When an odd polynomial is a one-one map on $\mathbb{R}$

Question

Let $f(x)=c_1x+c_3x^3+c_5x^5+\cdots+c_{2m+1}x^{2m+1}$, $c_1,c_3,c_5,\ldots,c_{2m+1} \in \mathbb{R}$, $m \in \mathbb{N}$, namely, $f$ is an odd polynomial over $\mathbb{R}$.

When such a polynomial is a one-one map on $\mathbb{R}$? Is it always one-one?

Example:

$f(x)=x+x^3$. If $f(a)=f(b)$ for $a,b \in \mathbb{R}$, then $a+a^3=b+b^3$, so $(a-b)+(a^3-b^3)=0$, and then, $(a-b)(1+(a^2+ab+b^2))=0$.

Therefore, $a-b=0$ or $1+(a^2+ab+b^2)=0$. In the first case $a=b$ and we are done, while in the second case, the discriminant is $-4-3a^2 < 0$, so there are no $a,b \in \mathbb{R}$ satisfying $1+(a^2+ab+b^2)=0$.

I guess that a general solution should be similar to the special case of the example: If $c_1a+c_3a^3+c_5a^5+\cdots+c_{2m+1}a^{2m+1}=c_1b+c_3b^3+c_5b^5+\cdots+c_{2m+1}b^{2m+1}$, then $(a-b)(c_1+c_3(a^4+a^3b+a^2b^2+ab^3+b^4)+\cdots)=0$, so (first case) $a-b=0$ and we are done, or (second case) $c_1+c_3(a^4+a^3b+a^2b^2+ab^3+b^4)+\cdots=0$, which I am not sure I know how to show that it is necessarily non-zero (it is an even polynomial).

Answer 1

That's not true in general. It is 1-1 iff its derivative never changes the sign. Counterexamples would be any of the antiderivatives of $g(x)=(x-1)(x+1)$. One the other hand, one of the sufficient conditions would be to have all of the coefficients of the same sign. Also requiring the coefficient of the top power monomial to be large(or small) enough with respect to other coefficients and to have the same sign as the constant term would be sufficient. Here the idea is the following: We know that if we go away from zero then any polynomial will be 1-1. Therefore forcing the polynomial to speed up around zero too(to make it 1-1) requires some dancing with the coefficient of the top power monomial.

The question of positivity of arbitrary polynomial is a difficult one and is the subject of real algebraic geometry. I suspect that even in the case of having only odd powers of monomials it is not an easy question.

Answer 2

I'm not 100% sure but I believe(another answer has now provided a proof) a non-constant polynomial forms a 1:1 mapping from the reals to the reals if and only if one of the following is true.

$$\frac{df}{dx}\ge 0 \space\space\forall x \in \mathbb{R} $$

$$\frac{df}{dx}\le 0 \space\space\forall x \in \mathbb{R} $$

Lets look a bit at the case presented in your question of polynomials with only odd powers of $x$ (note: I believe this is not the normal definition of "odd polynomial").

$\frac{df}{dx}=c_1+c_33x^2+c_55x^4+\cdots+c_{2m+1}(2m+1)x^{2m}$

The derivative only contains even powers. So the sign of each term in the derivative is always either zero or the same as the sign of it's coefficient.

• If all coefficients have the same sign then the function is one to one.
• If the first and last non-zero coefficients have different signs then the function is not one to one because for sufficiently large $x$ the largest power dominates and for sufficiently small $x$ the smallest (possibly zero) power dominates.

Of course that leaves unsolved that case were the first and last coefficients have the same sign but the intermediate coefficients have different signs.

Answer 3

The answer is no, ti is not always one-to-one.

Consider the following odd polynomial

$$P(x)=x(x-5)(x-10)= x^3-15x^2+50x$$

$$ P(0)=P(5)=P(10)=0$$

It is not one-to-one.

However, a function will be one-to-one if it's derivative is always positive or always negative.

Answer 4

I'm writing this answer in response to the discussion in the comments on Peter Green's answer.

Claim: A non-constant polynomial function $f\colon \mathbb{R}\to \mathbb{R}$ is one-to-one if and only if either $f'(x)\geq 0$ for all $x$ or $f'(x)\leq 0$ for all $x$.

Proof: Since $f'$ is a nonzero polynomial (this is where we use the fact that $f$ is nonconstant), it has only finitely many zeros $z_1<\dots<z_n$. And $f'$ is either strictly positive or strictly negative on each interval $(-\infty,z_1)$, $(z_1,z_2)$, $\dots$, $(z_{n-1},z_n)$, $(z_n,\infty)$.

If $f'(x)\geq 0$ for all $x$, then $f$ is strictly increasing on each of these intervals. It follows by continuity that $f$ is strictly increasing everywhere (e.g. $f(z_i) = \text{sup}\{f(x)\mid x\in (z_{i-1},z_i)\} > f(y)$ for all $y\in (z_{i-1},z_i)$). In particular, it is one-to-one. The case $f'(x)\leq 0$ for all $x$ is similar.

Conversely, suppose $f'$ is positive on some of these intervals and negative on others. Then there is some $z_i$ such that the sign of $f'$ is different on the intervals immediately to the left and right of $z_i$. Say $f'$ is positive on $(z_{i-1},z_i)$ and negative on $(z_i,z_{i+1})$, setting $z_0 = -\infty$ in case $i = 1$ and $z_{n+1} = \infty$ in case $i = n$. Then $f$ has a local maximum at $z_i$, which implies it is not one-to-one (in the other case, $f$ has a local mininum instead).

To argue this carefully, we can pick points $z_{i-1}<x<z_i<y<z_{i+1}$ in these intervals. Then $f(x) < f(z_i)$ since $f$ is increasing on $(z_{i-1},z_i)$ and $f(y)<f(z_i)$ since $f$ is decreasing on $(z_i,z_{i+1})$. If $f(x)<f(y)<f(z_i)$ (the other case is similar), then by the intermediate value theorem there is some $x<x'<z_i$ such that $f(x') = f(y)$. This shows that $f$ is not one-to-one.