AM@多元函数的近似和误差计算@二元函数泰勒公式
文章目录
abstract
- AM@多元函数的近似和误差计算@二元函数泰勒公式
一元泰勒公式
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f ( x ) f(x) f(x)= [ ∑ k = 1 n f ( k ) k ! ( x − x 0 ) k ] [\sum_{k=1}^{n}\frac{f^{(k)}}{k!}(x-x_0)^{k}] [∑k=1nk!f(k)(x−x0)k]+ R n ( x ) R_{n}(x) Rn(x)
- = f ( x 0 ) + f ′ ( x 0 ) ( x − x 0 ) + 1 2 ! f ′ ′ ( x 0 ) ( x − x 0 ) 2 + ⋯ + 1 n ! f ( n ) ( x 0 ) ( x − x 0 ) n f(x_0)+f'(x_0)(x-x_0)+\frac{1}{2!}f''(x_0)(x-x_0)^2+\cdots+\frac{1}{n!}f^{(n)}(x_0)(x-x_0)^{n} f(x0)+f′(x0)(x−x0)+2!1f′′(x0)(x−x0)2+⋯+n!1f(n)(x0)(x−x0)n+ R n ( x ) R_{n}(x) Rn(x)
- R n ( x ) R_{n}(x) Rn(x)= f ( n + 1 ) ( x 0 + θ ( x − x 0 ) ) ( n + 1 ) ! ( x − x 0 ) n + 1 \frac{f^{(n+1)}(x_0+\theta{(x-x_0)})}{(n+1)!}(x-x_0)^{n+1} (n+1)!f(n+1)(x0+θ(x−x0))(x−x0)n+1, θ ∈ ( 0 , 1 ) \theta\in{(0,1)} θ∈(0,1)
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利用一元函数的泰勒公式,我们可以用 n n n次多项式来逼近函数 f ( x ) f(x) f(x);且误差为当 x → x 0 x\to{x_0} x→x0时比 ( x − x 0 ) n (x-x_0)^{n} (x−x0)n高阶的无穷小
多元函数的近似和误差计算
- 对于多元函数来说,也有必要考虑多个变量的多项式(多元多项式)来近似表达一个给定的多元函数,必能具体地估算出误差
二元函数泰勒公式
- 设 z = f ( x , y ) z=f(x,y) z=f(x,y)在点 ( x 0 , y 0 ) (x_0,y_0) (x0,y0)的某一个邻域内连续且 ( n + 1 ) (n+1) (n+1)阶连续偏导数, ( x 0 + h , y 0 + k ) (x_0+h,y_0+k) (x0+h,y0+k)为此邻域内任意一点
- 二元函数泰勒公式解决的问题是:
- 把函数 f ( x 0 + h , y 0 + k ) f(x_0+h,y_0+k) f(x0+h,y0+k)近似地表达为 h = x − x 0 h=x-x_0 h=x−x0, k = y − y 0 k=y-y_0 k=y−y0的 n n n此多项式
- 由此产生的误差是当 ρ = h 2 + k 2 → 0 \rho=\sqrt{h^2+k^2}\to{0} ρ=h2+k2→0时比 ρ n \rho^{n} ρn高阶的无穷小
- 二元函数泰勒公式是一元函数泰勒公式的推广,其形式主要是将 n n n阶导数替换为多元 n n n阶偏导数的线性组合
二元泰勒定理
- 设
z
=
f
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z=f(x,y)
z=f(x,y)在点
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y
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(x_0,y_0)
(x0,y0)的某一邻域内连续,且具有
n
+
1
n+1
n+1阶来连续偏导数
- ( x 0 + h , y 0 + k ) (x_0+h,y_0+k) (x0+h,y0+k)为此邻域内任意一点
- 则
f
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f(x_0+h,y_0+k)
f(x0+h,y0+k)=
f
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f(x_0,y_0)
f(x0,y0)+
(
h
∂
∂
x
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k
∂
∂
y
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f
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x
0
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(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})f(x_0,y_0)
(h∂x∂+k∂y∂)f(x0,y0)+
1
2
!
(
h
∂
∂
x
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k
∂
∂
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2
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\frac{1}{2!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{2}f(x_0,y_0)
2!1(h∂x∂+k∂y∂)2f(x0,y0)+
⋯
\cdots
⋯+
1
n
!
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h
∂
∂
x
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k
∂
∂
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n
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\frac{1}{n!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{n}f(x_0,y_0)
n!1(h∂x∂+k∂y∂)nf(x0,y0)+
R
n
R_{n}
Rn,
θ
∈
(
0
,
1
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\theta\in(0,1)
θ∈(0,1)
(0)- 其中余项
R
n
R_{n}
Rn=
1
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n
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!
