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

Question Number 100624 by Dwaipayan Shikari last updated on 27/Jun/20

Find the value of \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}} \dots \infty using \cos function

Commented by MJS last updated on 27/Jun/20

why \cos ?

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

x^{2} = 2 x

x_{1} = 0 wrong

x_{2} = 2 is the solution

Commented by Dwaipayan Shikari last updated on 27/Jun/20

T h e r e w a s a m i s t a k e o n m y q u e s t i o n . N o w c a n y o u p r o v e i t

s i r u s i n g c o s f u n c t i o n ? ? ? ? I h a v e e d i t e d m y q u e s t i o n

Commented by Dwaipayan Shikari last updated on 27/Jun/20

S i r i h a v e f o u n d s o m e t h i n g

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}

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

1 + c o s \frac{π}{8} = 2 {c o s}^{2} \frac{π}{16}

\frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2}}} = c o s \frac{π}{16} = c o s \frac{π}{2^{4}}

S o \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 \dots}}}}}}} n = c o s \frac{π}{2^{n + 1}}

li \underset{n \to \infty}{m} \frac{1}{2} \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2 + \sqrt{2}}}}}} . . n = c o s (0) = 1

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

i s i t r i g h t ? ? ? ? ? ? ?

Answered by maths mind last updated on 27/Jun/20

u_{n + 1}^{2} = u_{n} + 2 \dots E

u_{0} = \sqrt{2}

w e c a n p r o o f t h i s w i t h e l o g i c

0 ⩽ u_{n} ⩽ 2 \dots e a s y t o o s e e

u_{n + 1} - u_{n} = \frac{u_{n} + 2 - u_{n}^{2}}{\sqrt{u_{n} + 2} + u_{n}} = \frac{- (u_{n} - 2) (u_{n} + 1)}{\sqrt{u_{n} + 2} + u_{n}} ⩾ 0

u_{n} ↾ i n c r e a s i n g

\Rightarrow s i n c e 0 ⩽ U_{n} ⩽ 2

\Rightarrow \exists (w_{n}) s u c h u_{n} = 2 c o s (w_{n}) p r o o f

w_{n} = \cos^{- 1} (\frac{u_{n}}{2}) w e l l d e f i n d s i n c e \frac{u_{n}}{2} \in [0, 1]

s i n c e u_{n} \in [0, 2] a n d i n c r a s i n g \Rightarrow w_{n} \in [0, \frac{π}{2}] \Rightarrow

w_{n} i s d e a c r i s i n g

E \Leftrightarrow 4 {c o s}^{2} (w_{n + 1}) = 2 + 2 c o s (w_{n}) \Rightarrow 2 {c o s}^{2} (w_{n + 1}) - 1 = c o s (w_{n})

c o s (2 x) = 2 {c o s}^{2} (x) - 1 \Rightarrow

E \Leftrightarrow c o s (2 w_{n + 1}) = c o s (w_{n})

\Rightarrow 2 w_{n + 1} = w_{n} b e c a u s e w_{n} \in [0, \frac{π}{2}]

\Rightarrow w_{n + 1} = \frac{w_{n}}{2} \Rightarrow w_{n} i s g e o m e t r i c s e q u e n c e s

\Rightarrow w_{n} = w_{0} {(\frac{1}{2})}^{n} w_{0} = \frac{π}{4}

u_{n} = 2 c o s (\frac{π}{4 . 2^{n}}) = 2 c o s (\frac{π}{2^{n + 2}})

Commented by ghiniboss last updated on 28/Jun/20

what aspect of maths is this ?

Commented by prakash jain last updated on 28/Jun/20

Two topics

Sequences and series

recurrence relation