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受欢迎的三角函数 >

3tanh^2(x)=5sech(x)+1

解答

3 tanh^{2} (x) = 5 sech (x) + 1

解答

x = ln (0.17157 \dots), x = ln (5.82842 \dots)

+1

度数

x = - 100.99797 \dots^{\circ}, x = 100.99797 \dots^{\circ}

求解步骤

3 tanh^{2} (x) = 5 sech (x) + 1

使用三角恒等式改写

3 tanh^{2} (x) = 5 sech (x) + 1

使用双曲函数恒等式:

tanh (x) = \frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 sech (x) + 1

使用双曲函数恒等式:

sech (x) = \frac{2}{e ^{x} + e ^{- x}}

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 \cdot \frac{2}{e ^{x} + e ^{- x}} + 1

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 \cdot \frac{2}{e ^{x} + e ^{- x}} + 1

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 \cdot \frac{2}{e ^{x} + e ^{- x}} + 1 : x = ln (0.17157 \dots), x = ln (5.82842 \dots)

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 \cdot \frac{2}{e ^{x} + e ^{- x}} + 1

使用指数运算法则

3 (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}})^{2} = 5 \cdot \frac{2}{e ^{x} + e ^{- x}} + 1

使用指数法则:

a^{b c} = (a^{b})^{c}

e^{- x} = (e^{x})^{- 1}

3 (\frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})^{2} = 5 \cdot \frac{2}{e ^{x} + ( e ^{x} ) ^{- 1}} + 1

3 (\frac{e ^{x} - ( e ^{x} ) ^{- 1}}{e ^{x} + ( e ^{x} ) ^{- 1}})^{2} = 5 \cdot \frac{2}{e ^{x} + ( e ^{x} ) ^{- 1}} + 1

用

e^{x} = u

改写方程式

3 (\frac{u - ( u ) ^{- 1}}{u + ( u ) ^{- 1}})^{2} = 5 \cdot \frac{2}{u + ( u ) ^{- 1}} + 1

解

3 (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{2} = 5 \cdot \frac{2}{u + u ^{- 1}} + 1 : u \approx 0.17157 \dots, u \approx 5.82842 \dots

3 (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{2} = 5 \cdot \frac{2}{u + u ^{- 1}} + 1

整理后得

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} = \frac{10 u}{u ^{2} + 1} + 1

乘以最小公倍数

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} = \frac{10 u}{u ^{2} + 1} + 1

找到

(u^{2} + 1)^{2}, u^{2} + 1

的最小公倍数:

(u^{2} + 1)^{2}

(u^{2} + 1)^{2}, u^{2} + 1

最小公倍数 (LCM)

计算出由出现在

(u^{2} + 1)^{2}

或

u^{2} + 1

中的因子组成的表达式

= (u^{2} + 1)^{2}

乘以最小公倍数=

(u^{2} + 1)^{2}

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} = \frac{10 u}{u ^{2} + 1} (u^{2} + 1)^{2} + 1 \cdot (u^{2} + 1)^{2}

化简

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} = \frac{10 u}{u ^{2} + 1} (u^{2} + 1)^{2} + 1 \cdot (u^{2} + 1)^{2}

化简

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2} : 3 (u^{2} - 1)^{2}

\frac{3 ( u ^{2} - 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}} (u^{2} + 1)^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ( u ^{2} - 1 ) ^{2} ( u ^{2} + 1 ) ^{2}}{( u ^{2} + 1 ) ^{2}}

约分:

(u^{2} + 1)^{2}

= 3 (u^{2} - 1)^{2}

化简

\frac{10 u}{u ^{2} + 1} (u^{2} + 1)^{2} : 10 u (u^{2} + 1)

\frac{10 u}{u ^{2} + 1} (u^{2} + 1)^{2}

分式相乘:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{10 u ( u ^{2} + 1 ) ^{2}}{u ^{2} + 1}

约分:

u^{2} + 1

= 10 u (u^{2} + 1)

化简

1 \cdot (u^{2} + 1)^{2} : (u^{2} + 1)^{2}

1 \cdot (u^{2} + 1)^{2}

乘以:

1 \cdot (u^{2} + 1)^{2} = (u^{2} + 1)^{2}

= (u^{2} + 1)^{2}

3 (u^{2} - 1)^{2} = 10 u (u^{2} + 1) + (u^{2} + 1)^{2}

3 (u^{2} - 1)^{2} = 10 u (u^{2} + 1) + (u^{2} + 1)^{2}

3 (u^{2} - 1)^{2} = 10 u (u^{2} + 1) + (u^{2} + 1)^{2}

解

3 (u^{2} - 1)^{2} = 10 u (u^{2} + 1) + (u^{2} + 1)^{2} : u \approx 0.17157 \dots, u \approx 5.82842 \dots

