Proof that for $\alpha > 0$ and $\tau \geq 0$ , P($s$) = $(s+\alpha+ e^{-s\tau})$ has no solution less than zero

Asked Apr 11 '15 at 12:04

Active Apr 11 '15 at 12:29

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P($s$) = $(s+\alpha+ e^{-s\tau})$

Proof that for $\alpha > 0$ and $\tau \geq 0$ the polynomial has no solution less than zero.

I am having difficulty proving that for $\alpha > 0$ and $\tau \geq 0$, the polynomial below doesn't lead to a real negative part (in case the solution was imaginary).

$s+\alpha+ e^{-s\tau} = 0$

edited Apr 11 '15 at 12:29

giorgiomugnaini

1,416

asked Apr 11 '15 at 12:04

MEssam

In case τ = 0, P(s) = s + α +1, which gives -α-1 is a negative solution. I guess you should modify the condition to τ > 0 – SiXUlm Apr 11 '15 at 13:14

Proof that for $\alpha > 0$ and $\tau \geq 0$ , P($s$) = $(s+\alpha+ e^{-s\tau})$ has no solution less than zero

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