Question

use newton's method to approximate SQRT of 11 to 5 decimal places

Answer 1

Answer on Question #58301 – Math – Algorithms | Quantitative Methods

Question

Use newton's method to approximate SQRT of 11 to 5 decimal places.

Solution

Finding SQRT(S) is the same as solving the equation

$$x ^ {2} - S = 0$$

Therefore, any general numerical root-finding algorithm can be used for solving (1). Let $f(x) = x^{2} - S$ , $f'(x) = 2x$ . In Newton's method we use the following recurrent equation

$$x _ {n + 1} = x _ {n} - \frac {f (x _ {n})}{f ^ {\prime} (x _ {n})}$$ $$x _ {n + 1} = x _ {n} - \frac {x _ {n} ^ {2} - S}{2 x _ {n}}$$ $$x _ {n + 1} = \frac {x _ {n} + \frac {S}{x _ {n}}}{2}$$

Using condition, here $S = 11$ , $n = 0,1,2,3,\ldots$ is a number of iteration step, and $x_{n}$ is the approximation for square root on the $n$ -th step of the iterative process. Substitute for $S = 11$ in formula (2):

$$x _ {n + 1} = \frac {x _ {n} + \frac {1 1}{x _ {n}}}{2}$$

Choose the zero-approximation, and let it be $x_0 = 3.2 < \sqrt{11}$ for instance.

Therefore, the calculations for next approximations will be following:

1. $x_{1} = \frac{3.2 + \frac{11}{3.2}}{2} \approx 3.31875$

2. $x_{2} = \frac{3.31875 + \frac{11}{3.31875}}{2} \approx 3.31663$

3. $x_{3} = \frac{3.31663 + \frac{11}{3.31663}}{2} \approx 3.31662$

The difference between $x_{3}$ and $x_{2}$ is less or equal to 0.00001. Thus, the required accuracy is attained and $x_{3}$ is the appropriate approximation

$$S Q R T (1 1) = \sqrt {1 1} \approx 3. 3 1 6 6 2$$

Answer: $\sqrt{11} \approx 3.31662$

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