(a) Assume that f(x) could be written as f(x) = (x-x₂)²q(x), where q(xr) 0 (i) Show that f(xr) = f'(xr) = 0 (ii) Show that µ(xr) = 0 but µ'(xr) ‡ 0. (iii) The results above show that finding the root of f(x) is equivalent to finding the root of μ(x). Hence, show that a modified version of the Newton-Raphson method that can be used in order to achieve quadratic convergence is f(x₁) f'(x₂) [f′(æi)]? – f(xi)f"(x) (iv) Use Eqs. (2.19) and (2.21) to obtain the root of ƒ(x) = e(x−¹) — x. (b) Solve (2.20) Xi+1 = Xi (2.21) (2.22) You might like to plot the function to see how it looks like. Start with an initial guess of xo 5. Which of the two methods give you faster convergence? The exact answer is r = 1. = x³ + 4x −9=0 (2.23) using both Eq. (2.19) and Eq. (2.21). Start with an initial guess of co = 5. (c) Compare the convergence and discuss the advantages and disadvantages of each schemes.
(a) Assume that f(x) could be written as f(x) = (x-x₂)²q(x), where q(xr) 0 (i) Show that f(xr) = f'(xr) = 0 (ii) Show that µ(xr) = 0 but µ'(xr) ‡ 0. (iii) The results above show that finding the root of f(x) is equivalent to finding the root of μ(x). Hence, show that a modified version of the Newton-Raphson method that can be used in order to achieve quadratic convergence is f(x₁) f'(x₂) [f′(æi)]? – f(xi)f"(x) (iv) Use Eqs. (2.19) and (2.21) to obtain the root of ƒ(x) = e(x−¹) — x. (b) Solve (2.20) Xi+1 = Xi (2.21) (2.22) You might like to plot the function to see how it looks like. Start with an initial guess of xo 5. Which of the two methods give you faster convergence? The exact answer is r = 1. = x³ + 4x −9=0 (2.23) using both Eq. (2.19) and Eq. (2.21). Start with an initial guess of co = 5. (c) Compare the convergence and discuss the advantages and disadvantages of each schemes.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:Exercise 2.11:
The formula for Newton-Raphson method is given by
Xi+1 = Xi
(2.19)
which can be used to find the root, xr, of a function, f(x). It has been been
shown in lectures that the Newton-Raphson method has quadratic convergence
providing that f'(x) ‡ 0. In cases where f'(x) = 0 one will not be able to
achieve quadratic convergence with Eq. (2.19). In order to achieve quadratic
convergence for cases where f'(x) = 0, we have to define another function, µ(x),
where
µ(x) =
=
f (x₂)
f'(xi)
-
f(x)
f'(x)
![(a) Assume that f(x) could be written as
f(x) = (x − x)²q(x),
where q(x) 0
(i) Show that f(x₁) = f'(x₁) = 0
(ii) Show that u(x₁) = 0 but µ'(x) ‡ 0.
µ(xr)
(iii) The results above show that finding the root of f(x) is equivalent
to finding the root of u(x). Hence, show that a modified version of
the Newton-Raphson method that can be used in order to achieve
quadratic convergence is
Xi+1 = Xi
f(i)f(i)
[f′(xi)]? – f(xi)f"(i)
m(xr)
(iv) Use Eqs. (2.19) and (2.21) to obtain the root of
(b) Solve
=
(2.20)
f(x) = e(x-¹) — x.
e(x−1)
(2.22)
You might like to plot the function to see how it looks like. Start with
an initial guess of xo 5. Which of the two methods give you faster
convergence? The exact answer is x = 1.
3
x³ + 4x − 9=0
(2.21)
(2.23)
using both Eq. (2.19) and Eq. (2.21). Start with an initial guess of xo = 5.
(c) Compare the convergence and discuss the advantages and disadvantages of
each schemes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12fdac56-4b52-4434-8b00-5772085cd8ad%2F5d7b56ec-d72a-44fd-bb5d-b20866f9cbe7%2F8jzmxdr_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(a) Assume that f(x) could be written as
f(x) = (x − x)²q(x),
where q(x) 0
(i) Show that f(x₁) = f'(x₁) = 0
(ii) Show that u(x₁) = 0 but µ'(x) ‡ 0.
µ(xr)
(iii) The results above show that finding the root of f(x) is equivalent
to finding the root of u(x). Hence, show that a modified version of
the Newton-Raphson method that can be used in order to achieve
quadratic convergence is
Xi+1 = Xi
f(i)f(i)
[f′(xi)]? – f(xi)f"(i)
m(xr)
(iv) Use Eqs. (2.19) and (2.21) to obtain the root of
(b) Solve
=
(2.20)
f(x) = e(x-¹) — x.
e(x−1)
(2.22)
You might like to plot the function to see how it looks like. Start with
an initial guess of xo 5. Which of the two methods give you faster
convergence? The exact answer is x = 1.
3
x³ + 4x − 9=0
(2.21)
(2.23)
using both Eq. (2.19) and Eq. (2.21). Start with an initial guess of xo = 5.
(c) Compare the convergence and discuss the advantages and disadvantages of
each schemes.
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