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For a function $f(x)$, it was given here with derivation that it has $E(x)$ as even part and $O(x)$ as odd part,

$$f(x)=E(x)+O(x)$$

why does this have to be always true? What is the proof?

Community
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Prabhpreet
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    The proof is in the derivation: you define those functions $f_e$ and $f_o$, check that they are respectively even and odd, and check that $f=f_e+f_o$. – icurays1 Aug 05 '13 at 06:36

1 Answers1

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If our function takes values in a ring in which $2$ has an inverse, we can always write:

$$f(x)=\underbrace{\frac{f(x)+f(-x)}{2}}_{E(x)}+\underbrace{\frac{f(x)-f(-x)}{2}}_{O(x)}$$

Jared
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    I love this trick. It can help a lot with simplifying proofs in some situations. – Cameron Williams Aug 05 '13 at 06:42
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    to remember how this works: given $f$, what is the laziest way to make an even function out of it? we just take $f(x) + f(-x)$. and the same sort of thing for odd, take $f(x) - f(-x)$. then we see that $2f(x) = (f(x) + f(-x)) + (f(x) - f(-x))$ – citedcorpse Aug 05 '13 at 06:53
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    only caveat is that $f(x)$ and $f(-x)$ must be defined – ratchet freak Aug 05 '13 at 08:47
  • @Jared Sorry I didn't get the proof. I wanted to ask whether all functions are equal to sum of even and odd functions of themselves. If so, how? – Prabhpreet Aug 06 '13 at 04:30
  • @Prabhpreet: Notice how in the above equation, the part labeled $E(x)$ is an even function: replacing $x$ with $-x$ doesn't change the expression at all. Similarly, the part labeled $O(x)$ is an odd function because replacing $x$ with $-x$ changes the sign of the expression. Now, also notice that by combining fractions, the RHS is just $f(x)$. So we've shown that $f(x)$ is indeed the sum of an even function and an odd function. – Jared Aug 06 '13 at 05:04