傅里叶级数部分难题解答.docx
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傅里叶级数部分难题解答.docx
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傅里叶级数部分难题解答
傅里叶级数部分难题解答
微积分(下)傅里叶级数部分难题解答
傅里叶级数部分难题解答
1((书中,第1题)三角多项式P301
n,0,,,,Tx,,,coskx,,sinkx,nkk2,1k
的傅里叶级数还是它自己吗,
,,Tx,,解:
是以为周期的函数,不妨在,,,,上展开成傅里叶级数.由于2,n
n,,,a1110,,,,coskx,,sinkxdx,dx,,a,Txdx,kk0n,,,,,,,,,,2,,1,k
,.(由于三角函数系的正交性)0
1对于由,m,1,m,n,,,a,Txcosmxdxmn,,,,
n,,,,110,,,,,.cosmxdx,coskx.cosmxdx,sinkx.cosmxdx,kk,,,,,,,,,,,,,2,,,1k
由于三角函数系的正交性,仅当时,k,m
cosmx.cosmxdx,,..,,,,
,其余而且,,,,coskx.cosmxdx,0k,m.sinkx.cosmxdx,01,k,m.,,,,,,
a,,.故mm
a,0.对于,m,m,n,同样由三角函数系的正交性知m
0,m,n,,,m即a,,m0,其他.,
1,m,n,,,m,,Tx同理,有所以,的傅里叶级数为b,,nm0,其他.,
n,a00,,,,,acosmx,bsinmx,,,cosmx,,sinmx(换记为),,mmmm22m1,1,m
nn,,00,,,,,,,,,coskx,,sinkx.Tx,,,coskx,,sinkx三角多项式,,kknkk22,1,1kk的傅里叶级数还是它自己.
1
微积分(下)傅里叶级数部分难题解答
4,,fx,sinx,x,,,,,,2((书中,第2题)将函数展开成傅里叶级数P301
,,11244解:
,,a,fxdx,sinxdx,sinxdx0,,,,,0,,,,,
,,,,2443!
!
3444222(第二个积分令,sinxdx,costdt,sinxdx,.,.,,,,,0004!
!
24,,,,,
t,x,)2
,114,,a,fxcosnxdx,sinxcosnxdxn,,,,,,,,
131,,2cos2xcos4xcosnxdx,,,,,,,,,422,,
a,0,由于三角函数系的正交性,仅当或时,此时n,2,n,4n
,1111a,,cos2x.cos2xdx,,;a,cos4x.cos4xdx,.24,,,,,,,,2288
4,,fx,sinx,x,,,,,,b,0(n,1,2,....)又由于是偶函数,故.n
311所以,,,,,fx,,cos2x,cos4x,x,,,,,.828
注意:
其实,聪明的同学还有更简单的做法:
21,cos2x311,,4,,fx,sinx,,...,,cos2x,cos4x,x,,,,,,,.既然,,2828,,
利用第1题的结论,它的傅里叶级数就是它自己,即:
311,,,,fx,,cos2x,cos4x,x,,,,,.828
,,,,,,3((书中,第3题)关于区间展开函数P302
1,x,0,,n1,,,1,,,,sgnx,0,x,0,为傅里叶级数,并由此证明,.,,n2,14n1,,,1,x,0.,
sgnx解:
注意到为奇函数,故显然
,11,a,sgnxdx,0;a,sgnx.cosnxdx,0;0n,,,,,,,,
,,122b,sgnx.sinnxdx,.sinnxdx,,cosnx|n,,0,0,,,,n
4,,n,1,3,5,...,2,,,.,,cosn,,1,n,,n,,0,n,2,4,6,..,
2
微积分(下)傅里叶级数部分难题解答
4111,,sgnx,bsinnx,sinx,sin3x,sin5x,sin7x,...,n,,,357,,n1,
41,,,,,sin2n,1x.,,,x,,.(*),2n,1,n1,
(*)中,令得:
x,,2
n1n1,,,,,414,1,,1,,,,,,,1,sin2n,1,所以,.,,,212n,,2n,1n2,14,n1,n1n1,,
,,4((书中,第4题)关于区间,,,,展开下列函数为傅里叶级数P302
(?
),,fx,x;
,,(?
)fx,cosax(为常数);a
,,(?
)fx,xsinx;
0,,,,0,x,(?
),fx,,,x,,x,,.,
,,fx解:
(?
)所给函数满足收敛定理的条件,它在整个数轴上连续,因此的傅
,,fx.里叶级数处处收敛于
,,b,0;n,1,2,?
