⑩天掌 Teaching Plan on Advanced Mathematics o 第八节一般周期函数的傅里叶级数 周期为2L周期函数的傅里叶级数 二、总结 tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 第八节 一般周期函数的傅里叶级数 一、周期为2L周期函数的傅里叶级数 二、总结
⑩天掌 Teaching Plan on Advanced Mathematics o 周期为2L周期函数的傅里叶级数 2兀 T=2l,0= 代入傅氏级数中 +>(a, cos nax +b sin nax) 2 定理设周期为的周期函数f(x)满足收敛 定理的条件则它的傅里叶级数展开式为 ∫(x0+∑(a11 ntx nTtr +b, sin -), 2 tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 周期为2L周期函数的傅里叶级数 T = 2l, . 2 T l = = 定理 定理的条件则它的傅里叶级数展开式 为 设周期为 的周期函数 满足收敛 , 2l f (x) ( cos sin ), 2 ( ) 1 0 l n x b l n x a a f x n n n + = + = ( cos sin ) 2 1 0 a n x b n x a n n + n + = 代入傅氏级数中
⑩天掌 Teaching Plan on Advanced Mathematics o 其中系数an,b为 nTtr f(x)cos",x,(n=0,1,2,…) nth f(x)sin",dx,(n=1,2,…) (1)如果f(x)为奇函数,则有 f(x)=∑bsin nTcr 其中系数b为b2 T x sin dx (n=1,2,…) tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 其中系数an , bn为 ( )cos , ( 0,1,2, ) 1 = = − dx n l n x f x l a l l n ( )sin , ( 1,2, ) 1 = = − dx n l n x f x l b l l n (1)如果f (x)为奇函数, 则有 ( ) sin , 1 = = n n l n x f x b ( )sin , 2 0 dx l n x f x l b b l n n 其中系数 为 = (n = 1,2, )
⑩天掌 Teaching Plan on Advanced Mathematics o (2)如果(x)为偶函数则有 f(x)="+∑ a. cOS nTo n=1 其中系数a为n-=2 nTx x)cOS (n=0,,2,…) 证明△,_πx ,一l≤x≤l→-兀≤z≤π 设f(x)=f()=F(z),F(z)以2m为周期 T F(3)=+2(an cos nz+b, sin nz), tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics (2)如果f (x)为偶函数, 则有 cos , 2 ( ) 1 0 = = + n n l n x a a f x dx l n x f x l a a l n n = 0 ( )cos 2 其中系数 为 (n = 0,1,2, ) 证明 , l x z 令 = − l x l − z , ( ) ( ) F(z), lz f x f = 设 = F(z)以2为周期. ( cos sin ), 2 ( ) 1 0 a nz b nz a F z n n = + n + =
⑩天掌 Teaching Plan on Advanced Mathematics o 其中 "F( z)coS naz 几=兀 z) SIn nza。 Tr z= F(=f(r) n f(x)=+2(an cosx+b, sin x) 2 其中an1=,,f(x)cos,xcr, n ∫(x)sin"xx tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics ( cos sin ) 2 ( ) 1 0 x l n x b l n a a f x n n n + = + = ( )sin . 1 ( )cos , 1 − − = = b F z nzdz a F z nzdz n 其中 n ( )sin . 1 ( )cos , 1 − − = = l l n l l n xdx l n f x l b xdx l n f x l 其中 a F(z) f (x) l x z = =
⑩天掌 Teaching Plan on Advanced Mathematics o 例1设f(x)是周期为4的周期函数,它在[-2,2)上的表 0-2≤x<0 达式为f(x)= 将其展成傅氏级数 k0≤x<2 解∵=2,满足狄氏充分条件 a=0+k=k, tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics k − 2 x y − 4 0 2 4 例 1 设 f (x)是周期为 4 的周期函数,它在[−2,2)上的表 达式为 − = 0 2 0 2 0 ( ) k x x f x , 将其展成傅氏级数. 解 l = 2, 满足狄氏充分条件. = + − 2 0 0 2 0 2 1 0 2 1 a dx kdx = k
⑩天掌 Teaching Plan on Advanced Mathematics o k cos xdx=o, (n 0 n k k. sin-xdx=( 1-cosn兀) 2 n 2k 当n=1,3,5, n几 0￥n=2,4,6, k2k.,πv 3πx1 STX .f(r) 2+元 (sin-+sin+in+…) 23 2 52 (-∞0<x<+;x≠0,+2,+4,) tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 2 0 2 cos 2 1 xdx n k = 0, = 2 0 2 sin 2 1 xdx n bn k (1− cos ) = n n k , 0 2,4,6, 1,3,5, 2 = = = n n n k 当 当 ) 2 5 sin 5 1 2 3 sin 3 1 2 (sin 2 2 ( ) + + + = + k k x x x f x (− x +; x 0,2,4, ) an = (n = 1,2, )
⑩天掌 Teaching Plan on Advanced Mathematics o 例2将函数f(x)=10-x(5<x<15)展开成傅氏级数 解作变量代换z=x-10, 5<x<15→-5<<5, f(x)=∫(z+10)=-z=F(z), 补充函数F(z)=-z(-5<z<5)的定义, 令F(-5)=5,然后将F(z)作周期延拓(T=10) 这拓广的周期函数满足收敛定理的条件 且展开式在(-5,5内收敛于F() tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 例 2 将函数 f (x) = 10 − x (5 x 15)展开成傅氏级数. 解 作变量代换 z = x −10, 5 x 15 −5 z 5, f (x) = f (z + 10)= −z = F(z), 补充函数 F(z) = −z (−5 z 5)的定义, 令 F(−5) = 5, 然后将F(z)作周期延拓(T = 10) 这拓广的周期函数满足收敛定理的条件, 且展开式在(−5, 5)内收敛于F(z)
⑩天掌 Teaching Plan on Advanced Mathematics o 0,(n=0,4,2,…) F(z) 5(-孔)9份4mz dz 2 50:510N 10 (n=1,2,) 10>(D) n几z 5 (-5<z<5 n n几 ∴10-x= 10y(-1) sin[-(x-10) 5 10(-1) nTC sinx.(5<x<15) tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics x y F(z) − 5 0 5 10 15 a = 0, (n = 0,1,2, ) n = − 5 0 2 ( )sin 5 2 dz n z b z n , 10 ( 1) = − n n (n = 1,2, ) , 5 sin 10 ( 1) ( ) 1 = − = n n n z n F z (−5 z 5) = − − − = 1 ( 10)] 5 sin[ 10 ( 1) 10 n n x n n x . 5 sin 10 ( 1) 1 = − = n n x n n (5 x 15)
⑩天掌 Teaching Plan on Advanced Mathematics o nTtr 另解 (10-x)c0s 5 2 s cos"-dx-- nTu rcos x=0,(n=1,2,…) 5 =2(10-x)d=0 b9(0-xSi ATx 10 15 d=(-1) 5 2,(n=12,…) 10÷(-1)"n 故f(x)=10-x=∑ sIn— H=1 5 (5<x<15) tianjin polytechnic dmivendity
Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 另解 = − 15 5 5 (10 )cos 5 1 dx n x an x = − 15 5 5 (10 )sin 5 1 dx n x bn x − = 15 5 15 5 5 cos 5 1 5 2 cos dx n x dx x n x = 0, = − 15 0 5 (10 ) 5 1 a x dx = 0, , 10 ( 1) n n = − (n = 1,2, ) = − = − = 1 5 sin 10 ( 1) ( ) 10 n n x n n 故 f x x (5 x 15) (n = 1,2, )
