Consider the functions $f(x)$, $g(x)$, $h(x)$, where $f(x)$ is neither odd nor even, $g(x)$ is even and $h(x)$ is odd. Is it possible for $f(x) + g(x)$ to be

  1. even;
  2. odd?

For the second case I can imagine for example $f(x) = x - 1$ and $g(x) = 1$. Then $f$ is neither even nor odd and $g$ is even but their sum is odd, hence it's possible to get odd function from the sum of neither odd nor even and even function.

It feels like $f(x) + g(x)$ can never be even, but I couldn't manage to prove that.

I've tried to do it the following way: Let $f(x) = - g(x) - h(x)$, which doesn't contradict the initial statement. Then we can express $g(x)$ and $h(x)$ and see whether the facts that they are either even or odd holds, but this always leads to valid equations:

$$ h(x) = \frac{f(-x) - f(x)}{2} \;\;\; \text{is an odd function} \\ g(x) = \frac{-f(x) - f(-x)}{2} \;\;\; \text{is an even function} $$

I'm stuck at that point.

How can I prove/disprove that $f(x) + g(x)$ may be even?

  • 2
    $\begingroup$ What is the purpose of $h$ $\endgroup$ – Jakobian Jul 10 '18 at 14:56
  • $\begingroup$ @Adam, I've used it in the definition of $f(x)$ to state that the sum of odd and even function is neither odd nor even $\endgroup$ – roman Jul 10 '18 at 15:08

Well, by definition $f(x)+g(x)$ is even if and only if $f(x)+g(x)=f(-x)+g(-x)$ for all $x$. If $g(x)$ is even, it satisfies $g(x)=g(-x)$ for all $x$. Inserting this into the previous equation gives $f(x)=f(-x)$ for all $x$, hence $f$ must be even.

Why did it work for odd? Here the definition requires $f(x)+g(x)=-f(-x)-g(-x)$. Since $g(x)$ is even, this transforms to $f(x)+g(x)=-f(-x)-g(x)$ i.e. $g(x) = -\frac 12 (f(x)+f(-x))$, which your example satisfies. In other words: if $f$ is not even, subtracting its even part [*] makes it odd.

[*] This refers to the fact that you can decompose every function $f(x)$ into its even part $f_e(x)$ and its odd part $f_o(x)$ such that $f(x) = f_e(x)+f_o(x)$, where $f_e(x) =\frac 12 (f(x)+f(-x))$ and $f_o(x) =\frac 12 (f(x)-f(-x))$.


If $f+g$ is even, then $(f+g)-g$ is the difference of even functions, hence even


f(x) = f(-x)

$g(x) \ne g(-x)$ for some x

z(x) = f(x) + g(x)

then $z(-x) = f(-x) + g(-x) \ne f(x) + g(x) $ for some x

  • 1
    $\begingroup$ You just wrote, at the bottom, that $f(-x) + g(-x) \neq f(-x) + g(-x)$. That's a contradiction, because for any statement $P$ (e.g., $P= f(-x) + g(-x)$) $P=P$ is in fact, true. I suspect you did not express what you intended to express. $\endgroup$ – amWhy Jul 10 '18 at 15:39

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