5. Use the definition to find the Taylor polynomial of degree n = 5 for f(x) = x2 - 2 sin(x) centered at x = 0. k f ( k ) ( 2 ) f (k) (0) f ( k ) (0 ) K ! termSolution: We know that Taylor polynomial of at x = Q is given by f ( x ) ~ ( a ) 1 "- 2 sink ) f ( x 1 = f ( *1 = x2- 2sinx f ( x ) = = ( fal ) = 2 ( x - 2 sinx ) = 2x - 2Cox do dx 7 ( x ) = = ( f ( x ) ) = = ( 2 x - 2 cos x ) = 2+ 2 sinx dx dx f ( x ) = d ( f ( x ) ) = d ( 2 + 2 sing ) = 2 cox f ( x ) = = ( f (x ) = 1 ( 2(ox ) = - 2 sinxf ( x ) = 2 ( f ( x 1 ) = d ( - 2 sinx ) = - 2 cox dx Now our table is K f ( x ) + (0 ) term K 0 f ( x ) = x - 2 sing f (0 ) = (@) - 2 sin(0) f ( 0 ) . .= 0 O. x = O = 0 I f ( * ) = 2x - 2casx f( 0 ) = 2(0) - 2108(0 ) -2 = -2 - 2X - 2 L IT 11 f ( 2 0 ) = 2 + 2 5 in (x ) | $ ( 0 ) = 2 + 2 sin ( 0 ) f ( 0 ) = 2 1 1 x = = 2+ 2(0 1 = 2 12 12 111 * ( * ) = 2 cos x pill f (0 ) = 260 (0 ) [WIN WI - = 2 3 IV f" ( x ) = - 2sin x * ( 0 ) = - 2sin (0 ) ( 10 _ O O. x= O = 0 14 v f (x1 = - 2405 x 7 (0 ) = - 2 cos ( 0 ) = - 2 15 = -1 .x = 60 60Now our taylor polynomial of oth degree is 3 f ( x ) u o ( xx ) + ( - 2 ) ( * ) + 1 ( x ) + + 0 ( x ) + 2 3 = ) f ( x ) u - 2x t x + 1 x 5 2 3