3 (u^{2} - 1)^{2} = 10 u (u^{2} + 1) + (u^{2} + 1)^{2}

展开

3 (u^{2} - 1)^{2} : 3 u^{4} - 6 u^{2} + 3

3 (u^{2} - 1)^{2}

(u^{2} - 1)^{2} = u^{4} - 2 u^{2} + 1

(u^{2} - 1)^{2}

使用完全平方公式:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = u^{2}, b = 1

= (u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

化简

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2} : u^{4} - 2 u^{2} + 1

(u^{2})^{2} - 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} - 2 \cdot 1 \cdot u^{2} + 1

(u^{2})^{2} = u^{4}

(u^{2})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

2 u^{2} \cdot 1 = 2 u^{2}

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} - 2 u^{2} + 1

= u^{4} - 2 u^{2} + 1

= 3 (u^{4} - 2 u^{2} + 1)

打开括号

= 3 u^{4} + 3 (- 2 u^{2}) + 3 \cdot 1

使用加减运算法则

+ (- a) = - a

= 3 u^{4} - 3 \cdot 2 u^{2} + 3 \cdot 1

化简

3 u^{4} - 3 \cdot 2 u^{2} + 3 \cdot 1 : 3 u^{4} - 6 u^{2} + 3

3 u^{4} - 3 \cdot 2 u^{2} + 3 \cdot 1

数字相乘:

3 \cdot 2 = 6

= 3 u^{4} - 6 u^{2} + 3 \cdot 1

数字相乘:

3 \cdot 1 = 3

= 3 u^{4} - 6 u^{2} + 3

= 3 u^{4} - 6 u^{2} + 3

展开

10 u (u^{2} + 1) + (u^{2} + 1)^{2} : 10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1

10 u (u^{2} + 1) + (u^{2} + 1)^{2}

(u^{2} + 1)^{2} : u^{4} + 2 u^{2} + 1

使用完全平方公式:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = u^{2}, b = 1

= (u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

化简

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2} : u^{4} + 2 u^{2} + 1

(u^{2})^{2} + 2 u^{2} \cdot 1 + 1^{2}

使用法则

1^{a} = 1

1^{2} = 1

= (u^{2})^{2} + 2 \cdot 1 \cdot u^{2} + 1

(u^{2})^{2} = u^{4}

(u^{2})^{2}

使用指数法则:

(a^{b})^{c} = a^{b c}

= u^{2 \cdot 2}

数字相乘:

2 \cdot 2 = 4

= u^{4}

2 u^{2} \cdot 1 = 2 u^{2}

2 u^{2} \cdot 1

数字相乘:

2 \cdot 1 = 2

= 2 u^{2}

= u^{4} + 2 u^{2} + 1

= u^{4} + 2 u^{2} + 1

= 10 u (u^{2} + 1) + u^{4} + 2 u^{2} + 1

乘开

10 u (u^{2} + 1) : 10 u^{3} + 10 u

10 u (u^{2} + 1)

使用分配律:

a (b + c) = ab + a c

a = 10 u, b = u^{2}, c = 1

= 10 u u^{2} + 10 u \cdot 1

= 10 u^{2} u + 10 \cdot 1 \cdot u

化简

10 u^{2} u + 10 \cdot 1 \cdot u : 10 u^{3} + 10 u

10 u^{2} u + 10 \cdot 1 \cdot u

10 u^{2} u = 10 u^{3}

10 u^{2} u

使用指数法则:

a^{b} \cdot a^{c} = a^{b + c}

u^{2} u = u^{2 + 1}

= 10 u^{2 + 1}

数字相加:

2 + 1 = 3

= 10 u^{3}

10 \cdot 1 \cdot u = 10 u

10 \cdot 1 \cdot u

数字相乘:

