AM@麦克劳林公式逼近以及误差分析 |
您所在的位置：网站首页 › 麦克劳林公式展开怎么记 › AM@麦克劳林公式逼近以及误差分析 |
|
n
f
(
n
)
(
x
)
f^{(n)}(x)
f(n)(x)
f
(
n
)
(
0
)
f^{(n)}(0)
f(n)(0)0
e
x
e^{x}
ex11
e
x
e^{x}
ex12
e
x
e^{x}
ex1
⋯
\cdots
⋯
⋯
\cdots
⋯
⋯
\cdots
⋯
n
n
n
e
x
e^{x}
ex1
n
+
1
n+1
n+1
e
x
e^{x}
ex
f
(
n
+
1
)
(
θ
x
)
f^{(n+1)}(\theta{x})
f(n+1)(θx)=
e
θ
x
e^{\theta{x}}
eθx e x e^{x} ex= 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 = 1 + x + 1 2 ! x 2 + ⋯ + 1 n ! x n 1+x+\frac{1}{2!}x^2+\cdots+\frac{1}{n!}x^{n} 1+x+2!1x2+⋯+n!1xn+ e θ x ( n + 1 ) ! x n + 1 \frac{e^{\theta{x}}}{(n+1)!}x^{n+1} (n+1)!eθxxn+1, θ ∈ ( 0 , 1 ) \theta\in(0,1) θ∈(0,1)(1)误差: ∣ R n ( x ) ∣ |R_{n}(x)| ∣Rn(x)∣= ∣ e θ x ( n + 1 ) ! x n + 1 ∣ |\frac{e^{\theta{x}}}{(n+1)!}x^{n+1}| ∣(n+1)!eθxxn+1∣ |
今日新闻 |
推荐新闻 |
| CopyRight 2018-2019 办公设备维修网 版权所有 豫ICP备15022753号-3 |