# How to prove whether a polynomial function is even or odd

We know that a function is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$. With this reasoning is it possible to prove that a polynomial function such as $f(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + ...+a_{2}x^2 + a_{0}$ is even or odd?

What do you suggest? How do we get started?

• Yes, it's even if all the powers are even ($(-x)^{m}=x^{m}$ if $m$ is even). Nov 26 '11 at 18:04
• Oh,ok.The degree of the polynomial says it all is it? Is there any other reasoning,axioms,theorems that we can use?
– alok
Nov 26 '11 at 18:08
• Not the degree (that's the power of the leading term)... If all the powers are even, as you have, it's even. If all the powers are odd, it's odd. Nov 26 '11 at 18:09
• Put it this way: all polynomials $p(x)$ that are even can be expressed in the form $q(x^2)$ for some polynomial $q(x)$, and all odd polynomials can be put into the form $x s(x^2)$, where $s(x)$ is some polynomial. Nov 26 '11 at 18:13
• Also you can break every polynomial as a sum of an even and odd polynomial (breaking to even and odd powers); if the even part is 0, the polynomial is odd, and if the odd part is 0, the polynomial is even. Nov 26 '11 at 18:21

Two polynomials have the same value at every real number if and only if they are identical (exact same coefficients in each and every power of $x$); this follows because a polynomial of degree $n\gt 0$ can have at most $n$ roots. If $f(x)$ and $g(x)$ are the same at every value of $x$, then $f-g$ has infinitely many roots, and so must be the zero polynomial.

So write $$f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x + a_0.$$

Then \begin{align*} f(-x) &= a_n(-x)^n + x_{n-1}(-x)^{n-1} + \cdots + a_1(-x) + a_0\\ &= a_n (-1)^nx^n + a_{n-1}(-1)^{n-1}x^{n-1} + \cdots + a_1(-1)x + a_0\\ &= \left((-1)^na_n\right)x^n + \left((-1)^{n-1}a_{n-1}\right)x^{n-1} + \cdots + \left(-a_1\right)x + a_0. \end{align*}

Now, $f(x)$ is even if and only if $f(x) = f(-x)$. That means that we need $$a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = (-1)^nx^n + (-1)^{n-1}a_{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$ That, in turn, means we need $$a_n = (-1)^{n}a_n,\quad a_{n-1}=(-1)^{n-1}a_{n-1},\ldots,\quad a_1 = -a_1, \quad a_0=a_0.$$

If $k$ is even, then $(-1)^k = 1$, so we automatically get $a_k=(-1)^ka_k$; that just says we need $a_k=a_k$, which is always true. If $k$ is odd, then $(-1)^k = -1$, so we need $a_k=-a_k$. This can only happen if $a_k=0$.

So $f(x)$ is an even function exactly when all odd terms to have coefficient $0$. So it must be a polynomial in which the only powers of $x$ that "show up" are even powers of $x$ (including $x^0$ which gives the constant term).

For $f(x)$ to be an odd function, we need $-f(x)=f(-x)$. That means that we need: $$-\Bigl(a_nx^n+a_{n-1}x^{n-1} + \cdots + a_1 x + a_0\Bigr) = (-1)^nx^n + (-1)^{n-1}x^{n-1}+\cdots +(-a_1)x + a_0.$$ That, in turn, means we need $$-a_n = (-1)^{n}a_n,\quad -a_{n-1}=(-1)^{n-1}a_{n-1},\ldots\quad, -a_1 = -a_1, \quad -a_0=a_0.$$ This time, if $k$ is odd, then we are asking for $-a_k = -a_k$ to be true, which it always is; and if $k$ is even we are asking for $-a_k = a_k$, which is true if and only if $a_k=0$.

So $f(x)$ is an odd function exactly when all the even coefficients are zero; that is, the only powers that "show up" are odd powers of $x$.

– alok
Nov 27 '11 at 2:45

Remember that $-1$ raised to an even power is $1$. With that in mind, we can factor out the $-1$ and see \begin{align*} f(-x) &= a_{2n}(-x)^{2n} + a_{2n-2}(-x)^{2n-2} + \cdots + a_2(-x)^2 + a_0\\ &= a_{2n}(-1)^{2n}x^{2n} + a_{2n-2}(-1)^{2n-2}x^{2n-2} + \cdots + a_2(-1)^2x^2 + a_0\\ &= a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + \cdots + a_2x^2 + a_0\\\\ &= f(x). \end{align*} Whenever all the powers of $x$ in a polynomial are even, the $-1$s will be "swallowed up" in this way, ensuring the polynomial itself is even.

You could use a similar approach to show that whenever all the powers of $x$ in a polynomial are odd, the polynomial itself is odd.

• Lovely.This is all i was looking for.
– alok
Nov 27 '11 at 2:44

let $f(x) = g(x^2)$ then $f(-x) = g((-x)^2) = g(x^2) = f(x)$.

For oddness let $f(x) = xg(x^2)$ then $f(-x) = -x g(x^2) = -f(x)$