10 \cdot 1 = 10

= 10 u

= 10 u^{3} + 10 u

= 10 u^{3} + 10 u

= 10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1

3 u^{4} - 6 u^{2} + 3 = 10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1

交换两边

10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1 = 3 u^{4} - 6 u^{2} + 3

将

3

para o lado esquerdo

10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1 = 3 u^{4} - 6 u^{2} + 3

两边减去

3

10 u^{3} + 10 u + u^{4} + 2 u^{2} + 1 - 3 = 3 u^{4} - 6 u^{2} + 3 - 3

化简

u^{4} + 10 u^{3} + 2 u^{2} + 10 u - 2 = 3 u^{4} - 6 u^{2}

u^{4} + 10 u^{3} + 2 u^{2} + 10 u - 2 = 3 u^{4} - 6 u^{2}

将

6 u^{2}

para o lado esquerdo

u^{4} + 10 u^{3} + 2 u^{2} + 10 u - 2 = 3 u^{4} - 6 u^{2}

两边加上

6 u^{2}

u^{4} + 10 u^{3} + 2 u^{2} + 10 u - 2 + 6 u^{2} = 3 u^{4} - 6 u^{2} + 6 u^{2}

化简

u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 3 u^{4}

u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 3 u^{4}

将

3 u^{4}

para o lado esquerdo

u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 3 u^{4}

两边减去

3 u^{4}

u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 - 3 u^{4} = 3 u^{4} - 3 u^{4}

化简

- 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 0

- 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 0

使用牛顿-拉弗森方法找到

- 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 0

的一个解:

u \approx 0.17157 \dots

- 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2 = 0

牛顿-拉弗森近似法定义

f (u) = - 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2

找到

f^{^{'}} (u) : - 8 u^{3} + 30 u^{2} + 16 u + 10

\frac{d}{d u} (- 2 u^{4} + 10 u^{3} + 8 u^{2} + 10 u - 2)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{4}) + \frac{d}{d u} (10 u^{3}) + \frac{d}{d u} (8 u^{2}) + \frac{d}{d u} (10 u) - \frac{d}{d u} (2)

\frac{d}{d u} (2 u^{4}) = 8 u^{3}

\frac{d}{d u} (2 u^{4})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{4})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 4 u^{4 - 1}

化简

= 8 u^{3}

\frac{d}{d u} (10 u^{3}) = 30 u^{2}

\frac{d}{d u} (10 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 10 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 10 \cdot 3 u^{3 - 1}

化简

= 30 u^{2}

\frac{d}{d u} (8 u^{2}) = 16 u

\frac{d}{d u} (8 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 8 \cdot 2 u^{2 - 1}

化简

= 16 u

\frac{d}{d u} (10 u) = 10

\frac{d}{d u} (10 u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 10 \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 10 \cdot 1

化简

= 10

\frac{d}{d u} (2) = 0

\frac{d}{d u} (2)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 8 u^{3} + 30 u^{2} + 16 u + 10 - 0

化简

= - 8 u^{3} + 30 u^{2} + 16 u + 10

令

u_{0} = 0

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 0.2 : Δ u_{1} = 0.2

f (u_{0}) = - 2 \cdot 0^{4} + 10 \cdot 0^{3} + 8 \cdot 0^{2} + 10 \cdot 0 - 2 = - 2

f^{^{'}} (u_{0}) = - 8 \cdot 0^{3} + 30 \cdot 0^{2} + 16 \cdot 0 + 10 = 10

u_{1} = 0.2

Δ u_{1} = ∣ 0.2 - 0 ∣ = 0.2

Δ u_{1} = 0.2

u_{2} = 0.17232 \dots : Δ u_{2} = 0.02767 \dots

f (u_{1}) = - 2 \cdot 0. 2^{4} + 10 \cdot 0. 2^{3} + 8 \cdot 0. 2^{2} + 10 \cdot 0.2 - 2 = 0.3968

f^{^{'}} (u_{1}) = - 8 \cdot 0. 2^{3} + 30 \cdot 0. 2^{2} + 16 \cdot 0.2 + 10 = 14.336

u_{2} = 0.17232 \dots

Δ u_{2} = ∣ 0.17232 \dots - 0.2 ∣ = 0.02767 \dots

Δ u_{2} = 0.02767 \dots

u_{3} = 0.17157 \dots : Δ u_{3} = 0.00074 \dots

f (u_{2}) = - 2 \cdot 0.17232 \dots^{4} + 10 \cdot 0.17232 \dots^{3} + 8 \cdot 0.17232 \dots^{2} + 10 \cdot 0.17232 \dots - 2 = 0.01017 \dots

f^{^{'}} (u_{2}) = - 8 \cdot 0.17232 \dots^{3} + 30 \cdot 0.17232 \dots^{2} + 16 \cdot 0.17232 \dots + 10 = 13.60704 \dots

