夢(mèng)開始的地方——Fourier級(jí)數(shù)與變換

2021-12-02 23:43 作者:子瞻Louis 0人讀過 | 我要投稿

這篇文章肯定是寫不下太多內(nèi)容的，因此我會(huì)寫一些以后會(huì)用到的一些東西，因?yàn)槟芰τ邢匏晕恼驴赡軙?huì)稍微有些簡略……

Euler公式

對(duì) $e%5E%7Bix%7D$ 展開為Maclaurin級(jí)數(shù)，有

$%5Cbegin%7Baligned%7De%5E%7Bix%7D%26%3D1%2Bix-%5Cfrac%7Bx%5E2%7D%7B2!%7D-i%5Cfrac%7Bx%5E2%7D%7B3!%7D%2B%5Cfrac%7Bx%5E4%7D%7B4!%7D-i%5Cfrac%7Bx%5E5%7D%7B5!%7D%2B%E2%80%A6%5C%5C%26%3D1-%5Cfrac%7Bx%5E2%7D%7B2!%7D%2B%5Cfrac%7Bx%5E4%7D%7B4!%7D-%E2%80%A6%2Bi%5Cleft(x-%5Cfrac%7Bx%5E3%7D%7B3!%7D%2B%5Cfrac%7Bx%5E5%7D%7B5!%7D-%E2%80%A6%5Cright)%5Cend%7Baligned%7D$

根據(jù)正弦余弦的Maclaurin級(jí)數(shù)，便有

$e%5E%7Bix%7D%3D%5Ccos%20x%2Bi%5Csin%20x$

這只是及其簡略且不嚴(yán)謹(jǐn)?shù)淖C明，不嚴(yán)謹(jǐn)在于這里并沒有指出它滿足交換求和次序的條件

它的詳細(xì)證明以及推廣什么的這里就不討論了，互聯(lián)網(wǎng)一搜就是一堆

Fourier級(jí)數(shù)

眾所周知，正弦函數(shù)和余弦函數(shù)都是周期為2π的連續(xù)函數(shù)，因此多個(gè)正弦函數(shù)和余弦函數(shù)的和仍然是周期函數(shù)

$%5Csin%20x%2B2%5Csin%203x%3D%5Csin%20(x%2B2n%5Cpi)%2B2%5Csin%203(x%2B2n%5Cpi)%0A$ ?

$%5Ccos%202x%2B%5Csin%204x%3D%5Ccos%202(x%2Bn%5Cpi)%2B%5Csin%204(x%2Bn%5Cpi)$

$%E2%80%A6%E2%80%A6$

那么反過來，是不是周期函數(shù)都可以寫成正弦函數(shù)與余弦函數(shù)的和呢？

Fourier就正好注意到了這點(diǎn)，他認(rèn)為，任何在一個(gè)周期內(nèi)只有有限個(gè)第一類間斷點(diǎn)的周期函數(shù)都可表為正余弦函數(shù)之和，即對(duì)周期函數(shù)f(t)，有

$f(t)%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Da_%7Bn%7D%5Ccos%20%5Cleft(%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%5Cright)%20%2Bb_%7Bn%7D%5Csin%20%5Cleft(%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%5Cright)$

其中T為f(t)的最小正周期，此即為經(jīng)典的Fourier三角級(jí)數(shù)

假設(shè)f(t)是處處連續(xù)的，且Fourier級(jí)數(shù)收斂于f(t)，我們來求解他的Fourier系數(shù)，利用歐拉公式，有

$%5Ccos%20x%3D%5Cfrac%7Be%5E%7Bix%7D%2Be%5E%7B-ix%7D%7D%7B2%7D%2C%5Csin%20x%3D%5Cfrac%7Be%5E%7Bix%7D-e%5E%7B-ix%7D%7D%7B2i%7D%3D-i%5Cfrac%7Be%5E%7Bix%7D-e%5E%7B-ix%7D%7D%7B2%7D$

代入到其中

$%5Cbegin%7Baligned%7Df(t)%26%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Ba_%7Bn%7D%7D%7B2%7D%5Cleft(e%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%2Be%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)-i%5Cfrac%7Bb_%7Bn%7D%7D%7B2%7D%5Cleft(e%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D-e%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)%5C%5C%26%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D%5Cfrac%7Ba_%7Bn%7D-ib_%7Bn%7D%7D%7B2%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%2B%5Cfrac%7Ba_%7Bn%7D%2Bib_%7Bn%7D%7D%7B2%7De%5E%7B-%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cend%7Baligned%7D$

看起來越變?cè)綇?fù)雜了，但我們可以做如下變換

$c_%7Bn%7D%3D%5Cleft%5C%7B%20%5Cbegin%7Barray%7D%7Brcl%7D%0A%5Cfrac%7Ba_%7Bn%7D-ib_%7Bn%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%0A%26%20n%EF%BC%9E0%20%5C%5C%20%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%20%26%20n%3D0%20%5C%5C%0A%5Cfrac%7Ba_%7B-n%7D%2Bb_%7B-n%7D%7D%7B2%7D%20%26%20%5Cmbox%7Bfor%7D%20%26%20n%EF%BC%9C0%0A%5Cend%7Barray%7D%5Cright.$

則

$f(t)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D$

到這一步后就看上去似乎無從下手了，

但如果我們用一指數(shù)函數(shù)乘以f(t)再對(duì)其在一個(gè)周期上進(jìn)行積分（這里選[-T/2,T/2]）

$%5Cbegin%7Baligned%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7D%26%3D%5Cleft(%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20int%7D%7BT%7D%7D%5Cright)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7D%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7D%5Cend%7Baligned%7D$

$%5Cbegin%7Baligned%7D%5CRightarrow%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7Ddt%26%3D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%5Cend%7Baligned%7D$