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∂
∂
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∂
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n
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1
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θ
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θ
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\frac{1}{(n+1)!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{n+1} f(x_0+\theta{h},y_0+\theta{h})
(n+1)!1(h∂x∂+k∂y∂)n+1f(x0+θh,y0+θh)
(0-1)
- 其中余项
R
n
R_{n}
Rn=
1
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+
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(
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∂
∂
x
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∂
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n
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\frac{1}{(n+1)!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{n+1} f(x_0+\theta{h},y_0+\theta{h})
(n+1)!1(h∂x∂+k∂y∂)n+1f(x0+θh,y0+θh)
记号
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(
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∂
∂
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m
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(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{m} f(x_0,y_0)
(h∂x∂+k∂y∂)mf(x0,y0)表示
∑
p
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\sum_{p=0}^{m}C_{m}^{p}h^{p}k^{m-p}\frac{\partial^{m}f}{\partial{x^{p}\partial{y}^{m-p}}}|_{(x_0,y_0)}
∑p=0mCmphpkm−p∂xp∂ym−p∂mf∣(x0,y0),例如
- ( h ∂ ∂ x + k ∂ ∂ y ) f ( x 0 , y 0 ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}}) f(x_0,y_0) (h∂x∂+k∂y∂)f(x0,y0)= h f x ( x 0 , y 0 ) + k f y ( x 0 , y 0 ) hf_{x}(x_0,y_0)+kf_{y}(x_0,y_{0}) hfx(x0,y0)+kfy(x0,y0)
- ( h ∂ ∂ x + k ∂ ∂ y ) 2 f ( x 0 , y 0 ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^2 f(x_0,y_0) (h∂x∂+k∂y∂)2f(x0,y0)= h 2 f x x ( x 0 , y 0 ) h^2f_{xx}(x_0,y_0) h2fxx(x0,y0)+ 2 h k f x y ( x 0 , y 0 ) 2hkf_{xy}(x_0,y_0) 2hkfxy(x0,y0)+ k 2 f y y ( x 0 , y 0 ) k^2f_{yy}(x_0,y_0) k2fyy(x0,y0)
- 总之,该记号展开后形如二项式定理展开: ( a + b ) m (a+b)^{m} (a+b)m= ∑ p = 0 m C n p a p b m − p \sum_{p=0}^{m}C_{n}^{p}a^{p}b^{m-p} ∑p=0mCnpapbm−p
- 此处规定偏导数运算符满足乘法对加法的分配律,偏导符号之间的乘积视为偏导重数(阶数)叠加或混合偏导数的阶数叠加
证明
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引入关于增量 t t t的函数 Φ ( t ) \Phi(t) Φ(t)= f ( x 0 + h t , y 0 + k t ) f(x_0+ht,y_0+kt) f(x0+ht,y0+kt), t ∈ [ 0 , 1 ] t\in[0,1] t∈[0,1],
(1)函数 f f f是二元函数,但 Φ ( t ) \Phi(t) Φ(t)是一元函数,若令 x = x ( t ) = x 0 + h t x=x(t)=x_0+ht x=x(t)=x0+ht(2-1), y = y ( t ) = y 0 + k t y=y(t)=y_0+kt y=y(t)=y0+kt(2-2),则 Φ ( t ) = f ( x ( t ) , y ( t ) ) \Phi(t)=f(x(t),y(t)) Φ(t)=f(x(t),y(t))(3),记 ( x 0 + h t , y 0 + k t ) (x_0+ht,y_0+kt) (x0+ht,y0+kt)为 P 0 P_{0} P0点坐标 -
x t ′ = h x_{t}'=h xt′=h
(4-1), y t ′ = k y_{t}'=k yt′=k(4-2) -
应用一元函数的泰勒公式证明
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Φ
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\Phi(0)
Φ(0)=
f
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f(x_0,y_0)
f(x0,y0)
(4-3) -
Φ
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\Phi(1)
Φ(1)=
f
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f(x_0+h,y_0+k)
f(x0+h,y0+k)
(4-4)
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Φ
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\Phi(0)
Φ(0)=
f
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f(x_0,y_0)
f(x0,y0)
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由多元复合函数求导法则(全导数公式):
-
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Φ ′ ( t ) \Phi'(t) Φ′(t)= f x ( x 0 + h t , y 0 + k t ) ⋅ h f_{x}(x_0+ht,y_0+kt)\cdot{h} fx(x0+ht,y0+kt)⋅h+ f y ( x 0 + h t , y 0 + k t ) ⋅ k f_{y}(x_0+ht,y_0+kt)\cdot{k} fy(x0+ht,y0+kt)⋅k= ( h ∂ ∂ x + k ∂ ∂ y ) f ( x 0 + h t , y 0 + k t ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}}) f(x_0+ht,y_0+kt) (h∂x∂+k∂y∂)f(x0+ht,y0+kt)
(5-1) -
