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Does there exist an even polynomial (i.e. $f(-x)=f(x)$ for all $x$) that has at least one odd power of $x$?

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In other words, is every even polynomial of the form $g(x^2)$ ? –  lhf Apr 24 at 18:10
    
@lhf yes, that's right. I assume the answer is yes but I'm not sure what a rigorous proof would be.. –  Superbus Apr 24 at 18:11

6 Answers 6

up vote 18 down vote accepted

Hint: You need $f(x) - f(-x)$ to be the zero polynomial.

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Why is the hint-only answer voted 16, while the proof answer is voted 5? –  ignis Apr 25 at 16:02
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Because it's much more valuable to look for the answer by yourself. –  mak Apr 25 at 17:00

The answer depends on which field you are working in. For example, over a field of characteristic $2$, every polynomial is even. Also, over a finite field $F_q$, we have that the polynomial $x^q - x$ is even (it vanishes for all $x\in F_q$) but its $x$ term has a non-zero coefficient.

However, over an infinite field of characteristic $\neq 2$, every even polynomial is a polynomial in $x^2$. This follows since $f(x) - f(-x) = 2g(x)$, where $g(x)$ consists of the odd powers of $x$ in $f$ and by assumption $2g(x)$ vanishes for all, and hence infinitely many, elements $x$. Thus $2g(x) = 0$ which implies $g(x) = 0$ (char $\neq 2$).

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+1 for the caveat about finite fields –  Mario Carneiro Apr 24 at 18:18
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Your comment about finite fields seems to be based on the identification of formal polynomials with polynomial functions. For the reason you describe, this is problematic in finite fields. –  Ryan Reich Apr 25 at 2:28
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I think the most natural definition of an even polynomial $P\in K[X]$ is one that does not change when $-X$ is substituted for$~X$. Then indeed in characteristic$~2$ all polynomials are even (the substitution is a no-op), but in other characteristics the even polynomials are exactly the ones without terms in odd degree, regardless of whether $K$ is finite or infinite. In particular $X^q-X$ is not even over $\Bbb F_q$ for an odd prime power$~q$. –  Marc van Leeuwen Apr 25 at 5:04

No. If $p(x)=a_nx^n+\cdots +a_1x+a_0$ then $$p(-x)=p(x)$$ $$a_0-a_1x+\cdots +(-1)^na_nx^n=a_0+a_1x+\cdots a_nx^n$$ Polynomials are identically equal exactly when their coefficients are equal, so $$a_1=-a_1\implies a_1=0,...$$ all coefficients of odd powers are $0$.

Alternatively, use induction. If $p$ is even, then $$-p'(-x)=p'(x)$$ so $p'$ is odd, and vice versa. Prove that constant and linear polynomials which are even and odd have only even and odd (respectively) terms, and you're done.

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Split $\,f(x)\,$ into its even part $\,f_0(x) = \sum f_{2k} x^{2k}$ and odd part $\,f_1(x) = \sum f_{2k+1} x^{2k+1}$

Thus $\,f(x) = f_0(x)+ f_1(x)\ $ where $\ f_0(-x) = f_0(x),\,$ and $\ f_1(-x) = -f_1(x)$

So $\ \ f(-x) = f_0(x)-f_1(x)\ $

So $\ \ f(-x) = f(x)\!\iff\! 2 f_1(x) = 0\!\iff\! f_1(x) = 0\ $ (assuming $\,2\,$ is cancellable).

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Alternate answer. The sum of two even functions is even. So if $p$ is an even polynomial, add scalar multiples of even powers of $x$ to eliminate the even powers in $p(x)$. You'll be left with a supposedly even polynomial that has only odd powers of $x$. But that makes what you have an odd function. So you have a function that is even and odd, and it's easy to prove such a function is the zero function. Therefore there were no odd terms left over after cancelling the even terms.

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Among smooth functions, the derivative of an even function is odd, and vice versa (and this is quite easy to prove). So if $p$ were even and there were an odd powered term, say $c_kx^k$, in a simplified expression for $p(x)$, then $\frac{d^k}{dx^k}p(x)$ would be an odd function (since $k$ is odd). But odd functions have $f(0)=0$, and this supposedly odd function would have a nonzero constant term $c_kk!$. So it's not possible.

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