Exam (elaborations)

Solutions for Classical and Modern Numerical Analysis, 1st Edition by Ackleh (All Chapters included)

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Classical and Modern Numerical Analysis 1e Ackleh

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Classical And Modern Numerical Analysis 1e Ackleh

Complete Solutions Manual for Classical and Modern Numerical Analysis, 1st Edition by Azmy S. Ackleh; Edward James Allen; R. Baker Kearfott; Padmanabhan Seshaiyer ; ISBN13: 9781420091571....(Full Chapters included)...1.Mathematical Review and Computer Arithmetic 2.Numerical Solution of Nonlinear E...

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Institution Classical and Modern Numerical Analysis 1e Ackleh
Course Classical and Modern Numerical Analysis 1e Ackleh

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INSTRUCTOR’S
ANSWER GUIDE FOR
Classical and Modern
Numerical Analysis:
Theory, Methods
and Practice

by
Azmy S. Ackleh
Edward J. Allen
R. Baker Kearfott
Padmanabhan Seshaiyer

** Immediate Download
** Swift Response
** All Chapters included

, Instructor’s Answer Guide
for
Classical and Modern Numerical Analysis:
Theory, Methods, and Practice
Azmy S. Ackleh, University of Louisiana at Lafayette
Edward J. Allen, Texas Tech University
R. Baker Kearfott, University of Louisiana at Lafayette
Padmanabhan Seshaiyer, George Mason University
January 4, 2010

Contents
Acknowledgments 2

Chapter 1 3

Chapter 2 16

Chapter 3 35

Chapter 4 58

Chapter 5 81

Chapter 6 91

Chapter 7 101

Chapter 8 122

Chapter 9 144

Chapter 10 155

1

,Chapter 1
1. (a) Let f (x) = ex , for x ∈ (−∞, 0]. We have f ∈ C 1 (−∞, 0] and, by
assumption, y ∈ (−∞, 0] with y ≤ x. We may therefore use the Mean
Value Theorem to obtain that there exists a ζ ∈ [y, x] ⊆ (−∞, 0], such
that f (x) − f (y) = eζ (x − y). Hence,
|ex − ey | = |eζ ||x − y| ≤ |e0 ||x − y| = |x − y|, ∀x, y ≤ 0.

(b) Let f (x) = xp , for x ≥ 0, and p ≥ 1. Then, f ∈ C 1 [0, ∞). Suppose
0 ≤ y ≤ x, The Mean Value Theorem then implies ∃ζ ∈ [y, x] ⊆
[0, ∞) such that
f (x) − f (y) = f (ζ)(x − y).
Since g(x) = xp−1 for x ≥ 0 and p ≥ 1 is an increasing function, we
have y p−1 ≤ ζ p−1 ≤ xp−1 . Thus,
py p−1 ≤ xp − y p ≤ pxp−1 (x − y)
for 0 ≤ y ≤ x and p ≥ 1.
(c) The solution is already available in the Appendix of the book.
(d) Since f ′ (x) 6= 0, ∀x ∈ (a, b), either f ′ (x) < 0 or f ′ (x) > 0, ∀x ∈ (a, b).
Since f ′ ∈ C[a, b], f is a continuous monotone function on [a, b].
Thus, f (x) can vanish at most one point in [a, b].
2. An answer is
x2 x4 x6
p(x) = 1 −
+ − .
3! 5! 7!
To see why, observe that, when x 6= 0, Taylor’s Theorem gives

x3 x5 x7 ξ9
sin(x) x− + − +
sinc(x) = = 3! 5! 7! 9! ,
x x
where ξ is between 0 and x. Thus,
ξ8
|sinc(x) − p(x)| ≤ .
9!
Then, −0.2 ≤ x ≤ 0.2 implies
ξ8 (0.2)8
≤ = 7.05 × 10−12 .
9! 9!

3. By Taylors Theorem,
Z x+h
h2 ′′ 1
f (x + h) = f (x) + hf ′ (x) + f (x) + (x + h − t)2 f ′′′ (t)dt,
2! 2! x
Z x−h
h2 1
f (x − h) = f (x) − hf ′ (x) + f ′′ (x) + (x − h − t)2 f ′′′ (t)dt.
2! 2! x

3

, Subtracting the second equation from the first, dividing by 2h, and sim-
plifying then gives
f (x+h)−f (x−h)
2h − f ′′ (x)
Z x+h Z x−h
1 2 ′′′
= (x + h − t) f (t)dt − (x − h − t)2 f ′′′ (t)dt
4h x x
1
≤ max |f ′′′ (t)|
4h x−h≤t≤x+h
Z x−h Z x−h
· (x − h − t)2 dt + (x − h − t)2 dt
x x
3

1 2h
= max |f ′′′ (t)| · = ch2 ,
4h x−h≤t≤x+h 3
1
where c = max |f ′′′ (t)|.
6 x−h≤t≤x+h
4. This is similar to Problem 3. By Taylor’s Theorem, since f has a contin-
uous fourth derivative, we have

h2 ′′ h3
f (x + h) = f (x) + hf ′ (x) + f (x) + f ′′′ (x)
2! 3!
Z
1 x+h
+ (x + h − t)3 f (4) (t)dt,
3! x
h2 (−h)3 ′′′
f (x − h) = f (x) + (−h)f ′ (x) + f ′′ (x) + f (x)
2! 3!
Z x−h
1
+ (x − h − t)3 f (4) (t)dt.
3! x

Combining these two equations according to the difference quotient gives

f (x + h) − 2f (x) + f (x − h)
− f ′′ (x)
h2
Z x+h Z x−h
1
= 2 (x + h − t)3 f (4) (t)dt + (x − h − t)3 f (4) (t)dt .
6h x x

4

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