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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.

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    $\begingroup$ Looks fine {}{}{}{}{}{} $\endgroup$ – hamam_Abdallah Apr 22 '18 at 18:08
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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If the polynomial has real coefficients and odd degree then it should be represented as

$$ p_{2n+1}(x) = (x+\alpha_0)\prod_{k=1}^n((x+\alpha_k)^2+\beta_k^2) $$

NOTE

The complex roots appear in conjugate complex pairs

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  • $\begingroup$ You are assuming what is needed to be proved. $\endgroup$ – Paramanand Singh Apr 23 '18 at 5:51
  • $\begingroup$ I am assuming that a $n$ degree polynomial has $n$ roots. It is a polynomial and not any other real function! $\endgroup$ – Cesareo Apr 23 '18 at 8:07
  • $\begingroup$ The fact that a polynomial of degree $n$ with real coefficients has $n$ real or complex roots is crucially dependent on the fact that a polynomial of odd degree with real coefficients has a real root. So you are already assuming what needs to be proved. $\endgroup$ – Paramanand Singh Apr 23 '18 at 8:46
  • $\begingroup$ I am assuming that it is a polynomial. A polynomial with odd degree has an odd number of roots. This is considered after Gauss a fundamental algebra result. $\endgroup$ – Cesareo Apr 23 '18 at 8:50
  • $\begingroup$ This is what I am trying to say, but unfortunately you don't get it. The result which you mention is called Fundamental Theorem of Algebra. The result can not be proved without using machinery of calculus and analysis and most proofs of this fundamental result depend on the fact that every polynomial of odd degree with real coefficients has a real root. I would suggest you to have a look at the proofs of the Fundamental Theorem of Algebra as mentioned in this blog post. $\endgroup$ – Paramanand Singh Apr 23 '18 at 9:47
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An odd degree polynomial can be written as $a_{2n+1} x^{2n+1}+p_{2n}(x)$ We know that $a_{2n+1} x^{2n+1}$ range, includes $p_{2n}(x)$ range because $x^{2n+1}$ range is $(-\infty, \infty)$. We know also that $a_{2n+1} x^{2n+1}+p_{2n}(x)$ is continuous then follows that there exists at least one point where the graphic of $a_{2n+1} x^{2n+1}+p_{2n}(x)$ crosses the $0$ value.

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