all derivatives of sinc are bounded by 1
Let us show that all derivatives of sinc are bounded by 1.
First of all, let us out that sinc(t)≤1 is
bounded by the Jordan’s inequality. To the derivatives, let
us write sinc as a Fourier integral,
sinc(t)=12∫1-1eixt𝑑x. |
Let k=1,2,…. Then
dkdtksinc(t)=12∫1-1(ix)keixt𝑑x. |
and
|dkdtksinc(t)| | = | |12∫1-1(ix)keixt𝑑x| | ||
≤ | 12∫1-1|(ix)keixt|𝑑x | |||
≤ | 12∫1-1|x|k𝑑x | |||
≤ | 12⋅2∫10|x|k𝑑x | |||
≤ | ∫10xk𝑑x | |||
≤ | 1k+1 | |||
< | 1. |
Title | all derivatives of sinc are bounded by 1 |
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Canonical name | AllDerivativesOfSincAreBoundedBy1 |
Date of creation | 2013-03-22 15:39:03 |
Last modified on | 2013-03-22 15:39:03 |
Owner | matte (1858) |
Last modified by | matte (1858) |
Numerical id | 10 |
Author | matte (1858) |
Entry type | Result |
Classification | msc 26A06 |