The recursion relation to solve x = e^-x using Newton-Raphson method is

A) $${{\text{x}}_{{\text{n}} + 1}} = {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$ B) $${{\text{x}}_{{\text{n}} + 1}} = {{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}$$ C) $${{\text{x}}_{{\text{n}} + 1}} = \left( {1 + {{\text{x}}_{\text{n}}}} \right)\frac{{{{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}{{1 + {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$ D) $${{\text{x}}_{{\text{n}} + 1}} = \frac{{{\text{x}}_{\text{n}}^2 - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}\left( {1 + {{\text{x}}_{\text{n}}}} \right) - 1}}{{{{\text{x}}_{\text{n}}} - {{\text{e}}^{ - {{\text{x}}_{\text{n}}}}}}}$$

Answer & Solution

Correct Answer: Option C

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The recursion relation to solve x = e-x using Newton-Raphson method...

The recursion relation to solve x = e-x using Newton-Raphson method is

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The recursion relation to solve x = e^-x using Newton-Raphson method is