微积分泰勒公式

即: f (x) = f (0) + f ′(0)x + f ′′(0) x2 + f ′′′(0) x3 + + f (n) (0) xn + f (n+1) (ξ ) xn+1.
2!
3!
n!
(n +1)!
(2) : 记Rn (x) =
f (n+1) (ξ )
(n +1)!
(
x
−
x0
) n +1
=
在近似计算或者理论分析中, 常用一个简单的函数近似地表示比较复杂的函数. 而多项式函数是最简单的一类函数.
2018/11/19
Edited by Lin Guojian
2
证 : 只证x ∈[x0, x0 + δ )的情形,其它情形类似可证.
构造F (z) = f (x) −[ f (z) + f ′(z)(x − z) + f ′′(z) (x − z)2 2!
2
x=0
2 x=0
2
2
则当n = 2k, k = 1,2,时,sin( n ⋅π ) = sin( 2k ⋅π ) = sin(kπ ) = 0.
2
2
当n = 2k −1, k = 1,2,时,
sin( n ⋅π ) = sin( 2k −1 ⋅π ) = sin(kπ − π ) = sin[(k −1)π + π ] = (−1)k−1 sin[π ] = (−1)k−1.
x0 )n+1
=
f
(n+1) (x0 ) + (n +1)!
o(1)
(x
−
x0 )n+1
=
f (n+1) (x0 ) (n +1)!
(x
−
x0 )n+1
+
o(1) (x (n +1)!
−
x0 )n+1
=
f
(n+1) ( x0 (n +1)!
)
(x
−
x0
) n +1
+
o[(x
−
x0 )n+1],
(x
→
问题(1) : 若f ′(x0) = 0,如何近似计算f(x) − f(x0).
问题(2) : 近似的精度不够,因为o(x − x0)仅仅是x − x0的高阶无穷 小量(x → x0),如何得到任意高阶的无穷小量o[(x − x0)n],n ≥ 1.
问题(3) : 这样近似只是定性的估计,没有给出具体的误差计算公式.
3! 5!
(2k −1)!
或 sin x = x − 1 x3 + 1 x5 + + (−1)k−1 x2k−1 + o(x2k−1), x → 0.
3! 5!
(2k −1)!
2018/11/19
Edited by Lin Guojian
8
例: 求f (x) = cos x带皮亚诺余项的麦克劳林公式.
)
(x
−
x0
)n
]
F′(z) = −[ f ′(z) + f ′′(z)(x − z) + (−1) ⋅ f ′(z) + f ′′′(z) (x − z)2 + f ′′(z) 2(x − z)(−1)
2!
2!
+ f (4) (z) (x − z)3 + f ′′′(z) 3(x − z)2 (−1)
f (n+1) (ξ ) =
f (n+1) (x0 )
⇒
lim[ f (n+1) (ξ ) −
x → x0
f (n+1) (x0 )] = 0 ⇒
f (n+1) (ξ ) =
f (n+1) (x0 ) + o(1), (x →
x0 ).
这时Rn (x) =
f (n+1) (ξ )
(n +1)!
(x
−
2!
3!
n!
(n +1)!
函数f (x)的n +1阶带皮亚诺余项的麦克劳林公式为:
f (x) = f (0) + f ′(0)x + f ′′(0) x2 + f ′′′(0) x3 ++ f (n) (0) xn + o(xn ).
2!
3!
n!
由e0 = 1与(ex )(n) = ex , ⇒ (ex )(n) = ex = 1,
f (n+1) (ξ ) = f (n+1) (x0 ) + o(1), x → x0 , x0 ∈[a, b].
其中x0 = a或x0 = b时, x → x0是指单侧极限过程.
这时因为: x0 < ξ < x,故x → x0时,ξ → x0.
⇒ lim x → x0
f (n+1) (ξ ) = lim ξ →x0
x0 ).
通常, 称o[(x − x0 )n+1], (x → x0 )为皮亚诺余项.
2018/11/19
Edited by Lin Guojian
6
例: 求f (x) = ex的n +1阶麦克劳林公式.
解 :函数f (x)的n +1阶带拉格朗日余项的麦克劳林公式为:
f (x) = f (0) + f ′(0)x + f ′′(0) x2 + f ′′′(0) x3 + + f (n) (0) xn + f (n+1) (ξ ) xn+1,
2! 3!
n!
2018/11/19
Edited by Lin Guojian
7
例: 求f (x) = sin x带皮亚诺余项的麦克劳林公式.
解 :由于sin 0 = 0, (sin x)(n) = sin(x + n π )⇒ (sin x)(n) = sin(x + n π ) = sin(n ⋅ π ) = sin( n ⋅π ),
f (x) =
f (x0 ) +
f ′(x0 )(x − x0 ) +
f
′′( x0 2!
)
(
x
−
x0
)2
+
+
f
(Байду номын сангаас
n) (x0 n!
)
(
x
−
x0
)
n
+
f (n+1) (ξ ) (n +1)!
(
x
−
x0
)n+1
其中ξ介于x0与x之间.
特别地, n = 0时,泰勒公式变为f (x) = f (x0 ) + f ′(ξ )(x − x0 ),即为拉格朗日公式.
x=0
x=0
且(ex )(n+1) = ex = eξ = eθx ,0 < θ < 1.
x=ξ
x=ξ
故 : ex = 1+ x + 1 x2 + 1 x3 + + 1 xn + eθx xn+1,0 < θ < 1.(拉格朗日余项)
2! 3!
n! (n +1)!
ex = 1+ x + 1 x2 + 1 x3 ++ 1 xn + o(xn ), x → 0.(皮亚诺余项)
对F (z)与G(z)在[x0, x]上应用柯西中值定理知 :
F ( x) G(x)
− −
F (x0 ) G(x0 )
=
F′(ξ ) G′(ξ )
,ξ
∈ (x0 ,
x).
− f (x) −[ f (x0 ) + 即
f ′(x0 )(x − x0 ) +
f
′′( x0 2!
)
(x
−
x0
)2
+
f
′′′( x0 3!
3!
3!
+ + f (n+1) (z) (x − z)n + f (n) (z) n(x − z)n−1(−1)] = − f (n+1) (z) (x − z)n.
n!
n!
n!
2018/11/19
Edited by Lin Guojian
3
再构造G(z) = (x − z)n+1,则G(x) = 0, G(x0 ) = (x − x0 )n+1. G′(z) = (n +1)(x − z)n (−1) = −(n +1)(x − z)n. 且G′(z)在(x0, x)内不为零.
2
2
2
2
2
因此(sin
x)(n)
x=0
=
0 (−1) k −1
n = 2k n = 2k −1
从而sin x = x + −1 x3 + 1 x5 + + (−1)k−1 x2k−1 + (−1)k x2k+1 +
3! 5!
(2k −1)!
(2k +1)!
= x − 1 x3 + 1 x5 + + (−1)k−1 x2k−1 + o(x2k ), x → 0.
2
2
2
2
当n = 2k, k = 1,2,时, cos( n ⋅π ) = cos( 2k ⋅π ) = cos(kπ ) = (−1)k