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I encountered a polynomial:

$$p_{10}=0.742134 + 32.583720 z + 345.639505 z^2 + 1369.404360 z^3 $$ $$+ 2400.069657 z^4 + 1996.926314 z^5 + 798.801952 z^6 + 147.695904 z^7 $$ $$+ 11.294899 z^8 + 0.274789 z^9 + 0.000907284 z^{10}$$

Numerical solution showed that all the zeros are real and negative:

z = -256.811, 
z = -27.684, 
z = -9.363, 
z = -4.326, 
z = -2.267, 
z = -1.241, 
z = -0.670, 
z = -0.334, 
z = -0.137, 
z = -0.0327.

Is there a method (without using numerical equation solver, and without using Sturm series) to prove that all the zeros of $p_{10}(z)$ are real and negative?

Thanks- mike

mike
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    Only negative roots seems simple since all coefficients (including the constant term) are positive. – Claude Leibovici Oct 08 '14 at 10:33
  • Yes. The real part of the zeros must be negative. – mike Oct 08 '14 at 10:35
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    I came across this from David Bevan after searching here "Kurtz (1992) proves that if all the $a_i$ are positive, and $a_{i}^{2}-4{{a}{i-1}}{{a}{i+1}}>0,,,,1\le i\le n-1$ , then all the roots are real (and thus negative) and distinct". – Paul Oct 08 '14 at 14:45
  • The coefficients in $p_{10}(z)$ are not log-concave enough to satisfy this relation. By the way, Hutchinson seemed to be the first one to prove this result. Please refer to "On a Remarkable Class of Entire Functions" Author(s): J. I. Hutchinson Source: Transactions of the American Mathematical Society, Vol. 25, No. 3 (Jul., 1923), pp.325-332 – mike Oct 08 '14 at 16:41

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