4

This is a theorem in Spivak's book "calculus"

Every equation $x^n + a_{n-1}x^{n-1} + \dots + a_0 = 0$ has a real root, if $n$ is odd.

However, I think his proof is overly complicated. Therefore, I would like to offer an alternative proof. It would be nice if someone can check if it correct.

Proof: Define $f: \mathbb{R} \to \mathbb{R}: x \mapsto x^n + a_{n-1}x^{n-1} + \dots + a_0$

Then for $x \neq 0$, we have

$$f(x) = x^n(1+ a_{n-1}/x + \dots + a_0/x^n)$$

and we see that $f(x) \to + \infty$ when $x \to +\infty$ and that $f(x)\to - \infty$ when $x \to - \infty$.

Hence, there is $M> 0$ such that $f(M) > 0$ and $f(-M) < 0$.

By applying IVT on $f\vert_{[-M,M]}$, we deduce that there is $x \in [-M,M]$ such that $f(x) = 0$, and this proves the result.

  • 3
    Your proof looks OK to me. What was Spivak's argument? – quasi Aug 09 '18 at 12:58
  • I will find a link to the proof. Give me a minut. –  Aug 09 '18 at 12:58
  • 1
    I would do it just like you. @Peter The OP gave this as easier proof, but did not say to believe in a proof even more easy. – drhab Aug 09 '18 at 13:01
  • https://math.stackexchange.com/questions/407211/spivak-calculus-chapter-7-theorem-9 Here is the first part of the proof. Spivak then shows that $|a_{n-1}/x + \dots + a_0/x^n| \leq 1/2$. If we choose $x_1 > 0$ satisfying (*) (see link) then $0 < (x_1)^n/2 \leq f(x_1)$. Similarly, we find $x_2 < 0$ with $f(x_2) < 0$ and we can then apply IVT. –  Aug 09 '18 at 13:04
  • @drhab Seems I misinterpreted the question. I thought THIS proof is considered overcomplicated. – Peter Aug 09 '18 at 13:05
  • 4
    The difference between Spivak's proof and yours is that Spivaks proves $f(x)\to+\pm\infty$ when $x\to\pm\infty$, or rather specifically exhibits the $M$, while you only throw it in there. –  Aug 09 '18 at 13:06
  • It is not hard to show that $x^n \to \pm\infty$ (n odd) when $x \to \pm \infty$, and that if $g(x) \to L > 0$, $h(x) \to \pm \infty$ then $h(x)g(x) \to \pm \infty$, which is the only thing that is necessary here. –  Aug 09 '18 at 13:07
  • 4
    That still doesn't change that your "simpler" proof is only the same proof hiding part of the argument. –  Aug 09 '18 at 13:08
  • 1
    Moreover, if somehow you were to obtain magically the limit property, then $M$ is only known to exists. On the other hand, Spivak's proof is an effective proof. It not only shows the existence, but how to effectively obtain such an $M$. As a consequence, it proves more. It gives an algorithm to compute an interval that contains a root. –  Aug 09 '18 at 13:11
  • Doesn't Spivak's proof take the same route, but just adds an explicit construction that gives a value for $|M|$ ? – gandalf61 Aug 09 '18 at 13:11
  • Ah that makes sense. If @floodbaharak is willing to summarise his comments in an answer, I will accept it. –  Aug 09 '18 at 13:15
  • @user582578. I don't know this book (except I've heard of it.) But is is easy to show that $f(z)\ne 0$ for any complex $z$ such that $|z|\geq 1+\max (|a_0|,...,|a_{n-1}|).$ In some cases this can be improved upon by a linear change of variable $z=Az'+B$. – DanielWainfleet Aug 10 '18 at 14:50

3 Answers3

2

Your proof using IVT is correct, in fact $x \in (-M, M)$.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
1

Your proof is correct. Some algebraists prefer a purely algebraic proof. Because the Galois group $Gal(\mathbb{C}|\mathbb{R})$ has order $2$ and it operates on the set of zeros of the polynomial all orbits must have length one or two and since the sum of the orbit lengths must be odd there must be at least one orbit of length one that contains the real root. However You need the IVT to prove the completeness of $\mathbb{C}$ so that this proof just "seems to be purely algebraic".

Peter Melech
  • 4,353
0

Yes, your proof is correct, though I would state that the IVT applies only because $f$ is continuous on $[-M,M]$ .

(It seems overkill to prove that a polynomial is continuous on any interval, but I'm sure you could!)

I think your proof is a nice formalisation of how a high-school student might understand that a polynomial of odd degree must have a real root: "the graph goes down on the left and up on the right so must cross the $x$-axis".

Malkin
  • 2,113