Chapter 12 - Sequences and Series - 12.6 Newton's Method - 12.6 Exercises - Page 652: 13

$$1.25$$

$$\eqalign{ & {x^2}{e^{ - x}} + {x^2} - 2 = 0;\,\,\,\,{\text{in the interval }}\left[ {0,3} \right] \cr & {\text{Let }}f\left( x \right) = {x^2}{e^{ - x}} + {x^2} - 2,{\text{ so that}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left( {{x^2}{e^{ - x}} + {x^2} - 2} \right) \cr & f'\left( x \right) = {x^2}\left( { - {e^{ - x}}} \right) + {e^{ - x}}\left( {2x} \right) + 2x \cr & f'\left( x \right) = - {x^2}{e^{ - x}} + 2x{e^{ - x}} + 2x \cr & {\text{Interval }}\left[ {0,3} \right]{\text{ then }}a = 0{\text{ and }}b = 3.{\text{ Check }}f\left( a \right){\text{ and }}f\left( b \right) \cr & f\left( 0 \right) = {\left( 0 \right)^2}{e^{ - \left( 0 \right)}} + {\left( 0 \right)^2} - 2 = - 2 < 0 \cr & f'\left( 3 \right) = {\left( 3 \right)^2}{e^{ - \left( 3 \right)}} + {\left( 3 \right)^2} - 2 \approx 7.44 < 0 \cr & f\left( 0 \right){\text{ and }}f\left( 3 \right){\text{ have opposite signs}}{\text{. Thus}}{\text{,}} \cr & {\text{There is a solution for the equation in the interval }}\left( {0,3} \right) \cr & {\text{As an initial guess }}{c_1} = 1.{\text{ we cannot use as initial value 0 because }}f'\left( 0 \right) = 0. \cr & {\text{Then}}{\text{. A better guess}}{\text{, }}{c_2},{\text{ can be found as follows }} \cr & {c_{n + 1}} = {c_n} - \frac{{f\left( {{c_n}} \right)}}{{f'\left( {{c_n}} \right)}} \cr & {c_2} = {c_1} - \frac{{f\left( {{c_1}} \right)}}{{f'\left( {{c_1}} \right)}} = 1 - \frac{{{{\left( 1 \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( 1 \right)}^2} - 2}}{{ - {{\left( 1 \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( 1 \right){e^{ - \left( 0 \right)}} + 2\left( 1 \right)}} \approx 1.2669 \cr & {\text{A third approximation}}{\text{, }}{c_3},{\text{ can now we found}}{\text{.}} \cr & {c_3} = 1.2669 - \frac{{{{\left( {1.2669} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( {1.2669} \right)}^2} - 2}}{{ - {{\left( {1.2669} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( {1.2669} \right){e^{ - \left( 0 \right)}} + 2\left( {1.2669} \right)}} \approx 1.2464 \cr & {\text{In the same way}} \cr & {c_4} = 1.2464 - \frac{{{{\left( {1.2464} \right)}^2}{e^{ - \left( 0 \right)}} + {{\left( {1.2464} \right)}^2} - 2}}{{ - {{\left( {1.2464} \right)}^2}{e^{ - \left( 0 \right)}} + 2\left( {1.2464} \right){e^{ - \left( 0 \right)}} + 2\left( {1.2464} \right)}} \approx 1.2463 \cr & \cr & {\text{Subsequent approximations yield no further accuracy}}{\text{, to the hundredth}}{\text{.}} \cr & {\text{Then}}{\text{. A solution for the given interval is }}x \approx 1.25 \cr} $$