# Proof that every odd polynomial has a real root - NOT a duplicate

I've done a search for this first because it is a common question, however I have yet to find one I follow/like.

I ask this because I actually have an idea for how I might do it, I just don't know how to write it.

Proof that every polynomial of odd degree has one real root <--look it is a duplicate (in title only) that question there has some good answers from a few perspectives (IVT, FTA) but not how I want to do it!

When I was doing my A-levels (before I ever questioned this stuff) I would sketch a lot of graphs (I still do but they all look the same) and one of the first things I'd do is look for asymptotes. Considering what a function does as $x\rightarrow-\infty$ and $x\rightarrow\infty$ was the first step.

An odd polynomial is easy because one extreme of x will have a different sign to the other, at some point it is intuitive/obvious that the value of the function will become almost entirely based on the highest power, that highest power term is a whole x times bigger than the previous term, so some tiny number as a coefficient doesn't really matter (because $x$ times bigger is HUGE).

I think there's a proof there. I want to show that way over at one end the polynomial is tending towards $+$ or $-\infty$ and the other end is the opposite of that. Then by the IVT it must be zero somewhere!

## Thoughts on proof

First of all I need to deal with the issue of going far enough that I've hit the "value of f is basically just that last term, everything else is tiny compared to it" point mentioned before. I don't really want to differentiate because this comes before differentiation in every text book. I could though. I go beyond all turning points (so the gradient does not change sign again) and I'm sorted!

Rather than finding turning points (which would be painful) I can instead just find something that is guaranteed to go beyond them eventually. A sequence that tends towards infinity comes to mind (and that sequence times $-1$ for the other). Then I'd be dealing with $f(x_n)$ and $f(-x_n)$ where $x_n=n$ or something nice.

I want to side-step the issue of finding exactly where it becomes monotonic at each side, to do that I need to work out what the limit of $f(x_n)$.

To do that I could use the ratio-test, I could show that $\frac{f}{x^n}$ tends towards $a_n$(the first coefficient). I can then pick an $\epsilon$ as close as I like to $a_n$ but I'd have to make sure this is beyond any turning points - back to that issue before. (I think - I've edited this twice, might be on to something...)

Not really sure where to go from there (which is why I've not invested any ink in this). Also I'm not sure how to properly write the last step, is $lim_{n\rightarrow\infty}(f(-x_n))<0$ and the other one being $>0$ enough? I could write $=-\infty$ for the limit, but I'm not sure how to write that showing the sign changes (I know that sounds silly but we can never have an infinity, only tend towards it. $x_n$ also has no limit, so I can't use the axiom of completeness)

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If you're going to use the IVT anyway, what's the problem with the usual proof? Since $\;\lim_{x\to -\infty}f(x)=-\infty\;,\;\;\lim_{x\to \infty}f(x)=\infty\;$ (or the other way around, of course), there are $\;a<b\;$ s.t. $\;f(a)f(b)<0\;$ , and we're done then ...Some things are so simple that trying to make them messier...we usually succeed. – DonAntonio Apr 6 '14 at 19:41
@DonAntonio $f(a)f(b)<0$? – Alec Teal Apr 6 '14 at 19:51
Every odd polynomial has a root; and polynomials without roots are very odd indeed! CONTRADICTION! Close the site, and everybody go home. Nothing to see here anymore. – Asaf Karagila Apr 6 '14 at 19:54
@Alec: $xy<0$ is a fancy way of saying "$x$ and $y$ are nonzero and have different signs" – Hurkyl Apr 6 '14 at 20:00
@AsafKaragila: Close the site! Ohhhhhhh Noooooo Br. Bill! – Robert Lewis Apr 6 '14 at 21:06

Certainly, "there's a proof there" as our OP Alec Teal put it, and he apparently has most of it, conceptually speaking at least; viz., in the sense that for $n$ odd, $x^n$ has oppsite signs as $x \to \pm \infty$. So, to flesh out a few remaining details, let us suppose that $p(x) \in \Bbb R[x]$ is of odd degree $n$ , writing the terms in order of descending degree, thus:

$p(x) = \sum_0^n p_{n - i} x^{n - i}, \; p_n \ne 0; \tag{1}$

we may furthermore assume that $p_n > 0$; otherwise we simply replace $p(x)$ by $-p(x)$.

