Determine-all-function-f-R-0-1-R-satisfying-the-functional-relation-f-x-f-1-1-x-2-1-2x-x-1-x-for-x-0-and-x-1-

Question Number 118150 by bobhans last updated on 15/Oct/20

Determine all function f : R ╲ {0, 1} \to R

satisfying the functional relation

f (x) + f (\frac{1}{1 - x}) = \frac{2 (1 - 2 x)}{x (1 - x)}; for x \neq 0 and x \neq 1

Commented by bemath last updated on 15/Oct/20

g o o d q u e s t i o n

Answered by bemath last updated on 15/Oct/20

l e t t i n g w = \frac{1}{1 - x}, t h e g i v e n f u n c t i o n a l

e q u a t i o n c a n b e w r i t t e n a s

f (x) + f (w) = 2 (\frac{1}{x} - w) \dots (i)

I f w e s e t t i n g z = \frac{1}{1 - w} t h e n x = \frac{1}{1 - z}

H e n c e w e a l s o h a v e t h e r e l a t i o n

f (w) + f (z) = 2 (\frac{1}{w} - z) \dots (i i)

a n d f (z) + f (x) = 2 (\frac{1}{z} - x) \dots (i i i)

a d d i n g (i) a n d (i i i) w e g e t

2 {f (x) + f (w) + f (z)} = 2 (\frac{1}{x} - x) - 2 w + \frac{2}{x}

u s i n g (i i) r e l a t i o n, t h i s r e d u c e s

t o \Rightarrow 2 f (x) = 2 (\frac{1}{x} - x) - 2 (w + \frac{1}{w}) + 2 (z + \frac{1}{z})

n o w u s i n g w + \frac{1}{w} = \frac{1}{1 - x} + 1 - x

z + \frac{1}{z} = \frac{x - 1}{x} + \frac{x}{x - 1}

t h u s w e g e t f (x) = \frac{x + 1}{x - 1}

Answered by bobhans last updated on 15/Oct/20

(1) replacing \frac{1}{1 - x} by x, gives

f (\frac{x - 1}{x}) + f (x) = \frac{2 (1 - 2 (\frac{x - 1}{x}))}{(\frac{x - 1}{x}) (1 - (\frac{x - 1}{x}))}

f (\frac{x - 1}{x}) + f (x) = \frac{2 x (2 - x)}{(x - 1)}

(2) replacing x by \frac{1}{1 - x}

f (\frac{1}{1 - x}) + f (\frac{1}{1 - (\frac{1}{1 - x})}) = \frac{2 (1 - 2 (\frac{1}{1 - x}))}{(\frac{1}{1 - x}) (1 - (\frac{1}{1 - x}))}

f (\frac{1}{1 - x}) + f (\frac{x - 1}{x}) = \frac{2 (1 - x) (1 + x)}{x}

now we have

(1) f (x) + f (\frac{1}{1 - x}) = \frac{2 - 4 x}{x - x^{2}}

(2) f (\frac{x - 1}{x}) + f (x) = \frac{4 x - {2 x}^{2}}{x - 1}

(3) f (\frac{1}{1 - x}) + f (\frac{x - 1}{x}) = \frac{2 - {2 x}^{2}}{x}

adding (1), (2) and (3) we get

2 [f (x) + f (\frac{1}{1 - x}) + f (\frac{x - 1}{x})] = \frac{- {4 x}^{3} + {6 x}^{2} + 6 x - 4}{x^{2} - x}

or f (x) + f (\frac{1}{1 - x}) + f (\frac{x - 1}{x}) = \frac{- {2 x}^{3} + {3 x}^{2} + 3 x - 2}{x^{2} - x}

substitute third equation, gives

f (x) = \frac{- {2 x}^{3} + {3 x}^{2} + 3 x - 2}{x^{2} - x} - \frac{2 - {2 x}^{2}}{x}

f (x) = \frac{- {2 x}^{3} + {3 x}^{2} + 3 x - 2 - (2 x - 2 - {2 x}^{3} + {2 x}^{2})}{x^{2} - x}

f (x) = \frac{x^{2} + x}{x^{2} - x} = \frac{x + 1}{x - 1} .

Commented by bemath last updated on 15/Oct/20

w a w \dots . g r e a t \dots

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