Φ ′ ′ ( t ) \Phi''(t) Φ′′(t)= h ( f x x ( x 0 + h t , y 0 + k t ) h + f x y ( x 0 + h t , y 0 + k t ) k ) h(f_{xx}(x_0+ht,y_0+kt)h+f_{xy}(x_0+ht,y_0+kt)k) h(fxx(x0+ht,y0+kt)h+fxy(x0+ht,y0+kt)k)+ k ( f y x ( x 0 + h t , y 0 + k t ) h + f y y ( x 0 + h t , y 0 + k t ) k ) k(f_{yx}(x_0+ht,y_0+kt)h+f_{yy}(x_0+ht,y_0+kt)k) k(fyx(x0+ht,y0+kt)h+fyy(x0+ht,y0+kt)k)= h 2 f x x ( x 0 + h t , y 0 + k t ) + 2 h k f x y ( x 0 + h t , y 0 + k t ) + k 2 f y y ( x 0 + h t , y 0 + k t ) h^2f_{xx}(x_0+ht,y_0+kt)+2hkf_{xy}(x_0+ht,y_0+kt)+k^2f_{yy}(x_0+ht,y_0+kt) h2fxx(x0+ht,y0+kt)+2hkfxy(x0+ht,y0+kt)+k2fyy(x0+ht,y0+kt)= ( h ∂ ∂ x + k ∂ ∂ y ) 2 f ( x 0 + h t , y 0 + k t ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^2 f(x_0+ht,y_0+kt) (h∂x∂+k∂y∂)2f(x0+ht,y0+kt)
(5-2) -
上述写法很重,借助式(3),(4-1,4-2);并引入偏导数算子 D x = ∂ ∂ x D_{x}=\frac{\partial}{\partial{x}} Dx=∂x∂可以简写为
- Φ ′ ( t ) \Phi'(t) Φ′(t)= ( f x h + f y k ) (f_{x}h+f_yk) (fxh+fyk)= ( h D x + k D y ) Φ ( t ) (hD_x+kD_{y})\Phi(t) (hDx+kDy)Φ(t)
- Φ ′ ′ ( t ) \Phi''(t) Φ′′(t)= h ( f x x h + f x y k ) h(f_{xx}h+f_{xy}k) h(fxxh+fxyk)+ k ( f y x h + f y y k ) k(f_{yx}h+f_{yy}k) k(fyxh+fyyk)= h 2 f x x + 2 h k f x y + k 2 f y y h^2f_{xx}+2hkf_{xy}+k^2f_{yy} h2fxx+2hkfxy+k2fyy= ( h D x + k D y ) 2 Φ ( t ) (hD_x+kD_{y})^{2}\Phi(t) (hDx+kDy)2Φ(t)
- ⋯ \cdots ⋯
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Φ
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n
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(
t
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\Phi^{(n)}(t)
Φ(n)(t)=
∑
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\sum_{p=0}^{m}C_{m}^{p}h^{p}k^{m-p}\frac{\partial^{m}f}{\partial{x^{p}\partial{y}^{m-p}}}|_{(x_0,y_0)}
∑p=0mCmphpkm−p∂xp∂ym−p∂mf∣(x0,y0)=
(
h
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k
D
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Φ
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(hD_x+kD_{y})^{n}\Phi(t)
(hDx+kDy)nΦ(t)
(5-n) -
Φ
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\Phi^{(n+1)}(t)
Φ(n+1)(t)=
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Φ
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(hD_x+kD_{y})^{n+1}\Phi(t)
(hDx+kDy)n+1Φ(t)
(5-(n+1))
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利用一元函数的Maclaurin公式:
- f ( x ) f(x) f(x)= [ ∑ k = 0 n f ( k ) ( 0 ) k ! x k ] [\sum_{k=0}^{n}\frac{f^{(k)}(0)}{k!}x^{k}] [∑k=0nk!f(k)(0)xk]+ f ( n + 1 ) ( θ x ) ( n + 1 ) ! x n + 1 \frac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1} (n+1)!f(n+1)(θx)xn+1= f ( 0 ) + f ′ ( 0 ) x + 1 2 ! f ′ ′ ( 0 ) x 2 f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2 f(0)+f′(0)x+2!1f′′(0)x2+ ⋯ \cdots ⋯+ 1 n ! f ( n ) ( 0 ) x n \frac{1}{n!}f^{(n)}(0)x^n n!1f(n)(0)xn+ f ( n + 1 ) ( θ x ) ( n + 1 ) ! x n + 1 \frac{f^{(n+1)}(\theta x)}{(n+1)!}x^{n+1} (n+1)!f(n+1)(θx)xn+1, ( θ ∈ ( 0 , 1 ) ) (\theta\in(0,1)) (θ∈(0,1))
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得 Φ ( 1 ) \Phi(1) Φ(1)= Φ ( 0 ) + Φ ′ ( 0 ) \Phi(0)+\Phi'(0) Φ(0)+Φ′(0)+ 1 2 ! Φ ′ ′ ( 0 ) + ⋯ \frac{1}{2!}\Phi^{''}(0)+\cdots 2!1Φ′′(0)+⋯+ 1 n ! Φ ( n ) ( 0 ) \frac{1}{n!}\Phi^{(n)}(0) n!1Φ(n)(0)+ 1 ( n + 1 ) ! Φ ( n + 1 ) ( θ ) \frac{1}{(n+1)!}\Phi^{(n+1)}(\theta) (n+1)!1Φ(n+1)(θ)
(6) -
将(4-3),(4-4),以及(5-1)直到(5-n)在 t = 0 t=0 t=0处的取值 Φ ′ ( 0 ) , ⋯ , Φ ( n ) ( 0 ) \Phi'(0),\cdots,\Phi^{(n)}(0) Φ′(0),⋯,Φ(n)(0),以及 Φ ( n + 1 ) ( θ ) \Phi^{(n+1)}(\theta) Φ(n+1)(θ)的值代入(6),得
-
f
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f(x_0+h,y_0+k)
f(x0+h,y0+k)
- = Φ ( 0 ) \Phi(0) Φ(0)+ ( h D x + k D y ) Φ ( 0 ) (hD_x+kD_{y})\Phi(0) (hDx+kDy)Φ(0)+ 1 2 ! ( h D x + k D y ) 2 Φ ( 0 ) \frac{1}{2!}(hD_x+kD_{y})^{2}\Phi(0) 2!1(hDx+kDy)2Φ(0)+ ⋯ \cdots ⋯+ 1 n ! ( h D x + k D y ) n Φ ( 0 ) \frac{1}{n!}(hD_x+kD_{y})^{n}\Phi(0) n!1(hDx+kDy)nΦ(0)+ R n R_{n} Rn
- =
f
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{f(x_0,y_0)}
f(x0,y0)+
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k
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f
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(hD_x+kD_{y}){f(x_0,y_0)}
(hDx+kDy)f(x0,y0)+
1
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2
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\frac{1}{2!}(hD_x+kD_{y})^{2}{f(x_0,y_0)}
2!1(hDx+kDy)2f(x0,y0)+
⋯
\cdots
⋯+
1
n
!