u_{3} = 0.17157 \dots

Δ u_{3} = ∣ 0.17157 \dots - 0.17232 \dots ∣ = 0.00074 \dots

Δ u_{3} = 0.00074 \dots

u_{4} = 0.17157 \dots : Δ u_{4} = 5.2738 E - 7

f (u_{3}) = - 2 \cdot 0.17157 \dots^{4} + 10 \cdot 0.17157 \dots^{3} + 8 \cdot 0.17157 \dots^{2} + 10 \cdot 0.17157 \dots - 2 = 7.16598 E - 6

f^{^{'}} (u_{3}) = - 8 \cdot 0.17157 \dots^{3} + 30 \cdot 0.17157 \dots^{2} + 16 \cdot 0.17157 \dots + 10 = 13.58789 \dots

u_{4} = 0.17157 \dots

Δ u_{4} = ∣ 0.17157 \dots - 0.17157 \dots ∣ = 5.2738 E - 7

Δ u_{4} = 5.2738 E - 7

u \approx 0.17157 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 2 u ^{4} + 10 u ^{3} + 8 u ^{2} + 10 u - 2}{u - 0.17157 \dots} = - 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots

- 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots \approx 0

使用牛顿-拉弗森方法找到

- 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots = 0

的一个解:

u \approx 5.82842 \dots

- 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots

找到

f^{^{'}} (u) : - 6 u^{2} + 19.31370 \dots u + 9.65685 \dots

\frac{d}{d u} (- 2 u^{3} + 9.65685 \dots u^{2} + 9.65685 \dots u + 11.65685 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{3}) + \frac{d}{d u} (9.65685 \dots u^{2}) + \frac{d}{d u} (9.65685 \dots u) + \frac{d}{d u} (11.65685 \dots)

\frac{d}{d u} (2 u^{3}) = 6 u^{2}

\frac{d}{d u} (2 u^{3})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{3})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 3 u^{3 - 1}

化简

= 6 u^{2}

\frac{d}{d u} (9.65685 \dots u^{2}) = 19.31370 \dots u

\frac{d}{d u} (9.65685 \dots u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.65685 \dots \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 9.65685 \dots \cdot 2 u^{2 - 1}

化简

= 19.31370 \dots u

\frac{d}{d u} (9.65685 \dots u) = 9.65685 \dots

\frac{d}{d u} (9.65685 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 9.65685 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 9.65685 \dots \cdot 1

化简

= 9.65685 \dots

\frac{d}{d u} (11.65685 \dots) = 0

\frac{d}{d u} (11.65685 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 6 u^{2} + 19.31370 \dots u + 9.65685 \dots + 0

化简

= - 6 u^{2} + 19.31370 \dots u + 9.65685 \dots

令

u_{0} = 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = - 0.26120 \dots : Δ u_{1} = 1.26120 \dots

f (u_{0}) = - 2 \cdot 1^{3} + 9.65685 \dots \cdot 1^{2} + 9.65685 \dots \cdot 1 + 11.65685 \dots = 28.97056 \dots

f^{^{'}} (u_{0}) = - 6 \cdot 1^{2} + 19.31370 \dots \cdot 1 + 9.65685 \dots = 22.97056 \dots

u_{1} = - 0.26120 \dots

Δ u_{1} = ∣ - 0.26120 \dots - 1 ∣ = 1.26120 \dots

Δ u_{1} = 1.26120 \dots

u_{2} = - 2.59994 \dots : Δ u_{2} = 2.33873 \dots

f (u_{1}) = - 2 (- 0.26120 \dots)^{3} + 9.65685 \dots (- 0.26120 \dots)^{2} + 9.65685 \dots (- 0.26120 \dots) + 11.65685 \dots = 9.82895 \dots

f^{^{'}} (u_{1}) = - 6 (- 0.26120 \dots)^{2} + 19.31370 \dots (- 0.26120 \dots) + 9.65685 \dots = 4.20267 \dots

u_{2} = - 2.59994 \dots

Δ u_{2} = ∣ - 2.59994 \dots - (- 0.26120 \dots) ∣ = 2.33873 \dots

Δ u_{2} = 2.33873 \dots

u_{3} = - 1.52768 \dots : Δ u_{3} = 1.07225 \dots

f (u_{2}) = - 2 (- 2.59994 \dots)^{3} + 9.65685 \dots (- 2.59994 \dots)^{2} + 9.65685 \dots (- 2.59994 \dots) + 11.65685 \dots = 86.97662 \dots