.,,fx注意到为偶函数,故显然而n
,22,,a,fx.dx,x.dx,,;0,,00,,
,22,,a,fx.cosnxdx,x.cosnxdxn,,00,,
,,,2122,,,,,0,cosnx,,,,,.xdsinnxxsinnxsinnxdx||,,,,00,,00,,,nn,,nn,,
4,n,,1,3,5,...,2,2,,,cosn,,1,.n,,2n,,n,0,2,4,6,..,
,,fx所以,的傅里叶级数展开式为
,a4111,,0x,,acosnx,,cosx,cos3x,cos5x,cos7x,...,n222,,,22357,,n1,
3
微积分(下)傅里叶级数部分难题解答
,41,,cos2n,,,,,1x.,,,x,,.,22,,,2n,1n,1
(?
)所给函数满足收敛定理的条件,它在整个数轴上连续,因此,,的傅里fx,,叶级数处处收敛于fx.
,,b,0;n,1,2,?
.,,注意到fx为偶函数,故显然而n
,,2222,,a,fx.dx,cosax.dx,sinax,sina,;|0,,000,,aa,,
,22,,a,fx.cosnxdx,cosax.cosnxdxn,,00,,
21,,,,,,,cosn,ax,cosn,axdx,0,2
,11,sin,,n,ax,sin,,n,ax,,,,||00,,,,,,n,an,a
n,,,111,,,,,a,,sin,,n,an,a,,,
1nn,,,,a,,,a,,,,12a2sin1,,,,..sina,,,n,1,2,?
2222,n,a,na,
,,fx所以,的傅里叶级数展开式为
,aanx2sin,1cos,,,1n,,,,,,x,,.ax,,,,,cos1,,,,22,a2na,1,,n,
,,fx(?
)所给函数满足收敛定理的条件,它在整个数轴上连续,因此的傅里
,,b,0;n,1,2,?
.,,,,fx.fx叶级数处处收敛于注意到为偶函数,故显然而n
,,222,,,,a,fx.dx,xsinx.dx,,x.dcosx0,,,000,,,
,2,,,,,,2;xcosxcosxdx|,0,,0,,,
,,122,,a,fxcosx.dx,xsinx.cosxdx,xsin2xdx1,,,000,,,
,,111,,,,,,x.dcos2x,,xcos2x,cos2xdx,,;|,,00,0,,,,,222
,22,,a,fx.cosnxdx,xsinx.cosnxdxn,,00,,
4
微积分(下)傅里叶级数部分难题解答
21,,,,,,,x.sinn,1x,sinn,1xdx,0,2
,,11,xd,,cosn,1x,xd,,cosn,1x,,,,,,00,,,,,,n,1n,1
,,1,,,,,xcos,,n,1x,cos,,n,1xdx,,|,00,,,n,1
,1,,,,,xcos,,n,1x,cos,,n,1xdx,,|,00,,,n,1
11,22,,n1nn,,,,,1,,,n,2,3,?
,1,,,,,,1.22,,n,1n,1n,1n,1,,
,,所以,fx的傅里叶级数展开式为
n1,,1,1,,,,xsinx,1,cosx,2cosnx.,,,x,,.,22n,1n2,
,,111,(?
),,,,a,fx.dx,fx.dx,x.dx,;0,,,,00,,,2,
,11,,a,fx.cosnxdx,x.cosnxdxn,,,0,,,
,,,1111,,,,,0,cosnx,,,,,.xdsinnxxsinnxsinnxdx||,,,,00,,00,,,nn,,nn,,
2,n,,1,3,5,...,1,2,,,cosn,,1,.n,,2n,,n,0,2,4,6,..,
,11,,b,fx.sinnxdx,x.sinnxdxn,,,0,,,
,,11,,,,,,,,,.xdcosnxxcosnxcosnxdx|,,0,,00,,,,nn
n,1,11,,,11,,,,,cosn,,sinnx,.,,n,1,2,?
,cosn,,|,,0nnn,n,,
n1,,,,2cos2n,1x,1,,,,所以,,,,,fx,,,sinnx.,,,x,,.,,24n,,,2n,1n1n1,,
,,,,,,fx,x,,x0,x,,5((书中,第1题)将函数分别展开成正弦级数与P307
余弦级数.
,,fx解:
(一)对作奇延拓及周期延拓.
