1
$\begingroup$

Question: Show that every polynomial of odd degree with real coefficients has at least one real root.

Proof: $f(x)$ is of the form:

$f(x) = a_nx^n +a_{n-1}x^{n-1}+.....+a_2x^2 + a_1x +a_0$

$f(x)$ is an odd degree polynomial and is hence continuous on $\mathbb{R}$.

Then, if $f(x)$ has a positive leading coefficient

$\implies \lim_{x \to \infty}f(x)=\infty$ and $\lim_{x \to -\infty}f(x)=-\infty$

$\implies \exists \alpha,\beta \in\mathbb{R}$ with $ \alpha > \beta$ such that:

$ \alpha>0$ and $f(\alpha)>0$ and $\beta <0$ and $f(\beta)<0$

Now, consider the restriction of $f(x)$ onto the interval $I:=[\beta,\alpha]$

The restriction of $f$ on $I$ is also continuous on $I$

Then, since $f(\beta)<0<f(\alpha)$, we have by the Location of roots theorem that:

$ \exists$ a number $c \in (\beta,\alpha)$ such that $f(c) =0$

Can anyone please verify this proof and let me know if it is correct and /or if I'm missing out on something?

Note: I've missed out on the case with negative leading coefficient as I believe that too will be done in a similar fashion should this proof be alright.

Thank you.

$\endgroup$
  • 1
    $\begingroup$ Looks fine {}{}{}{}{}{} $\endgroup$ – hamam_Abdallah Apr 22 '18 at 18:08
  • 1
    $\begingroup$ I don't spot any issues, but it can likely be shortened in a few places. For example, it isn't really necessary to have $\alpha>0$ and $\beta<0$, all you need is to show the existence of $\alpha$ and $\beta$ where $f(\alpha)>0$ and $f(\beta)<0$. In this case, you need to make $I$ the smallest interval containing $\alpha$ and $\beta$, if you want a formula, $I=[\min\{\alpha,\beta\},\max\{\alpha,\beta\}]$. For the negative leading coefficient, observe that $f(x)=0$ iff $-f(x)=0$ and use the proof you already have. $\endgroup$ – Michael Burr Apr 22 '18 at 18:11
  • $\begingroup$ Can you elaborate on the leading coefficient condition a bit? I fail to understand how I can carry forward/ use the fact that $f(x)=0 \iff -f(x)=0$ to show this $\endgroup$ – A.Asad Apr 22 '18 at 18:14
  • 1
    $\begingroup$ I would say that you should really prove the limits you use. If you want to avoid cases for different signs you can note that you can divide by the (non-zero) leading coefficient without changing the zeros of the polynomial, and if you do this first you can deal with the case of leading coefficient equal to $1$. $\endgroup$ – Mark Bennet Apr 22 '18 at 18:21
  • 2
    $\begingroup$ I have no complaints about your proof although you can improve it following Michael's suggestions (because $f$ and $-f$ have the same roots, you can just say "without loss of generality, we can assume $a_n > 0$") However I despair about the terminology you have been taught: why oh why oh why have educators decided to introduce the name "location of roots theorem" for a special case of the intermediate value theorem? $\endgroup$ – Rob Arthan Apr 22 '18 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.