If $\displaystyle \sum_{r=0}^{21}f(\frac{r}{11}+2x)=constant $ for all real x and f(x) is periodic ,then what could be period of f(x)
Hint you know that cos is periodic function right it is periodic with period $\displaystyle 2\pi$ right
see this
$\displaystyle h(x)=cosx = cos (x+2\pi) = cos(x+4\pi) = cos(x+6\pi ) = cos(x+8\pi) $
so h(x) periodic with period $\displaystyle 2\pi$
ok $\displaystyle cos(2x) $ periodic with period ?? $\displaystyle \pi $ right
$\displaystyle cos(2x) = cos(2x+\pi) = cos (2x+2\pi ) = cos(2x+3\pi)= ...$
dose this make a sense .
did you got it
$\displaystyle f(2x)+f(2x+\frac{1}{11}) + f(2x+\frac{2}{11})+f(2x+\frac{3}{11})+...f(2x+\fra c{21}{11})=c ...(1)$
$\displaystyle f(2x+\frac{1}{11})+f(2x+\frac{2}{11})+f(2x+\frac{3 }{11})+....f(2x+\frac{22}{11}) = c...(2) $
find the subtraction of (2)-(1)
$\displaystyle f(2x)-f(2x+2)=0 $
$\displaystyle f(2x)=f(2x+2) $
so f(2x) is peroid with period 2 and
f(x) is periodic with period 4 right
is there a different between the periods of f(x) and f(2x)
ok lats do you said replace x with x/2
$\displaystyle f(x)+f(x+\frac{1}{11})+f(x+\frac{2}{11})+f(x+\frac {3}{11})+...+f(x+\frac{21}{11})=c $
sub instead of x x+1/11
$\displaystyle f(x+\frac{1}{11})+f(x+\frac{2}{11})+f(x+\frac{3}{1 1})+f(x+\frac{4}{11})...+f(x+\frac{22}{11})=c$
$\displaystyle f(x)=f(x+2)$ ??
ok sub instead of x 3x/2
$\displaystyle f(3x)+f(3x+\frac{1}{11})+f(3x+\frac{2}{11})+...+f( 3x+\frac{21}{11})=c$
instead of 3x 3x+1\11
$\displaystyle f(3x+\frac{1}{11})+f(3x+\frac{2}{11})+...f(3x+\fra c{22}{11})=c$
$\displaystyle f(3x)=f(3x+2)$ ??
so the periodic dose not change for all values of x so we can say
$\displaystyle f(nx)=f(nx+2)..n\in R$ right