Delta 與 Fourier

寫著寫著突然覺得不對勁, 近日忙...等哪天有空了再想想辦法看能不能讓這幾個概念自洽

如果有dalao知道如何證明 $[fleft( t ight)=sumlimits_{n=-infty }^{infty }{{{e}^{inomega t}}frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{-inomega {t}}}d{t}}}]$ 的話還望賜教.

如果ω和t都是離散的, 就可以推導出來, 這個和固體物理中晶格震動的簡正坐標的處理思路是一致的, 固體物理裡面那個一維晶格進行的幺正變換得到簡正坐標的過程初學看起來像是個傅里葉級數展開, 現在看來好像又不是這麼回事了, 因為傅里葉級數展開作用的函數是連續函數.

我就總覺得傅立葉展開介於之間會不會是晶格震動這種情況取了一個長波極限,等有空了就準備從這方面入手推導一下 .

Part.I - δ函數^[1]的數學表達式

傅里葉變換: $[gleft( omega ight)=frac{1}{sqrt{2pi }}int_{-infty }^{infty }{fleft( t ight){{e}^{-iomega t}}dt}]$

傅里葉積分: $[fleft( t ight)=frac{1}{sqrt{2pi }}int_{-infty }^{infty }{gleft( omega ight){{e}^{iomega t}}domega }]$

將變換式代入積分式:

$[egin{align} & fleft( t ight)=frac{1}{sqrt{2pi }}int_{-infty }^{infty }{gleft( omega ight){{e}^{iomega t}}domega } \ & =frac{1}{sqrt{2pi }}int_{-infty }^{infty }{left[ frac{1}{sqrt{2pi }}int_{-infty }^{infty }{fleft( {{t}} ight){{e}^{-iomega {t}}}d{t}} ight]{{e}^{iomega t}}domega } \ & =int_{-infty }^{infty }{fleft( {{t}} ight)left[ frac{1}{2pi }int_{-infty }^{infty }{{{e}^{iomega left( t-{t} ight)}}domega } ight]d{t}} \ end{align}]$

這種選擇性實際上就是是δ函數的定義, 所以我們可以說:

$[delta left( t-{t} ight)=frac{1}{2pi }int_{-infty }^{infty }{{{e}^{iomega left( t-{t} ight)}}domega }]$

Part.II - 離散的δ函數:Kronecker delta^[2]

克羅內克符號的定義: $[{{delta }_{t{t}}}=left{ egin{align} & 1 t={t} \ & 0 t e {t} \ end{align} ight.]$

我們將證明他的表達式為 $[{{delta }_{t{t}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{inomega left( t-{t} ight)}}}]$

離散形式的取值也是離散的^[3]: $[t=frac{2pi }{Nomega }p {其中} pin Z]$

當 $[t={t}+mfrac{2pi }{omega }]$ 其中時:

$[{{delta }_{t{t}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{inomega left( t-{t} ight)}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{0}}}=1]$

當 $[t e {t}+nfrac{2pi }{omega }]$ 其中時^[4]: $[egin{align} & {{delta }_{t{t}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{inomega left( t-{t} ight)}}}=frac{1}{N+1}frac{{{e}^{-ifrac{N}{2}omega left( t-{t} ight)}}left( 1-{{e}^{iNomega left( t-{t} ight)}} ight)}{1-{{e}^{iomega left( t-{t} ight)}}} \ & =frac{1}{N+1}frac{{{e}^{-ifrac{N}{2}omega left( t-{t} ight)}}}{1-{{e}^{ifrac{2pi left( p-{p} ight)}{N}}}}left( 1-{{e}^{i2pi left( p-{p} ight)}} ight)=0 \ end{align}]$

綜上所述在 $[tin left( -frac{pi }{omega },frac{pi }{omega } ight)]$ 或者說 $[tin left( -frac{T}{2},frac{T}{2} ight)]$ 時^[5]:

$[{{delta }_{t{t}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{inomega left( t-{t} ight)}}}=left{ egin{align} & 1 t={t} \ & 0 t e {t} \ end{align} ight.]$

Part.III - Kronecker delta與傅里葉展開^[6]

$[fleft( t+T ight)=fleft( t ight), T=frac{2pi }{omega }]$

周期函數的傅里葉變換: $[{{g}_{n}}=frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( t ight){{e}^{-inomega t}}dt}]$ ------[1]

周期函數的傅里葉展開: $[fleft( t ight)=sumlimits_{n=-infty }^{infty }{{{e}^{inomega t}}{{g}_{n}}}]$ -------------[2]

仿照Part. I

$[egin{align} & fleft( t ight)=sumlimits_{n=-infty }^{infty }{{{e}^{inomega t}}frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{-inomega {t}}}d{t}}} \ & =frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight)sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}} \ & =fleft( t ight)frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}} \ & =fleft( t ight)frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}} \ end{align}]$

莫非... $[frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}=1]$ 真的成立?

分與兩種情況, 前者不難看出發散了,是無窮個相加.

