Question

Find a positive root of cos x =x2 by using Newton Raphson
method.

Answer 1

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I will approximate cosine by Taylor series, 1 - x²/2. Solving the quadratic, I get an approximate answer, x0 = .73205. There are other ways of getting a starting seed: graph, trial values, bisection, etc. The better the seed, the fewer iterations.

Then I set up Newton’s method

f(x) = x - cos(x)

f’(x) = 1 + sin(x)

x(n) = x(n-1) - f(x(n-1))/f’(x(n-1)

x1 = 0.739 096 140

x2 = 0.739 085 1332