Period

**pankaj** · June 11th 2009, 08:22 AM

If $\sum_{r=0}^{21}f(\frac{r}{11}+2x)=constant$ for all real x and f(x) is periodic ,then what could be period of f(x)

**Amer** · June 11th 2009, 10:21 AM

Originally Posted by pankaj

If $\sum_{r=0}^{21}f(\frac{r}{11}+2x)=constant$ 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 $2\pi$ right

see this

$h(x)=cosx = cos (x+2\pi) = cos(x+4\pi) = cos(x+6\pi ) = cos(x+8\pi)$

so h(x) periodic with period $2\pi$

ok $cos(2x)$ periodic with period ?? $\pi$ right

$cos(2x) = cos(2x+\pi) = cos (2x+2\pi ) = cos(2x+3\pi)= ...$

dose this make a sense .

did you got it

**pankaj** · June 11th 2009, 11:08 AM

Taking example is allright but we need to generalize .

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

I am going to bed.It is midnight here.

**Amer** · June 11th 2009, 11:38 AM

Originally Posted by pankaj

If $\sum_{r=0}^{21}f(\frac{r}{11}+2x)=constant$ for all real x and f(x) is periodic ,then what could be period of f(x)

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

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

find the subtraction of (2)-(1)

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

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

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

**pankaj** · June 11th 2009, 03:52 PM

Originally Posted by Amer

[tex]

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

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

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

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

**Amer** · June 11th 2009, 11:01 PM

Originally Posted by pankaj

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

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

ok lats do you said replace x with x/2

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

sub instead of x x+1/11

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

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

ok sub instead of x 3x/2

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

instead of 3x 3x+1\11

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

$f(3x)=f(3x+2)$ ??

so the periodic dose not change for all values of x so we can say

$f(nx)=f(nx+2)..n\in R$ right