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Consider Newton's Method for calculating √a​ .

(a) Show that the Newton's Method formula is given by xₖ+₁ = 21​ (xₖ​ + a/xₖ),x 0​ >0.
(b) Prove that x k ​≥ a​ for k=1,2,⋯, and the sequence {x₁​ ,x₂​ ,⋯} is decreasing.

User Pdb
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Final answer:

Newton's method for calculating the square root of a number involves iteratively improving the approximate value using a specific formula. The sequence of approximations obtained by this method is always decreasing and the approximations converge to the exact square root.

Step-by-step explanation:

Newton's Method for calculating the square root of a number is given by the formula: xₖ₊₁ = ½(xₖ + a/xₖ). To prove this formula, we start by assuming that √a is equal to xₖ. Then, we solve for a new value, xₖ₊₁, which is a better approximation of √a. We repeat this process until we reach a desired level of accuracy.

To prove that xₖ ≥ a for k=1,2,..., we can start by assuming that xₖ ≥ a for some value of k. We then substitute this assumption into the formula xₖ₊₁ = ½(xₖ + a/xₖ) and try to prove that xₖ₊₁ ≥ a. By simplifying the expression, we can show that xₖ₊₁ - a = (xₖ² - a) / (2xₖ) ≥ 0. Since xₖ² - a ≥ 0 and xₖ > 0, we can conclude that xₖ₊₁ - a ≥ 0, which means xₖ₊₁ ≥ a.

The sequence {x₁, x₂, ...} is decreasing because each successive term, xₖ₊₁, is closer to √a than the previous term, xₖ. This can be proven by comparing xₖ and xₖ₊₁, and showing that their difference, xₖ - xₖ₊₁, is always positive.

User Koders
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