Solve-for-f-x-f-x-1-f-x-3-

Question Number 1295 by prakash jain last updated on 20/Jul/15

Solve for f (x)

f (x + 1) = {[f (x)]}^{3}

Commented by Rasheed Ahmad last updated on 20/Jul/15

(R a s h e e d S o o m r o)

f (x) = \pm 1

V e r i f i c a t i o n :

1) f (x) = 1 \Rightarrow f (x + 1) = 1 \land {[f (x)]}^{3} = {(1)}^{3} = 1

2) f (x) = - 1 \Rightarrow f (x + 1) = - 1 \land {[f (x)]}^{3} = {(- 1)}^{3} = - 1

Commented by Rasheed Soomro last updated on 20/Jul/15

{L e t}^{'} s f i r s t c o n s i d e r f o r f (x) t o b e p o l y n o m i a l .

T h e n

f (x) = a o r f (x) = a x + b o r {a x}^{2} + b x + c \dots \dots \dots . .

1) f (x) = a = {a x}^{0}

f (x + 1) = {[f (x)]}^{3} \dots \dots \dots \dots \dots . (i)

f (x + 1) = a {(x + 1)}^{0} = a \dots \dots \dots (i i)

a n d {[f (x)]}^{3} = a^{3} \dots \dots \dots \dots \dots \dots . (i i i)

f r o m (i), (i i) a n d (i i i)

a = a^{3} \Rightarrow a^{2} = 1 \Rightarrow a^{3} - a = 0 \Rightarrow a = 0 \lor a^{2} = 1 \Rightarrow a = 0, 1, - 1

H e n c e f (x) = 0, \pm 1

f (x) = a x + b a n d f (x) = {a x}^{2} + b x + c b o t h w i l l g i v e s a m e

r e s u l t t h o u g h c a l c u l a t i o n w i l l b e c o m p l i c a t e d .

I t h i n k f o r f (x) t o b e p o y n o m i a l t h e r e e x i s t o n l y t h e s e t h r e e

s o l u t i o n s .

Answered by prakash jain last updated on 20/Jul/15

Use another technique to remove power .

F (x) = \log_{q} f (x)

F (x + 1) = 3 F (x)

F (x) = 3^{x}

\log_{q} f (x) = 3^{x}

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

f (x + 1) = q^{3^{x + 1}} = q^{3^{x}} \cdot q^{3^{x}} \cdot q^{3^{x}} = {(q^{3^{x}})}^{3} = {(f (x))}^{3}

Commented by Rasheed Ahmad last updated on 20/Jul/15

M a r v e l l o u s! I {c o u l d n}^{'} t e v e n t h i n k

o f i t . H o w e v e r I w a n t t o k n o w h o w

y o u d e t e r m i n e F (x) = 3^{x} ?