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Question Number 163942 by Rasheed.Sindhi last updated on 12/Jan/22

f (2 x) - 2 {[f (x)]}^{2} + 1 = 0

f (x) = ?

Answered by mr W last updated on 12/Jan/22

x \in R

s o {i t}^{'} s a l s o v a l i d f o r x \in N, s a y x = n,

f (2 n) = 2 {(f (n))}^{2} - 1

a_{2 n} = 2 a_{n}^{2} - 1

? ? ? ? s i m i l a r t o \cos (2 θ) = 2 {(\cos θ)}^{2} - 1 ? ? ? ?

l e t a_{n} = \cos θ_{n}

\cos θ_{2 n} = 2 \cos^{2} θ_{n} - 1 = \cos (2 θ_{n})

\Rightarrow θ_{2 n} = 2 θ_{n}

θ_{2} = 2 θ_{1}

θ_{4} = 2^{2} θ_{1}

\dots

θ_{2^{n}} = 2^{n} θ_{1}

g e n e r a l l y

θ_{n} = n θ_{1}

w i t h θ_{1} = c o n s t a n t = c

f (n) = a_{n} = \cos θ_{n} = \cos (n θ_{1}) = \cos (n c)

o r

\Rightarrow f (x) = \cos (c x)

e x a m p l e : f (1) = 0 \Rightarrow c = \frac{π}{2}

\Rightarrow f (x) = \cos (\frac{x π}{2})

Commented by Rasheed.Sindhi last updated on 13/Jan/22

V F i n e S i r!

Commented by Rasheed.Sindhi last updated on 13/Jan/22

You can't use 'macro parameter character #' in math mode

Commented by mr W last updated on 13/Jan/22

t h a n k s s i r!

s i n c e \cos 2 x = 2 \cos^{2} x - 1 a n d

\cosh 2 x = 2 \cosh^{2} x - 1, w e h a v e

g e n e r a l l y t w o p o s s i b l i t i e s :

f (x) = \cos (c x) o r f (x) = \cosh (c x) .

b u t d e p e n d i n g o n i n i t i a l c o n d i t i o n,

t h e r e i s o n l y o n e s o l u t i o n s u i t a b l e .

Commented by Rasheed.Sindhi last updated on 13/Jan/22

T h a n k s a l o t s i r!

T h a n k s a l o t \boldsymbol s i r!