Is-there-f-R-R-such-that-f-x-0-f-x-3-3f-x-2-

Question Number 6482 by prakash jain last updated on 28/Jun/16

Is there f : R \to R such that f (x) \neq 0

f (x^{3}) = 3 f (x^{2})

Commented by Yozzii last updated on 29/Jun/16

f (x^{3}) = 3 f (x^{2})

x = 0 \Rightarrow f (0) = 3 f (0) \Rightarrow f (0) = 0 .

x = 1 \Rightarrow f (1) = 3 f (1) \Rightarrow f (1) = 0

x = - 1 \Rightarrow f (- 1) = 3 f (1) = 0

x = 0, \pm 1 a r e s o m e r o o t s o f f (x) = 0, i f e v e r

f (x) = 0 a s f (x) i s n o t e v e r y w h e r e z e r o .

f (x) = \underset{i = 0}{\sum^{\infty}} a (i) x^{i} = a (0) + a (1) x + a (2) x^{2} + a (3) x^{3} + \dots

{a (i)}_{i = 0}^{\infty} i s a r e a l s e q u e n c e

3 f (x^{2}) = 3 a (0) + 3 a (1) x^{2} + 3 a (2) x^{4} + 3 a (3) x^{6} + 3 a (4) x^{8} + 3 a (5) x^{10} + 3 a (6) x^{12} \dots

f (x^{3}) = a (0) + a (1) x^{3} + a (2) x^{6} + a (3) x^{9} + a (4) x^{12} + \dots

f (x^{3}) = 3 f (x^{2}) \Rightarrow a (i) = 0 f o r a l l i \in Z^{⩾} \Rightarrow f (x) = 0 .

\Rightarrow f (x) \neq \underset{i = 0}{\sum^{\infty}} a (i) x^{i}

Answered by nburiburu last updated on 29/Jun/16

f (x^{k}) = \frac{\partial^{k} (x^{k})}{{(\partial x)}^{k}} .

f (x^{3}) = {(x^{3})}^{‴} = {(3 x^{2})}^{″} = {(6 x)}^{'} = 6

f (x^{2}) = {(x^{2})}^{″} = {(2 x)}^{'} = 2

f (x^{3}) = 3 f (x^{2})