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If $f$ is to be written as a sum of the even function $E$ and the odd function $O$,
$E=\dfrac{f(x) + f(-x)}{2} \quad$ and
$O=\dfrac{f(x)-f(-x)}{2}$
obviously works.

I get a bit confused though because in a previous question I was asked to investigate if the sum of two functions $g$ and $h$ is even, odd or neither depending on if $g$ and $h$ is even or odd. I came to the conclusion that (with the subscript denoting if function is even or odd) $g_e(x) + h_o(x) $ is neither even nor odd assuming no function is the zero function:

$k(x) = g_e(x) + h_o(x) \implies k(-x) = g_e(-x) + h_o(-x) = g_e(x) - h_o(x) \quad$($k$ not even)
$k(x) = g_e(x) + h_o(x) \implies -k(-x) = -g_e(-x) - h_o(-x) = -g_e(x) + h_o(x)\quad$($k$ not odd)

So does the two facts:

1.) The sum of an even function $g_e$ and an odd function $h_o$ is neither even nor odd unless $g_e(x)=0$ or $h_o(x)=0$.

2.) Any function, even, odd or netheir, can be written as the sum of an even and an odd function.

mean that $E(x)=0$ or $O(x)=0$ if $f$ is even or odd?
Or am I just totally wrong about everything.

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    $\begingroup$ You got everything right! $\endgroup$ – Zardo Jul 20 '15 at 8:25
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You are actually right. However, it would be better to add something to fact number 2:

Every function can be written uniquely as the sum of an even function and an odd function.

Another way to say the same thing (using linear algebra terminology) is: The vector space consisting of functions $\mathbb{R}\to\mathbb{R}$ is the direct sum of the space of even functions and the space of odd functions.

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The function $k(x)=0$ statistifies both $k(x)=-k(-x)$ and $k(x)=k(-x)$, so it is even and odd. This is the only function that is both even and odd.

If $f$ is originally even or odd, substituting the even or odd definition will give you that $O(x)=0$ or $E(x)=0$.

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