至於這種情況:

$[egin{align} & int_{-frac{T}{2}}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}=T \ & Rightarrow int_{0}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}+int_{-frac{T}{2}}^{0}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}=T \ & Rightarrow int_{0}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}-int_{0}^{-frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}}=T \ end{align}]$

兩邊對求導試試:

$[frac{1}{2}sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-frac{T}{2} ight)}}}+frac{1}{2}sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t+frac{T}{2} ight)}}}=1]$

$[egin{align} & frac{1}{2}sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-frac{T}{2} ight)}}}+frac{1}{2}sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t+frac{T}{2} ight)}}} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{2}left( {{e}^{inomega left( t-frac{T}{2} ight)}}+{{e}^{inomega left( t+frac{T}{2} ight)}} ight)} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{2}left( {{e}^{inomega left( t-frac{pi }{omega } ight)}}+{{e}^{inomega left( t+frac{pi }{omega } ight)}} ight)} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{2}left( {{e}^{-inpi }}+{{e}^{inpi }} ight)}{{e}^{inomega t}}=sumlimits_{n=-infty }^{infty }{{{e}^{inomega t}}cos npi } \ end{align}]$

無解, 評論區有提到兩次運用分部積分, 試了一下:

$[egin{align} & fleft( t ight)=frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight)sumlimits_{n=-infty }^{infty }{{{e}^{inomega left( t-{t} ight)}}}d{t}} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}}d{t}}} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{-inomega }int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight)d{{e}^{inomega left( t-{t} ight)}}}} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{-inomega }left[ left. fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}} ight|_{-frac{T}{2}}^{frac{T}{2}}-int_{-frac{T}{2}}^{frac{T}{2}}{frac{dfleft( {{t}} ight)}{dt}{{e}^{inomega left( t-{t} ight)}}d{t}} ight]} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{inomega }int_{-frac{T}{2}}^{frac{T}{2}}{frac{dfleft( {{t}} ight)}{dt}{{e}^{inomega left( t-{t} ight)}}d{t}}} \ end{align}]$

可以計算 $[sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{inomega }int_{-frac{T}{2}}^{frac{T}{2}}{{{e}^{inomega left( t-{t} ight)}}dfleft( {{t}} ight)}}]$ 或 $[sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{{{n}^{2}}{{omega }^{2}}}int_{-frac{T}{2}}^{frac{T}{2}}{frac{dfleft( {{t}} ight)}{dt}d{{e}^{inomega left( t-{t} ight)}}}}]$

前者沒什麼意義, 又變回去了:

$[egin{align} & sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{inomega }int_{-frac{T}{2}}^{frac{T}{2}}{{{e}^{inomega left( t-{t} ight)}}dfleft( {{t}} ight)}} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{inomega }left[ left. fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}} ight|_{-frac{T}{2}}^{frac{T}{2}}+inomega int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}}d{t}} ight]} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}}d{t}}} \ end{align}]$

至於後者:

$[egin{align} & sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{{{n}^{2}}{{omega }^{2}}}int_{-frac{T}{2}}^{frac{T}{2}}{frac{dfleft( {{t}} ight)}{dt}d{{e}^{inomega left( t-{t} ight)}}}} \ & ext{=}sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{{{n}^{2}}{{omega }^{2}}}left[ left. frac{dfleft( {{t}} ight)}{dt}{{e}^{inomega left( t-{t} ight)}} ight|_{-frac{T}{ ext{2}}}^{frac{T}{2}}-int_{-frac{T}{2}}^{frac{T}{2}}{frac{{{d}^{2}}fleft( {{t}} ight)}{d{{t}^{2}}}{{e}^{inomega left( t-{t} ight)}}d{t}} ight]} \ & =sumlimits_{n=-infty }^{infty }{frac{1}{T}frac{1}{{{n}^{2}}{{omega }^{2}}}int_{-frac{T}{2}}^{frac{T}{2}}{frac{{{d}^{2}}fleft( {{t}} ight)}{d{{t}^{2}}}{{e}^{inomega left( t-{t} ight)}}d{t}}} \ end{align}]$

這說明 $[sumlimits_{n=-infty }^{infty }{int_{-frac{T}{2}}^{frac{T}{2}}{fleft( {{t}} ight){{e}^{inomega left( t-{t} ight)}}d{t}}}=sumlimits_{n=-infty }^{infty }{frac{1}{{{n}^{2n}}{{omega }^{2n}}}int_{-frac{T}{2}}^{frac{T}{2}}{frac{{{d}^{2n}}fleft( {{t}} ight)}{d{{t}^{2n}}}{{e}^{inomega left( t-{t} ight)}}d{t}}}]$

...I dont really...eh?

其實從[2]出發通過正交性會很容易推導出[1]:

$[egin{align} & fleft( t ight)=sumlimits_{n=-infty }^{infty }{{{e}^{inomega t}}{{g}_{n}}} \ & Rightarrow fleft( t ight){{e}^{-i{n}omega t}}=sumlimits_{n=-infty }^{infty }{{{e}^{ileft( n-{n} ight)omega t}}{{g}_{n}}} \ & Rightarrow int_{-frac{T}{2}}^{frac{T}{2}}{fleft( t ight){{e}^{-i{n}omega t}}dt}=int_{-frac{T}{2}}^{frac{T}{2}}{sumlimits_{n=-infty }^{infty }{{{e}^{ileft( n-{n} ight)omega t}}{{g}_{n}}}dt} \ end{align}]$

由於 $[left{ left. {{e}^{inomega t}} ight|nin Z ight}]$ 是一組完備正交函數所以有: $[egin{align} & frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( t ight){{e}^{-i{n}omega t}}dt}=sumlimits_{n=-infty }^{infty }{{{g}_{n}}frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{{{e}^{ileft( n-{n} ight)omega t}}dt}} \ & Rightarrow {{g}_{n}}=frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{fleft( t ight){{e}^{-i{n}omega t}}dt} \ end{align}]$

也就是得到了[1], 但這裡只說明有 $[{{delta }_{{n}n}}=frac{1}{T}int_{-frac{T}{2}}^{frac{T}{2}}{{{e}^{ileft( n-{n} ight)omega t}}dt}]$ 是一個平凡的結論.

能不能得到 $[{{delta }_{t{t}}}=frac{1}{N+1}sumlimits_{n=-frac{N}{2}}^{frac{N}{2}}{{{e}^{inomega left( t-{t} ight)}}}=left{ egin{align} & 1 t={t} \ & 0 t e {t} \ end{align} ight.]$ 呢?