狄利克雷核

${\displaystyle D_{n}(x)=\sum _{k=-n}^{n}e^{ikx}=1+2\sum _{k=1}^{n}\cos(kx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}.}$

应用

{\displaystyle {\begin{aligned}(D_{n}*f)(x)&=\int _{-\pi }^{\pi }f(y)D_{n}(x-y)\,dy=\int _{-\pi }^{\pi }f(y)\left(\sum _{k=-n}^{n}e^{ik(x-y)}\right)\,dy=\int _{-\pi }^{\pi }\left(\sum _{k=-n}^{n}f(y)e^{-iky}\right)e^{ikx}\,dy\\&=2\pi \sum _{k=-n}^{n}{\hat {f}}(k)e^{ikx}\end{aligned}}}

${\displaystyle {\hat {f}}(k)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)e^{-ikx}\,dx}$

${\displaystyle f}$ 的第${\displaystyle k}$ 个傅里叶系数。需要特别注意，在傅里叶级数上下文中采用的卷积定义，有时会加上了特有的系数${\textstyle {\frac {1}{2\pi }}}$ ，从而将上式表达为：

${\displaystyle (D_{n}*f)(x)={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(y)D_{n}(x-y)\,dy=\sum _{k=-n}^{n}{\hat {f}}(k)e^{ikx}}$

核的L1范数

${\displaystyle \|D_{n}\|_{L^{1}}=\Omega (\log n)}$

与周期狄拉克δ函数的关系

${\displaystyle \sum _{m=-\infty }^{\infty }e^{\pm i\omega mT}={\frac {2\pi }{T}}\sum _{k=-\infty }^{\infty }\delta (\omega -2\pi k/T)={\frac {1}{T}}\sum _{k=-\infty }^{\infty }\delta (\xi -k/T)}$

${\displaystyle {\mathcal {F}}\left[D_{n}(2\pi x)\right](\xi )={\mathcal {F}}^{-1}\left[D_{n}(2\pi x)\right](\xi )=\int _{-\infty }^{\infty }D_{n}(2\pi x)e^{\pm i2\pi \xi x}dx=\sum _{k=-n}^{+n}\delta (\xi -k)\equiv \mathrm {comb} _{n}(\xi )}$
${\displaystyle {\mathcal {F}}\left[\mathrm {comb} _{n}\right](x)={\mathcal {F}}^{-1}\left[\mathrm {comb} _{n}\right](x)=\int _{-\infty }^{\infty }\mathrm {comb} _{n}(\xi )e^{\pm i2\pi \xi x}d\xi =D_{n}(2\pi x)}$

${\displaystyle \mathrm {comb} _{n}(x)}$ ${\displaystyle n\rightarrow \infty }$ 时成为了周期${\displaystyle T=1}$ 狄拉克采样函数英语Dirac comb${\displaystyle \,\operatorname {\text{Ш}} }$ ，它在傅里叶变换下保持不变：${\textstyle {\mathcal {F}}[\operatorname {\text{Ш}} ]=\operatorname {\text{Ш}} }$ 。因此${\displaystyle D_{n}(2\pi x)}$ ${\displaystyle n\rightarrow \infty }$ 时也必定收敛为${\displaystyle \,\operatorname {\text{Ш}} }$

${\displaystyle f*(\Delta )=f}$

${\displaystyle \Delta (x)\sim \sum _{k=-\infty }^{\infty }e^{ikx}=\left(1+2\sum _{k=1}^{\infty }\cos(kx)\right).}$

三角恒等式的证明

${\displaystyle \sum _{k=-n}^{n}e^{ikx}={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}}$

${\displaystyle \sum _{k=0}^{n}ar^{k}=a{\frac {1-r^{n+1}}{1-r}}.}$

${\displaystyle \sum _{k=-n}^{n}r^{k}=r^{-n}\cdot {\frac {1-r^{2n+1}}{1-r}}.}$

${\displaystyle {\frac {r^{-n-1/2}}{r^{-1/2}}}\cdot {\frac {1-r^{2n+1}}{1-r}}={\frac {r^{-n-1/2}-r^{n+1/2}}{r^{-1/2}-r^{1/2}}}.}$

r = eix 时就有：

${\displaystyle \sum _{k=-n}^{n}e^{ikx}={\frac {e^{-(n+1/2)ix}-e^{(n+1/2)ix}}{e^{-ix/2}-e^{ix/2}}}={\frac {-2i\sin((n+1/2)x)}{-2i\sin(x/2)}}}$

${\displaystyle \lim \limits _{x\to 2k\pi }{\frac {\sin((n+1/2)x)}{\sin(x/2)}}=2n+1}$

狄利克雷核的性质

• 狄利克雷核是一个三角多项式，因此是无穷阶可导的周期函数；
• 狄利克雷核是偶函数
• 狄利克雷核的平均值是1；
• 在正无穷处的平均值为：
${\displaystyle \|D_{n}\|_{1}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }|D_{n}(t)|dt={\frac {4}{\pi ^{2}}}\ln n+O(1)}$

来源

• Andrew M. Bruckner, Judith B. Bruckner, Brian S. Thomson: Real Analysis. ClassicalRealAnalysis.com 1996, ISBN 0-13-458886-X, S.620 (vollständige Online-Version (Google Books))
• Podkorytov, A. N. (1988), "Asymptotic behavior of the Dirichlet kernel of Fourier sums with respect to a polygon". Journal of Soviet Mathematics, 42(2): 1640–1646. doi: 10.1007/BF01665052
• Levi, H. (1974), "A geometric construction of the Dirichlet kernel". Transactions of the New York Academy of Sciences, 36: 640–643. doi: 10.1111/j.2164-0947.1974.tb03023.x
• Hazewinkel, Michiel (编), Dirichlet kernel, 数学百科全书, Springer, 2001, ISBN 978-1-55608-010-4
• Dirichlet-Kernel[失效链接] at PlanetMath[永久失效链接]