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Newton’s method for a square root · Part 3 of 3

You compute 10\sqrt{10} with Newton’s method on f(x)=x2−10f(x) = x^2 - 10, which gives the iteration xk+1=12(xk+10/xk)x_{k+1} = \tfrac{1}{2}\left(x_k + 10/x_k\right), starting from x0=3x_0 = 3.

You reuse the solver on g(x)=(x−3)2g(x) = (x - 3)^2, whose root at 3 is a double root, starting from x0=4x_0 = 4. How does it converge?

  1. AQuadratically, as for the square root
  2. BLinearly: the error halves on every step
  3. CIt diverges, because g′g' is zero at the root
  4. DIt lands on the root in one step, because Newton’s method is exact on any quadratic polynomial

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Newton’s method for a square root, part 3 of 3 · QuantMax