0
$\begingroup$

What are odd and even functions?

An interpretation with an example will be appreciated.

Thanks in advance.

3 Answers 3

1

Let $f:A\to\Bbb R$ be defined on a set $A\subset\Bbb R$ s.t $-a\in A\iff a\in A$ ($A$ is symmetric wrt. origin).

We have two conditions $f(-x)=-f(x)$ and $f(-x)=f(x)$. So, $f(x)=0$ for all $x\in A$.

  • 0
    So, is this the only case? – 2017-02-27
  • 0
    As you can see. You can trivially verify that $f\equiv 0$ is both even and odd. I have checked necessity. – 2017-02-27
  • 0
    Thank you. Now it is clearer. – 2017-02-27
1

Functions who are odd and even have to satisfy two properties $$(even) \qquad f(x)=f(-x)$$ $$(odd) \qquad f(x)=-f(-x)$$

from which you can deduce that, for all $x$, $f(x)=-f(x)$ which, in most contexts, means that $f(x)=0$. (for example, if you are in the reals, $0$ is the only odd and even function)

  • 0
    Thanks for your answer. So I understand from this that the only case that a function is both odd and even is when $f(x)=0$? – 2017-02-27
  • 0
    No, when $f(x)=0$. $x$ is not an even function, because $x \neq -x$ – 2017-02-27
  • 1
    sorry, it was a typo! – 2017-02-27
  • 0
    @Hamad If a function is both even and odd it must satisfy both of the above properties/equations sum those two equations and see what you get. – 2017-02-27
0

And odd function is a function $f$ satisfying

$$\forall x,\quad f(x)=-f(-x).$$

Examples.

  • $f(x)=\sin(x)$

enter image description here

  • $f(x)=x^3$

enter image description here

Remark: and odd function $f$ always satisfies $f(0)=0$, and it's graph have a central-symmetry with the origin.

And even function is a function $f$ satisfying

$$\forall x,\quad f(x)=f(-x).$$

Examples.

  • $f(x)=\cos(x)$

enter image description here

  • $f(x)=x^2$

enter image description here

Remark: the graph of an even function is symmetric to the line $x=0$.

Edit.

If the function is odd and even, you can deduce from what I have previously said about the symmetries of the graph that $f(x)=0$ for all $x$.

You can also deduce that from an algebraic point of view using $f(x)=-f(-x)=f(-x)$, so $f(-x)=0$ for all $x$, so $f(x)=0$ everywhere.

  • 0
    Thanks for your answer, much appreciated, but I meant when the function is both odd **and** even. – 2017-02-27
  • 0
    @Hamad My mistake, I edited at the end, to provide an answer using the geometric properties of the graph of such a function. – 2017-02-27