= J F(z) n b not SIN d 5 2 50510Ni5 10 =(-1) (n=1,2,) nTt (2=10∑)sinm,(5<z<5 T 5 10 10-x ∑ sin["(x-10) 5 n= 10(-1)”兀 sIn--x (5<x<15) T nx 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)