5
微积分(下)傅里叶级数部分难题解答
a,0(n,0,1,2,...).n
,,222,,,b,x,xsinnxdx,2xsinnxdx,xsinnxdxn,,,000,,
,2其中,,,,xdcosnx2xsinnxdx,,00n
,,212,,,,,,,cosn,,sinnx,,,xcosnxcosnxdx||,,,00,,0,,nnn,,,22,n,1,,,,cosn,,,1.nn
,,,222222,,,,,,,xsinnxdx,xdcosnxxcosnx2xcosnxdx|,,,0,,000,,,,,nn
,241,,n,,,,1,0,cosnx|2,,0nn,n,,
,2424nnnn,,,,,,,,,,,,,,,,,,,1,,1,1111.33nnn,n,
242,n,1nnb111故,,,,,,,,,,1,,,,,n3nnn,
8,n,,1,3,5,...,4,n311,,,,,.,,,n,,3n,,n,0,2,4,6,..,
81,,,,fx,sin2n,1x,,0,x,,所以,.,3,,,2n,1n,1
,,fx
(二)对作偶延拓及周期延拓.
b,0(n,1,2,...);n
2,,,1,22,x.dx,;,,,,a,fx.dx,x,,x.dx0,,,0003,,,
,,222,,,a,x,xcosnxdx,2xcosnxdx,xcosnxdxn,,,000,,
,,,22,,,,其中,,,xdsinnxxsinnxsinnxdx2xcosnxdx|,,,0,,000,,nn
2122,,n,0,cosnx,,,,,,,cosn,,1,,1,1.|22,,0nnnn,,
,2222,,,xcosnxdx,,xdsinnx,,00,,n
6
微积分(下)傅里叶级数部分难题解答
,22,,,,,xsinnx2xsinnxdx|,0,,0,,,n
,444n,,,,,,,,,,,,xcosnxcosnxdx,,,cosn,,01.|222,0,,0,,,nnn,
2424nnnn,1,,故a11n,1,2,?
,1,,,,,,,,,,,,1,1,,,,,,,,1.n2222nnnn
2,,,11n1n,,,,,,,fx,,21,,1cosnx,,,4,1cosnx所以,,,226nnn1,n1,2,,11,n1,,,,4,1cosnx,,0,x,,,,4cos2n,,,1x,,22n6,,2n,1n1,n1,
,,,,6((书中,第2题)设ft是周期为,高为的锯齿形波,它在0,2,的P3072,h
h函数表示式为,试把它展开成傅里叶级数.,,,ftt2,
2,22,2,1hht1.,tdt,,h,,解:
;n,1,2,?
,,a,ftdt|0,2,00022,,2,,
2,2,2,1h1h,,,,a,ftcosntdt,t.cosntdt,tdsinntn,,,2000,,,,22n
2,2,hh,,,,,,tsinntsinntdt,,0,0,0.|2,2,,00,,,n,2n22,2,2,1h1h,,,,b,ftsinntdt,t.sinntdt,,tdcosntn,,,2000,,,,22n
2,2,h,,,,,tcosntcosntdt|,2,,00,,,2n
hh,,n,1,2,?
,,,,,2,cos2n,,0.2n,2n,
hh1,,ft,,sinnt.,,x,2k,,k,0,,1,,2,?
所以,,2nn1,
x,,,0,2,fx,7((书中,第3题)在展开函数为傅里叶级数.,,P3072,,fx解:
对作周期延拓.
2,2,2,,1,x11112;,,,,a,fxdx,dx,,..,,x,0|0,,000,222,,
2,2,11x,,,,a,fxcosnxdxcosnxdx,n,,00,2,
2,1,,,,,,,xdsinnx,02n,
2,2,1,,,,,,,,xsinnxsinnxdx|,,,00,,2n,
7
微积分(下)傅里叶级数部分难题解答
1;,,,0,0,0n,2
2,2,11x,,,,b,fxsinnxdxsinnxdx,n,,00,2,
2,1,,,,,,,,xdcosnx,02n,
2,2,1,,,,,,,,,xcosnxcosnxdx|,,,00,,2n,
11,,,,,cos2n,,;n,1,2,?
