高數(shù)課件1757年數(shù)學(xué)家在研究太陽引起的攝動(dòng)時(shí)大膽

u當(dāng)o非正弦周期函數(shù):矩形波u(t) 當(dāng)u當(dāng)o0tt不同頻率正弦波逐sint,1sin3t,1sin5t,1sin7t, u4sinu4(sint1sin3t u4(sint1sin3t1sin5t u4(sint1sin3t1sin5t1sin7t u4(sint1sin3t1sin5t1sin7t1sin9t u(t)4(sint1sin3t1sin5t1sin7tL

(7t,t二、三角級(jí)數(shù)三角函數(shù)系的正三角f(t)A0Ansin(ntn A0(AnsinncosntAncosnsinnt

A

Asin,

Acos

t

aa0

(ancosnxbnsin

三角級(jí)三角函數(shù)系1,cosx,sinx,cos2x,sin2xLcosnx,sinnx任意兩個(gè)不同函數(shù)在[,]上的積分等于零cosnxdx1dsinnx 1sinnxdx n

m

m mcosmxcosnxdx

msinmxcosnxdx (其中mn1,2,L)cosmxcosnx1[cos(mn)xcos(mn) mncosmxcosnxdx2[cos(mnxcos(mn1sin(mn)

sin(mn)x2

m m

(m,n1,2,3L,km

cosmxcosnxdx

cos

時(shí) 1cos2mxdx120 cosmxcosnxdxmsinmxsinnx1[cos(mn)xcos(mn)m,sinmxcosnx [sin(mn)xsin(mn)212aa02

(ancosnxbnsin

三角三、函數(shù)展開 級(jí)問題:1.若能展開,ai,bi是什么 2.展開的條件是什么若有fx2

(akcoskxbksin求a0

kf(x)dx

a0dx

bsin

k

cos 0dx

coskxdx

sinkxdx

0

k1a01

f(

k

fxfxa02k coskxbsinkk f(x)cosnxdx2

cos coskxcosnxdx sinkxcos

k k

nxdx

1cos2nxdxan

n f(x)cos

(n1,2,3,L) fxfxa02k coskxbsinkk

f(x)sinnxdx2

sin [akcoskxsinnxdxbksinkxsin

sin

nxdx

(1cos2nx)dxb(2 bn n 11bn f(x)sin

(n1,2,3,L)傅里

f(x)cos

(n0,1,2,L)

f(x)sin

(n1,2,L)

系

f(x)cos

(n0,1,2,L) f(x)sin

(n1,2,L)

f(x)cos (n0,1,2,L) 2f(x)sin

(n1,2,L)

(acosnxbsin級(jí) 問題

fx)條件?a02

(ancosnxbnsin

aa02

(ancosnxbnsin 級(jí)雷(Dirichlet)充分條件(收斂定至多只有有限個(gè)極值點(diǎn),則f(x)的 當(dāng)xfx)的連續(xù)點(diǎn)時(shí),級(jí)數(shù)收斂于fx);當(dāng)xfx)fx0fx0);2當(dāng)x為端點(diǎn)xf(0f(0).2回憶：判斷一個(gè)函數(shù)能展成泰勒級(jí)數(shù)定理2 f(x)在點(diǎn)x0的泰勒級(jí)數(shù),在U(x0)內(nèi)收斂于f(x)在U(x0)內(nèi)limRn(x)0.n

fx)Ux0M0x(

R,

R),恒

f(n)(

M(n0,1,2,L)fx)在x0Rx0R)內(nèi)可展開成點(diǎn)x0注意:函數(shù)展開成 a0

(a cosnxbsin

u1以2uu(t)Em 0t

t

將其展開

解所給函數(shù)滿 雷充分條件在點(diǎn)xk(k0,1,2,L)處不連續(xù)收斂于

EmEm(Em) 2

(ancosnxbnsin當(dāng)xk時(shí),收斂于f( 和函數(shù)圖象u u(t)cos011 011(Em)cosntdt

1 1 cosntdt

cosntdt

(n0,1,2,L) u(t)sin

( )sinntdt

sin

2Em(1cosn)2Em[1(1)n

n2k1,k1,2,L

(2k

n2k,k1,2,L(t)

4Em sin(2nn1(2n(t;t0,,2,L

aa級(jí) 0

(ancosnxbnsin

f(x)cos

(n0,1,2,L) f(x)sin (n1,2,L) 注意作法

對(duì)于非周期函數(shù),如果函fx)只在周期延拓(T F(x)f( (,端點(diǎn)處收斂于1[f(0f(2函數(shù)的周期性若f(x)在[-ππ上有定義，則f(x可以開拓成f(x+2nπ)=f(x) -π≤x≤π.aa)當(dāng)f(0)f(0)時(shí)就有f(＋0)ff(＋0f(－0)故在x，處以x(2k1)都是如此開拓周期函 y

b)b)當(dāng)f(0)f(0)時(shí)則開拓的周期函數(shù)f(x)連續(xù)。y o1)先進(jìn)行周期延拓(T2)構(gòu)成F2)再將F(x)限制在(,)內(nèi)，此時(shí)F(x) f(端點(diǎn)處收斂于1[f(0f(22fx)

x

0x解所給函數(shù)滿足 拓廣的周期函數(shù)的 氏級(jí)數(shù)展開式在[,收斂于fx f(x)

x

2a0 (acosnxbsin2a 0x

f(x)dx 0xdx

f(x)cosnxdx

0(x)cosnxdx

xcos0 0

(x)dsinnx1

xdsin 1[xsin

0sinnxdx]

