The value of x that satisfies the relation x | vedclass

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Question

$x = \frac{1}{{1 + x}}$      $\Rightarrow$      $x^2 + x - 1 = 0$

$x = \frac{{ - 1 + \sqrt 5 }}{2}$   or&nb

Vedclass · Accepted Answer

$2\, sin18^&deg;$

Vedclass · Answer

$x = \frac{1}{{1 + x}}$      $\Rightarrow$      $x^2 + x - 1 = 0$

$x = \frac{{ - 1 + \sqrt 5 }}{2}$   or  $\frac{{ - 1 - \sqrt 5 }}{2}$ (rejected, think ! )

hence $x = \left( {\frac{{\sqrt 5  - 1}}{4}} \right)\,\cdot\,2 = 2 \, sin18^&deg;$

The value of $x$ that satisfies the relation $x = 1 - x + x^2 - x^3 + x^4 - x^5 + ......... \infty$

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