Alexander's question via email about Newton's Method

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SUMMARY

The discussion focuses on applying Newton's Method to solve the equation $\displaystyle \mathrm{e}^{1.2\,x} = 1.5 + 2.5\cos^2{\left( x \right) }$ with an initial estimate of $\displaystyle x_0 = 1$. The equation is rewritten as $\displaystyle f\left( x \right) = \mathrm{e}^{1.2\,x} - 1.5 - 2.5\cos^2{\left( x \right) } = 0$. The derivative is calculated as $\displaystyle f'\left( x \right) = 1.2\,\mathrm{e}^{1.2\,x} + 5\sin{\left( x \right) }\cos{\left( x \right) }$. After three iterations, the approximate solution is $\displaystyle x_3 = 0.81797$.

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  • Knowledge of calculus, specifically derivatives
  • Familiarity with exponential and trigonometric functions
  • Experience with Computer Algebra Systems (CAS)
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  • Study the convergence properties of Newton's Method
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  • Explore the implementation of Newton's Method in programming languages like Python
  • Investigate the use of Computer Algebra Systems (CAS) for solving complex equations
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Mathematicians, engineering students, and anyone interested in numerical methods for solving equations will benefit from this discussion.

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Alexander asks:

Apply three iterations of Newton's Method to find an approximate solution of the equation

$\displaystyle \mathrm{e}^{1.2\,x} = 1.5 + 2.5\cos^2{\left( x \right) } $

if your initial estimate is $\displaystyle x_0 = 1 $.

What solution do you get?
 

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@Prove It answers:

Newton's Method solves an equation of the form $\displaystyle f\left( x \right) = 0 $, so we need to rewrite the equation as

$\displaystyle \mathrm{e}^{1.2\,x} - 1.5 - 2.5\cos^2{\left( x \right) } = 0 $

Thus $\displaystyle f\left( x \right) = \mathrm{e}^{1.2\,x} - 1.5 - 2.5\cos^2{\left( x \right) }$.

Newton's Method is: $\displaystyle x_{n+1} = x_n - \frac{f\left( x_n \right) }{f'\left( x_n \right) } $

We will need the derivative, $\displaystyle f'\left( x \right) = 1.2\,\mathrm{e}^{1.2\,x} + 5\sin{\left( x \right) }\cos{\left( x \right) } $.I have used my CAS to do this problem:

View attachment 9644

View attachment 9645

So after three iterations the root is approximately $\displaystyle x_3 = 0.81797 $.
 
Last edited:

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