Feedback Math Example 4

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Example 4

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Newton's method for finding 2\sqrt{2} uses xn+1=12(xn+2xn)x_{n+1} = \frac{1}{2}\left(x_n + \frac{2}{x_n}\right). Starting from x0=1x_0=1, compute x1,x2,x3x_1, x_2, x_3 and compare to 2โ‰ˆ1.41421\sqrt{2}\approx1.41421.

Solution

  1. 1
    x1=12(1+2/1)=1.5x_1 = \frac{1}{2}(1+2/1) = 1.5.
  2. 2
    x2=12(1.5+2/1.5)=12(1.5+1.3333)=1.4167x_2 = \frac{1}{2}(1.5+2/1.5) = \frac{1}{2}(1.5+1.3333)=1.4167.
  3. 3
    x3=12(1.4167+2/1.4167)โ‰ˆ12(1.4167+1.4118)=1.41421x_3 = \frac{1}{2}(1.4167+2/1.4167)\approx\frac{1}{2}(1.4167+1.4118)=1.41421. Already accurate to 5 decimal places.

Answer

x1=1.5,x2โ‰ˆ1.4167,x3โ‰ˆ1.41421โ‰ˆ2x_1=1.5, x_2\approx1.4167, x_3\approx1.41421\approx\sqrt{2}
Newton's method is a feedback iteration with quadratic convergence โ€” the number of correct decimal places roughly doubles each step. This makes it extremely efficient for root-finding.

About Feedback

Feedback occurs when the output of a system influences its future input โ€” positive feedback amplifies changes; negative feedback stabilizes them.

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