数学分析讲义 第四版 (刘玉琏 傅沛仁 著) 高等教育出版社

1、 y > x , f (y > f (x; y < x , f (y < f (x. ,x (x0 , x0 + (x x , x + x , f (x (x x , x + x . x, y (x0 , x0 + , x < y. , (xi xi , xi + xi ,i 1, 2, · · · , n, x1 = x, xn = y x1 < x2 · · · < xn , x, y , n x, y i=1 (xi xi , xi + xi . ai (xi , xi+1 , i =

2、1, 2, · · · , n 1, ai (xi xi , xi + xi (xi+1 xi+1 , xi+1 + xi+1 . , f (x < f (a1 < f (a2 · · · < f (an1 < f (y , f (x (x0 , x0 + . §3.2 7, y = f (x (x0 , x0 + , . , x = (y , y0 = f (x0 §5.2 4,y (f (x0 , f (x0 + , (x = x = (y , . (f (x0 , f (x0 + 1

3、. f (x y = 0, y0 + y (f (x0 , f (x0 + , , x = (y0 + y (y0 = 0. x0 + x (x0 , x0 + , f (x0 + x = y0 + y, y 0 y 0. 1 1 (y0 + y (y0 f (x0 + x f (x0 lim = lim y 0 x0 f (x0 + x f (x0 y f (x0 + x f (x0 f (x0 x = lim = , x0 f (x0 3 f (x0 + x f (x0 f (x0 f (x0 + x x (y0 = f (x0 . f (x0 3 f (x = 0 12. : f (x

4、n ,a k (k n · · · = f (k1 (a = 0, f (k1 (a = 0. = a f (x = 0 k (k n, f (x = (x ak g (x, , g (a = 0. ,m = 1, 2, · · · , k 1, k, 0 f (m (x = Cm (x ak (m g (x 1 +Cm (x ak (m1 g (x + · · · (l1 +Cm (x ak (ml+1 g (l1 (x (l +Cm (x ak (ml g (l (x (m + · 

5、3; · + Cm (x ak g (m (x. f (a = f (a = g (x n k (1 m k 1, ,(1 xa , f (a = f ( a = · · · = f (k1 (a = 0. n=k ,(1 k !g (x, f (k (a = k !g (a = 0 21 xa , = 9 n f (x = xa n f (n (x f (n1 (x (x an + (x an1 n! (n 1! , f (k1 (x f (k (x k (x a + (x ak1 + ··· + k! (k 1! + &

6、#183; · · + f (a. f (a = f ( a = · · · = f (k1 (a = 0, f (k (a = 0, f (x = f (n (x f (n1 (x f (k (x (x an + (x an1 + · · · + (x ak n! (n 1! k! f (k (a f (k+1 (x f (n (x + (x a + · · · + (x a(nk k! (k + 1! (n! f k (a = k! = (x ak = (x ak g (x. f

7、(n (x f (k (a f (k+1 (x + (x a + · · · + (x a(nk k! (k + 1! (n! 0. ,a f (x = 0 k . 1 dn 2 13. : Pn = n (x 1n 2 n! dxn y = (x2 1n , y = 2nx(x2 1n1 , n+1 u(x = y u(k (x = y (k+1 v (x = x2 + 1 v (x = 2x v (x = 2 nk , g (a = Pn (x = (1 x2 Pn (x 2xPn (x + n(n + 1Pn (x = 0. (x2 1y = 2nxy. .

8、 u(x = y u(k (x = y (k v (x = 2xn v (x = 2n (1 0 y (n+2 (x2 1 + C 1 y (n+1 2x + C 2 y (n · 2 = C 0 y (n+1 · 2xn + C 1 y (n · 2n Cn n n n+1 n+1 (x2 1y (n+2 + 2(n + 1xy (n+1 + n(n + 1y (n = 2nxy (n+1 + 2n(n + 1y (n = 0, (x2 1y (n+2 + 2xy (n+1 n(n + 1y (n = 0 y = (x2 1n , Pn (x = 1 dn 2 1 (x 1n = n = Pn (x n n 2 n! dx 2 n! y