If we re-write $p(x)$ by "factoring out" $p_n x^n$, we find

$p(x) = p_n x^n (1 + \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i}) = p_n x^n (1 + \dfrac{p_{n - 1}}{p_n} x^{-1} + \ldots + \dfrac{p_0}{p_n} x^{-n}); \tag{2}$

we have

$\vert \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i} \vert \le \sum_1^n \vert \dfrac{p_{n - i}}{p_n}x^{-i} \vert = \sum_1^n \vert \dfrac{p_{n - i}}{p_n} \vert \vert x \vert^{-i}, \tag{3}$

and picking any $\epsilon > 0$, we may choose $\vert x \vert$ sufficiently large that $\vert \dfrac{p_{n - i}}{p_n} \vert \vert x \vert^{-i} < \epsilon / n$ for all $i \ge 1$. Then (3) yields

$\vert \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i} \vert < \epsilon, \tag{4}$

and so

$\vert (1 + \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i}) - 1 \vert = \vert \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i} \vert < \epsilon, \tag{5}$

or

$-\epsilon < (1 + \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i}) - 1 < \epsilon, \tag{6}$

or

$1 - \epsilon < (1 + \sum_1^n \dfrac{p_{n - i}}{p_n}x^{-i}) < 1 + \epsilon; \tag{7}$

choosing $\epsilon < 1$ now shows that, for $\vert x \vert$ sufficiently large, the sign of $1 + \sum_1^n \frac{p_{n - i}}{p_n}x^{-i}$ is positive; thus by (2) the sign of $p(x)$ and that of $x^n$ agree, since $p_n > 0$. $p(x)$ is thus negative for $x < 0$ and positive for $x > 0$ if $\vert x \vert$ is sufficiently large; the intermediate value theorem now takes care of the rest, there is an $x_0 \in \Bbb R$ with $p(x_0) = 0$. QED

Now there's an old saw retold one more time!

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

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Hey, I just want to say I enjoyed reading this. It does say "no thanks" style comments but this wasn't just an answer, I liked reading it, thanks. – Alec Teal Apr 6 '14 at 22:49
@Alec Teal: where does it say that? Anyway, the comment cops are usually a little slow on the uptake, so we get away with lots! ;-)! But I'm gratified that you enjoyed it. "Thanks" for the '"thanks"! Regards, RKL – Robert Lewis Apr 6 '14 at 23:01
OK, let's go flag a few comments for thanking and rethanking... ;-) – vonbrand Apr 6 '14 at 23:27
@vonbrand: Get those grateful bastards! – Robert Lewis Apr 6 '14 at 23:28
Wait! There is one thing, why $\le$, why not just $<$ for eq 4 onwards? (seems more effort to type, unless you hate the shift key!) Or is it just "why not" – Alec Teal Apr 7 '14 at 20:29

Suppose that $f(x) = a_n x^n + ... + a_1 x + a_0$ is a polynomial, $n$ is odd, and $a_n \neq 0$. Wlog, we take $a_n > 0$ (else replace $f$ with $-f$). First we need to show that $f(x) \to \infty$ as $x \to \infty$. For this, set $a = \min \{a_{n-1}, ..., a_0\}$. If $a \ge 0$, then there's nothing to show, so we may assume that $a < 0$. So $f(x) = a_n x^n + ... + a_0 \ge a_n x^n + a(x^{n-1} + ... + 1) \ge a_nx^n + nax^{n-1}$ if $x \ge 1$. But $a_n x^n + n ax^{n-1}= x^{n-1}(a_n x + na) \to \infty$ as $x \to \infty$ clearly. Thus $f(x) \to \infty$ as $x \to \infty$.

Make a similar argument for $x \to -\infty$ (here is the point that the fact that $n$ is odd will be necessary)

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The usual argument is to factor out the most significant term:

\begin{align} \lim_{x \to +\infty} a_n x^n + \ldots + a_0 x^0 &=\lim_{x \to +\infty} x^n (a_n x^0 + a_{n-1} x^{-1} + \ldots + a_0 x^{-n}) \\&= \left( \lim_{x \to +\infty} x^n \right) \left( \lim_{x \to +\infty} a_n x^0 + a_{n-1} x^{-1} + \ldots + a_0 x^{-n} \right) \\&= +\infty \cdot \left(a_n + a_{n-1} \cdot 0 + \ldots + a_0 \cdot 0 \right) \\&= a_n \cdot (+\infty) \\&= \operatorname{sgn}(a_n) \cdot (+\infty) \end{align}

We can break the limit into a product of limits, because the two individual limits exist and have values where the product is defined. The two individual limits are continuous at $+\infty$, and so we can evaluate by plugging in.

