(n=2,3.4…) f(x)sin 5(7x+2) b,=F/(x)sin 2丌nx (n+1)丌 (n-1)丌 SIn (n=2,3,4 4)-2-5-.w+(+3)mar (n+1)z (n-1)x2 cos cOs CoS not 2 n+1 4 4 sin(n-1) 1(n+1)x1( 4-snot。 8.设f(x)在-z,n]上可积或绝对可积,证明 (1)若对于任意x∈[-z,n],成立f(x)=f(x+n),则a21=bn1=0 (2)若对于任意x∈[-z,],成立f(x)=-f(x+),则a2n=b2n=0 证(1)an1=1∫2/(0-)h f(x)cos(2n-1)xdx+=f(x)cos(2n-1).xdx 「。(0)cs2n-1)-(2n-1)1+(x)o2n-1)oh(=x+) 0,(n=1,2,3 b, f(x)sin(2n-1)xdx f()sin(2n-1)xdx+- f(x)sin(2n-1)xdx 10-(21+(xm0)U=x =0,(n=1,2,3 (2)an=1 f∫(x)cos(2nx)dh f(x)cos(2nx)dx+_Jo/(x)cos(2nx)dx f(ocos(2nt-2nz ) dt+-f(x)cos(2nx)dx (t=x+r) 1,2,3,( n = 2,3, 4,"), 1 b = 0 2 2 ( )sin T x f x dx T T π ∫ 5(7 2) 8 π π + = ， bn = 0 2 2 ( )sin T nx f x dx T T π ∫ 5 1 ( 1) 1 ( 1) sin sin 2 1 4 1 4 n n n n π π π ⎡ + − ⎤ = − ⎢ ⎥ ⎣ + − ⎦ ， ( n = 2,3, 4,")。 f x( ) ∼ t ω t π ω π π sin 8 35 4 5 cos 4 5 (2 2) 4 5 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − + + n t n n n n n n ω π π π cos 1 2 4 ( 1) cos 1 1 4 ( 1) cos 1 1 2 5 2 ∑ 2 ∞ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − + − − − + + + n t n n n n n ω π π π sin 4 ( 1) sin 1 1 4 ( 1) sin 1 1 2 5 2 ∑ ∞ = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − − + + + 。 ⒏ 设 f (x)在[−π ,π ]上可积或绝对可积，证明： ⑴ 若对于任意 x ∈[−π ,π ]，成立 f (x) = f (x + π ) ，则a b 2 1 n n − = 2 −1 = 0； ⑵ 若对于任意 x ∈[−π ,π ]，成立 f (x) = − f (x + π ) ，则a b 2 2 n n = = 0 . 证 (1) 2 1 n a − = 1 f ( ) x n cos(2 1)x π π −π − ∫ dx 0 0 1 1 f ( ) x n cos(2 1)xdx f (x) cos(2n 1)x π π π −π = − + ∫ ∫ − dx 0 0 1 1 f t( ) cos[(2n 1)t (2n 1) ]dt f (x) cos(2n 1)xdx (t x ) π π π π π π = − − − + − ∫ ∫ = + = 0 , ( n =1, 2,3,…), 2 1 n b − = 1 f ( ) x n sin(2 1)x π π −π − ∫ dx 0 0 1 1 f ( ) x n sin(2 1)xdx f (x n )sin(2 1)x π π dx π − π = − + ∫ ∫ − 0 0 1 1 f t( )sin[(2n 1)t (2n 1) ]dt f (x)sin(2n 1)xdx (t x ) π π π π π π = − − − + − ∫ ∫ = + = 0 , ( n =1, 2,3,…)。 (2) 2n a = 1 f ( ) x n cos(2 x) π π ∫ −π dx 0 0 1 1 f ( ) x n cos(2 x)dx f ( ) x cos(2nx) π π π −π = + ∫ ∫ dx 0 0 1 1 f t( ) cos(2nt 2n )dt f (x) cos(2nx)dx (t x ) π π π π π π = − − + = + ∫ ∫ = 0 , ( n =1, 2,3,…), 7