f^{^{'}} (u_{2}) = - 6 (- 2.59994 \dots)^{2} + 19.31370 \dots (- 2.59994 \dots) + 9.65685 \dots = - 81.11583 \dots

u_{3} = - 1.52768 \dots

Δ u_{3} = ∣ - 1.52768 \dots - (- 2.59994 \dots) ∣ = 1.07225 \dots

Δ u_{3} = 1.07225 \dots

u_{4} = - 0.74271 \dots : Δ u_{4} = 0.78497 \dots

f (u_{3}) = - 2 (- 1.52768 \dots)^{3} + 9.65685 \dots (- 1.52768 \dots)^{2} + 9.65685 \dots (- 1.52768 \dots) + 11.65685 \dots = 26.57243 \dots

f^{^{'}} (u_{3}) = - 6 (- 1.52768 \dots)^{2} + 19.31370 \dots (- 1.52768 \dots) + 9.65685 \dots = - 33.85150 \dots

u_{4} = - 0.74271 \dots

Δ u_{4} = ∣ - 0.74271 \dots - (- 1.52768 \dots) ∣ = 0.78497 \dots

Δ u_{4} = 0.78497 \dots

u_{5} = 0.58655 \dots : Δ u_{5} = 1.32927 \dots

f (u_{4}) = - 2 (- 0.74271 \dots)^{3} + 9.65685 \dots (- 0.74271 \dots)^{2} + 9.65685 \dots (- 0.74271 \dots) + 11.65685 \dots = 10.63096 \dots

f^{^{'}} (u_{4}) = - 6 (- 0.74271 \dots)^{2} + 19.31370 \dots (- 0.74271 \dots) + 9.65685 \dots = - 7.99758 \dots

u_{5} = 0.58655 \dots

Δ u_{5} = ∣ 0.58655 \dots - (- 0.74271 \dots) ∣ = 1.32927 \dots

Δ u_{5} = 1.32927 \dots

u_{6} = - 0.48314 \dots : Δ u_{6} = 1.06969 \dots

f (u_{5}) = - 2 \cdot 0.58655 \dots^{3} + 9.65685 \dots \cdot 0.58655 \dots^{2} + 9.65685 \dots \cdot 0.58655 \dots + 11.65685 \dots = 20.23988 \dots

f^{^{'}} (u_{5}) = - 6 \cdot 0.58655 \dots^{2} + 19.31370 \dots \cdot 0.58655 \dots + 9.65685 \dots = 18.92109 \dots

u_{6} = - 0.48314 \dots

Δ u_{6} = ∣ - 0.48314 \dots - 0.58655 \dots ∣ = 1.06969 \dots

Δ u_{6} = 1.06969 \dots

u_{7} = 8.32623 \dots : Δ u_{7} = 8.80938 \dots

f (u_{6}) = - 2 (- 0.48314 \dots)^{3} + 9.65685 \dots (- 0.48314 \dots)^{2} + 9.65685 \dots (- 0.48314 \dots) + 11.65685 \dots = 9.47094 \dots

f^{^{'}} (u_{6}) = - 6 (- 0.48314 \dots)^{2} + 19.31370 \dots (- 0.48314 \dots) + 9.65685 \dots = - 1.07509 \dots

u_{7} = 8.32623 \dots

Δ u_{7} = ∣ 8.32623 \dots - (- 0.48314 \dots) ∣ = 8.80938 \dots

Δ u_{7} = 8.80938 \dots

u_{8} = 6.72569 \dots : Δ u_{8} = 1.60054 \dots

f (u_{7}) = - 2 \cdot 8.32623 \dots^{3} + 9.65685 \dots \cdot 8.32623 \dots^{2} + 9.65685 \dots \cdot 8.32623 \dots + 11.65685 \dots = - 392.91766 \dots

f^{^{'}} (u_{7}) = - 6 \cdot 8.32623 \dots^{2} + 19.31370 \dots \cdot 8.32623 \dots + 9.65685 \dots = - 245.48993 \dots

u_{8} = 6.72569 \dots

Δ u_{8} = ∣ 6.72569 \dots - 8.32623 \dots ∣ = 1.60054 \dots

Δ u_{8} = 1.60054 \dots

u_{9} = 6.00490 \dots : Δ u_{9} = 0.72078 \dots

f (u_{8}) = - 2 \cdot 6.72569 \dots^{3} + 9.65685 \dots \cdot 6.72569 \dots^{2} + 9.65685 \dots \cdot 6.72569 \dots + 11.65685 \dots = - 95.03936 \dots

