# Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 17

$${\text{1}}{\text{.414}}$$

#### Work Step by Step

\eqalign{ & {\text{We can note that }}\sqrt 2 {\text{ is a solution of the equation }}{x^2} - 2 = 0.{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^2} - 2,\,\,\,\,\,\,f'\left( x \right) = 2x \cr & {\text{Since 1 < }}\sqrt 2 < 2,{\text{ we can use }}{c_1} = 1{\text{ as the first approximation}}{\text{. }} \cr & {\text{Then using Newton's method we have}}: \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{{{\left( 1 \right)}^2} - 2}}{{2\left( 1 \right)}} = 1.5 \cr & {c_3} = {c_2} - \frac{{f\left( {{c_2}} \right)}}{{f'\left( {{c_2}} \right)}} = 1.5 - \frac{{{{\left( {1.5} \right)}^2} - 2}}{{2\left( {1.5} \right)}} = 1.4166 \cr & {\text{In the same way}} \cr & {c_4} = 1.4166 - \frac{{{{\left( {1.4166} \right)}^2} - 2}}{{2\left( {1.4166} \right)}} = 1.4142 \cr & {c_5} = 1.4142 - \frac{{{{\left( {1.4142} \right)}^2} - 2}}{{2\left( {1.4142} \right)}} = 1.4142 \cr & {\text{Since }}{c_3} = {c_4} = 1.4142,{\text{ to the nearest thousand}}{\text{, }}\sqrt 2 = {\text{1}}{\text{.4142}} \cr}

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