[xsinnx

sin00 (cosn0

n2k1,k1,2,L(2k1)2 2[(1)n

n2k,k1,2,

f(x)dx

f(x)

x

0x f(x)sinnxdx 0xsinnxdx xsinnxdx 1an1

f(x)cosnxdx

(2k1)2

n2k1,k1,2,

n2k,k1,2,L所求函數(shù)的傅氏展f(x)

(2n

cos(2n1)

(x當(dāng)x0時(shí),f(0

28

111 利用傅氏展開式求設(shè)1111L

1

111L

2), 111L 1111L

2

22 26

一、奇函數(shù)和偶函數(shù) 級(jí) 定理(1)當(dāng)周期為2的奇函數(shù)f(x)展開 anb

f(x)sin

(n0,1,2,L)(n1,2,L) (2)當(dāng)周期為2的偶函數(shù)f(x)展開 里葉級(jí)

f(x)cos (n0,1,2,L) bn (n1,2,L)(1設(shè)fx)是奇函數(shù)1an1

f(x)cosnxdx奇函

(n0,1,2,3,L)1bn1

f(x)sinnxdx0偶函 0

f(x)sinnxdx(n1,2,3,L)同理可證

定理證畢f(xié)x)為奇函數(shù),傅氏級(jí)數(shù)bnsin fx)a0

ancos稱為例1fx)是周期為2的周期函數(shù)它在[)fxx,fx展開成解所給函數(shù)滿足雷充分條件因?yàn)閒(0f(0)故在x(2k1)(k0,1,2,L)處不連續(xù)收斂于f(0f(0)() 在連續(xù)點(diǎn)xx(2k1))處收斂于fx),Qx(2k1)fx)是以2為周期的奇函數(shù)y圖象

f(x)cosnxdx20bn20

f(x)sin

2xsin00

(n0,1,2,L) f(x) (ancosnxbnsin 220bn f(x)sinnxdx0

2xsin00 xcosnx

xcos

sinnx

n0n

cosnxdx]

2cosn2(1)n1 (n1,2,L)n

f(x)cosnxdxf(x)2(sinx1sin2x1sin3xL

n

sinnx.(x;x,3,L)y2(sinx1sin2x1sin3x1sin4x1sin5x) yy2將周期函數(shù)u(t其中E是正常數(shù).

Esin

展 氏級(jí)數(shù)解所給函數(shù)滿 數(shù)軸上連續(xù)Qu(t)為偶函數(shù)bn(n1,2,L)

E a u(t)dt2Esintdt4E 2將周期函數(shù)u(t其中E是正常數(shù).

Esin

展 氏級(jí)數(shù)an

2u(t)cos 00

2Esintcos00E0[sin(n1)tsin(nEEcos(n1)tcos(n1)t

n n

(n

當(dāng)n2k[(2k)21]

(k1,2,L)

當(dāng)n2ka

u(t)dt

a1

2u(t)costdt00

2Esintcostdt00an

2u(t)cos00

當(dāng)n2k[(2k)21]

當(dāng)n2ku(t) 0a1costa2cos2ta3cos3ta4cos4t2u(t)4E(11cos2t1cos4t1cos6tL 2E[1

(x) n14n2二、函數(shù)展開成正弦級(jí)數(shù)或余弦非周期函數(shù)的周期性設(shè)fx)定義在[0]上延拓成以2為周期的函數(shù)Fx).f 0x 且Fx2FFx

g( x則有如下兩種

奇延偶延拓奇延拓 g(x)f((則Fxf(

0xxx

fx)的傅氏正弦f(x)bnsin (0x偶延拓 g(x)f(0x 則Fxf f

0xxfx)的傅氏f(x)a02

ancos

(0x3fx)x10x)分別展開成解(1)求正弦級(jí)數(shù) 對(duì)f(x)進(jìn)行奇延拓2bn2

f(x)sinnxdx

2(x1)sin0000 2xcosnxn

sinn

cosn

2(1cosncosn)

2 2

當(dāng)n1,3,5,L當(dāng)n2,4,6,L x12[(2)sinxsin2x1(2)sin3xsin4x1(2)sin5 fxx10x

(0xx0處級(jí)數(shù)和為0，不代表原函

f(4 值(2)求余弦級(jí)數(shù) 對(duì)f(x)進(jìn)行偶延拓a (x1)dx

2xsin

cos

sinnxan0(x1)cosnxdx

4 (cosn1) 4

當(dāng)n2,4,6,Ln2

n2

當(dāng)n1,3,5,Lx1

14(cosx1cos3x1cos5xL

(0x基本概系；雷分條件傅氏級(jí)數(shù)的意義——整體傅氏展1、正弦和余弦級(jí)數(shù)的基本內(nèi)容2、注意幾個(gè)問題不是只有周期函數(shù)才能 氏級(jí)數(shù)在[0,]上,展成周期