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k
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Φ
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\frac{1}{n!}(hD_x+kD_{y})^{n}\Phi(0)
n!1(hDx+kDy)nΦ(0)+
R
n
R_{n}
Rn
(8),这就是公式(0)
- 其中
R
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R_{n}
Rn=
1
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!
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\frac{1}{(n+1)!}(hD_x+kD_{y})^{n+1}\Phi(\theta)
(n+1)!1(hDx+kDy)n+1Φ(θ)=
1
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!
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\frac{1}{(n+1)!}(hD_x+kD_{y})^{n+1}f(x_0+\theta{h},y_0+\theta{k})
(n+1)!1(hDx+kDy)n+1f(x0+θh,y0+θk)
(9), θ ∈ ( 0 , 1 ) \theta\in{(0,1)} θ∈(0,1) - 还可以进一步简写方程(8),令
D
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D=hD_x+kD_{y}
D=hDx+kDy
(9-1),则 f ( x 0 + h , y 0 + k ) f(x_0+h,y_0+k) f(x0+h,y0+k)=- f ( x 0 , y 0 ) + D f ( x 0 , y 0 ) + 1 2 ! D 2 f ( x 0 , y 0 ) + ⋯ + 1 n ! D n f ( x 0 , y 0 ) + R n f(x_0,y_0)+D{f(x_0,y_0)}+\frac{1}{2!}D^{2}{f(x_0,y_0)}+\cdots+\frac{1}{n!}D^{n}{f(x_0,y_0)}+R_{n} f(x0,y0)+Df(x0,y0)+2!1D2f(x0,y0)+⋯+n!1Dnf(x0,y0)+Rn= [ ∑ k = 0 n 1 k ! D k f ( x 0 , y 0 ) ] + R n [\sum_{k=0}^{n}\frac{1}{k!}D^{k}{f(x_0,y_0)}]+R_{n} [∑k=0nk!1Dkf(x0,y0)]+Rn
- R n = 1 ( n + 1 ) ! D n + 1 f ( x 0 + θ h , y 0 + θ k ) R_{n}=\frac{1}{(n+1)!}D^{n+1}f(x_0+\theta{h},y_{0}+\theta{k}) Rn=(n+1)!1Dn+1f(x0+θh,y0+θk), θ ∈ ( 0 , 1 ) \theta\in{(0,1)} θ∈(0,1)
-
f
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f(x_0+h,y_0+k)
f(x0+h,y0+k)
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二元泰勒公式
- 上述公式(0)称为** f ( x , y ) f(x,y) f(x,y)在点 ( x 0 , y 0 ) (x_0,y_0) (x0,y0)的 n n n阶泰勒公式**
- 而 R n R_{n} Rn的表达式(0-1)称为Lagrange余项
公式的应用
- 公式(0)的左端是
f
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h
,
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k
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f(x_0+h,y_0+k)
f(x0+h,y0+k),这个形式和一元泰勒公式有所不同
- 如果我们要近似某个二元函数 f ( x , y ) f(x,y) f(x,y),要怎么用公式(0)解决?
