taylor dice que f (x) = f (x0 ) + f 0 (x0 )(x f (4) (x0 ) (x 4! x0 )4 + f 00 (x0 ) (x 2 f (5) (x0 ) (x x0 )5 5! f (3) (x0 ) (x 3! f (6) (x0 ) (x x0 )6 6! x0 )2 + x0 ) + + x0 )3 + en nuestro caso: cos(x) = cos( =2) cos( =2)(x 4! sen( =2)(x cos( =2)(x 2 =2) =2)4 sen( =2)(x 5! = (x =2)5 =2) + (x cos( =2)(x 6! =2)3 3! (x 1 =2)6 =2)5 5! =2)2 + sen( =2)(x 3! =2)3 +