The recurrence relation to solve x = e -x using Newton-Raphson me

The recurrence relation to solve x = e -x using Newton-Raphson me
| The recurrence relation to solve x = e-x using Newton-Raphson method is

A. <span class="math-tex">\({x_{n + 1}} = {e^{ - {x_n}}}\)</span>

B. <span class="math-tex">\({x_{n + 1}} = {X_n} - {e^{ - {x_n}}}\)</span>

C. <span class="math-tex">\({x_{n + 1}} = \left( {1 + {x_n}} \right)\frac{{{e^{ - {x_n}}}}}{{1 + {e^{ - {x_n}}}}}\)</span>

D. <span class="math-tex">\({x_{n + 1}} = \frac{{{x_n}{e^{ - {x_n}}}}}{{1 + {e^{ - {x_n}}}}}\)</span>

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Right Answer is: C

SOLUTION

Explanation:

Let f(x) = x – e-x then f(x) = 1 + e-x

Newton Raphson iteration formula is given by:

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

\({x_{n + 1}} = {x_n} - \frac{{{x_n} - {e^{ - {x_n}}}}}{{1 + {e^{ - {x_n}}}}}\)

\(\therefore {x_{n + 1}} = \left( {1 + {x_n}} \right)\frac{{{e^{ - {x_n}}}}}{{1 + {e^{ - {x_n}}}}}\)

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