we obtain the relation
3
(x n ) − 5
(1) x n+1 = xn − 2
.
3 (xn )
(b) Perform three iterations of the Newton algorithm as given in (1.1(a)), starting
0 = 2.at x (3)
Applying the formula in equation (1), we have:
Iteration 1: 3
(x 0) − 5 23 − 5
x 1 = x0 − 2 = 2 − 3(2)2 = 1.75
3 (x0)
Iteration 2:
3
(x 1) − 5 (1.75)3 − 5
x 2 = x1 − 2 = 1.75 − 2
= 1.7109
3 (x1) 3(1.75)
Iteration 3:
(x 2)3 − 5 (1.7109)3
−5
x 3 = x2 − 2 = 1.7109 − 2
= 1.7100
3 (x2) 3(1.7109)
x2 y2
(1.2) The ellipse defined by + = 1 and the parabola defined by y =2 x− 2x − 4 intersect at four
16 9
distinct points.
(a) Write the Newton formula (recurrence relation) in the form of a system of equations, to(6)
approxi-
mate any of the intersection points.
We define
x2 y2
F (x, y) = 216 + 9 − 1
x − 2x − y − 4
The Jacobian is
x 2y
J F (x, y) = 8 9
2x − 2 −1
and its inverse is
1 −1 − 2y
(J F (x, y))−1 = − 4 x
9 .
x
8 + 9 y(x − 1) −2x + 2 8
The Newton’s formula is given by
x n+1 x −1
= n − (J F (x n , yn )) F (x n , yn )
yn+1 yn
[TURN OVER]
, 2 COS2633
OCTOBER/NOVEMBER 2014
and if we let D n = det JF (x n , yn ) = − x8n − 4
(x n − 1), we obtain
y
9 n
" #
x 2n y n2
x n+1 xn 1 −1 − 2y9n + − 1
= − 16 9
yn+1 yn Dn −2x n + 2 x8n x 2n − 2x n − y n − 4
x 2n y2
xn 1 −1 16 + 9n − 1 − 2y9n x 2n − 2x n − y n − 4
= − 2 2
yn D n (−2x n + 2) x n + y n − 1 + x n x 2n − 2x n − y n − 4
16 9 8
and therefore the required recurrence relation is
h 2
i
y2
x n+1 = x n − D1n −1 x16 n
+ 9n − 1 − 2y9n x 2n − 2x n − y n − 4
h i
x 2n y2
yn+1 = y n − D1n (−2x n + 2) 16 + 9n − 1 + x8n x 2n − 2x n − y n − 4
(b) Perform two iterations of the Newton algorithm as given in (1.2(a)), starting 0, y0)with
= (4,
(x1). (4)
The results are as follows:
Iteration 1: We have D0 = − x80 − 49 y0 (x 0 − 1) = − 48 − 49 × 1 (4 − 1) = −11
6
. Then
h i
x2 y2
x 1 = x 0 − D10 −1 160 + 90 − 1 − 2y90 x 20 − 2x 0 − y 0 − 4
h i
6 42 2
= 4 + 11 −1 16 + 19 − 1 − 29 42 − 8 − 1 − 4
118
= = 3.5758
33
h i
x 20 y 02
y1 = y 0 − D10 (−2x 0 + 2) 16
+ 9
− 1 + x80 x 20 − 2x 0 − y 0 − 4
h i
6 42 2
= 1 + 11 (−2 × 4 + 2) 16 + 19 − 1 + 48 42 − 8 − 1 − 4
16
= 11 = 1.4545
118
Iteration 2: We have D1 = − x81 − 4
y (x 1
9 1
− 1) = − 33
− 4
9
× 16
11
118
33
− 1 = − 471
223
. Then
8
h i
1 x 21 y 12 2y1
x2 =x 1 − D1 −1 16 + 9 −1 − 9
x 21 − 2x 1 − y 1 − 4
" ! #
118 2 16 2 32
118 118 2
= 33
+ 223
471
−1 33
+ 11
−1 − 11
33
−2× 118
33
− 16
11
−4
16 9 9
717
= 203 = 3.5320
h i
x 21 y 12
y2 =y 1 − 1
D1 (−2x 1 + 2) 16
+ 9
− 1 + x81 x 21 − 2x 1 − y 1 − 4
" 2 2
! #
118 16 118
16 223 118 33 11 33 118 2 118 16
= 11
+ 471
−2 × 33
+2 + −1 + 33
−2× 33
− 11
−4
16 9 8
1133
= 804
= 1.4092
(c) What is the error of this process, given the fact that the exact solution is (3 .5316, 1.4088)?
(1)
The error of the process is
p −4
2 + (1.4088 − 1.4092)
e = (3.5316 − 3.5320) 2 = 5.6569 × 10
[17]
QUESTION 2
h i
..
(2.1) By means of
Gaussian Elimination on a larger augmented matrix
A . I where I is the identity (7)
matrix of appropriate size, and by using exact calculation, find the inverse of the matrix
1 2 3
A = 2 3 4 .
3 4 1
(2.2) The SOR algorithm, with relaxation parameter ω for solving a linear system Ax = b is defined by the
formula
i−1
X Xn
ω
x [k]
i
[k−1]
= (1 − ω)xi + bi − aij x [k]
j − aij x [k−1]
j
.
aii
j=1 j=i+1
(a) Perform two iterations of the SOR algorithm to solve the system (4)
1 2 x 8
=
3 1 y 9
with ω = 0.5 starting with0(x
, y0) = (1, 1).
The results are as follows:
Iteration 1: ω
x1 = (1 − ω)x +
[b1 − a 12y0]
0
a11
0.5
= (1 − 0.5)1 + [8 − 2 × 1]
7
1
= 2
[TURN OVER]
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through EFT, credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying this summary from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller LOVELY01. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy this summary for R53,93. You're not tied to anything after your purchase.
