Let f(x) = 2x3 – 0.0064x2 – 1.5x + 6 (a) Indicate that f(x) = 0 has a root in the interval [-3, 0] with |bo – ao| = 1. (b) Examine which method is less iteration (i) to find the root in Q2 (a) by using Secant and Newton-Raphson method. Stop iteration if |f(xi)| < 0.0005 ori>7.
Let f(x) = 2x3 – 0.0064x2 – 1.5x + 6 (a) Indicate that f(x) = 0 has a root in the interval [-3, 0] with |bo – ao| = 1. (b) Examine which method is less iteration (i) to find the root in Q2 (a) by using Secant and Newton-Raphson method. Stop iteration if |f(xi)| < 0.0005 ori>7.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Q2 Let f(x) = 2x3 – 0.0064x2 – 1.5x + 6
(a) Indicate that f(x) = 0 has a root in the interval [-3, 0] with |bo – aol = 1.
(b) Examine which method is less iteration (i) to find the root in Q2 (a) by using Secant
and Newton-Raphson method. Stop iteration if |f(x)| < 0.0005 or i > 7.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bd5699c-9f23-4bc6-98a2-fb84f0f61568%2Fcc7d93fb-c59a-482f-a968-d92276b6e0ed%2Fm7o6jn4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q2 Let f(x) = 2x3 – 0.0064x2 – 1.5x + 6
(a) Indicate that f(x) = 0 has a root in the interval [-3, 0] with |bo – aol = 1.
(b) Examine which method is less iteration (i) to find the root in Q2 (a) by using Secant
and Newton-Raphson method. Stop iteration if |f(x)| < 0.0005 or i > 7.
![LIST OF FORMULA
a, +b,
i = 0,1,2,...
2
Bisection method:
Secant method:
Xi+2
, i = 0,1,2 ..
f(xi+1) - f(x)
.....
Newton-Raphson method: x = x,
f(x,)
.i = 0,1,2....
f"(x,)
f.0 = fi,i = 0,1,..n
fU-1] - fj-1
Newton's interpolation:
%3D
j = 1,2,..,n
%3D
Xi+j - Xị
P,(x) = f0 + f"(x-x)+ f"(x-x, )(x- x)++f"(x-x, )(x-).-(x-x)
Lagrange interpolation: P,(x) = L,(x)f, for k 0,1,2,3,.,n with L(x)= -*,)
1-6 (x, -x)
i=0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5bd5699c-9f23-4bc6-98a2-fb84f0f61568%2Fcc7d93fb-c59a-482f-a968-d92276b6e0ed%2Fqsx2sit_processed.jpeg&w=3840&q=75)
Transcribed Image Text:LIST OF FORMULA
a, +b,
i = 0,1,2,...
2
Bisection method:
Secant method:
Xi+2
, i = 0,1,2 ..
f(xi+1) - f(x)
.....
Newton-Raphson method: x = x,
f(x,)
.i = 0,1,2....
f"(x,)
f.0 = fi,i = 0,1,..n
fU-1] - fj-1
Newton's interpolation:
%3D
j = 1,2,..,n
%3D
Xi+j - Xị
P,(x) = f0 + f"(x-x)+ f"(x-x, )(x- x)++f"(x-x, )(x-).-(x-x)
Lagrange interpolation: P,(x) = L,(x)f, for k 0,1,2,3,.,n with L(x)= -*,)
1-6 (x, -x)
i=0
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