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Question Number 33375 by rahul 19 last updated on 15/Apr/18

I f f : R \to R i s a n o d d f u n c t i o n s u c h

t h a t :

a) f (1 + x) = 1 + f (x) .

b) x^{2} f (\frac{1}{x}) = f (x), x \neq 0 .

T h e n f i n d f (x) ?

Commented by prof Abdo imad last updated on 15/Apr/18

f i s o d d \Rightarrow f (0) = 0 \Rightarrow f (1 + 0) = 1 + f (0) \Rightarrow f (1) = 1

f (2) = f (1 + 1) = 1 + f (1) = 2

l e t s u p p o s e f (n) = n \Rightarrow f (n + 1) = 1 + f (n) = 1 + n \Rightarrow

\forall n \in N f (n) = n i f n \in Z^{-} n = - m

f (n) = f (- m) = - f (m) = - m = n \Rightarrow

\forall n \in Z f (n) = n

f (1) = f (\frac{n}{n}) = n f (\frac{1}{n}) = 1 \Rightarrow f (\frac{1}{n}) = \frac{1}{n}

f (\frac{p}{n}) = f (\frac{1}{n} + . . + \frac{1}{n}) (p \times) = p f (\frac{1}{n}) = \frac{p}{n}

\Rightarrow \forall x \in Q f (x) = x b u t Q i s d e n s e i n s i d e R \Rightarrow

\forall x \in R f (x) = x w e h a v e f o r x \neq o

x^{2} f (\frac{1}{x}) = x^{2} . \frac{1}{x} = x = f (x) c o n d . b i s v e r i f i e d .

Commented by rahul 19 last updated on 15/Apr/18

t h a n k y o u s i r!