How to Solve a Nonlinear Equation Using Newton's Method?

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SUMMARY

The discussion focuses on solving the nonlinear equation 2.5x^3 = 2cos(x) + 8 using Newton's Method. The equation is rearranged into the form f(x) = 0, resulting in f(x) = 2cos(x) + 8 - 2.5x^3. The user successfully applies Newton's Method, yielding approximate solutions x1 = 1.508306, x2 = 1.48451, and x3 = 1.48416, confirming the effectiveness of the method for this problem.

PREREQUISITES
  • Understanding of Newton's Method for root-finding
  • Familiarity with calculus concepts, specifically derivatives
  • Ability to manipulate equations into the form f(x) = 0
  • Basic knowledge of trigonometric functions, particularly cosine
NEXT STEPS
  • Study the derivation and application of Newton's Method in detail
  • Learn about the convergence criteria for Newton's Method
  • Explore alternative methods for solving nonlinear equations, such as the Bisection Method
  • Investigate the role of numerical analysis in solving complex equations
USEFUL FOR

Students in mathematics or engineering courses, educators teaching numerical methods, and anyone interested in solving nonlinear equations using computational techniques.

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Homework Statement


I've attached the question


Homework Equations


x(n+1) = x(n) - f(x(n)) / f '(x(n))


The Attempt at a Solution


2.5x^3 =2cos(x) + 8
okay so this has to be in the form of f(x) = ... but there is no y variable for me to make it into that form. once I've gotten it into that form, it should be easy but its the manipulation that's got me stuck.
0 =2cos(x) + 8 - 2.5x^3
 

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You want it in the form of f(x) = 0, and you have the equation 0 = 2cos(x) + 8 - 2.5x^3, so f(x) = 2cos(x) + 8 - 2.5x^3
 
oh i didnt know that, i got an answer of x1=1.508306, x2=1.48451 and x3=1.48416 looks right
 

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