2015/034) Given $f(x-1) + f(x+1) = \sqrt{3} * f(x)$. What is the period of f(x)

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we need to have 2 term $f(x)$ and $f(y)$ where $y = x + t$

now $f(x-1) + f(x+1) = \sqrt{3} f(x)\cdots (1)$

putting $x-1 = y$

$f(y) + f(y+2) = \sqrt{3} f(y+1)\cdots (2)$ 


hence $f(y+2) + f(y+4) = \sqrt{3} f(y+3)\cdots (3)$  ( from 1 putting y = x - 3)

add (2) and 3 to get


$f(y ) + 2 f(y+2) + f(y+4) = \sqrt{3}(f(y+1) + f(y+3) = 3 f(y+2)$

so $f(y) + f(y+4) = f(y+2)$

putting y = x and y = x +3 we get 2 equations

$f(x) + f(x+4) = f(x+2)$
$f(x+4) + f(x+8) = f(x+6)$

add to get $f(x) + 2 f(x+4) + f(x+8) = f(x+2) + f(x+6) = f(x+4)$

so $f(x) + f(x+4) + f(x+8) = 0 \cdots(4)$

changing x to x + 4 we get

$f(x+ 4) + f(x+8) + f(x+12) = 0$ 0 and subtacting (4) from above

$f(x+12) = f(x) = 0$

so $f(x+12) = f(x)$

so period is 12 or a factor of 12

it can be checked that period is not 2 4 3 or 6.

hence it is 12

refer to https://in.answers.yahoo.com/question/index?qid=20111105051143AAcISWZ