I want to prove that every real polynomial of odd degree has at least one real root, using the intermediate value theorem.

Let $P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0$ for each $a_i \in \mathbb{R}$ and $n \in \mathbb{N}$.

By the fundamental theorem of algebra I know that $P(x)$ has exactly $2n+1$ complex roots, so

$P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})$ for each $r_i \in \mathbb{C}$

I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since $2n+1$ is odd there is at least $1$ root that is imaginary part $0$ and thus real.

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    $\begingroup$ We do not need the Fundamental Theorem of Algebra at all. If we decide to use it, we do not need the IVT. $\endgroup$ – André Nicolas Feb 25 '14 at 7:44
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    $\begingroup$ $andre my book says that I must do this proof with the fundamental theorem of algebra and IVT, that is why I have included this. $\endgroup$ – terrible at math Feb 25 '14 at 7:44
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    $\begingroup$ Do you mean you should supply two proofs, one with the Fundamental Theorem and another with IVT? $\endgroup$ – Hagen von Eitzen Feb 25 '14 at 7:45
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    $\begingroup$ @terribleatmath That's impossible unless you accidentally prove the result twice in the same proof. $\endgroup$ – Hagen von Eitzen Feb 25 '14 at 7:47
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    $\begingroup$ Show that $P(x) \to + \infty$ as $x \to + \infty$ and $P(x) \to - \infty$ as $x \to - \infty$. $\endgroup$ – copper.hat Feb 25 '14 at 7:49

Method of FTA: $$P(\overline z)=\sum_{k=0}^{2n+1}a_k\overline z^k=\sum_{k=0}^{2n+1}\overline a_k\overline{z^k}=\sum_{k=0}^{2n+1}\overline{a_kz^k}=\overline{\sum_{k=0}^{2n+1}a_kz^k}=\overline{P(z)}$$ which states $z$ is a root for $P(z)=0$ iff its complex conjugate $\bar z$ is. According to FTA, there are odd number of roots for a polynomial of odd degree. That implies there must be one single root $z$ satisfying $z=\bar z$, hence the real root.

Method of IVT:

$$\frac{P(x)}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_k\frac{x^k}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_kx^{k-(2n+1)}$$ For any $\varepsilon>0$, there exists $N>0$ such that for all $|x|>N$, $\left|\sum_{k=0}^{2n}a_kx^{k-(2n+1)}\right|<\varepsilon$. Hence for $x>N$, we have $P(x)>x^{2n+1}-\varepsilon x^{2n+1}>0$ and similarly for $x<-N$, we have $P(x)<0$. Then IVT implies there exists some $y$ such that $P(y)=0$.

  • $\begingroup$ Hello @Shuchang. Could you please elaborate a little more on how for any $\varepsilon>0$, there is a $N>0$ such that for all $\left|x\right|>N$ we have $\left| \sum_{k=0}^{2n} a_k x^{k-(2n+1)} \right|<\varepsilon$? $\endgroup$ – AttractorNotStrangeAtAll Nov 20 '20 at 12:55

Indeed it is true that all proofs of the fundamental theorem of algebra need some piece of analysis. Even the most algebraic proof of FTA (Euler, Gauß II) relies on the fact that all odd-degree real polynomials have at least one real root.

First consider the case of relatively large positive $x$. Assuming $x\ge 1$ as provisional lower bound, then $1\le x^k\le x^{2n}$ for $0\le k\le 2n$ and the value of the polynomial is bounded below by $$ P(x)\ge x^{2n+1}-\sum_{k=0}^{2n}|a_k|x^k\ge x^{2n+1}-x^{2n}\sum_{k=0}^{2n}|a_k| =x^{2n}\left(x-\sum_{k=0}^{2n}|a_k|\right) $$

We can now try to push the last expression on the right into positive territory by increasing the lower bound for $x$. At the Lagrange root bound $$ R=\max\left(1,\sum_{k=0}^{2n}|a_k|\right), $$ the right side for $x\ge R$ gives a non-negative bound. Increasing the lower bound to $x\ge 2R$ will result in $$ x≥2R \implies P(x)\ge (2R)^{2n}\cdot R\ge 2^{2n}>0. $$

The same reasoning can be applied to $-P(-x)=x^{2n+1}-a_{2n}x^{2n}+a_{2n-1}x^{2n-1}\mp...-a_0$, so that

$$x≤-2R \implies P(x)≤-(2R)^{2n}\cdot R≤-2^{2n}<0.$$

In total one obtains $$ P(-2R)≤-(2R)^{2n}\cdot R ≤ -2^{2n}<0<2^{2n}≤(2R)^{2n}\cdot R≤P(2R) $$ which allows to apply the intermediate value theorem for $P$ concluding for a real root of $P$ inside $(-2R, 2R)$, but really already inside $(-R,R)$.

