PlanetMath: all derivatives of sinc are bounded by $1$

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all derivatives of sinc are bounded by $1$

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Let us show that all derivatives of $\operatorname{sinc}$ are bounded by .

First of all, let us point out that $\operatorname{sinc}(t)\le 1$ is bounded by the Jordan's inequality. To bound the derivatives, let us write $\operatorname{sinc}$ as a Fourier integral,

$\displaystyle \operatorname{sinc}(t) = {1\over 2} \int_{-1}^1 e^{ixt} \, dx.$

Let $k=1,2,\ldots$ . Then

$\displaystyle {d^k\over dt^k} \operatorname{sinc}(t) = {1\over 2} \int_{-1}^1 (ix)^k e^{ixt} \, dx.$

and

$\displaystyle \left\vert {d^k\over dt^k} \operatorname{sinc}(t) \right\vert$	$\displaystyle =$	$\displaystyle \left\vert {1\over 2} \int_{-1}^1 (ix)^k e^{ixt} \, dx \right\vert$
	$\displaystyle \le$	$\displaystyle {1\over 2} \int_{-1}^1 \vert (ix)^k e^{ixt} \vert \, dx$
	$\displaystyle \le$	$\displaystyle {1\over 2} \int_{-1}^1 \vert x \vert^k \, dx$
	$\displaystyle \le$	$\displaystyle {1\over 2} \cdot 2 \int_0^1 \vert x \vert^k \, dx$
	$\displaystyle \le$	$\displaystyle \int_0^1 x^k \, dx$
	$\displaystyle \le$	$\displaystyle \frac{1}{k+1}$
	$\displaystyle <$	$\displaystyle 1.$

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Cross-references: Fourier integral, Jordan's inequality, bounded, derivatives
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This is version 7 of all derivatives of sinc are bounded by $1$

, born on 2006-02-02, modified 2006-09-09.
Object id is 7583, canonical name is AllDerivativesOfSincAreBoundedBy1.
Accessed 2201 times total.

Classification:

AMS MSC:

26A06 (Real functions :: Functions of one variable :: One-variable calculus)

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