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Thread: Period

  1. #1
    Senior Member pankaj's Avatar
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    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)
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by pankaj View Post
    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
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  3. #3
    Senior Member pankaj's Avatar
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    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.
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  4. #4
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by pankaj View Post
    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
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  5. #5
    Senior Member pankaj's Avatar
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    Quote Originally Posted by Amer View Post
    [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
    $\displaystyle f(x)=f(x+2)$
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  6. #6
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by pankaj View Post
    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 $

    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
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  7. #7
    Senior Member pankaj's Avatar
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    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 $\displaystyle a\neq 0$
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