# The polynomial whose roots are all real

Suppose $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0\in \mathbb{R}[x]$ is a polynomial whose roots are all real where $a_n=1$. We want to show that

The polynomial $g(x)=\sum_{i=0}^{n} \binom{n}{i}a_ix^i\in \mathbb{R}[x]$ has only real roots.

This problem is the Exercise 5 of the first section of chapter 5 in the page 257 of the book "Algebra" by T.T.Moh.

I have tried a couple of days to solve it out, but fail in end.

Thanks.

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And what is the title of the section? And what are the preceeding theorems? – Berci Sep 28 '12 at 13:28
an induction argument based on the case $p(0) = 0$ suggests $\sum_{i=0}^{n} i \binom{n}{i}a_ix^i$ has all real zeros. it would be nice if you could then finish taking a derivative but it seems to go the wrong way – mike Sep 28 '12 at 14:00
In the book, the sum defining $g$ seems to be until $n-1$. – Davide Giraudo Sep 28 '12 at 14:36
@Davide Giraudo, there is an errata about the book on his homepage. math.purdue.edu/~ttm/book.html – wxu Sep 28 '12 at 15:06
@Berci The title of that section is "Algebraically closed field", the preceeding theorem is "the fundamental theorem of algebra". – wxu Sep 28 '12 at 16:22

Perhaps there's an easier way, but here's a sketch using some general facts:

For any polynomial $f$ it follows from Rolle's theorem that the derivative $f'$ has at most as many non-real zeros as $f$ itself. (If $f$ is of degree $n$ and has $k$ real zeros (and hence $n-k$ nonreal zeros), then $f'$ is of degree $n-1$ and has at least $k-1$ real zeros, since $f'$ has at least one zero between any two adjacent zeros of $f$; hence $f'$ has at most $(n-1)-(k-1) = n-k$ nonreal zeros.)

More generally (for any real number $\alpha$), $f'-\alpha f$ has at most as many non-real zeros as $f$ itself; this can be shown by counting zeros of $e^{-\alpha x} f(x)$ and its derivative using Rolle's theorem as above. (If $\alpha \neq 0$, then $f'-\alpha f$ is of degree $n$ instead of $n-1$, so one also has to take into account that $e^{-\alpha x} f(x) \to 0$ at either $+\infty$ or $-\infty$ in order to find at least one more real zero of its derivative.)

In other words, the operator $D-\alpha$ (where $D=d/dx$ and $\alpha\in\mathbf{R}$) doesn't increase the number of non-real zeros of a polynomial that it acts upon. Thus, if $p(x)=c (x-\alpha_1) \dots (x-\alpha_n)$ has only real zeros, then the operator $p(D)=c (D-\alpha_1) \dots (D-\alpha_n)$ doesn't increase the number of non-real zeros; in particular, it preserves the property of having only real zeros.

Another operation which preserves the property of having only real zeros is "reversing the coefficients", i.e., mapping $f(x) = \sum_0^n a_k x^k$ to $x^n f(1/x) = \sum_0^n a_k x^{n-k}$. (This is because the zeros of the new polynomial are exactly the reciprocals of the zeros of the original polynomial, except that the root $x=0$ may appear or disappear or change its multiplicity.)

Now let $p$ be your polynomial with only real zeros. Then the reversed polynomial $q(x)=\sum_0^n a_k x^{n-k}$ has only real zeros too, so the operator $q(D)$ doesn't increase the number of non-real zeros. Apply this operator to the polynomial $x^n$ (which has no non-real zeros). Then the resulting polynomial $r(x)=q(D)x^n=\sum_0^n b_k D^{n-k} x^n = n! \sum_0^n \frac{b_k}{k!} x^k$ has no non-real zeros. The conclusion is that we can divide the coefficients by $k!$ without ruining the property of having only real zeros. Now reverse $r(x)$ and perform our newly-discovered operation again; this has the effect of further dividing the coefficients by $(n-k)!$ while preserving the reality of the zeros, and then we're done.

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Great! The proof works for me, though there is something not clear (for me), but we can fix it. For exmaple, we only need that if a polynomial $f(x)$ has only real zeros, then $D-\alpha$ acts on $f$ has only real zeros, and the reversing the coefficients has the property that if a real polynomial $f(x)$ has only real zeros then the "reversing" has only real zeros or a nonzero constant. And all we need can be satisfied in this proof. – wxu Sep 29 '12 at 14:09
Sorry, but I don't think I understand what your problem is... – Hans Lundmark Sep 29 '12 at 14:24
When $x^n$ "reverse" will become $x^n(1/x)^n=1$, so the number of real zeros count multiplicity will change! And I donot know why $D-\alpha$ dosen't increase the number of non-real zeros in general, but I know that if $f$ has $n$ real zeros, then applying rolle's theorem will get that $\alpha f- f^{\prime}$ has at least $n-1$ real zeros, since the degree of $\alpha f-f^{\prime}$ is $n$(here we assume $\alpha \neq 0$), so we in fact get that $\alpha f- f^{\prime}$ has $n$ real zeros. – wxu Sep 29 '12 at 14:41
I just want to say maybe there is something not very rigorous in your proof but it doesn't matter for me and your proof is wonderful! Thanks! – wxu Sep 29 '12 at 14:50
OK, I think I see your point now. I will try to clarify the answer. – Hans Lundmark Sep 29 '12 at 14:51