Timeline for Bounding derivative of a function

Current License: CC BY-SA 3.0

10 events

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Jan 10, 2013 at 7:19	history	edited	ACL	CC BY-SA 3.0	correct spelling of Bernstein
Nov 29, 2012 at 6:24	comment	added	Neeks		In the case I am dealing with, only real zeros of $a(t)$ are of interest. But you brought out interesting possibility. Now taking a simple example, $a(t)=1+\mu\sin\omega t$, with $0<\mu<1$ we have,\newline $\Big\|\frac{a'(t)}{a(t)}\Big\|=\frac{\mu\omega\cos\omega t}{1+\mu\sin\omega t}\leq\frac{\mu\omega}{\sqrt{1-\mu^2}}, \forall t$. So above it can be bounded, though a simple example. Can we have a bound for sum of harmonic sinusoids and generalize it. But polynomials are also entire function and hence band-limited but the bandwidth is too large for it. Please correct me.
Nov 29, 2012 at 2:14	answer	added	Alexandre Eremenko		timeline score: 1
Nov 29, 2012 at 2:07	comment	added	Alexandre Eremenko		You did not tell me what sort of estimate you want. There is no uniform estimate, of course: $a$ can have complex zeros as close as you wish to the real line, condition $a>0$ does not help, and $a'/a$ can be arbitrarily large at some points.
Nov 28, 2012 at 22:51	comment	added	Neeks		Thanks Alexandre, I have corrected the notation and the question.
Nov 28, 2012 at 22:50	history	edited	Neeks	CC BY-SA 3.0	added 40 characters in body
Nov 28, 2012 at 22:48	history	edited	Yemon Choi	CC BY-SA 3.0	retagged
Nov 28, 2012 at 22:45	history	edited	Neeks	CC BY-SA 3.0	corrected the error in variable notation.; deleted 4 characters in body
Nov 28, 2012 at 22:42	comment	added	Alexandre Eremenko		Please edit your question for English and math. What is $a(t)$? is this the same as $x(t)$? What sort of estimate do you want? $x(t)$ can be zero at some points where $x'(t)/x(t)=\infty$.
Nov 28, 2012 at 22:28	history	asked	Neeks	CC BY-SA 3.0