Depending on what you already know and are comfortable with (e.g. some people are allergic to working with $+\infty$), you may need to prove the facts I used, or use alternative methods.

e.g. the right limit can be done by repeatedly applying the limit laws for addition and multiplication, and the limit form can be seen by proving the lemma

If $\lim f = +\infty$ and $\lim g = L > 0$, then $\lim fg = +\infty$.

(and similarly with the other signs). I think you are thinking these lines, and you're close to the key: what you need to realize is that you can constrain $x$ to ensure $L - \epsilon < g(x) < L + \epsilon$ and $f(x) > N$ for any positive $\epsilon, N$ you choose, and so $f(x)g(x) > (L - \epsilon) N$.

Taking your starting point isn't all that different: your method of proving

$$\lim_{x \to +\infty} \frac{f(x)}{x^n} = a_n$$

is essentially the same thing as my right hand limit. And to get what you want:

$$\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} x^n \cdot \frac{f(x)}{x^n} = (+\infty) \cdot a_n$$

The extended real numbers (which do include $\pm \infty$) also satisfy the completeness axiom; in fact, they satisfy a slightly stronger version: every subset of the extended real numbers has a least upper bound. For example, the least upper bound of the empty set is $-\infty$, and the least upper bound of $\mathbf{R}$ is $+\infty$.

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I hope this would help:

We first check a particular case, $P(x)= x^n +a_{n-1}x^{n-1}+ \ldots+a_0$. We shall show that the polynomial contain a real root. The basic idea is that the leader term is what determine the behavior of $P(x)$ for $|x|$ bigger. Since $P(x)= x^n \left(1+\frac{a_{n-1}}{x}+\ldots \frac{a_0}{x^n}\right)$ when $x\not = 0$, the first thing to do is to estimate what is inside the parenthesis:

$$\left|\frac{a_{n-1}}{x}+\ldots \frac{a_0}{x^n} \right|\le\frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x^2|}+\ldots+\frac{|a_{0}|}{|x^n|}$$

Now let $|x|> N= \max \bigg(1,2n|a_{n-1}|,\ldots, 2n |a_0|\bigg)$. So we have $\frac{|a_{n-i}|}{|x^i|}\le \frac{|a_{n-i}|}{|x|}< \frac{1}{2n}$

$$\underbrace{\frac{|a_{n-1}|}{|x|}+\frac{|a_{n-2}|}{|x^2|}+\ldots+\frac{|a_{0}|}{|x^n|}}_\text{n terms} < \frac{1}{2}$$

Hence

$$\frac{1}{2}<1+\frac{a_{n-1}}{x}+\ldots \frac{a_0}{x^n}$$

Let $x>0$ and $x>N$. Thus

$$0<\frac{x^n}{2}<x^n \left(1+\frac{a_{n-1}}{x}+\ldots \frac{a_0}{x^n}\right)= P(x)$$

in other words for $x> N$ we must have $P(x)>0$

Now let $y<0$ and $|y|> N$. Thus $y^n<0$ since $n$ is odd. So

$$0>\frac{y^n}{2}>y^n \left(1+\frac{a_{n-1}}{y}+\ldots \frac{a_0}{y^n}\right)= P(y)$$

Then for $y<-N$ we must have $P(y)<0$. Thus in $[y,x]$ we have that $P(y)<0<P(x)$ and so by the IVT there is a $c\in (y,x)$ such that $P(c)=0$ as desired.

For the general case if $P(x)= a_nx^n +a_{n-1}x^{n-1}+ \ldots+a_0$, where $a_n\not= 0$. Define $F(x)=x^n +b_{n-1}x^{n-1}+ \ldots+b_0$, where $b_j=a_j/a_n$. So there is a closed interval $[y,x]$ such that $F(y)<0<F(x)$. Now if $a_n>0$ so $P(y)=a_nF(y)<0<a_nF(x)=P(x)$. If $a_n<0$ so $P(x)=a_nF(x)<0<a_nF(y)=P(y)$. Thus in any case $P(x)P(y)<0$.

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