if-f-g-are-functions-of-R-R-not-constant-such-for-all-x-y-R-2-f-x-y-f-x-f-y-g-x-g-y-g-x-y-f-x-g-y-g-x-f-y-if-f-0-0-then-proof-os-disproof-that-x-R-f-x-2-g-x-2-1-

Question Number 636 by 123456 last updated on 17/Feb/15

i f f, g a r e f u n c t i o n s o f R \to R

n o t c o n s t a n t s u c h f o r a l l (x, y) \in R^{2}

{\begin{cases} f (x + y) = f (x) f (y) - g (x) g (y) \\ g (x + y) = f (x) g (y) + g (x) f (y) \end{cases}

i f f^{'} (0) = 0 t h e n p r o o f o s d i s p r o o f

t h a t \forall x \in R, {[f (x)]}^{2} + {[g (x)]}^{2} = 1

Commented by prakash jain last updated on 16/Feb/15

g (x + y) = f (x) g (y) - g (x) f (y)

g (x + x) = f (x) g (x) - g (x) f (x)

g (2 x) = 0 ?

do you mean

g (x + y) = f (x) g (y) + g (x) f (y) ?

If

g (x + y) = f (x) g (y) - g (x) f (y)

g (2 x) = f (x) g (x) - g (x) f (y) = 0

Contradicts assumption f (x), g (x) are

not constant .

Commented by prakash jain last updated on 17/Feb/15

Squaring and adding

{[f (x + y)]}^{2} + {[g (x + y)]}^{2} = ({[f (x)]}^{2} + {[g (x)]}^{2}) ({[f (y)]}^{2} + {[g (y)]}^{2})

If u (x) = {[f (x)]}^{2} + {[g (x)]}^{2}

u (x + y) = u (x) u (y) \Rightarrow u (x) = e^{k x}

{[f (x)]}^{2} + {[g (x)]}^{2} = e^{k x}

Answered by prakash jain last updated on 17/Feb/15

From comments

{[f (x)]}^{2} + {[g (x)]}^{2} = e^{k x}

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

f (2 x) = 2 {[f (x)]}^{2} - e^{k x}

Differenting both sides

2 f^{'} (2 x) = 4 f (x) f^{'} (x) - {k e}^{k x}

put x = 0

2 f^{'} (0) = 4 f (0) f^{'} (0) - k

Given f^{'} (0) = 0

0 = 0 - k \Rightarrow k = 0

Hence

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