- 首先选择一个点 ( x 0 , y 0 ) (x_0,y_0) (x0,y0),例如常见的简单取值为 ( x 0 , y 0 ) = ( 0 , 0 ) (x_0,y_0)=(0,0) (x0,y0)=(0,0);此时 f ( x 0 + h , y 0 + k ) f(x_0+h,y_0+k) f(x0+h,y0+k)改写为 f ( h , k ) f(h,k) f(h,k),这个函数和 f ( x , y ) f(x,y) f(x,y)是同一个函数,只是自变量字母不同,但是映射规则同为 f f f,求得 f ( h , k ) f(h,k) f(h,k)的近似函数后,直接将 h , k h,k h,k分别替换为 x , y x,y x,y即可
- 分析公式(0)可知,二元函数的泰勒展开就是求各阶偏导数的过程,容易法线,二元泰勒公式需要计算的偏导数比一元泰勒公式在同阶展开的情况下求导次数要多得多,例如,不计算余项的情况下,展开到3阶时,二元泰勒公式就要计算4个不同的3阶偏导数,而一元情形只需要计算一个3阶导数即可
二元二阶情形
Peano型余项(TODO)
-
f ( x , y ) f(x,y) f(x,y)= f ( x 0 , y 0 ) f(x_0,y_0) f(x0,y0)+ f x ( x 0 , y 0 ) ( x − x 0 ) + f y ( x 0 , y 0 ) ( x − x 0 ) f_{x}(x_0,y_0)(x-x_0)+f_{y}(x_0,y_0)(x-x_0) fx(x0,y0)(x−x0)+fy(x0,y0)(x−x0)+ 1 2 ! [ D x x f ∣ ( x 0 , y 0 ) ( x − x 0 ) 2 + 2 D x y f ∣ ( x 0 , y 0 ) ( x − x 0 ) ( y − y 0 ) + D y y f ∣ ( x 0 , y 0 ) ( y − y 0 ) 2 ] + o ( ρ 2 ) \frac{1}{2!}[D_{xx}f|_{(x_0,y_0)}(x-x_0)^2+2D_{xy}f|_{(x_0,y_0)}(x-x_0)(y-y_0)+D_{yy}f|_{(x_0,y_0)}(y-y_0)^2]+o(\rho^2) 2!1[Dxxf∣(x0,y0)(x−x0)2+2Dxyf∣(x0,y0)(x−x0)(y−y0)+Dyyf∣(x0,y0)(y−y0)2]+o(ρ2)
-
其中 ρ = ( x − x 0 ) 2 + ( y − y 0 ) 2 \rho=\sqrt{(x-x_0)^2+(y-y_0)^2} ρ=(x−x0)2+(y−y0)2,上述公式称为 f ( x , y ) f(x,y) f(x,y)在 P 0 ( x 0 , y 0 ) P_{0}(x_0,y_0) P0(x0,y0)处带有Peano型余项的二元二阶泰勒公式
-
此外还有Lagrange型余项
近似误差
- 由二元函数的泰勒公式(0)可知,式(0)右端的 h , k h,k h,k的 n n n次多项式(二元 n n n次泰勒多项式)来近似表达函数 f ( x 0 + h , y 0 + k ) f(x_0+h,y_0+k) f(x0+h,y0+k)时,其误差为 ∣ R n ∣ |R_{n}| ∣Rn∣
- 由假设,函数的各
(
n
+
1
)
(n+1)
(n+1)阶偏导数都连续,所以它们的绝对值在点
(
x
0
,
y
0
)
(x_0,y_0)
(x0,y0)的某一邻域内都不超过某一正常数
M
M
M
- 对于
R
n
R_{n}
Rn=
1
(
n
+
1
)
!
(
h
∂
∂
x
+
k
∂
∂
y
)
n
+
1
f
(
x
0
+
θ
h
,
y
0
+
θ
k
)
\frac{1}{(n+1)!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{n+1} f(x_0+\theta{h},y_0+\theta{k})
(n+1)!1(h∂x∂+k∂y∂)n+1f(x0+θh,y0+θk)=
1
(
n
+
1
)
!
∑
p
=
0
n
+
1
C
n
+
1
p
h
p
k
n
+
1
−
p
∂
n
+
1
f
∂
x
p
∂
y
n
+
1
−
p
∣
(
x
0
+
θ
h
,
y
0
+
θ
k
)
\frac{1}{(n+1)!} \sum_{p=0}^{n+1}C_{n+1}^{p}h^{p}k^{n+1-p} \frac{\partial^{n+1}f}{\partial{x^{p}\partial{y}^{n+1-p}}}|_{(x_0+\theta{h},y_0+\theta{k})}
(n+1)!1∑p=0n+1Cn+1phpkn+1−p∂xp∂yn+1−p∂n+1f∣(x0+θh,y0+θk)
(10) -
T
=
∂
n
+
1
f
∂
x
p
∂
y
n
+
1
−
p
∣
(
x
0
+
θ
h
,
y
0
+
θ
k
)
T=\frac{\partial^{n+1}f}{\partial{x^{p}\partial{y}^{n+1-p}}}|_{(x_0+\theta{h},y_0+\theta{k})}
T=∂xp∂yn+1−p∂n+1f∣(x0+θh,y0+θk)
(11)在 ( x 0 , y 0 ) (x_0,y_0) (x0,y0)邻域内不超过 M M M,从 T T T而放大为 ∑ p = 0 n + 1 C n + 1 p h p k n + 1 − p M \sum_{p=0}^{n+1}C_{n+1}^{p}h^{p}k^{n+1-p}M ∑p=0n+1Cn+1phpkn+1−pM= M ( h + p ) n + 1 M(h+p)^{n+1} M(h+p)n+1(12),由 h + p ⩽ ∣ h ∣ + ∣ p ∣ h+p\leqslant{|h|+|p|} h+p⩽∣h∣+∣p∣, T T T进一步放大为 M ( ∣ h ∣ + ∣ p ∣ ) n + 1 M(|h|+|p|)^{n+1} M(∣h∣+∣p∣)n+1- Note:逆用二项式定理: ∑ p = 0 n + 1 C n + 1 p h p k n + 1 − p \sum_{p=0}^{n+1}C_{n+1}^{p}h^{p}k^{n+1-p} ∑p=0n+1Cn+1phpkn+1−p= ( h + p ) n + 1 (h+p)^{n+1} (h+p)n+1
- 对于
R
n
R_{n}
Rn=
1
(
n
+
1
)
!
