f-nx-f-x-n-f-x-f-0-0-f-x-

Question Number 3789 by 123456 last updated on 21/Dec/15

f (n x) = f (x + n) - f (x)

f (0) = 0

f (x) = ?

Commented by Rasheed Soomro last updated on 21/Dec/15

f (n x) = f (x + n) - f (x)

- - - -

x = 0

f (0) = f (n) - f (0)

f (n) = 0

n i s a n a r b i t r a r y c o n s t a n t

S o

f (x) = 0 ? ? ?

Commented by Rasheed Soomro last updated on 21/Dec/15

f (n x) = f (x + n) - f (x)

n = 1

f (x) = f (1 + n) - f (x)

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

x = 0 i n a b o v e

f (0) = \frac{f (1 + 0)}{2} = 0

f (1) = 0

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

f (2) = 0

f (3) = 0

⋮

f (x) = 0

Commented by 123456 last updated on 21/Dec/15

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

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

n cannot be treated as variable since

it is a constant for all x .

f o r e x a m p l e

g (x) = x - n

g (n) = 0 b u t g (x) \neq 0

Commented by Rasheed Soomro last updated on 21/Dec/15

T h a n k S!

Commented by Rasheed Soomro last updated on 21/Dec/15

f o r x = 0, f (n x) = f (x + n) - f (x) \Rightarrow f (n) = 0

W h a t i s m e a n t b y d e f i n i t i o n f (n) = 0

{C o u l d n}^{'} t w e u s e t h i s d e f i n i t i o n f o r

n = 1, n = 4, n = a n y v a l u e

{C o u l d n}^{'} t w e s a y f (6) = 0, f (a n y v a l u e) = 0

I f y e s t h e n n i s j u s t l i k e a v a r i a b l e ?

Answered by Rasheed Soomro last updated on 21/Dec/15

f (n x) = f (x + n) - f (x) \dots \dots \dots \dots \dots \dots . (i)

f (0) = 0

f (x) = ?

- - - - - - - -

n = 1 i n (i)

f (x) = f (x + 1) - f (x) \Rightarrow 2 f (x) = f (x + 1)

x \to x - 1 i n a b o v e

f (x) = 2 f (x - 1)

f (0) = 0 [g i v e n]

f (1) = 2 f (0) = 2 (0) = 0

f (2) = 2 f (1) = 0

f (3) = 2 f (2) = 0

f (x) = 0

T h i s c a n b e p r o v e d b y m a t h e m a t i c a l i n d u c t i o n .