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Prove if the function $f: \mathbb{R} \to \mathbb{R}$ is a polynomial function of odd degree, then $f(\mathbb{R}) = \mathbb{R}.$

We know a polynomial, $f(x)=a_nx^n +a_{n−1}x^{n−1} ...a_1x+a_0$ with real coefficients is continuous. Also, $\mathbb{R}$ is connected now since $\mathbb{R}$ is connected then $f(\mathbb{R})$ is connected, thus we can apply the intermediate value theorem. Now any polynomial of odd degree has at least one real root since every polynomial, with real coefficients, has as many roots as it has degrees. Also, any complex roots are paired with their complex conjugates thus they take up an even number of roots. So, an odd degree polynomial has only an even number of roots that can be complex therefore at least one root must be real. Therefore, there exists $p,q\in \mathbb{R}$, $p<q$ such that $f(p)<0$ and $f(q)>0$. Thus we can then choose $p\in \mathbb{R}$ such that $f(p) \ge 0,$ $f(p)\le 0$ and obtain any $f(\mathbb{R})\in \mathbb{R}$.

Similarly, suppose that the leading coefficient of $f$ is positive then, $\lim_{x\to \infty} f(x) = \infty$, and $\lim_{x\to -\infty} f(x) = -\infty$. So, any arbitrarily large or small value can be attained. Due to the intermediate value theorem and since $f$ is continuous, any value between such a large value $M$ and a small value $m$ is attained. Hence, $f(\mathbb{R}) = \mathbb{R}$. And a similar argument follows if the leading coefficient of $f$ is negative. Therefore, $f(\mathbb{R}) = \mathbb{R}$.

Is this correct?

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  • $\begingroup$ "since every polynomial, with real coefficients, has as many roots as it has degrees" is very wrong. To begin with every polynomial has just one degree, but it can have any natural number of roots (or even infinitely many if one allows the zero polynomial). But even with the degree itself instead of the "number of degrees", it is not correct. The only correct statement vaguely resembling this is that for a nonzero polynomial $P$, the number of complex roots counted with their multiplicity is equal to $\deg P$. But (non-real) complex roots are not useful to this problem. $\endgroup$ Nov 3, 2013 at 10:04
  • $\begingroup$ @MarcvanLeeuwen Can you explain a bit more? $\endgroup$
    – user104235
    Nov 3, 2013 at 20:10
  • $\begingroup$ There are two points to my remark, having to do with formulation. In mathematics one does not say a quadratic polynomial has two degrees; there is only one degree, which happens to be$~2$. (Similarly $\Bbb R^3$ does not have three dimensions, but dimension$~3$.) $\endgroup$ Nov 4, 2013 at 6:05
  • $\begingroup$ The other point is mostly due to my not continuing to read the next sentence. But if you say a real polynomial of odd degree$~d$ has at least one real root because it has $d$ roots, then it seems natural to think you are talking about $d$ real roots (arguments saying something must exist because their number is odd are not uncommon); you apparently mean complex roots instead (and counted with multiplicity), so I would advise saying that explicitly in the first sentence. $\endgroup$ Nov 4, 2013 at 6:16
  • $\begingroup$ @MarcvanLeeuwen Got it. Thank you! $\endgroup$
    – user104235
    Nov 4, 2013 at 8:39

2 Answers 2

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That is correct, but all you really need is the last part : Assume without loss of generality that the leading coefficient is positive (else apply this result to $-f$), and note that $$ \lim_{x\to +\infty} f(x) = +\infty, \text{ and } \lim_{x\to -\infty} f(x) = -\infty $$ Then, for any $y\in \mathbb{R}$, there is $M > 0$ such that $$ x > M \Rightarrow f(x) > y, \text{ and } x < -M \Rightarrow f(x) < y $$ Hence, by intermediate value theorem, there is $x_0 \in [-M,M]$ such that $f(x_0) = y$.

This is true for all $y \in \mathbb{R}$, and hence $f(\mathbb{R}) = \mathbb{R}$

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every odd degree polynomial over $\mathbb{R}$ has a root in $\mathbb{R}$....

we want to prove that for each $b\in \mathbb{R}$ there exist some $a\in \mathbb{R}$ such that $f(a)=b$

Set $f(x)=b$and consider $g(x)=f(x)-b$ this $g(x)$ is an odd degree polynomial and hence has a real root

thus we have $a\in \mathbb{R}$ such that $g(a)=0$ i.e., $f(a)-b=0$ i.e., $f(a)=b$

I was expecting that the hint which i have given is more that sufficient...

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  • $\begingroup$ please see the edit.. $\endgroup$
    – user87543
    Nov 3, 2013 at 10:29
  • $\begingroup$ you are welcome! $\endgroup$
    – user87543
    Nov 4, 2013 at 8:53
  • $\begingroup$ This is a valid answer, +1. However if you think of it, you can see that it is hard to imagine proving that every odd degree real polynomial has a root without at some point arguing that their polynomial functions must take on all real values, invoking the intermediate value theorem. Or using something stronger still, such as that all nonconstant complex polynomials have a root (as far as I know all proofs of that fact ultimately use that odd degree real polynomials have a real root). So in this sense there is an implicit circularity to this answer. $\endgroup$ Nov 4, 2013 at 9:29
  • $\begingroup$ Yes Yes.. I also feel the same... $\endgroup$
    – user87543
    Nov 4, 2013 at 9:33
  • $\begingroup$ @PraphullaKoushik I am looking over your proof and I'm wondering how did you establish that $g(x)$ is an odd degree polynomial? $\endgroup$
    – user104235
    Nov 5, 2013 at 4:21

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