f-2x-2-f-x-2-1-0-f-x-

Question Number 1307 by Rasheed Ahmad last updated on 21/Jul/15

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

f (x) = ?

Commented by Rasheed Soomro last updated on 22/Jul/15

(R a s h e e d A h m a d)

A c t u a l l y t h i s e q u a t i o n i s d i s g u i s e d f o r m o f t h e t r i g o n o m e t r i c

i d e n t i t y g i v e n b e l o w :

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

S o f (x) = c o s x i s e x p e c t e d s o l u t i o n f o r m e .

I t i s a m a t t e r o f i n t e r e s t f o r m e t h a t t h e r e e x i s t o t h e r

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

A n y w a y T H A N K S a n d A P P R E C I A T I O N S f o r t h e

p a r t i c i p a n t s .

Commented by 123456 last updated on 21/Jul/15

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

f (0) = x

- 2 x^{2} + x + 1 = 0

Δ = {(1)}^{2} - 4 (- 2) (1) = 1 + 8 = 9

x = \frac{- (1) \pm \sqrt{9}}{2 (- 2)} = \frac{- 1 \pm 3}{- 4} = \frac{1 \mp 3}{4}

x_{1} = \frac{1 - 3}{4} = - \frac{2}{4} = - \frac{1}{2}

x_{2} = \frac{1 + 3}{4} = \frac{4}{4} = 1

f (0) = - \frac{1}{2} \lor f (0) = 1

Commented by Rasheed Soomro last updated on 22/Jul/15

f (x) = \frac{1}{2} (b^{a x} + \frac{1}{b^{a x}}) \dots \dots \dots . (i)

f (x) = \frac{b^{a x} + b^{- a x}}{2} \dots \dots \dots \dots \dots . (i i)

I t h i n k t h e s e t w o a n s w e s a r e s a m e .

A n s w e r f (x) = \cosh x a n d a n s w e r (i i) h a s s i m i l a r i t y . A l t h o u g h

t h e y a r e n o t s a m e :

f (x) = \cosh x = \frac{e^{x} + e^{- x}}{2}

f (x) = \frac{b^{a x} + b^{- a x}}{2}

I t h i n k l a t t e r i s g e n e r a l f o r m a n d f o r m e r i s s p e c i a l c a s e .

t a k i n g b = e a n d a = 1 i n l a t t e r (a s a a n d b a r e a r b i t r a r y c o n s t a n t s) f o r m e r c a n b e a c h i e v e d .

o n e c a n a c h i e v e f o r m e r .

A l s o n o t e t h a t w e m a y t a k e x \in C i n t h e s e t w o s o l u t i o n s .

Commented by 123456 last updated on 23/Jul/15

\cos x = \frac{e^{i x} + e^{- i x}}{2}

a^{b x} = {(a^{b})}^{x} \equiv c^{x}, c = a^{b}

\cosh x = \frac{e^{x} + e^{- x}}{2}

also

\cosh (x i) = \cos x

\cos (x i) = \cosh x

f (x) = \frac{b^{a x} + b^{- a x}}{2}

= \frac{{(e^{\ln b})}^{a x} + {(e^{\ln b})}^{- a x}}{2}

= \frac{e^{a x \ln b} + e^{- a x \ln b}}{2}

= \cosh (a x \ln b)

a \ln b = k

f (x) = \cosh (k x)

Commented by prakash jain last updated on 23/Jul/15

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

In general when addition in domain results

in multiplication in range then function

is exponential . The extra 1 on the right

side suggests that exponent and its reciprocal

are multiplied together .

\cosh (k x) is general solution .

Other form of solution \frac{b^{a x} + b^{- a x}}{2} is also reducible

to \cosh (k x) .

Answered by prakash jain last updated on 21/Jul/15

f (x) = \frac{1}{\sqrt{2}} (\frac{2^{a x}}{\sqrt{2}} + \frac{1}{\sqrt{2} \cdot 2^{a x}})

2 {(f (x))}^{2} = \frac{2^{2 a x}}{2} + \frac{1}{2 \cdot 2^{2 a x}} + 1

f (2 x) = \frac{1}{\sqrt{2}} (\frac{2^{2 a x}}{\sqrt{2}} + \frac{1}{\sqrt{2} \cdot 2^{2 a x}}) = \frac{2^{2 a x}}{2} + \frac{1}{2 \cdot 2^{2 a x}}

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

f (x) = \frac{1}{\sqrt{2}} (\frac{b^{a x}}{\sqrt{2}} + \frac{1}{\sqrt{2} \cdot b^{a x}}) will also satisfy .

Commented by prakash jain last updated on 21/Jul/15

f (x) = \frac{1}{2} (b^{a x} + \frac{1}{b^{a x}})

Commented by 112358 last updated on 21/Jul/15

f (x) = c o s x o r f (x) = c o s h x

a l s o s a t i s t f y t h e e q u a t i o n .

Commented by prakash jain last updated on 21/Jul/15

Thanks . After working thru one solution

with base 2, I see any base b can be used .

f (x) = \frac{b^{a x} + b^{- a x}}{2}

Commented by 112358 last updated on 21/Jul/15

N i c e .

Commented by Rasheed Soomro last updated on 22/Jul/15

\frac{1}{\sqrt{2}} (\frac{b^{a x}}{\sqrt{2}} + \frac{1}{\sqrt{2} \cdot b^{a x}}) c a n b e r e d u c e d i n t o \frac{b^{a x} + b^{- a x}}{2} .

O v e r a l l t w o a n s w e r s h a v e c o m e .

f (x) = c o s x a n d f (x) = \frac{b^{a x} + b^{- a x}}{2}

O t h e r a n s w e r s i n c l u d e i n f o r m e r o r i t s s p e c i a l c a s e s i n c l u d i n g

f (x) = \cosh x .

Commented by Rasheed Soomro last updated on 22/Jul/15

N o t e : S o m e c o m m e n t s o n a n s w e r s a r e m i s t a k e n l y i n c l u d e d i n

{Q u e s t i o n}^{'} s c o m m e n t s . p l e a s e s e e t h e r e .