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Solve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditio

Solve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditio

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter29: Dc Generators
Section: Chapter Questions
Problem 5RQ: What are interpoles, and what is their purpose?
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Solve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditio
Transcribed Image Text:Solve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditio
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