Exponents in Odd and Even Functions I was hoping someone could show or explain why it is that a function of the form $f(x) = ax^d + bx^e + cx^g+\cdots $  going on for some arbitrary length will be an odd function assuming $d, e, g$ and so on are all odd numbers, and likewise why it will be even if $d, e, g$ and so on are all even numbers.  Furthermore, why is it if say, $d$ and $e$ are even but $g$ is odd that $f(x)$ will then become neither even nor odd?
Thanks.

It’s just a matter of checking the definitions of even function and odd function. Let’s look at a simple case that displays all of the possible behaviors of the more general case: consider $f(x)=ax^m+bx^n$. 


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*Suppose that $m$ and $n$ are even, say $m=2k$ and $n=2\ell$; then $$(-x)^m=(-x)^{2k}=\left((-x)^2\right)^k=\left(x^2\right)^k=x^{2k}=x^m\;,$$ and similarly $(-x)^n=x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=ax^m+bx^n=f(x)\;,$$ and by definition $f$ is an even function.

*Now suppose that $m$ and $n$ are odd, say $m=2k+1$ and $n=2\ell+1$. Then $$(-x)^m=(-x)^{2k+1}=(-x)^{2k}(-x)=x^{2k}(-x)=-x^{2k+1}=-x^m\;,$$ and similarly $(-x)^n=-x^n$, so $$f(-x)=a(-x)^m+b(-x)^n=-ax^m-bx^n=-\left(ax^m+bx^n\right)=-f(x)\;,$$ and by definition $f$ is an odd function.

*Finally, suppose that $m$ is odd and $n$ is even. (The argument is the same if $m$ is even and $n$ is odd, just reversing the rôles of $m$ and $n$.) Then by what we’ve already seen we know that $$f(-x)=a(-x)^m+b(-x)^n=-ax^m+bx^n\;,$$ which is neither $f(x)$ nor $f(-x)$. Thus, $f$ is neither even nor odd.
Nothing essentially different happens if there are more than two terms.
In fact the names even function and odd function come from the behavior of even and odd powers, respectively.
