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Greetings,

I am currently working on a paper that involves an upper bound of the largest root of a polynomial. With the help of the Mean Value Theorem, I believe a colleague and I have proved the following: if f^(n)(p) > 0 (the n-th derivative evaluated at "p", for n=0,1,2..d, where d is the degree of the polynomial), then p>r, where r is the largest root of the polynomial.

The only question I have is the following: is this well-known or has been cited/proven elsewhere? It just seems basic enough that someone in the past has referenced it in a paper/Lemma. Though we could be mistaken and have a flaw in our own proof...

thanks,

Mike

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Doesn't this basically just say that if all the coefficients are positive, then the roots are all negative? (by shifting $p$ to $0.$)? If so, this predates the Descartes' rule of signs... –  Igor Rivin Aug 10 '11 at 20:02
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It sounds like a consequence of Rolle's Theorem to me. You might reword it as a request for a reference. If it is true and published, you may get a reference. Otherwise (though I doubt it) the result is new or wrong and you may find out which anyway without getting a reference. I suggest you add the reference-request tag yourself, and don't worry in print about whether there is a mistake: if there is, it will be found. Gerhard "Ask Me About System Design" Paseman, 2011.08.10 –  Gerhard Paseman Aug 10 '11 at 20:04
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up vote 2 down vote accepted

For me, it is a folklore result for people interested in Descartes' rule of signs, even though I am not aware of any paper stating it explicitly. It follows from Descartes' rule of signs.

Theorem (Descartes' rule of signs). The number of positive roots of a real polynomial is not larger than the number of sign changes in the sequence of its coefficients.

You can prove the following generalization:

Lemma. Let $P$ be a degree-$d$ polynomial and $r\in\mathbb R$. The number of sign changes in the sequence of the $P^{(n)}(r)$ for $n=0$ to $d$ is an upper bound on the number of roots of P larger than $r$.

Proof. Let $Q(x)=P(x+r)$. Then $Q^{(n)}(0)/n!$ is the coefficient of $x^n$ in $Q$. Therefore, the number of signs changes in the sequence $(P^{(n)}(r))_n$ equals the number of sign changes in the sequence of coefficients of $Q$. By Descartes' rule of signs, this number of sign changes upper bounds the number of positive roots of $Q$. $\square$

Remark. It is a consequence of Rolle's Theorem since Rolle's Theorem is used to prove Descartes' rule of signs.

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Didn't Descartes predate Rolle? –  Igor Rivin Aug 10 '11 at 21:28
    
Historically, perhaps. Foundationally, not in my calculus text (the one where analytic geometry is in a separate text). Gerhard "Priority Matters In Dessert Lines" Paseman, 2011.08.10 –  Gerhard Paseman Aug 11 '11 at 3:20
    
You are right, Descartes' rule of signs appears in La Géométrie in 1637, and Rolle proved his theorem in 1691. I should have written "Rolle's Theorem can be used to prove Descartes' rule of signs." What I meant is that those results are very close to each other, thus your two initial comments are right. –  Bruno Aug 11 '11 at 8:24
    
Should the coefficient be represented as $\frac{Q^{(n)}(0)}{n!}$? Regardless, the proof still works. It's been so long since I used it that I forgot about Descartes' Rule of Signs. Thank you very much for all the input. –  Mike Aug 11 '11 at 21:36
    
You're of course right, I forgot to divide by $n!$. –  Bruno Aug 12 '11 at 7:49
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