# all derivatives of sinc are bounded by $1$

Let us show that all derivatives of $\mathrm{sinc}$ are bounded by $1$.

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

$$\mathrm{sinc}(t)=\frac{1}{2}{\int}_{-1}^{1}{e}^{ixt}\mathit{d}x.$$ |

Let $k=1,2,\mathrm{\dots}$. Then

$$\frac{{d}^{k}}{d{t}^{k}}\mathrm{sinc}(t)=\frac{1}{2}{\int}_{-1}^{1}{(ix)}^{k}{e}^{ixt}\mathit{d}x.$$ |

and

$\left|{\displaystyle \frac{{d}^{k}}{d{t}^{k}}}\mathrm{sinc}(t)\right|$ | $=$ | $\left|{\displaystyle \frac{1}{2}}{\displaystyle {\int}_{-1}^{1}}{(ix)}^{k}{e}^{ixt}\mathit{d}x\right|$ | ||

$\le $ | $\frac{1}{2}}{\displaystyle {\int}_{-1}^{1}}|{(ix)}^{k}{e}^{ixt}|\mathit{d}x$ | |||

$\le $ | $\frac{1}{2}}{\displaystyle {\int}_{-1}^{1}}{|x|}^{k}\mathit{d}x$ | |||

$\le $ | $\frac{1}{2}}\cdot 2{\displaystyle {\int}_{0}^{1}}{|x|}^{k}\mathit{d}x$ | |||

$\le $ | ${\int}_{0}^{1}}{x}^{k}\mathit{d}x$ | |||

$\le $ | $\frac{1}{k+1}$ | |||

$$ | $1.$ |

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 |