MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The formula in

gives upper and lower bounds x1 and x2 for the roots of a polynomial all of whose roots are real. Where can I find a proof ? Unfortunately, Wikipedia does not give any sources.

share|cite|improve this question
Pretty sure sure there should be something in Marsden's Geometry of Polynomials. – J. M. Nov 2 '10 at 12:33

Actually I happened to write a short section on wikipedia about the classic bounds here. To locate the zeros of a polynomial you can also use the Rouché theorem. Also, ymay find interesting material e.g. in the book by Pólya and Szegő.

rmk 1: note that the above bounds of course hold for complex roots as well.

rmk 2: the spirit of the square root formula given in the link is: assume you have a real monic polynomial $p$ of degree $n$, and you just know $a_{n-1}$ and $a_{n-2}$. These numbers put a constraint (Nebenbedingung) on the size of the roots. Remember that $-a_ {n-1}= -\sum_k z_k$ and $a_{n-2}=\sum_ {j < k} z_ j z_ k$ are elementary symmetric functions of the roots $z_k:=x_k+ i y_k$. In particular, for the real parts of the roots we have $$\sum_{k=1}^n x_k =-a_{n-1}$$ and $$\sum_{k=1}^n x_k^2 \leq a_{n-1}^2-2a_{n-2}$$

Within these constraints, the maximum, resp. the minimum, of a real root is found maximizing, resp. minimizing $x_1;$ the method of Lagrange multipliers should easily yield (I didn't check) to (a generalization of) the given result.

share|cite|improve this answer
Hello Pietro. Thank you for the answer, but I'm afraid it doesn't really apply to my question about the specific formula for the case of polynomials that have only real zeros. General formulas (for any complex roots) are much weaker than the considered formula. – Yakob Nov 2 '10 at 13:04
OK, this one ( $1+\max_k |a_k|$ ) was quoted at the end of the link. Actually, in the link there is a sketch of the proof for the square root formula (I'm adding some remark on this above). – Pietro Majer Nov 2 '10 at 13:30
Above I assumed that the polynomial has real coefficients, but it may have some non-real roots. In this case, $a_{n-1}$ and $a_{n-2}$ are sufficient to bound the real parts of the roots, and give some generalization of the formula in the link. However, the formula there must assume that all roots are real, or some other hypothesis, otherwise it can't be correct: note that a polynomial with $a_{n-1}=a_{n-2}=0$ may certainly have non zero real roots. – Pietro Majer Nov 2 '10 at 15:26
The assumption really is that all roots are real. Pietro, thank you for your answers. What I originally wanted to know is, where the theorem comes from. Does anyone know the original source ? – Yakob Nov 2 '10 at 16:02
@Yakob: As I commented already, look at Marsden's book. I'm far away from my copy so I can't check, but if it's there, there should be a reference to where it first appeared. – J. M. Nov 2 '10 at 16:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.