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Thread: Even and odd parts of a piecewise defined function

  1. #1
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    Even and odd parts of a piecewise defined function

    Hello,

    I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$

    We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.

    So for even/odd to mean anything $x$ should live in an interval which is symmetric with respect to 0. For otherwise there would exist some $x_0$ such that either $f(x)\neq f(-x)$ or $f(-x) \neq -f(x)$ simply because the function isn't defined on appropriate values.

    I tried just naively shoving function values into the formula. So $f_e(x)=1/2 (f(x)+f(-x))=(e^{-x}+x^2)$ but since $f(-x)=1/2(e^{-(-x)}+(x)^2)$ this function is not even. So this can't be correct.

    Is there some mistake?
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  2. #2
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    Re: Even and odd parts of a piecewise defined function

    use the formula more carefully

    $f_e(x) = \begin{cases}\dfrac{x^2 + e^x}{2} &x < 0 \\\dfrac{e^{-x}+x^2}{2} &0 < x \end{cases}$
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  3. #3
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    Re: Even and odd parts of a piecewise defined function

    Quote Originally Posted by bkbowser View Post
    I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$
    We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.
    What is the exact formulation of $f(-x)$ you are using.

    i.e. $f(-x)=\begin{cases}? & x<0 \\ ?? & x>0.\end{cases}$

    FYI This is a very important idea: Any function is sum of an even function and an odd function.
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  4. #4
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    Re: Even and odd parts of a piecewise defined function

    Quote Originally Posted by bkbowser View Post
    Hello,

    I'm supposed to find the even and odd parts of the following $$f(x)=\begin{cases}x^2 & x<0 \\ e^{-x} & x>0.\end{cases}$$

    We have a formula, $f(x)=f_e(x)+f_o(x)=1/2 (f(x)+f(-x))+1/2(f(x)-f(-x))$.

    So for even/odd to mean anything $x$ should live in an interval which is symmetric with respect to 0. For otherwise there would exist some $x_0$ such that either $f(x)\neq f(-x)$ or $f(-x) \neq -f(x)$ simply because the function isn't defined on appropriate values.

    I tried just naively shoving function values into the formula. So $f_e(x)=1/2 (f(x)+f(-x))=(e^{-x}+x^2)$ but since $f(-x)=1/2(e^{-(-x)}+(x)^2)$ this function is not even. So this can't be correct.

    Is there some mistake?
    The "even part" of a function is $\frac{f(x)+ f(-x)}{2}$.

    If x< 0 then -x> 0 so for x< 0, $f(x)= x^2$ and $f(-x)= e^{-(-x)}= e^x$. If x< 0, $\frac{f(x)+ f(-x)}{2}= \frac{x^2+ e^x}{2}$. If x> 0, x< 0 for $f(x)= e^{-x}$ and $f(-x)= (-x)^2= x^2$ so $\frac{f(x)+ f(-x)}{2}= \frac{e^{-x}+ x^2}{2}.
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  5. #5
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    Re: Even and odd parts of a piecewise defined function

    Quote Originally Posted by HallsofIvy View Post
    The "even part" of a function is $\frac{f(x)+ f(-x)}{2}$.

    If x< 0 then -x> 0 so for x< 0, $f(x)= x^2$ and $f(-x)= e^{-(-x)}= e^x$. If x< 0, $\frac{f(x)+ f(-x)}{2}= \frac{x^2+ e^x}{2}$. If x> 0, x< 0 for $f(x)= e^{-x}$ and $f(-x)= (-x)^2= x^2$ so $\frac{f(x)+ f(-x)}{2}= \frac{e^{-x}+ x^2}{2}.
    Sorry, it may just me but I cannot read the above.
    I still ask how this is defined: $f(-x)=\begin{cases}? & x<0 \\ ?? & x>0.\end{cases}$
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