天津工业大学数学系：《高等数学》课程电子教案（PPT课件）第十一章 无穷级数（11.7）傅里叶级数

⑩天掌 Teaching Plan on Advanced Mathematics o 第七节傅里叶级数 、三角级数三角函数系的正交性 函数展开成傅里叶级数 、正弦级数和余弦级数 tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 第七节 傅里叶级数 二、函数展开成傅里叶级数 三、正弦级数和余弦级数 一、三角级数 三角函数系的正交性

⑩天掌 Teaching Plan on Advanced Mathematics o 、三角级数三角函数系的正交性 1.三角级数 ∫(t)=A+∑Asin(mof+qn)谐波分析 =1 A0+2(A, sin (p, cos not+ An, cos (n sin nat) A0=Ao, a,=A, sin(m, b,=A, cos n5 ot=x au+∑( a. cos nx+ h sinx)三角级数 2 tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 一、三角级数 三角函数系的正交性 = +   +   =1 0 ( ) sin( ) n n n f t A A n t 1.三角级数 谐波分析 = +    +    =1 0 ( sin cos cos sin ) n n n n n A A n t A n t +  +  =1 0 ( cos sin ) 2 n an nx bn nx a , 2 0 0 A a 令 = sin , n An n a =  cos , n An n b =  t = x, 三角级数

⑩天掌 Teaching Plan on Advanced Mathematics o 2.三角函数系的正交性 三角函数系 l,cosx,sinx,c0s2x,sin2x,… cos r, SInr,… 正交: 任意两个不同函数在[-m上的积分等于零 cos ndx=0 sin nxd=0, (n=1,2,3,…) tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 2.三角函数系的正交性 1,cos x,sin x,cos 2x,sin2x, cos nx,sinnx,  [ , ] . : 任意两个不同函数在 上的积分等于零 正交 −  cos = 0, −   nxdx sin = 0, −   nxdx 三角函数系 (n = 1,2,3, )

⑩天掌 Teaching Plan on Advanced Mathematics o 0,m≠n sin mx sinnxdx TC. =n cos mx cos nrds/o n≠n 7. =n sin mx cos ndx=0.(其中m,n=1,2,) tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics , , 0, sin sin     =  =   − m n m n mx nxdx , , 0, cos cos     =  =   − m n m n mx nxdx sin cos = 0.   − mx nxdx (其中m,n = 1,2, )

⑩天掌 Teaching Plan on Advanced Mathematics o 、函数展开成傅里叶级数 问题:1若能展开,a1,b是什么 2展开的条件是什么? 1.傅里叶系数 若有f(x)=+∑( a, cos kx+ b, sin hx) k=1 1)求a f(x)dx= dx+E(a, cos kx+bk sin kx)]dx k=1 tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 二、函数展开成傅里叶级数 问题: 1.若能展开, ai ,bi 是什么? 2.展开的条件是什么? 1.傅里叶系数 = +  +  =1 0 ( cos sin ) 2 ( ) k ak kx bk kx a 若有 f x (1) . 求a0 dx a kx b kx dx a f x dx k k k [ ( cos sin )] 2 ( ) 1 0  =  +   +  −  =  −  −

⑩天掌 Teaching Plan on Advanced Mathematics o f(x)dx 2 (2)求an f(x) cos ndx= o cos nrdc +ela. cos kx cos ndx +bs sin k cos ndx] tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 2 , 2 0 =   a   =  − a f (x)dx 1 0 dx a kxdx b kxdx a k k k k cos sin 2 1 1 0    −   = −  = − = + +       (2) . n 求a  =   −  − nxdx a f x nxdx cos 2 ( )cos 0 [ cos cos sin cos ] 1     −  −  = + a kx nxdx + bk kx nxdx k k

⑩天掌 Teaching Plan on Advanced Mathematics o a. cos2md=an冗 f∫(x) cos nro(n=1,2,3,…) (3)求bn ∫(x) sinned sinned +Ela[ cos kx sin ndxb sin kr sin/xx l =b,T, f(x)simn(n=1,2,3,…) T tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics =   − an nxdx 2 cos = , an   =  − a f x nxdx n ( )cos 1 (n = 1,2,3, ) (3) . 求bn   =  − bn f (x)sinnxdx 1 (n = 1,2,3, ) − − =     nxdx a f x nxdx sin 2 ( )sin 0 [ cos sin sin sin ] 1     −  −  = + a kx nxdx + bk kx nxdx k k = , bn

⑩天掌 Teaching Plan on Advanced Mathematics o 傅里叶系数 f∫(x) cos ndx,(n=0,1,2,…) f(x) sin ndx,(n=1,2,… T ∫(x) cos ndx,(n=0,2,) 或 兀0 ∫(x) sinned,(n tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics      =  = =  =    −  − ( )sin , ( 1,2, ) 1 ( )cos , ( 0,1,2, ) 1   b f x nxdx n a f x nxdx n n n      =  = =  =     2 0 2 0 ( )sin , ( 1,2, ) 1 ( )cos , ( 0,1,2, ) 1   b f x nxdx n a f x nxdx n n n 或 傅里叶系数

⑩天掌 Teaching Plan on Advanced Mathematics o 傅里叶级数 +∑(anC0sHx+ b sinn) H=1 问题: f(x)条件?0+∑ (a, cos nx+ b sin nx) tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 傅里叶级数 +  +  =1 0 ( cos sin ) 2 n an nx bn nx a 问题: +  +  =1 0 ( cos sin ) 2 ( ) ? n an nx bn nx a f x 条件

⑩天掌 Teaching Plan on Advanced Mathematics o 2.狄利克雷( Dirichlet)充分条件收敛定理) 设∫(x)是以2π为周期的周期函数如果它满足条件:在 个周期内连续或只有有限个第一类间断点,并且至多只有 有限个极值点则f(x)的傅里叶级数收敛,并且 (1)当x是f(x)的连续点时级数收敛于f(x) (2)当x是f(x)的间断点时收敛于 f(x-0)+f(x+0) 2 (3)当x为端点x=±7时收敛于(兀+0)+/(=0 2 tianjin polytechnic dmivendity

Tianjin Polytechnic University Teaching Plan on Advanced Mathematics 2.狄利克雷(Dirichlet)充分条件(收敛定理) 设 f (x)是以2 为周期的周期函数.如果它满足条件:在 一个周期内连续或只有有限个第一类间断点,并且至多只有 有限个极值点,则 f (x)的傅里叶级数收敛,并且 (1) 当x 是 f (x)的连续点时,级数收敛于f (x) ; (2)当x是 f ( x)的间断点时,收敛于 2 f (x − 0) + f (x + 0) ; (3) 当x 为端点x = 时,收敛于 2 f (− + 0) + f ( − 0)