1. ## Period

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)

2. Originally Posted by pankaj
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

3. Taking example is allright but we need to generalize .

If we replace $\displaystyle x$ by $\displaystyle x+\frac{1}{22}$ then see what happens.

I am going to bed.It is midnight here.

4. Originally Posted by pankaj
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)
$\displaystyle f(2x)+f(2x+\frac{1}{11}) + f(2x+\frac{2}{11})+f(2x+\frac{3}{11})+...f(2x+\fra c{21}{11})=c ...(1)$

$\displaystyle f(2x+\frac{1}{11})+f(2x+\frac{2}{11})+f(2x+\frac{3 }{11})+....f(2x+\frac{22}{11}) = c...(2)$

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

5. Originally Posted by Amer
[tex]

$\displaystyle f(2x)-f(2x+2)=0$

$\displaystyle f(2x)=f(2x+2)$

so f(2x) is peroid with period 2 and
f(x) is periodic with period 4 right

Actually if you replace $\displaystyle x$ by $\displaystyle \frac{x}{2}$ then we have
$\displaystyle f(x)=f(x+2)$

6. Originally Posted by pankaj
Actually if you replace $\displaystyle x$ by $\displaystyle \frac{x}{2}$ then we have
$\displaystyle f(x)=f(x+2)$

is there a different between the periods of f(x) and f(2x)

ok lats do you said replace x with x/2

$\displaystyle f(x)+f(x+\frac{1}{11})+f(x+\frac{2}{11})+f(x+\frac {3}{11})+...+f(x+\frac{21}{11})=c$

$\displaystyle f(x+\frac{1}{11})+f(x+\frac{2}{11})+f(x+\frac{3}{1 1})+f(x+\frac{4}{11})...+f(x+\frac{22}{11})=c$

$\displaystyle f(x)=f(x+2)$ ??

ok sub instead of x 3x/2

$\displaystyle f(3x)+f(3x+\frac{1}{11})+f(3x+\frac{2}{11})+...+f( 3x+\frac{21}{11})=c$

$\displaystyle f(3x+\frac{1}{11})+f(3x+\frac{2}{11})+...f(3x+\fra c{22}{11})=c$
$\displaystyle f(3x)=f(3x+2)$ ??
$\displaystyle f(nx)=f(nx+2)..n\in R$ right
7. there is a well known proven result that if period of $\displaystyle f(x)$ is $\displaystyle T$,then period of $\displaystyle f(ax+b)$ is given by $\displaystyle \frac{T}{|a|}$ where $\displaystyle a\neq 0$