Solution to Computational Math II
Assignment 3
Department of Applied Mathematics
I. Section 7.1
17.Establish a formula of the form
00
fn ≈ [Afn+3 + Bfn+2 + Cfn+1 + Dfn ]/h2
Here fn+i = f (xn + ih).
Solution I:
By Taylor expansion, we have
h2 00
h3 000
f (xn ) + f (xn ) + O(h4 )
2
6
4h3 000
0
00
f (xn ) + O(h4 )
fn+2 = f (xn ) + 2hf (xn ) + 2h2 f (xn ) +
3
9h2 00
9h3 000
0
fn+3 = f (xn ) + 3hf (xn ) +
f (xn ) +
f (xn ) + O(h4 )
2
2
From the problem, we have the following linear system:


A+B+C +D =0
A = −1






 3A + 2B + C = 0
 B=4
⇒
 9A + 4B + C = 2
C = −5






27A + 8B + C = 0
D=2
0
fn+1 = f (xn ) + hf (xn ) +
Hence, we have the form
00
fn ≈ [−fn+3 + 4fn+2 − 5fn+1 + 2fn ]/h2 .
Solution II: Using equations 01-03, we have
fn+1 + fn+2 − fn+3 − fn = −2f 00 (xn )h2 + O(h2 ),
which implies
f 00 (xn ) ≈
So, A = D = 12 and B = C = − 12 .
II. Section 7.2
8. Find the formula
Z
fn+1 + fn+2 − fn+3 − fn
.
−2h2
1
f (x)dx ≈ A0 f (0) + A1 f (1)
0
(0.1)
(0.2)
(0.3)
2
that is exact for all functions of the form f (x) = aex + b cos(πx/2).
Solution:
First, we integral the function on [0,1], then
Z
1
Z
1
f (x)dx =
0
0
1
1
2b
2
[aex + b cos(πx/2)]dx = aex +
sin(πx/2) = a(e − 1) + b.
π
π
0
0
And we have f (0) = a + b, f (1) = ae. Then we have the equation
a(e − 1) +
2
b = A0 (a + b) + A1 ae = (A0 + A1 e)a + A0 b.
π
Then we got the linear system
(
(
A0 = π2
A0 + A1 e = e − 1
⇒
A1 = 1 −
A0 = π2
1
e
−
2 .
eπ
31. Determine the minimum number of subintervals need to approximate
Z 2
2
(x + e−x )dx
1
to an accuracy of at least
1
2
×
10−7
using the trapezoid rule.
Solution:
f (x) = x + e
√
−x2
00
2 −x2
⇒ f (x) = 4x e
− 2e
−x2
00
00
x ∈ [1, 2] ⇒ max f = f (
x∈[1,2]
3
6
) = 4e− 2 .
2
By the accuracy of trapezoid rule, we know the the error is
00
−
f (ξ)
(b − a)h2
12
where b = 2, a = 1, ξ ∈ [1, 2]. As we want the accuracy of at least
1
2
× 10−7 , then
00
f (ξ)
1
(b − a)h2 | ≤ × 10−7
12
2
1 −3
(2 − 1)2
1
⇔ 4e 2 · (2 − 1) ·
≤ × 10−7
2
12
n
2
r
2 3
2 −3
⇔ n2 ≥ e− 2 · 107 ⇔ n ≥
e 2 · 107 ≈ 1219.6
3
3
|−
Hence, the minimal number of subintervals is 1220.
III. Section 7.2
R1
R1
16. We intend to use 0 p(x)dx as an estimated of 0 f (x)dx, where p is a polynomial of degree
n that interpolates f at nodes x0 , x1 , . . . , xn in [0, 1]. Assume that |f (n+1) (x)| < M on [0, 1] what
3
R1
R1
upper bound can be given for the error| 0 f (x)dx − 0 p(x)dx| if nothing is known about the location of the nodes? Can you find the best upper bound?
Solution:
By Interpolation polynomial Error Theorem, we have the error term
n
f (x) − p(x) =
Y
1
f (n+1) (ξx ) (x − xi ), ξx ∈ [0, 1], x ∈ [0, 1].
(n + 1)!
i=0
Hence the upper bound at least can be
# Z 1 "
Z 1
Z 1
n
Y
1
(n+1)
=
p(x)dx
f
(ξ
)
f
(x)dx
−
(x
−
x
)
dx
x
i
(n + 1)!
0
0
0
i=0
Z 1
n
Y
1
(n+1)
f
(ξ
)
(x
−
x
)
≤
x i dx
(n
+
1)!
0
i=0
Z 1
M
1dx (as|x − xi | ≤ 1)
≤
(n + 1)! 0
M
=
.
(n + 1)!
Now, we want to have the best upper bound, then we can choose Chebyshev points to interpolate
f (x). Then we use interval change, x = 12 + 12 t, t ∈ [−1, 1] and let xi = 12 + 12 ti , i = 0, 1, . . . , N, where
n
Q