(
h
∂
∂
x
+
k
∂
∂
y
)
n
+
1
f
(
x
0
+
θ
h
,
y
0
+
θ
k
)
\frac{1}{(n+1)!}(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{n+1} f(x_0+\theta{h},y_0+\theta{k})
(n+1)!1(h∂x∂+k∂y∂)n+1f(x0+θh,y0+θk)=
1
(
n
+
1
)
!
∑
p
=
0
n
+
1
C
n
+
1
p
h
p
k
n
+
1
−
p
∂
n
+
1
f
∂
x
p
∂
y
n
+
1
−
p
∣
(
x
0
+
θ
h
,
y
0
+
θ
k
)
\frac{1}{(n+1)!} \sum_{p=0}^{n+1}C_{n+1}^{p}h^{p}k^{n+1-p} \frac{\partial^{n+1}f}{\partial{x^{p}\partial{y}^{n+1-p}}}|_{(x_0+\theta{h},y_0+\theta{k})}
(n+1)!1∑p=0n+1Cn+1phpkn+1−p∂xp∂yn+1−p∂n+1f∣(x0+θh,y0+θk)
- 于是有如下误差估计式:
- R n = 1 ( n + 1 ) ! T ⩽ 1 ( n + 1 ) ! M ( ∣ h ∣ + ∣ p ∣ ) n + 1 R_{n}=\frac{1}{(n+1)!}T \leqslant{\frac{1}{(n+1)!}M(|h|+|p|)^{n+1}} Rn=(n+1)!1T⩽(n+1)!1M(∣h∣+∣p∣)n+1
-
∣
R
n
∣
⩽
M
(
n
+
1
)
!
(
∣
h
∣
+
∣
k
∣
)
n
+
1
|R_{n}|\leqslant{\frac{M}{(n+1)!}}(|h|+|k|)^{n+1}
∣Rn∣⩽(n+1)!M(∣h∣+∣k∣)n+1=
M
(
n
+
1
)
!
ρ
n
+
1
(
∣
h
∣
ρ
+
∣
k
∣
ρ
)
n
+
1
\frac{M}{(n+1)!}\rho^{n+1}(\frac{|h|}{\rho}+\frac{|k|}{\rho})^{n+1}
(n+1)!Mρn+1(ρ∣h∣+ρ∣k∣)n+1
⩽
\leqslant
⩽
M
(
n
+
1
)
!
(
2
)
n
+
1
ρ
n
+
1
\frac{M}{(n+1)!}(\sqrt{2})^{n+1}\rho^{n+1}
(n+1)!M(2)n+1ρn+1
(13),- 其中 ρ = h 2 + k 2 \rho=\sqrt{h^2+k^2} ρ=h2+k2
- Note:令
∣
h
∣
ρ
=
cos
α
\frac{|h|}{\rho}=\cos\alpha
ρ∣h∣=cosα;
∣
k
∣
ρ
=
sin
α
\frac{|k|}{\rho}=\sin\alpha
ρ∣k∣=sinα,则
cos
α
+
sin
α
\cos\alpha+\sin\alpha
cosα+sinα=
2
sin
(
α
+
π
4
)
⩽
2
\sqrt{2}{\sin{(\alpha+\frac{\pi}{4})}}\leqslant{\sqrt{2}}
2sin(α+4π)⩽2
- 由 2 2 cos α + 2 2 sin α \frac{\sqrt{2}}{2}\cos\alpha+\frac{\sqrt{2}}{2}\sin\alpha 22cosα+22sinα= sin ( α + π 4 ) \sin(\alpha+\frac{\pi}{4}) sin(α+4π)
- 即 2 2 ( sin α + cos α ) \frac{\sqrt{2}}{2}(\sin\alpha+\cos\alpha) 22(sinα+cosα)= sin ( α + π 4 ) \sin(\alpha+\frac{\pi}{4}) sin(α+4π),两边同时乘以 2 \sqrt{2} 2,得 sin α + cos α \sin\alpha+\cos\alpha sinα+cosα= 2 sin ( α + π 4 ) \sqrt{2}\sin(\alpha+\frac{\pi}{4}) 2sin(α+4π)
- 或者由辅助角公式,直接得到 cos α + sin α \cos\alpha+\sin\alpha cosα+sinα= 2 sin ( α + ϕ ) \sqrt{2}\sin(\alpha+\phi) 2sin(α+ϕ), tan ϕ = 1 \tan{\phi=1} tanϕ=1,即 ϕ = π 2 \phi=\frac{\pi}{2} ϕ=2π
- 由式(13)可知,误差 ∣ R n ∣ |R_{n}| ∣Rn∣是当 ρ → 0 \rho\to{0} ρ→0时比 ρ n \rho^{n} ρn高阶的无穷小
二元Lagrange中值公式
- 当
n
=
0
n=0
n=0时,公式(0)改写为
f
(
x
0
+
h
,
y
0
+
k
)
f(x_0+h,y_0+k)
f(x0+h,y0+k)=
f
(
x
0
,
y
0
)
+
R
0
f(x_0,y_0)+R_{0}
f(x0,y0)+R0=
f
(
x
0
,
y
0
)
f(x_0,y_0)
f(x0,y0)+
h
f
x
(
x
0
+
θ
h
,
y
0
+
θ
k
)
hf_{x}(x_0+\theta{h},y_0+\theta{k})
hfx(x0+θh,y0+θk)+
k
f
y
(
x
0
+
θ
h
,
y
0
+
θ
k
)
kf_{y}(x_0+\theta{h},y_0+\theta{k})
kfy(x0+θh,y0+θk)
(14), ( θ ∈ ( 0 , 1 ) ) (\theta\in(0,1)) (θ∈(0,1))
推论
- 若函数
f
(
x
,
y
)
f(x,y)
f(x,y)的偏导数
f
x
(
x
,
y
)
f_{x}(x,y)
fx(x,y),
f
y
(
x
,
y
)
f_{y}(x,y)
fy(x,y)在某一区域内都恒等于0,则函数
