Recall that Newton’s1 method often provides a good approximation to a root of an equation...

Recall that Newton’s1 method often provides a good approximation to a root of an equation f(x) = 0: An initial guess x0 at a root is improved by calculating a sequence x1, x2, x3,… given by

The same idea can be used for a vector field f: U ⊂ ℝn → ℝn. Suppose we have a point x0 that is close to a solution of the equation f(x) = 0. Define a sequence of points x1, x2,… by

 xi+1 = xi − Jf(xi)−1 f(xi), i = 0, 1, 2 (3.5.12)

It can be shown that if Jf(x) is continuous and invertible at x = a, then there is an open set U containing a such that for x0 ∈ U, the sequence given by (3.5.12) converges to a.

(a) Let f(x, y) = (x2 + y2 − 1, 3xy − 1) and x0 = (0.333, 1). Use (3.5.12) to obtain the next three points in the sequence of approximation for the roots of f(x, y) = 0.

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(b) Compute x1, x2, and x3 if x0 = (1.5, 0) and verify that f(x3) ≈ 0.

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(c) Find two other solutions to f(x, y) = 0 either by Newton’s method or other means.

1Isaac Newton (1642−1727), British mathematician and physicist, was one of the inventors of differential and integral calculus and is regarded as one of the greatest minds of all time.