Prove that odd polynomials have at least one root.

Question

Intuitively this is easy for me to understand, but I don't know how to start the proof. Can someone help me?

Prove that any polynomial of the form $x^n+\sum_{i<n}a_{i}x_{i}$, with $a_{i}\in \mathbb{R}$ and $n$ odd, has at least one real root.

Answer 1

Use Intermediate Value Theorem.

Since $\lim_{x\to\infty}f(x)=\infty$, and $\lim_{x\to-\infty}f(x)=-\infty$, in particular there exists $(a,b)$ such that $f(b)>0$ and $f(a)<0$.

By IVT, there exists $c\in (a,b)$ such that $f(c)=0$.

Answer 2

If $f(x) = x^n+\sum_{i<n}a_{i}x^{i} $, (you wrote $x_i$; I assume you meant $x^i$) then for large enough $x$, $f(x) > 0$.

To show this, write $f(x) = x^n(1+\sum_{i<n}a_{i}x^{i-n}) $. We want to choose $x$ large enough so that $\sum_{i<n}|a_{i}|x^{i-n} < 1$. This will be true if $|a_{i}|x^{i-n} < \frac1n $ or $x^{n-i} > n|a_{i}| $ or $x > (n|a_{i}|)^{1/(n-i)} $.

Therefore, if $x > \max_{i<n} (n|a_{i}|)^{1/(n-i)} $, $f(x) > 0$.

Note: There are other, better bounds that can be worked out. This is an easy one.

This is true whether $n$ is even or odd.

If $n$ is odd, then $x^n < 0$ for $x < 0$. Therefore, if $x < - \max_{i<n} (n|a_{i}|)^{1/(n-i)}$, then, as before, $\sum_{i<n}|a_{i}||x|^{i-n} < 1$ so that $|\sum_{i<n}a_{i}x^{i-n}| \le \sum_{i<n}|a_{i}x^{i-n}| < 1$ so that $1+\sum_{i<n}a_{i}x^{i-n} > 0$.

Therefore, if $x < - \max_{i<n} (n|a_{i}|)^{1/(n-i)}$, then $f(x) = x^n(1+\sum_{i<n}a_{i}x^{i-n}) < 0 $.

Since $f(x) > 0$ for all large enough $x$ and $f(x) < 0$ for all negatively large $x$ (with $|x| > \max_{i<n} (n|a_{i}|)^{1/(n-i)}$ working in both directions), since $f(x)$ is continuous, there has to be an $x_0$ such that $f(x_0) = 0$.