MHB Alexander's question via email about Newton's Method

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Alexander inquires about applying Newton's Method to solve the equation e^(1.2x) = 1.5 + 2.5cos²(x) starting with an initial estimate of x₀ = 1. The equation is reformulated to f(x) = e^(1.2x) - 1.5 - 2.5cos²(x) = 0. The derivative is calculated as f'(x) = 1.2e^(1.2x) + 5sin(x)cos(x). After three iterations using Newton's Method, the approximate root found is x₃ = 0.81797.
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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 $.
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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