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.

If the exponents are all odd, then $f(x)$ is the sum of odd functions, and hence is odd.
If the exponents are all even, then $f(x)$ is the sum of even functions, and hence is even.
As far as your last question, the sum of an odd function and even function is neither even nor odd.

Proof: Sum of Odd Functions is Odd:
Given two odd functions $f$ and $g$.   Since they are odd functions $f(-x) = -f(x)$, and $g(-x) = -g(x)$.
Hence:
\begin{align*}
f(-x) + g(-x) &= -f(x) - g(x) \\
              &= -(f+g)(x) \\
\implies (f+g) & \text{ is odd if $f$ and $g$ are odd.}
\end{align*}
Proof: Sum of Even Functions is Even:
Given two even functions $f$ and $g$, then $f(-x) = f(x)$, and $g(-x) = g(x)$.
Hence:
\begin{align*}
f(-x) + g(-x) &= f(x) + g(x) \\
\implies (f+g) &\text{ is even if $f$ and $g$ are even.}
\end{align*}
Proof: Sum of an odd function and even function is neither odd or even.
If $f$ is odd, and $g$ is even
\begin{align*}
f(-x) + g(-x) &= -f(x) + g(x) \\
              &= -(f-g)(x) \\
\implies (f+g)&\text{ is neither odd or even if $f$ is even, $g$ is odd.}
\end{align*}
