Q 79 Use Newton鈥檚 method to derive ... [FREE SOLUTION]

Chapter 7: Q 79 (page 593)

Use Newton鈥檚 method to derive the recursion formula
$x_{k + 1} = \frac{1}{2} (x_{k} + \frac{a}{x_{k}})$
for approximating $\sqrt{a}$ .

Short Answer

Expert verified

We can derive the recursion formula as following :-

$x_{k + 1} = x_{k} - \frac{{x_{k}}^{2} - a}{2 x_{k}} 鈬� x_{k + 1} = \frac{2 {x_{k}}^{2} - {x_{k}}^{2} + a}{2 x_{k}} 鈬� x_{k + 1} = \frac{{x_{k}}^{2} + a}{2 x_{k}} 鈬� x_{k + 1} = \frac{1}{2} (\frac{{x_{k}}^{2} + a}{x_{k}}) 鈬� x_{k + 1} = \frac{1}{2} (x_{k} + \frac{a}{x_{k}})$

Step by step solution

Step 1. Given Information

We have to derive a recursion formula by using Newton's method to approximate the value of $\sqrt{a}$ .

Step 2. Derivation of recursion formula

We have to approximate the value of $\sqrt{a}$ .

Let:-

$x = \sqrt{a} 鈬� x^{2} = a 鈬� x^{2} - a = f (x) S a y$

localid="1654268190999" $鈬� f (x) = x^{2} - a$

Then:-

$f' (x) = 2 x$

We have the following newton's formula to approximate a value of a function:-

$x_{k + 1} = x_{k} - \frac{f (x_{k})}{f' (x_{k})}$

Put the above values, then we have:-

localid="1654498543121" $\begin{array}{rcl} x_{k + 1} & = & x_{k} - \frac{{x_{k}}^{2} - a}{2 x_{k}} \\ = & \frac{2 {x_{k}}^{2} - {x_{k}}^{2} + a}{2 x_{k}} \\ = & \frac{{x_{k}}^{2} + a}{2 x_{k}} \\ = & \frac{1}{2} (\frac{{x_{k}}^{2} + a}{x_{k}}) \\ = & \frac{1}{2} (x_{k} + \frac{a}{x_{k}}) \end{array}$

This is the required formula.

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91影视

Use Newton鈥檚 method to derive the recursion formula
$x_{k + 1} = \frac{1}{2} (x_{k} + \frac{a}{x_{k}})$
for approximating $\sqrt{a}$ .

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Derivation of recursion formula

One App. One Place for Learning.

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91影视

Use Newton鈥檚 method to derive the recursion formulaxk+1=12xk+axkfor approximatinga.

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Derivation of recursion formula

One App. One Place for Learning.

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Use Newton鈥檚 method to derive the recursion formula
$x_{k + 1} = \frac{1}{2} (x_{k} + \frac{a}{x_{k}})$
for approximating $\sqrt{a}$ .