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Thread: Extension of sin function

  1. #1
    Senior Member bugatti79's Avatar
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    Extension of sin function

    Folks,

    Given $\displaystyle f(x)=sin x, \pi \le x \le 2\pi$ and 0 otherwise and
    $\displaystyle F(x)=\begin{cases}f(x), x \ge 0 \\ -f(-x), x \le 0 \\ \end{cases}$

    How do I complete this set up for, indicated by question mark


    $\displaystyle F(x)=\begin{cases}f(x)=sin x, \pi \le x \le 2\pi \\ f(x)=0, -2\pi \le x \le -\pi \\ -f(-x)=?, \pi \le x \le 2\pi \\ -f(-x)=0, -2\pi \le x \le -\pi \end{cases}$

    Ie, I want to use the same method that was applied to the following
    $\displaystyle F(x)=\begin{cases}f(x)=xe^{-x}, x\ge 0 \\-f(-x)=-(-x)e^{-(-x)}, x \ge 0 \\ \end{cases}=\begin{cases}xe^{-x}, x\ge 0 \\ xe^x, x \le 0 \\ \end{cases}$

    thanks
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  2. #2
    MHF Contributor

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    What you are writing simply doesn't make sense. F(x)= -f(-x) only for x<0 so you cannot have "$\displaystyle F(x)= -f(-x)$" for $\displaystyle \pi\le x\le 2\pi$. If $\displaystyle 0\le x\le\pi$ then F(x)= 0. If $\displaystyle \pi\le x\le 2\pi$ then F(x)= sin(x). If $\displaystyle 2\pi\le x$, F(x)= 0. For negative values of x, reverse those: if $\displaystyle -\pi\le x\le 0$ then $\displaystyle 0\le -x\le \pi$ so F(x)= -f(-x)= 0. If $\displaystyle -2\pi\le x\le -\pi$ then $\displaystyle \pi\le -x\le 2\pi$ so F(x)= -f(-x)= -sin(-x)= sin(x). If $\displaystyle x\le -2\pi$ then $\displaystyle x\ge 2\pi$ so F(x)= -f(-x)= 0.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    What you are writing simply doesn't make sense. F(x)= -f(-x) only for x<0 so you cannot have "$\displaystyle F(x)= -f(-x)$" for $\displaystyle \pi\le x\le 2\pi$.
    Ok, you say F(x)=-f(-x) only for x<0. This apply for all situations or only in the context of what was given in this example?
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