Find-the-value-of-2-2-2-2-2-2-by-cos-function-

Question Number 99568 by Dwaipayan Shikari last updated on 22/Jun/20

F i n d t h e v a l u e o f \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 +}}}}}} \dots \infty b y c o s f u n c t i o n

Commented by bemath last updated on 22/Jun/20

x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{\dots}}}}}

x^{2} = 2 + x

x^{2} - x - 2 = 0

(x - 2) (x + 1) = 0

{\begin{cases} x = 2 s o l u t i o n \\ x = - 1 r e j e c t e d \end{cases}

Commented by Dwaipayan Shikari last updated on 22/Jun/20

T h a n k y o u s i r . C a n y o u p r o v e i t {u s i n g}^{'} {c o s}^{'} f u n c t i o n ?

Commented by Dwaipayan Shikari last updated on 22/Jun/20

1 + c o s \frac{π}{4} = 2 {c o s}^{2} \frac{π}{8}

\sqrt{\frac{1}{2} (1 + \frac{1}{\sqrt{2}})} = c o s \frac{π}{8} = c o s \frac{π}{2^{3}} = \frac{1}{2} \sqrt{2 + \sqrt{2}}

A s w e p r o c c e e d \sqrt{\frac{1}{2} (1 + \frac{1}{2} \sqrt{2 + \sqrt{2}})} = c o s \frac{π}{16} = c o s \frac{π}{2^{4}} = \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}}

s o \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}}} \dots n t e r m s = c o s \frac{π}{2^{n + 1}}

\underset{n \to \infty}{} l i m c o s \frac{π}{2^{n + 1}} = c o s (0) = 1

S o \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 +}}}}}}} \dots \infty = 1

s o \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 +}}}}}}}} \dots . \infty = 2

Commented by Dwaipayan Shikari last updated on 22/Jun/20

S i r, i h a v e f o u n d t h i s w a y w h i l e e x p e r i m e n t i n g . I s i t t r u e ?

Answered by mahdi last updated on 21/Jun/20

\sqrt{2 + \sqrt{2 + \sqrt{2 + \dots}}} = A \Rightarrow

\sqrt{2 + A} = A \Rightarrow A^{2} = A + 2 \Rightarrow A^{2} - A - 2 = 0

\Rightarrow A = \frac{1 \pm \sqrt{5}}{2} \overset{A > 0}{\Rightarrow} A = \frac{1 + \sqrt{5}}{2}

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