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)1(12 ??? ???????????,6,4,2,0,5,3,1,4nnn ?.02 11, ???? 級(jí)數(shù)收斂于(間斷點(diǎn))當(dāng) ?kx? ])12s i n (1213s i n31s i n 4)(),2,1,0( ???????????????xkkxxxfkkx??當(dāng) (連續(xù)點(diǎn)) xy s in4??)3s i n31( s i n4 xxy ?? ?)5s i n513s i n31( s i n4 xxxy ??? ?)7s i n715s i n513s i n31( s i n4 xxxxy ???? ?)9s i n917s i n715s i n513s i n31( s i n4 xxxxxy ????? ?注:泰勒級(jí)數(shù)是局部逼近,而傅里葉級(jí)數(shù)是全局逼近。 xy0 ??? ?2?2??? ????? dxxfa )(10 ?? ??? ????? 00 1)(1 x dxdxx,??解:所給函數(shù)滿足狄利克雷充分條件 . ? ????? nx d xxfa n c o s)(1 ?? ??? ????? 00 c o s1c o s)(1 n xd xxn xd xx)1( c o s22 ???? nn ]1)1[(22 ???? nn???????????,6,4,2,0,5,3,1,42nnn ?? ????? nx d xxfb n s i n)(1 ?? ??? ????? 00 s i n1s i n)(1 n xd xxn xd xx,0?)( ????? x所求函數(shù)的傅氏展開(kāi)式為 ),3,2,1( ??n.c os2)(.s i n)(101nxaaxfnxbxfnnnn???????余弦級(jí)數(shù)有常數(shù)項(xiàng)和余弦項(xiàng)的級(jí)數(shù)是只含是偶函數(shù),則其傅立葉如果正弦級(jí)數(shù)數(shù)是只含有正弦項(xiàng)的葉級(jí)是奇函數(shù),則它的傅立定理:如果注意以上兩個(gè)例子的結(jié)果,容易證明 x? ??oy例 3. 上的表達(dá)式為 將 f (x) 展成傅 里 葉級(jí)數(shù) . 解 : ?????? xxfa d)(10??? 0 dc o s1 ?? xxnx??? ??? xnxxfa n dc o s)(1??? 0 d1 ?? xx ?? ????????0221 x????? ??????? ?? 02c o ss i n1nnxnnxx??2c o s1nn??2 3? ?? ??? 3