,,,,2nn,
1,,fx,sinnx.,,所以,0,x,2,,n,n1
TT,,,,,,,,8((书中,第4题)设ft是周期为TT,0的周期函数,它在内P307,,22,,
的函数表示式为
T2,,Esint,0,t,,,,,,m,2T,ft,,,,试把它展开成傅里叶级数.,T0,,,t,0.,2,,
TTT2E121,,m222,,cos,t,Esin,tdt解:
,,a,ftdt,,|m0T,,00,TTT,,,2
2
2,,,.T,,2E2E2ET,,,mmTm,,,1,cos;,,1,cos,,2T2,2,,,,,,T.,,T,,
TT1nt22,nt,22,Etdt.cossin.cos,,a,ftdt,mnT,,0,TTTT2
22
TE,,nn2,2,,,,,m2,,,,ttdtsinsin,,,,,,,,,0TTT,,,,,,
,,,,TT,,,,EEnn12,12,,,,,mm22,,,,,,ttcoscos,,,,,,,,,,||00nn22,,TTTT,,,,,,,,,,,,,,,,,,,,TT
8
微积分(下)傅里叶级数部分难题解答
,,,,EE,,,,2nT2nT,,,,,,mm,1,cos,,.,,1,cos,,.,,,,,,,,,,,,,,,2n2nT2T2,,,,,,,,,,,,,,,,,,,T,T,,,,,,TT,,,,
,,E,,22nT,,,,m,1,cos,,.,,,,,,,22nTT2,,,,,,,,,,,T,,TT,,
,,E,,2,2n,T,,m,1,cos,,.,,,,,,,22nTT2,,,,,,,,,,,T,,TT,,
EEmm,,,1,cosn,,,,1,cosn,,,,,2n,121,n,,
E1,mn,.,2,4,6,...,EE112,,,nmm2,,,1,cosn,,,1,,1,,,,,.,,n1,2,,2n,11,n,,21,n,,,n,0,1,3,5,..,
TT1t22,t,22,Etdt.sinsin.sin,,b,ftdt,m1T,,0,TTTT2
22
4t,1cosTTtt22,2,2T22,Edt,Edtsin.sinmm,,00T2TTT
TEEE4,TtTT,,,,mmm2,,sin,t,,sin2,,;|,,,,0T4T242T,,,,,,
TT1nt12,nt,22,Etdt.sinsin.sin,,b,ftdt,mnT,,0,TTTT2
22
TE,,nn2,2,,,,,m2,,,,ttdtcoscos,,,,,,,,,0TTT,,,,,,
,,,,TT,,,,EEnn12,12,,,,,mm22,,,,ttsinsin,,,,,,,,,,||00nn22,,TTTT,,,,,,,,,,,,,,,,,,,,TT
,,,,EE,,,,2nT,2n,T,,,,mm,sin,,.,,0,sin,,.,,1,,,,,,,,,,,,,,2n2nT2T2,,,,,,,,,,,,,,,,,,,T,T,,,,,,TT,,,,
9
微积分(下)傅里叶级数部分难题解答
,,E,,22nT,,,,m,sin,,.,,0,,,,,,22nTT2,,,,,,,,,,,T,,TT,,
,,E,,2,2n,T,,m,sin,,.,,0,,,,,,22nTT2,,,,,,,,,,,T,,TT,,
EEmm,sinn,,0.,sinn,n,2,3,?
.,,,,21,n21,n,,
,,所以,ft的傅里叶级数展开式为
EEtnt2,14,mm,,fx,,sin,cos.,2TT2n,41,n1,
,,,,,,9((书中fx,,,,0,,,第5题)设在上为可积的奇函数且在上有P307
,,b,,fx,0.fx求证:
(其中为的傅里叶系数).b,kbkk1
提示:
,,sinkx,ksinx0,x,,.
,证明:
(一)先用数学归纳法证明:
对于,有(*),,k,Nsinkx,ksinx0,x,,.
成立。
事实上:
当时,(*)式显然以等式形式成立;k,1
假设当时,(*)式成立,则当时,k,nk,n,1
,,sink,1x,sinkx.cosx,coskx.sinx
所以(由归纳假设),,sink,1x,sinkx.cosx,coskx.sinx
,,,ksinx.1,1.sinx,k,1sinx.
(二)由公式
2,,b,fxsinkxdxk,0,
,22所以(由(*)式),,,,b,fxsinkxdx,fxsinkxdxk,,00,,
,22,,,,,k.fxsinxdx,kb.,,,fx.ksinxdx1,,,,00,,,,
b,0.,,,,,,fx0,2,k,1,2,?
10((书中,第6题)设在上单调减少,求证:
P308k
b,,fx(其中为的傅里叶系数).k
10
微积分(下)傅里叶级数部分难题解答
2,1证明:
(令),,b,fxsinkxdxkx,tk,0,
k22k,n,1t1t,,,,fsintdt,fsintdt,,,,,,,,,,2,10n,k,k,kk,,,,,1n
,,)(再令u,t,2n,1,
k2,,,,12n,1,u,,,,,fsin,,2n,1,,udu,,,,0kk,,,,1n
k2,,,12n,1,,u,,fsinudu,,,,,0kk,,,,1n
k22,,,,,,,,12n1u2n1u,,,,,,,,,,,,fsinudufsinudu,,,,,,,,,0,kkk,,,,,,1n,,
2,,,2n,1,,u,,fsinudu其中(令)x,u,,,,,,k,,
,,,,2n,1,x,,,,,,fsinx,,dx,,,0k,,
,2n,,x,2n,u,,,,fsinxdxfsinudu,,,,,,,,,,00kk,,,,
k,,,,,,,,12,,n1u2nu,,,,,,b,,所以ffsinudu,,,,,,k,,,,0kkk,,,,,,1n,,,,
,,,,fx0,2,注意到,在上单调减少,故
,,,,2n,1,,u2n,,u,,,,f,f,0,,,,,kk,,,,
b,0.,,k,1,2,?
所以,k
11
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