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 inequalityMathworldPlanetmath. To the derivatives, let us write sinc as a Fourier integral,

sinc(t)=12-11eixt𝑑x.

Let k=1,2,. Then

dkdtksinc(t)=12-11(ix)keixt𝑑x.

and

|dkdtksinc(t)| = |12-11(ix)keixt𝑑x|
12-11|(ix)keixt|𝑑x
12-11|x|k𝑑x
12201|x|k𝑑x
01xk𝑑x
1k+1
< 1.
Title all derivatives of sinc are bounded by 1
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