Answer to Question #260266 in Real Analysis for sam

Question

Check the convergence of the sequence defined by 𝑢𝑛+1 = 1 2 (𝑢𝑛 + 𝑎 𝑢𝑛 ) , 𝑎 > 0. Note that this is the sequence associated with finding the square root of a number 𝑎 > 0 by the Newton’s method

Answer 1

Let the sequence $u_n$ converges to $b.$ .

Then,

$$u_{n+1}=\frac{1}{2}(u_n+\frac{a}{u_n})$$

$$b=\frac{1}{2}(b+\frac{a}{b})\\$$

Solving this equation,we get

$$b=\frac{1}{2}(b+\frac{a}{b})\\$$

$$2b^2=b^2+a$$ $$b^2=a$$

$$b=\pm\sqrt{a}$$

Given that $a>0.$ Then $b>0=>b=\sqrt{a}.$

Therefore the sequence defined by

$$u_{n+1}=\frac{1}{2}(u_n+\frac{a}{u_n}), a>0$$

converges to $\sqrt{a}.$