# What is the reason for naming a function odd or even [duplicate]

We say that a function is called odd if $$f(-x)=-f(x)\\ (1)$$ and a function is called even if $$f(-x)=f(x)\\\\\\ (2)$$ But why do we call them odd and even. It feels a very peculiar choice of naming. After a bit of work what I saw is this: $$\begin{array}{ | m{1cm} | m{1cm}| m{5em}| } \hline f(x) & g(x) & f(g(x)) \{\text{even/odd}\} \\ \hline even & even & even \\ \hline \hline even & odd & even \\ \hline \hline odd & even & even \\ \hline \hline odd & odd & odd \\ \hline \end{array}$$

Now if we think functions as numbers in $\Bbb{N}$then whatever happens in the above table hold true if we think of composition of functions as multiplication of numbers in $\Bbb{N}$. But it does not feel right to take multiplication as the same as composition of functions. So is there any other reason why we have named functions which goes according to the properties $(1)$ and $(2)$ as odd and even respectively.

Note: I have written a table which does not seem to process. So if someone can kindly edit the question and fix the table it would be very good. Thankyou

• My guess would be that it has something to do with $x^n$, and more generally the degree of polynomials May 17, 2016 at 17:02
• Here's an experiment you can try: write down examples of odd polynomials and even polynomials that you know, and note carefully what the exponents look like. May 17, 2016 at 17:03
• Actually, the taylor series expansions (power series) for even functions contain only even degree polynomials, and similarly taylor expansions for odds have only odds. May 17, 2016 at 17:04
• If you have the requisite knowledge a really cool to try to prove that odd functions only have odd powers as a part of their taylor series and the same for even functions. May 17, 2016 at 17:07
• One fun thing is that if you look at the pointwise product $fg$ where $f$ and $g$ are either even or odd, it looks like addition - i.e. if both are even or odd, the product is even. If one is even and the other is odd, the product is odd. (Though multiplication and composition don't interact quite right to make this look like the integers - composition isn't right-distributive over multiplication) May 17, 2016 at 17:19

A polynomial functions with every term of an odd power is an odd function. A polynomial functions with all term of an even power is an even function.

And odd function * an odd function is an even function (like an odd number + an even number)

And even function * an even function is an even function (like an even number + an even number)

Does that really seem peculiar to you? Why do we call numbers divisible by two "even" and those that aren't "odd". "even" means.... well, "balanced", level, symmetric etc. An even number splits perfectly in half so "even" makes sense. Likewise if f(-x) = f(x) it's graph is a perfect mirror image: reflective and the same in either direction... i.e. "even". Odd also means self-reflective but off kilter. An odd number splits in to halves with a center unbalanced core. An odd function f(-x) = -f(x) rotates and fits onto itself but isn't reflexive. The negativity influences it. It's an odd weighting factor.

But the best mnemonic and probably the origin of the term is the function $f(x) = x^n$ which is even or odd precisely when $n$ is odd or even.

... or it may have something to do with transformations. If you rotate something by $k\pi$ radians. If $k$ is odd that an odd rotation and the function is upside down. If $k$ is even those are even rotation and the function is back to its cheery rightside up.

This is a really cool reason. Let's think about the set of polynomials $f(x) = x^n$. If $n$ is even then $f(-x) = f(x)$ and if n is odd $f(-x) = -f(x)$.

If you look at other function like $\sin$ the maclaurin polynomial for that function has only has odd powers and $\cos$ only has even powers so $cos$ is even and $sin$ is odd.

This holds for all even and odd functions. The maclaurin polynomial of an even function only has even powers and the maclaurin polynomial for an odd power only contains odd powers.

• Some additional precision is necessary: one should look at the Taylor series centered at $0$ (i.e. Maclaurin). May 17, 2016 at 17:06
• Thanks for that. I'll fix it. May 17, 2016 at 17:07