f^{^{'}} (u_{8}) = - 6 \cdot 6.72569 \dots^{2} + 19.31370 \dots \cdot 6.72569 \dots + 9.65685 \dots = - 131.85466 \dots

u_{9} = 6.00490 \dots

Δ u_{9} = ∣ 6.00490 \dots - 6.72569 \dots ∣ = 0.72078 \dots

Δ u_{9} = 0.72078 \dots

u_{10} = 5.83735 \dots : Δ u_{10} = 0.16754 \dots

f (u_{9}) = - 2 \cdot 6.00490 \dots^{3} + 9.65685 \dots \cdot 6.00490 \dots^{2} + 9.65685 \dots \cdot 6.00490 \dots + 11.65685 \dots = - 15.19941 \dots

f^{^{'}} (u_{9}) = - 6 \cdot 6.00490 \dots^{2} + 19.31370 \dots \cdot 6.00490 \dots + 9.65685 \dots = - 90.71934 \dots

u_{10} = 5.83735 \dots

Δ u_{10} = ∣ 5.83735 \dots - 6.00490 \dots ∣ = 0.16754 \dots

Δ u_{10} = 0.16754 \dots

u_{11} = 5.82845 \dots : Δ u_{11} = 0.00890 \dots

f (u_{10}) = - 2 \cdot 5.83735 \dots^{3} + 9.65685 \dots \cdot 5.83735 \dots^{2} + 9.65685 \dots \cdot 5.83735 \dots + 11.65685 \dots = - 0.73089 \dots

f^{^{'}} (u_{10}) = - 6 \cdot 5.83735 \dots^{2} + 19.31370 \dots \cdot 5.83735 \dots + 9.65685 \dots = - 82.05068 \dots

u_{11} = 5.82845 \dots

Δ u_{11} = ∣ 5.82845 \dots - 5.83735 \dots ∣ = 0.00890 \dots

Δ u_{11} = 0.00890 \dots

u_{12} = 5.82842 \dots : Δ u_{12} = 0.00002 \dots

f (u_{11}) = - 2 \cdot 5.82845 \dots^{3} + 9.65685 \dots \cdot 5.82845 \dots^{2} + 9.65685 \dots \cdot 5.82845 \dots + 11.65685 \dots = - 0.00201 \dots

f^{^{'}} (u_{11}) = - 6 \cdot 5.82845 \dots^{2} + 19.31370 \dots \cdot 5.82845 \dots + 9.65685 \dots = - 81.59922 \dots

u_{12} = 5.82842 \dots

Δ u_{12} = ∣ 5.82842 \dots - 5.82845 \dots ∣ = 0.00002 \dots

Δ u_{12} = 0.00002 \dots

u_{13} = 5.82842 \dots : Δ u_{13} = 1.88507 E - 10

f (u_{12}) = - 2 \cdot 5.82842 \dots^{3} + 9.65685 \dots \cdot 5.82842 \dots^{2} + 9.65685 \dots \cdot 5.82842 \dots + 11.65685 \dots = - 1.53818 E - 8

f^{^{'}} (u_{12}) = - 6 \cdot 5.82842 \dots^{2} + 19.31370 \dots \cdot 5.82842 \dots + 9.65685 \dots = - 81.59797 \dots

u_{13} = 5.82842 \dots

Δ u_{13} = ∣ 5.82842 \dots - 5.82842 \dots ∣ = 1.88507 E - 10

Δ u_{13} = 1.88507 E - 10

u \approx 5.82842 \dots

使用长除法

Eq u a t i o n 0 : \frac{- 2 u ^{3} + 9.65685 \dots u ^{2} + 9.65685 \dots u + 11.65685 \dots}{u - 5.82842 \dots} = - 2 u^{2} - 2.00000 \dots u - 2.00000 \dots

- 2 u^{2} - 2.00000 \dots u - 2.00000 \dots \approx 0

使用牛顿-拉弗森方法找到

- 2 u^{2} - 2.00000 \dots u - 2.00000 \dots = 0

的一个解:

u \in R

无解

- 2 u^{2} - 2.00000 \dots u - 2.00000 \dots = 0

牛顿-拉弗森近似法定义

f (u) = - 2 u^{2} - 2.00000 \dots u - 2.00000 \dots

找到

f^{^{'}} (u) : - 4 u - 2.00000 \dots

\frac{d}{d u} (- 2 u^{2} - 2.00000 \dots u - 2.00000 \dots)

