lease write a Pyhton code for NR iteration solution of the previous function. Implementation of the derivative should be an approximation through limit theorem.
lease write a Pyhton code for NR iteration solution of the previous function. Implementation of the derivative should be an approximation through limit theorem.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Question
fx = 3 - x**3 - x**5 - 2.7183**-x
NR iteration on the addion picture part ı uploaded example
with a precision of 4digits and the inital guess of 2.0000
Please write a Pyhton code for NR iteration solution of the previous function. Implementation of the derivative should be an approximation through limit theorem.
![Xo = 5.0000
f (xo)
= 5
f'(xo)
f (5)
= 5
f'(5)
121.272
X1 = X0
= 3.4009
75.8391
f (x1)
= 3.4009
f'(x1)
f(3.4009)
35.0884
X2 = X1
= 3.4009
= 2.3637
f'(3.4009)
33.8305
f(x2)
f'(x2)
f(x3)
f'(x3)
f (2.3637)
f'(2.3637)
f(1.7841)
f'(1.7841)
9.7069
X3 = X2 -
= 2.3637
= 2.3637
= 1.7841
16.7468
1.8859
X4 = X3
= 1.7841
= 1.7841
= 1.6038
10.4596
f(x4)
f'(x4)
f(1.6038)
0.1584
X5 = X4 –
= 1.6038 -
= 1.6038
= 1.5856
f'(1.6038)
8.7145
f (x5)
f'(xs)
f(1.5856)
f'(1.5856)
0.0016
X6 = X5
= 1.5856 -
= 1.5856
= 1.5855
8.5423](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6f97bcc-2b22-49d4-9090-30bfde25b931%2F0e671537-4350-4246-9a3e-fd18b438e4cb%2F0lc150h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Xo = 5.0000
f (xo)
= 5
f'(xo)
f (5)
= 5
f'(5)
121.272
X1 = X0
= 3.4009
75.8391
f (x1)
= 3.4009
f'(x1)
f(3.4009)
35.0884
X2 = X1
= 3.4009
= 2.3637
f'(3.4009)
33.8305
f(x2)
f'(x2)
f(x3)
f'(x3)
f (2.3637)
f'(2.3637)
f(1.7841)
f'(1.7841)
9.7069
X3 = X2 -
= 2.3637
= 2.3637
= 1.7841
16.7468
1.8859
X4 = X3
= 1.7841
= 1.7841
= 1.6038
10.4596
f(x4)
f'(x4)
f(1.6038)
0.1584
X5 = X4 –
= 1.6038 -
= 1.6038
= 1.5856
f'(1.6038)
8.7145
f (x5)
f'(xs)
f(1.5856)
f'(1.5856)
0.0016
X6 = X5
= 1.5856 -
= 1.5856
= 1.5855
8.5423
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