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Question Number 6247 by Yozzii last updated on 20/Jun/16

S h o w t h a t, \forall x \in R, t h e s e q u e n c e {f (n)}_{0}^{\infty}

d e f i n e d b y f (0) = c o s x, f (1) = s i n (c o s x) a n d

f (n) = {\begin{cases} s i n (c o s (f (n - 2))) i f n ⩾ 3 i s o d d \\ c o s (s i n (f (n - 2))) i f n ⩾ 2 i s e v e n, \end{cases}

c o n v e r g e s t o a l i m i t l \in (0.6, 0.7) .

I f y o u c a n, d e t e r m i n e t h e e x a c t v a l u e

o f l .

Commented by Yozzii last updated on 20/Jun/16

f (0) = c o s x

f (1) = s i n (c o s x)

f (2) = c o s (s i n (c o s x))

f (3) = s i n (c o s (s i n (c o s x)))

f (4) = c o s (s i n (c o s (s i n (c o s x))))

f (5) = s i n (c o s (s i n (c o s (s i n (c o s x)))))

\dots

\lim_{n \to \infty} f (n) = ?

G r a p h i c a l l y, t h e c u r v e s f_{n} (x) f l a t t e n

t o t h e l i n e y \approx 0.695 f o r s u f f i c i e n t l y

l a r g e n . I a m y e t t o f i g u r e o u t h o w

t o s h o w t h i s b e h a v i o u r f o r m a l l y .

Commented by prakash jain last updated on 20/Jun/16

The value will become constant and take

value u if for some u

\sin (\cos u) = u

\cos (\sin u) = u

Then the value will be a constant .

However

\sin (\cos x) = x \Rightarrow x = . 695

\cos (\sin x) = x \Rightarrow x = . 768

Also for \cos (. 695) = . 768

So the graph should show 2 lines

. 695 and . 768

Commented by Yozzii last updated on 20/Jun/16

T h e r e s u l t o f h a v i n g t w o v a l u e s o f t h e l i m i t

u f r o m t h e t w o r e c u r r e n c e f o r m u l a e

i n d i c a t e s t h a t n o l i m i t e x i s t s ?

Commented by Yozzii last updated on 21/Jun/16

u = s i n (c o s u)

u = c o s (s i n u)

\Rightarrow 2 u = s i n (c o s u) + c o s (s i n u)

u t u r n s o u t t o b e a p p r o x i m a t e l y 0.731

a c c o r d i n g t o t h i s e q u a t i o n .

- - - - - - - - - - - - - - - - - - - - - - - -

{s i n}^{- 1} u = c o s u

{c o s}^{- 1} u = s i n u

\Rightarrow {({s i n}^{- 1} u)}^{2} + {({c o s}^{- 1} u)}^{2} = 1 (*)

W o l f r a m g i v e s c o m p l e x s o l u t i o n s o f u

f o r (*) .

{s i n}^{- 1} u = 0.5 π - {c o s}^{- 1} u

j = {c o s}^{- 1} u \Rightarrow {(0.5 π - j)}^{2} + j^{2} = 1

2 j^{2} - π j + 0.25 π^{2} - 1 = 0

j = \frac{π \pm \sqrt{π^{2} - 4 (2) (\frac{π^{2}}{4} - 1)}}{4}

j = \frac{π \pm \sqrt{- π^{2} + 8}}{4} = \frac{π \pm i \sqrt{π^{2} - 8}}{4}

∴ u = c o s (\frac{π \pm i \sqrt{π^{2} - 8}}{4})

Commented by prakash jain last updated on 21/Jun/16

u = \sin (\cos u)

u = \cos (\sin u)

Both equation are not simulatenoeus .

u = . 695 w i l l s t a t i f y o n e e q u a t i o n .

Commented by Yozzii last updated on 21/Jun/16

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