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d u} (2 u^{2}) - \frac{d}{d u} (2.00000 \dots u) - \frac{d}{d u} (2.00000 \dots)

\frac{d}{d u} (2 u^{2}) = 4 u

\frac{d}{d u} (2 u^{2})

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d u} (u^{2})

使用幂法则:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 2 \cdot 2 u^{2 - 1}

化简

= 4 u

\frac{d}{d u} (2.00000 \dots u) = 2.00000 \dots

\frac{d}{d u} (2.00000 \dots u)

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 2.00000 \dots \frac{d u}{d u}

使用常见微分定则:

\frac{d u}{d u} = 1

= 2.00000 \dots \cdot 1

化简

= 2.00000 \dots

\frac{d}{d u} (2.00000 \dots) = 0

\frac{d}{d u} (2.00000 \dots)

常数微分:

\frac{d}{d x} (a) = 0

= 0

= - 4 u - 2.00000 \dots - 0

化简

= - 4 u - 2.00000 \dots

令

u_{0} = - 1

计算

u_{n + 1}

至

Δ u_{n + 1} < 0.000001

u_{1} = 2.53131 E - 13 : Δ u_{1} = 1

f (u_{0}) = - 2 (- 1)^{2} - 2.00000 \dots (- 1) - 2.00000 \dots = - 2

f^{^{'}} (u_{0}) = - 4 (- 1) - 2.00000 \dots = 1.99999 \dots

u_{1} = 2.53131 E - 13

Δ u_{1} = ∣ 2.53131 E - 13 - (- 1) ∣ = 1

Δ u_{1} = 1

u_{2} = - 0.99999 \dots : Δ u_{2} = 0.99999 \dots

f (u_{1}) = - 2 \cdot 2.53131 E - 1 3^{2} - 2.00000 \dots \cdot 2.53131 E - 13 - 2.00000 \dots = - 2.00000 \dots

f^{^{'}} (u_{1}) = - 4 \cdot 2.53131 E - 13 - 2.00000 \dots = - 2.00000 \dots

u_{2} = - 0.99999 \dots

Δ u_{2} = ∣ - 0.99999 \dots - 2.53131 E - 13 ∣ = 0.99999 \dots

Δ u_{2} = 0.99999 \dots

u_{3} = 1.26743 E - 12 : Δ u_{3} = 1.00000 \dots

f (u_{2}) = - 2 (- 0.99999 \dots)^{2} - 2.00000 \dots (- 0.99999 \dots) - 2.00000 \dots = - 1.99999 \dots

f^{^{'}} (u_{2}) = - 4 (- 0.99999 \dots) - 2.00000 \dots = 1.99999 \dots

u_{3} = 1.26743 E - 12

Δ u_{3} = ∣ 1.26743 E - 12 - (- 0.99999 \dots) ∣ = 1.00000 \dots

Δ u_{3} = 1.00000 \dots

u_{4} = - 0.99999 \dots : Δ u_{4} = 0.99999 \dots

f (u_{3}) = - 2 \cdot 1.26743 E - 1 2^{2} - 2.00000 \dots \cdot 1.26743 E - 12 - 2.00000 \dots = - 2.00000 \dots

f^{^{'}} (u_{3}) = - 4 \cdot 1.26743 E - 12 - 2.00000 \dots = - 2.00000 \dots

u_{4} = - 0.99999 \dots

Δ u_{4} = ∣ - 0.99999 \dots - 1.26743 E - 12 ∣ = 0.99999 \dots

Δ u_{4} = 0.99999 \dots

u_{5} = 5.32463 E - 12 : Δ u_{5} = 1.00000 \dots

f (u_{4}) = - 2 (- 0.99999 \dots)^{2} - 2.00000 \dots (- 0.99999 \dots) - 2.00000 \dots = - 1.99999 \dots

f^{^{'}} (u_{4}) = - 4 (- 0.99999 \dots) - 2.00000 \dots = 1.99999 \dots

u_{5} = 5.32463 E - 12

Δ u_{5} = ∣ 5.32463 E - 12 - (- 0.99999 \dots) ∣ = 1.00000 \dots

Δ u_{5} = 1.00000 \dots

u_{6} = - 0.99999 \dots : Δ u_{6} = 0.99999 \dots

f (u_{5}) = - 2 \cdot 5.32463 E - 1 2^{2} - 2.00000 \dots \cdot 5.32463 E - 12 - 2.00000 \dots = - 2.00000 \dots