f
(
x
,
y
)
f(x,y)
f(x,y)在该区域内为一常数
- Note:用处处相等的方式证明
例
-
求函数 f ( x , y ) f(x,y) f(x,y)= ln ( 1 + x + y ) \ln{(1+x+y)} ln(1+x+y)在点 ( 0 , 0 ) (0,0) (0,0)的3阶泰勒公式
-
根据公式(0), n = 3 n=3 n=3,需要分别计算:方程组
(15)-
f x = f y = 1 1 + x + y f_{x}=f_{y}=\frac{1}{1+x+y} fx=fy=1+x+y1;则 f x ∣ ( 0 , 0 ) f_{x}|_{(0,0)} fx∣(0,0)= f y ∣ ( 0 , 0 ) f_{y}|_{(0,0)} fy∣(0,0)= 1 1 1
(15-1) -
f x x = f x y = f y y f_{xx}=f_{xy}=f_{yy} fxx=fxy=fyy= − 1 ( 1 + x + y ) 2 -\frac{1}{(1+x+y)^2} −(1+x+y)21,则 f x x ∣ ( 0 , 0 ) = f x y ∣ ( 0 , 0 ) = f y y ∣ ( 0 , 0 ) f_{xx}|_{(0,0)}=f_{xy}|_{(0,0)}=f_{yy}|_{(0,0)} fxx∣(0,0)=fxy∣(0,0)=fyy∣(0,0)= − 1 -1 −1
(15-2) -
∂ 3 f ∂ x p ∂ y 3 − p \frac{\partial^{3}f}{\partial{x^{p}\partial{y}^{3-p}}} ∂xp∂y3−p∂3f= 2 ! ( 1 + x + y ) 3 \frac{2!}{(1+x+y)^3} (1+x+y)32!, ( p = 0 , 1 , 2 , 3 ) (p=0,1,2,3) (p=0,1,2,3);则 f x x x ∣ ( 0 , 0 ) f_{xxx}|_{(0,0)} fxxx∣(0,0)= f x x y ∣ ( 0 , 0 ) f_{xxy}|_{(0,0)} fxxy∣(0,0)= f x y y ∣ ( 0 , 0 ) f_{xyy}|_{(0,0)} fxyy∣(0,0)= f y y y ∣ ( 0 , 0 ) f_{yyy}|_{(0,0)} fyyy∣(0,0)=2
(15-3) -
∂ 4 f ∂ x p ∂ y 4 − p \frac{\partial^{4}f}{\partial{x^{p}\partial{y}^{4-p}}} ∂xp∂y4−p∂4f= − 3 ! ( 1 + x + y ) 4 -\frac{3!}{(1+x+y)^4} −(1+x+y)43!, ( p = 0 , 1 , 2 , 3 , 4 ) (p=0,1,2,3,4) (p=0,1,2,3,4);
(15-4) -
D f ( 0 , 0 ) D{f(0,0)} Df(0,0)= ( h ∂ ∂ x + k ∂ ∂ y ) f ( 0 , 0 ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})f(0,0) (h∂x∂+k∂y∂)f(0,0)= [ h f x + k f y ] ∣ ( 0 , 0 ) [hf_x+kf_{y}]|_{(0,0)} [hfx+kfy]∣(0,0)= h + k h+k h+k
-
D 2 f ( 0 , 0 ) D^2f(0,0) D2f(0,0)= ( h ∂ ∂ x + k ∂ ∂ y ) 2 f ( 0 , 0 ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^2f(0,0) (h∂x∂+k∂y∂)2f(0,0)= [ h 2 f x x + 2 h k f x y + k 2 f y y ] ∣ ( 0 , 0 ) [h^2f_{xx}+2hkf_{xy}+k^2f_{yy}]|_{(0,0)} [h2fxx+2hkfxy+k2fyy]∣(0,0)= − ( h + k ) 2 -(h+k)^2 −(h+k)2
-
D 3 f ( 0 , 0 ) D^{3}f(0,0) D3f(0,0)= ( h ∂ ∂ x + k ∂ ∂ y ) 3 f ( 0 , 0 ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^3f(0,0) (h∂x∂+k∂y∂)3f(0,0)= [ h 3 f x x x + 3 h 2 k f x x y + 3 h k 2 f x y y + k 3 f y y y ] ∣ ( 0 , 0 ) [h^3f_{xxx}+3h^2kf_{xxy}+3hk^2f_{xyy}+k^3f_{yyy}]|_{(0,0)} [h3fxxx+3h2kfxxy+3hk2fxyy+k3fyyy]∣(0,0)= 2 ( h + k ) 3 2(h+k)^3 2(h+k)3
-
Note: D f ( 0 , 0 ) Df(0,0) Df(0,0)= D f ∣ ( 0 , 0 ) Df|_{(0,0)} Df∣(0,0)= [ ( h ∂ ∂ x + k ∂ ∂ y ) f ] ( 0 , 0 ) [(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})f]_{(0,0)} [(h∂x∂+k∂y∂)f](0,0)
-
又 f ( 0 , 0 ) = 0 f(0,0)=0 f(0,0)=0,将 h = x , k = y h=x,k=y h=x,k=y代入,3阶泰勒公式得
- ln ( 1 + x + y ) \ln(1+x+y) ln(1+x+y)= x + y − 1 2 ( x + y ) 2 + 1 3 ( x + y ) 3 + R 3 x+y-\frac{1}{2}(x+y)^2+\frac{1}{3}(x+y)3+R_3 x+y−21(x+y)2+31(x+y)3+R3
- 其中
R
3
=
1
4
!
