# I Spivak calculus, page 123 - if n is odd, then the 'n'th degree polynomial equation f(x) has a root

1. Apr 3, 2016

### Alpharup

In chapter 7 of Spivak calculus, it is proved that if n is odd, then the 'n'th degree polynomial equation f(x) has a root. I do understand what goes into the proof and can follow steps easily.
But, my question is
1.How did they think of a proof like that?
2.By trial and error, did they find the set of values of x where the conditions for Intermediate Value theorem are satisfied?

2. Apr 3, 2016

### andrewkirk

One knows from experience with polynomials that, if they have odd order, they go to $+\infty$ as $x\to\infty$ and to $-\infty$ as $x\to -\infty$. The manipulations in the proof are simply a way of formalising that so that, given a set of polynomial coefficients, we get an expression for how big $|x|$ has to be for the polynomial to be certainly negative at $-x$ and positive at $x$. The approach taken is to separate the polynomial into a part that is constant and a part that shrinks to zero as $|x|\to\infty$, because that allows us to quantify how big $|x|$ needs to be for us to be sure that the constant part wins, so that the sign of the polynomial value will be the same as the sign of $x^n$ which is the same as the sign of $x$ (because $n$ is odd).
No. The proof doesn't have to find the value, and doesn't do so. It is a proof that the root exists. It says nothing about how to find it. To find the value, one would use Numerical Analysis techniques, unless a factorisation suggested itself.

3. Apr 5, 2016

### mathwonk

in my opinion writing the proof as mike does makes it look mysterious when actually it is easy. the point is just to show that f(x) goes to + infinity as x does, and f(x) goes to - infinity as x does. hence it must equal zero somewhere. but he is being super pedantic and actually showing how to express the points where f(x) is positive and negative in terms of the coefficients. really, by the factorization he used near the beginning, pulling out a factor of x^n, he is reducing to the case of the polynomial x^n.

he could have made it look easier by proving as lemmas that result for x^n, and then also proving that c/x^k goes to zero for any constant c and any exponent k ≥ 1.

4. Apr 5, 2016

### HallsofIvy

Staff Emeritus
I don't know if this is what you would consider "super-pedantic" but the fact that f is a polynomial of odd degree does NOT immediately imply that "f(x) goes to + infinity as x does, and f(x) goes to - infinity as x does. It depends upon the leading coefficient. If the leading coefficient is positive then this is true. If the leading coefficient is negative then "f(x) goes to - infinity as x goes to + infinity" and "f(x) goes to + infinity as x goes to - infinity". Of course, in either case, f must have a 0 somewhere.

5. Apr 6, 2016

### Alpharup

This is really logical...

6. Apr 7, 2016

### mathwonk

quite right Halls for the general case. however if you read the cited section in Spivak (theorem 9, chapter 7, in my edition), you will see that he considers only monic polynomials f(x), which is why i said what i did, since with Spivak's definition of f(x), i.e. leading coefficient = 1, hence positive, it is correct.

7. Apr 7, 2016

### HallsofIvy

Staff Emeritus
Thanks for letting me know that.

8. Apr 7, 2016

### mathwonk

I should have been more self contained, but after a moments reflection I decided to save writing. Unfortunately it made my comment understandable only by people actually reading spivak. Of course, as Halls explains, we also have the theorem not just in the monic case, but for all odd degree real polynomials.

Last edited: Apr 7, 2016