f^{^{'}} (u_{5}) = - 4 \cdot 5.32463 E - 12 - 2.00000 \dots = - 2.00000 \dots

u_{6} = - 0.99999 \dots

Δ u_{6} = ∣ - 0.99999 \dots - 5.32463 E - 12 ∣ = 0.99999 \dots

Δ u_{6} = 0.99999 \dots

u_{7} = 2.15534 E - 11 : Δ u_{7} = 1.00000 \dots

f (u_{6}) = - 2 (- 0.99999 \dots)^{2} - 2.00000 \dots (- 0.99999 \dots) - 2.00000 \dots = - 1.99999 \dots

f^{^{'}} (u_{6}) = - 4 (- 0.99999 \dots) - 2.00000 \dots = 1.99999 \dots

u_{7} = 2.15534 E - 11

Δ u_{7} = ∣ 2.15534 E - 11 - (- 0.99999 \dots) ∣ = 1.00000 \dots

Δ u_{7} = 1.00000 \dots

u_{8} = - 0.99999 \dots : Δ u_{8} = 0.99999 \dots

f (u_{7}) = - 2 \cdot 2.15534 E - 1 1^{2} - 2.00000 \dots \cdot 2.15534 E - 11 - 2.00000 \dots = - 2.00000 \dots

f^{^{'}} (u_{7}) = - 4 \cdot 2.15534 E - 11 - 2.00000 \dots = - 2.00000 \dots

u_{8} = - 0.99999 \dots

Δ u_{8} = ∣ - 0.99999 \dots - 2.15534 E - 11 ∣ = 0.99999 \dots

Δ u_{8} = 0.99999 \dots

u_{9} = 8.64686 E - 11 : Δ u_{9} = 1.00000 \dots

f (u_{8}) = - 2 (- 0.99999 \dots)^{2} - 2.00000 \dots (- 0.99999 \dots) - 2.00000 \dots = - 1.99999 \dots

f^{^{'}} (u_{8}) = - 4 (- 0.99999 \dots) - 2.00000 \dots = 1.99999 \dots

u_{9} = 8.64686 E - 11

Δ u_{9} = ∣ 8.64686 E - 11 - (- 0.99999 \dots) ∣ = 1.00000 \dots

Δ u_{9} = 1.00000 \dots

无法得出解

解为

u \approx 0.17157 \dots, u \approx 5.82842 \dots

u \approx 0.17157 \dots, u \approx 5.82842 \dots

验证解

找到无定义的点（奇点）:

u = 0

取

3 (\frac{u - u ^{- 1}}{u + u ^{- 1}})^{2}

的分母，令其等于零

u = 0

取

5 \frac{2}{u + u ^{- 1}} + 1

的分母，令其等于零

u = 0

以下点无定义

u = 0

将不在定义域的点与解相综合:

u \approx 0.17157 \dots, u \approx 5.82842 \dots

u \approx 0.17157 \dots, u \approx 5.82842 \dots

代回

u = e^{x}

，求解

x

解

e^{x} = 0.17157 \dots : x = ln (0.17157 \dots)

e^{x} = 0.17157 \dots

使用指数运算法则

e^{x} = 0.17157 \dots

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (0.17157 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (0.17157 \dots)

x = ln (0.17157 \dots)

解

e^{x} = 5.82842 \dots : x = ln (5.82842 \dots)

e^{x} = 5.82842 \dots

使用指数运算法则

e^{x} = 5.82842 \dots

若

f (x) = g (x)

，则

ln (f (x)) = ln (g (x))

ln (e^{x}) = ln (5.82842 \dots)

使用对数计算法则:

ln (e^{a}) = a

ln (e^{x}) = x

x = ln (5.82842 \dots)

x = ln (5.82842 \dots)

x = ln (0.17157 \dots), x = ln (5.82842 \dots)

x = ln (0.17157 \dots), x = ln (5.82842 \dots)

作图

流行的例子

tan (θ) = \frac{5}{7}

(cot(x)-1)(sqrt(3)tan(x)+1)=0

(cot (x) - 1) (3 tan (x) + 1) = 0

cos(2θ-pi/2)=(sqrt(2))/2

cos (2 θ - \frac{π}{2}) = \frac{2}{2}

tan^3(x)+tan^2(x)-3tan(x)=3

tan^{3} (x) + tan^{2} (x) - 3 tan (x) = 3

cos (x) = \frac{7}{9}