The sum of two odd functions

(a) is always an even function

(b) is always an odd function

(c) is sometimes odd and sometimes even

(d) may be neither odd nor even

The answer provided is b.

Here (another Q) the answers seems intuitive and I am able to prove that the sum of two odd functions is always odd. using this - $-f(-x)-g(-x)=-(f+g)(-x)$

I have a function that gives $0$ always yet is the sum of two odd functions:

$f(x) = \sin(x) + \sin(\pi + x)$

Does this not serve as a counterexample for the property? Why?

  • $\begingroup$ @Babai Nope. 1. $h(x) = 0$ is an odd function ($-1 \times 0 = 0$). 2. $g(x) = -x +c $ is not an odd function. $\endgroup$ – stochasticboy321 Oct 14 '15 at 11:24
  • $\begingroup$ @Babai: $g(x)$ is odd only for $c=0$. And in any case $f\equiv 0$ is an odd function (and very special because it is even too) $\endgroup$ – gammatester Oct 14 '15 at 11:25
  • $\begingroup$ Yes, my bad. $f(x)=0$ is odd and even both and $g(x)= -x+c$ is not odd. $\endgroup$ – Babai Oct 14 '15 at 11:27
  • $\begingroup$ @gammatester The property holds true because f(x) = 0 is considered both even and "odd" and my function is also even and odd so, does not serve as a counter example. Correct? $\endgroup$ – Aditya Guru Oct 25 '15 at 15:53

A function $f$ defined over the reals is odd if, for all $x$, $f(x)=-f(-x)$.

Does this hold for the constant zero function? Yes, it does.

The constant zero function is also even, so what? Nothing in the definition of even or odd function suggests that a function cannot be both.

One might exclude the constant zero function from the even and odd functions, but so doing we wouldn't have the theorem saying that every function over the reals is the sum of an even and an odd function (which are uniquely determined). The statement would become unnecessarily complicated.

The situation is similar to the complex numbers: a complex number $z$ is real if (and only if) $z=\bar{z}$ (complex conjugate); it is purely imaginary if (and only if) $z=-\bar{z}$.

It happens that $0$ satisfies both definitions, just like the constant zero function happens to be both even and odd. And, like $0$ is the only real and purely imaginary complex number, the constant zero function is the only function that is even and odd.

There are other similar cases: an $n\times n$ matrix can be both symmetric and antisymmetric (it would be the zero matrix); a relation can be both an equivalence relation and an order relation (it would be the identity relation).


Hint: $f(x)+g(x)=(f+g)(x)$ Therefore $-f(-x)-g(-x)=-(f+g)(-x)$

  • $\begingroup$ I have reiterated now in my Q that I do not need this proof. Anyways thanks for the try. $\endgroup$ – Aditya Guru Oct 25 '15 at 15:46

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