(
h
∂
∂
x
+
k
∂
∂
y
)
4
f
(
θ
h
,
θ
k
)
∣
h
=
x
,
k
=
y
R_{3}=\frac{1}{4!} (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{4}f(\theta{h},\theta{k})|_{h=x,k=y}
R3=4!1(h∂x∂+k∂y∂)4f(θh,θk)∣h=x,k=y=
−
1
4
(
x
+
y
)
4
(
1
+
θ
x
+
θ
y
)
4
-\frac{1}{4}\frac{(x+y)^4}{(1+\theta{x}+\theta{y})^{4}}
−41(1+θx+θy)4(x+y)4,
θ
∈
(
0
,
1
)
\theta\in(0,1)
θ∈(0,1)
- ξ 1 = θ h \xi_1=\theta{h} ξ1=θh, ξ 2 = θ k \xi_2=\theta{k} ξ2=θk,
- 为例简化书写,令 D x = ∂ ∂ x D_{x}=\frac{\partial}{\partial{x}} Dx=∂x∂; D y = ∂ ∂ y D_{y}=\frac{\partial}{\partial{y}} Dy=∂y∂
-
(
h
∂
∂
x
+
k
∂
∂
y
)
4
(h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{4}
(h∂x∂+k∂y∂)4=
(
h
D
x
+
k
D
y
)
4
(hD_{x}+kD_y)^{4}
(hDx+kDy)4=
C
4
0
(
h
D
x
)
4
+
C
4
1
(
h
D
x
)
3
(
k
D
y
)
C_{4}^{0}(hD_{x})^4+C_{4}^{1}(hD_{x})^3(kD_y)
C40(hDx)4+C41(hDx)3(kDy)+
C
4
2
(
h
D
x
)
2
(
k
D
y
)
2
C_{4}^{2}(hD_{x})^{2}(kD_{y})^2
C42(hDx)2(kDy)2+
C
4
3
(
h
D
x
)
(
k
D
x
)
3
C_{4}^{3}(hD_{x})(kD_{x})^{3}
C43(hDx)(kDx)3+
C
4
4
(
k
D
y
)
4
C_{4}^{4}(kD_{y})^{4}
C44(kDy)4
(16) - 由(15-4), ( h ∂ ∂ x + k ∂ ∂ y ) 4 f ( x , y ) (h\frac{\partial}{\partial{x}}+k\frac{\partial}{\partial{y}})^{4} f(x,y) (h∂x∂+k∂y∂)4f(x,y)= − 3 ! ( 1 + x + y ) 4 ( C 4 0 h 4 + C 4 1 h 3 k + C 4 2 h 2 k 2 + ⋯ + C 4 4 k 4 ) -\frac{3!}{(1+x+y)^4}(C_4^{0}h^{4}+C_4^1h^{3}k+C_4^{2}h^2k^2+\cdots+C_{4}^{4}k^{4}) −(1+x+y)43!(C40h4+C41h3k+C42h2k2+⋯+C44k4)= − 3 ! ( 1 + x + y ) 4 ( h + k ) 4 -\frac{3!}{(1+x+y)^4}(h+k)^4 −(1+x+y)43!(h+k)4
- R 3 = 1 4 ! ( − 3 ! ( 1 + θ h + θ k ) 4 ( h + k ) 4 ) ∣ h = x , k = y R_{3}=\frac{1}{4!}(-\frac{3!}{(1+\theta{h}+\theta{k})^4} (h+k)^4)|_{h=x,k=y} R3=4!1(−(1+θh+θk)43!(h+k)4)∣h=x,k=y= − 1 4 ( x + y ) 4 ( 1 + θ x + θ y ) 4 -\frac{1}{4}\frac{(x+y)^4}{(1+\theta{x}+\theta{y})^{4}} −41(1+θx+θy)4(x+y)4, θ ∈ ( 0 , 1 ) \theta\in(0,1) θ∈(0,1)
-
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