ti are Chebyshev points. Then we have x − x(i) = 21 (t − t(i)), and we know that | (t − ti )| = 21n .
i=0
Hence, the best bound can be
Z 1
Z
f
(x)dx
−
0
0
1
# Z 1 "
n
Y
1
(n+1)
f
(ξx ) (x − xi ) dx
p(x)dx = 0 (n + 1)!
i=0
Z "
#
n
1
Y
1
(t − ti )
=
f (n+1) (ξx )
dx
0 (n + 1)!
2
i=0
1
M
2n+1 (n + 1)! 2n
M
= 2n+1
2
(n + 1)!
≤
IV. Section 7.3
7. a) Find a formula of the form
Z
1
xf (x)dx ≈
0
n
X
Ai f (xi )
i=0
with n = 1 that is exact for all polynomials of degree 3.
Solution:
We want to the formula is exact for all polynomials of degree 3, by the theorem of Gaussian
4
Quadratue let q(x) = x2 + bx + c, q0 (x) = 1, q1 (x) = x desperately, we need to q(x) is w-orthogonal
q0 , q1 where w(x) = x from the problem. Hence ,we obtain that
Z 1
Z 1
1 4 b 3 c 2 1 1 b
c
2
(x + bx + c) · 1 · xdx = x + x + x = + + = 0,
q(x)p0 (x)w(x)dx =
4
3
2 0 4 3 2
0
0
Z
1
1
Z
(x2 + bx + c) · x · xdx =
q(x)p1 (x)w(x)dx =
0
0
− 65
Solve that we have b =
and c =
3
. Substitute to q(x)
10 √
√
6− 6
and x1 = 6+10 6 .
10
Solve that, we obtain the roots x0 =
f0 (x) = 1 and f1 (x) = x. Then we have
(
A0 + A1√=
√
6− 6
10 A0
1 5 b 4 c 3 1 1 b
c
x + x + x = + + = 0.
5
4
3 0 5 4 3
+
1
2
6+ 6
10 A1
=
1
3
⇒
= x2 + bx + c = x2 − 6/5x + 3/10 = 0.
Now we want to solve A0 and A1 Let
(
A0 =
A1 =
√
9− 6
36
√
9+ 6
36
Hence, the formula is
Z
0
1
√
9− 6
f
xf (x)dx ≈ A0 f (x0 ) + A1 f (x1 ) =
36
√ !
√
6− 6
9+ 6
f
+
10
36
V. Section 7.3
22. Show how the Gaussian quadrature rule
r !
Z 1
5
8
5
3
f (x)dx ≈ f −
+ f (0) + f
9
5
9
9
−1
√ !
6+ 6
.
10
r !
3
5
Rb
can be used for a f (x)dx. Apply this result to evaluate
R4
(b) 0 sint t dt
Solution:
Using change of intervals, we have
t=
Then we have
Z 4
0
sin t
dt =
t
Z
4−0
0+4
x+
= 2x + 2 where x ∈ [−1, 1].
2
2
1
sin 2x + 2
2dx
−1 2x + 2
"
!
r !
5
3
8
5
=2
·f 2· −
+ 2 + · f (2) + · f
9
5
9
9
≈ 2 × (0.5369 + 0.4041 − 0.062)
≈ 1.758
VII. Section 3.2
r
2×
3
+2
5
!#
5
23Perform two iterations of Newton’s method on these systems:
(a) Starting with (0,1)
(
4x21 − x22 = 0
4x1 x22 − x1 = 1
Solution:
First, we have f1 (x) = 4x21 − x22 , f2 (x) = 4x1 x22 − x1 − 1, then we derive Jacobian matrix
!
8x1
−2x2
J=
4x22 − 1 8x1 x2
As x(0) = (0, 1)T , we have f (x(0) ) = (−1, −1)T . Then we need to solve J(x(0) )h(0) = −f (x(0) ).
!
!
!
!
1
1
0 −2
1
3
h=
⇒h=
⇒ x(1) = x(0) + h(0) = 31
3 0
1
− 12
2
Then we have
8
3
0
!
−1
4
3
h=
7
− 36
1
!
⇒h=
5
24
3
4
!
⇒ x(2) = x(1) + h(1) =
VIII. Section 3.2
14.
( Use Newton method to solve the nonlinear system.
4y 2 + 4y + 52x = 19
a)
169x2 + 3y 2 + 111x − 10y = 10
Solution:
The Jacobian matrix is
!
52
8y + 4
J=
338x + 111 6y − 10
Use the Matlab code, we have the solution at tolerance τ = 1e − 9:
(
x = 0.1342
y = 1.3043
13
24
5
4
!
6
Appendix
code for 14.a)
clear a l l ;
clc ;
% define i n i t i a l point .
x0 = [ 0 ; 1 ] ;
N=100;
x=x0 ;
% define tolerence ;
t a o=1e −9;
for i =1:N;
J =[52 8∗ x ( 2 ) ; 338∗ x (1)+111 6∗ x ( 2 ) − 1 0 ] ;
f =−[4∗x (2)ˆ2+4∗ x (2)+52∗ x (1) −19;
169∗ x (1)ˆ2+3∗ x (2)ˆ2+111∗ x (1) −10∗ x ( 2 ) − 1 0 ] ;
h=J\ f ;
xs=x+h ;
i f norm( xs−x)<1e −8;
break ;
else
x=xs ;
end ;
end ;