Are there some nice results characterizing which polynomial functions (reals to reals) are injective (for example a necessary and sufficient condition)? Obviously all polynomials of degree $1$ are injective, and for degree ${}>1$ such a polynomial must have odd degree and at at most one root, but that is hardly sufficient.
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3The polynomial must be monotone, so derivative everywhere nonnegative or everywhere nonpositive. – hardmath Jan 05 '15 at 02:22
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Sorry for that stupid mistake. I removed it. – math_lover Jan 05 '15 at 03:04
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@hardmath that is a nessecary and sufficient condition so it looks like this was a stupid question after all! – math_lover Jan 05 '15 at 03:18
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To add to hardmath's comment, we can actually find algebraic conditions for the derivative not changing signs by using Sturm's theorem to find algebraic conditions for which the derivative has no roots (and then we can take the closure of the resulting set, to allow for double roots, which do not change the sign) – Milo Brandt Jan 05 '15 at 03:26
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A (maybe small) subclass of them are those with all odd powers with all coefficients the same sign. – David P Jan 05 '15 at 03:43
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A continuous function $f:\mathbb{R}\to \mathbb{R}$ is injective (1-1) if and only if it is monotone (strictly increasing or strictly decreasing). If $f$ is a (real) polynomial, this means the derivative $f'$ is nonnegative or nonpositive, respectively, as well as not identically zero.
This is only possible if $f$ has odd degree (equiv. $f'$ has even degree), but of course this is only a necessary condition. A sufficient condition would be (as Meelo comments) that $f'$ has no real roots. A necessary and sufficient condition would be that any real roots of $f'$ have even multiplicity.
In exact arithmetic these conditions can be investigated using Sturm sequences. In particular the derivative $f'$ would factor into $P(x)Q^2(x)$, where $P(x)$ is square-free and has no real roots, if and only if $f$ is monotone.
hardmath
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