其中，積分與和式可以交換次序是因?yàn)樵摷?jí)數(shù)的和收斂

假設(shè)該積分絕對(duì)值收斂，這樣我們便可以對(duì)其每一項(xiàng)討論

n=k時(shí)

$%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%3D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B0%7Ddt%3Dc_%7Bn%7DT$

n≠k時(shí)

$%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Dc_%7Bn%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7Ddt%3Dc_%7Bn%7D%5Cfrac%7BT%7D%7B2%5Cpi%20i(n-k)t%7De%5E%7B%5Cfrac%7B2%5Cpi%20i(n-k)t%7D%7BT%7D%7D%5Cvert_%7B-T%2F2%7D%5E%7BT%2F2%7D%3D0$

于是我們得到

$%5Cbegin%7Baligned%7Dc_%7Bk%7D%26%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)e%5E%7B-%5Cfrac%7B2%5Cpi%20ikt%7D%7BT%7D%7Ddt%5C%5C%26%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt-i%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Csin%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt%5Cend%7Baligned%7D$

根據(jù) $c_n$ 的定義對(duì)比實(shí)部與虛部又可得：

$a_%7Bk%7D%3Dc_n%2Bc_%7B-n%7D%3D%5Cfrac%7B2%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt$

$b_%7Bk%7D%3D-i(c_n-c_%7B-n%7D)%3D%5Cfrac%7B2%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(t)%5Csin%5Cleft(%7B%5Cfrac%7B2%5Cpi%20kt%7D%7BT%7D%7D%5Cright)dt$

這樣就能得到周期函數(shù)的Fourier級(jí)數(shù)展開了，

$f(t)%3D%5Cfrac%7Ba_%7B0%7D%7D%7B2%7D%2B%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7Da_%7Bn%7D%5Ccos%5Cleft(%7B%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%7D%5Cright)%2Bb_%7Bn%7D%5Csin%20%5Cleft(%7B%5Cfrac%7B2%5Cpi%20nt%7D%7BT%7D%7D%5Cright)$

為 $f(t)$ 的Fourier三角級(jí)數(shù)，同樣有一下更優(yōu)美形式的三角級(jí)數(shù)

$f(t)%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B%7B-%5Cfrac%7B2%5Cpi%20nx%7D%7BT%7D%7D%7Ddx%5Cright)e%5E%7B%7B%5Cfrac%7B2%5Cpi%20nx%7D%7BT%7D%7D%7D$

Fourier積分

看到這里的同學(xué)不難發(fā)現(xiàn)Fourier級(jí)數(shù)僅僅局限于對(duì)周期函數(shù)展開，那么非周期函數(shù)要怎么辦呢？

對(duì)此，我們可以將一個(gè)非周期連續(xù)函數(shù)看為周期為無窮大的周期函數(shù)

令 $%5Cfrac%20nT%3D%5Comega_n$ ?，則 $%5Cfrac1T%3D%5Cfrac%7Bn%2B1%7DT-%5Cfrac%20nT%3D%5CDelta%5Comega$

$%5Cbegin%7Baligned%7D%5CRightarrow%20f(t)%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cfrac%7B1%7D%7BT%7D%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B-%5Cfrac%7B2%5Cpi%20inx%7DT%7Ddx%5Cright)e%5E%7B%5Cfrac%7B2%5Cpi%20in%20t%7DT%7D%5C%5C%26%3D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-T%2F2%7D%5E%7BT%2F2%7Df(x)e%5E%7B-2%5Cpi%20i%5Comega_n%20x%7Ddx%5Cright)e%5E%7B2%5Cpi%20i%5Comega_n%20t%7D%5CDelta%20%5Comega%5Cend%7Baligned%7D$

令 $T%5Crightarrow%20%2B%5Cinfty$ ?，則 $%5CDelta%20%5Cxi%20%5Crightarrow%200%5E%7B%2B%7D$

$f(t)%3D%5Clim_%7B%5CDelta%5Comega%20%5Cto0%5E%7B%2B%7D%7D%5Csum_%7Bn%3D-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7Df(x)e%5E%7B2%5Cpi%20i%5Comega_n%20x%7Ddx%5Cright)e%5E%7B2%5Cpi%20i%5Comega_n%20t%7D%5CDelta%20%5Comega$

運(yùn)用小學(xué)二年級(jí)學(xué)過的微積分知識(shí)，可知右邊為一黎曼和，于是可以將級(jí)數(shù)寫成積分

$f(t)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Cleft(%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7Df(t)e%5E%7B-2%5Cpi%20i%5Comega%20t%7Ddt%5Cright)e%5E%7B2%5Cpi%20i%5Comega%20t%7Dd%5Comega$

這樣，我們就得到了Fourier積分，其中

$%5Cmathcal%20F%5C%7Bf(t)%5C%7D(%5Comega)%3D%5Chat%20f(%5Comega)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%7Bf%7D(t)e%5E%7B-2%5Cpi%20i%5Comega%20t%7Ddt$

稱為f(t)的Fourier變換，且其逆變換為

$%5Cmathcal%20F%5E%7B-1%7D%5C%7B%5Chat%20f(%5Comega)%5C%7D(t)%3Df(t)%3D%5Cint_%7B-%5Cinfty%7D%5E%7B%5Cinfty%7D%5Chat%7Bf%7D(%5Comega)e%5E%7B2%5Cpi%20i%5Comega%20t%7D%5Cmathrm%20d%5Comega$

那么本篇文章就到此結(jié)束了

今后呢也是會(huì)不定期更新一些文章的

拜拜~

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夢(mèng)開始的地方——Fourier級(jí)數(shù)與變換

Euler公式

Fourier級(jí)數(shù)

Fourier積分