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Consider an odd-degree polynomial. How to prove that it starts a value that has different sign from its end value?


$$ \lim_{x\to -\infty} f(x) \lim_{x\to+\infty } f(x)<0 $$

Please don't use calculus (limit, derivative, integral) in the proof.

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What do you mean by "polynomial starts at a value"? – Dan Shved Apr 30 '13 at 5:59
Do you mean to show that if $f(x)$ is an odd polynomial then $f(x )$ approaches negative infinity to the left and positive infinity to the right? – Jeremy Apr 30 '13 at 6:02
(if its leading coefficient is positive) – Hurkyl Apr 30 '13 at 6:04
@Jeremy: Yes if its leading coefficient is positive as Hurkyl said in his comment. – kiss my armpit Apr 30 '13 at 6:16
Consider the extended statement: For any $n$, polynomials of degree $2n+1$ "start and end at different signs" while polynomials of degree $2n$ "start and end at the same sign". Formalize "start at a negative sign" as: There exists a number $C$ such the $f(x)<0$ for any $x<C$. Formalize similar notions in a similar fashion. Now prove the (extended) statement by induction on $n$. Sometimes, proving a stronger statement is easier than proving a weaker one, when using induction. – user3533 Apr 30 '13 at 6:42
up vote 5 down vote accepted

Fix a polynomial $f(x) \in \mathbb{R}[x]$ of odd degree. Then $f$ has an even number of non-real roots (because roots appear in conjugate pairs), hence an odd number of real roots (counting multiplicities).

Suppose that $f(x)$ has a root of multiplicity $m$ at $x_0$. It's easy to show that $f$ changes sign at $x_0$ if and only if $m$ is odd. As well, $f$ may only change sign at these points, by continuity. It follows that $f$ has an odd number of sign changes (once again, by a parity argument).

But this is precisely to say that $$\lim_{x \to \infty} f(x) \qquad \text{and} \qquad \lim_{x \to -\infty} f(x)$$ have different signs (of course, both will be infinite).

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If your polynomial is odd it is of the form : $a_0 + a_1x + \ldots a_nx^n$ where $n$ is odd. Now we can write this as $x^n (a_0/x^n + \ldots a_n)$. Now for $|x|$ sufficiently large we see that we can make $(a_0/x^n + \ldots a_{n-1}/x)$ arbitrarily small i.e smaller than the leading coefficient $a_n$ thus we see that taking $\lim x \to \infty$ we must have $x^n (a_0/x^n + \ldots a_n) = + \infty$ and since the degree $n$ is odd we get $\lim x \to -\infty (a_0/x^n + \ldots a_n) = - \infty $. As we for each $x$ are multiplying something negative $x^n$ with something positive $(a_0/x^n + \ldots a_n)$

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I see you asked for a proof without limits, however your question is formulated using limits hence I find it hard to answer it without using limits. Maybe you meant that the proof should not use any theorems conserning limits? – Oliver E. Anderson Apr 30 '13 at 6:45
I am sorry if this wasn't the answer you were looking for. – Oliver E. Anderson Apr 30 '13 at 6:51
Thank you it is a good answer as well. – kiss my armpit Apr 30 '13 at 7:17

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