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 dmivendityTianjin 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, )