  • $\begingroup$ I'm curious on what you mean by your first assertion. I'm feel that many proofs of the FTA doesn't require the fact that all odd degree polynomials have at least one real root. For example, off the top of my head, the proof using Liouville's theorem doesn't seem to require this fact. $\endgroup$ – EuYu Feb 25 '14 at 8:24
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    $\begingroup$ This is then an almost purely analytical proof. If one tries to prove the FT of Algebra with almost exclusively algebraic means one sees that it is impossible, you need properties of real numbers, continuous functions and the intermediate value theorem in some fashion. In the maximum principle of harmonic functions these basic facts are buried some layers deep. $\endgroup$ – Lutz Lehmann Feb 25 '14 at 8:49
  • $\begingroup$ I split the first sentence to reflect that. All proofs need calculus, the minimal amount of calculus is required for the Euler-Gauß II proof using only the roots of odd-degree polynomials and complex square roots. $\endgroup$ – Lutz Lehmann Feb 25 '14 at 8:55
  • $\begingroup$ @LutzL hey i put a bounty on this question because I am desperate for an answer, I am almost close -- I will give it to you if you can answer this for me. I ended up proving that a polynomial $ax^3 + bx^2 + cx + d$ always has at least one real root. Can i extend this to ALL odd degree polynomials ? $\endgroup$ – terrible at math Feb 28 '14 at 1:40
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    $\begingroup$ If it were that easy, then the FTA would not be such a famous theorem with 100 years of failed or incomplete proof attempts. If a root is known, then factoring by deflation a la Horner-Ruffini is easy. Proving the existence of the root is non-trivial. $\endgroup$ – Lutz Lehmann Feb 28 '14 at 3:19

Let $p(x)=a_0 + a_1 x + \dots + a_n x^n$ a polynomial $p: \mathbb{R} \to \mathbb{R}$ with $n$ odd and $a_n \neq 0$. Suppose $a_n >0$. We can write $p(x) = a_n x^n \cdot r(x)$ with $$r(x) = \frac{a_0}{a_n} \cdot \frac{1}{x^n} + \frac{a_1}{a_n} \cdot \frac{1}{x^{n-1}}+ \cdots + \frac{a_{n-1}}{a_n} \cdot \frac{1}{x} + 1.$$

So, we have $$\lim_{x \to +\infty} r(x) = \lim_{x \to -\infty} = 1.$$ Thus $$\lim_{x \to +\infty} p(x) = \lim_{x \to +\infty} a_n x^n = +\infty$$ and $$\lim_{x \to -\infty} p(x) = \lim_{x \to -\infty} a_n x^n = -\infty$$ because $n$ is odd. Therefore the interval $p(\mathbb{R})$ is ilimited inferior and superiorly, i.e. $p(\mathbb{R}) = \mathbb{R}$. This means that $p: \mathbb{R} \to \mathbb{R}$ is sujective. In particular, there is $c \in \mathbb{R}$ that $p(c)=0$.

For a polynomial with even degree, take $p(x)=x^2+1$. This have not real roots.

  • $\begingroup$ Are you allowed to use limit properties if one sequence diverges? e.g. look at the "Product Rule" here: math24.net/properties-limits. it requires that the limits exist in order to conclude something about the product. $\endgroup$ – makansij Apr 28 '19 at 2:10

Let $f(x)$ be a polynomial of odd degree function. Then $f(x)\to +\infty$ as $x\to +\infty$ and $f(x)\to -\infty$ as $x \to -\infty$, or vice versa, depending on whether the leading coefficient is positive or negative. Hence, there are $a, b \in \mathbb{R}$ such that $f(a) < 0$ and $f(b)> 0$. Now IVT applies to give an $x \in [a,b]$ such that $f(x) = 0$. Do you think this is a valid prove?


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