Sigmoid函数导数推导详解
Sigmoid函数导数推导详解
- 在逻辑回归中,Sigmoid函数的导数推导是一个关键步骤,它使得梯度下降算法能够高效地计算。
1. Sigmoid函数定义
首先回顾Sigmoid函数的定义:
g(z)=11+e−zg(z) = \frac{1}{1 + e^{-z}}g(z)=1+e−z1
2. 导数推导过程
-
从Sigmoid函数出发:
g(z)=11+e−zg(z) = \frac{1}{1 + e^{-z}}g(z)=1+e−z1 -
令u=1+e−zu = 1 + e^{-z}u=1+e−z,则g(z)=u−1g(z) = u^{-1}g(z)=u−1
-
使用链式法则:
dgdz=dgdu⋅dudz=−u−2⋅(−e−z)=e−z(1+e−z)2\frac{dg}{dz} = \frac{dg}{du} \cdot \frac{du}{dz} = -u^{-2} \cdot (-e^{-z}) = \frac{e^{-z}}{(1 + e^{-z})^2}dzdg=dudg⋅dzdu=−u−2⋅(−e−z)=(1+e−z)2e−z -
现在,我们将其表示为g(z)g(z)g(z)的函数:
e−z1+e−z=1−11+e−z=1−g(z)\frac{e^{-z}}{1 + e^{-z}} = 1 - \frac{1}{1 + e^{-z}} = 1 - g(z)1+e−ze−z=1−1+e−z1=1−g(z) -
因此:
g′(z)=11+e−z⋅e−z1+e−z=g(z)⋅(1−g(z))g'(z) = \frac{1}{1 + e^{-z}} \cdot \frac{e^{-z}}{1 + e^{-z}} = g(z) \cdot (1 - g(z))g′(z)=1+e−z1⋅1+e−ze−z=g(z)⋅(1−g(z))
3. 代码实现
import numpy as np
import matplotlib.pyplot as plt
def sigmoid(z):
return 1 / (1 + np.exp(-z))
def sigmoid_derivative(z):
return sigmoid(z) * (1 - sigmoid(z))
z = np.linspace(-10, 10, 100)
plt.figure(figsize=(10, 6))
plt.plot(z, sigmoid(z), label="Sigmoid function")
plt.plot(z, sigmoid_derivative(z), label="Sigmoid derivative")
plt.xlabel("z")
plt.ylabel("g(z)")
plt.title("Sigmoid Function and its Derivative")
plt.legend()
plt.grid(True)
plt.show()

4. 导数性质分析
- 最大值:当g(z)=0.5g(z) = 0.5g(z)=0.5时,导数达到最大值0.250.250.25
- 对称性:导数在z=0z=0z=0时最大,随着∣z∣|z|∣z∣增大而迅速减小
- 非负性:导数始终非负,因为0<g(z)<10 < g(z) < 10<g(z)<1
5. 导数形式的重要型
- 在逻辑回归的梯度下降中,需要计算损失函数对参数的导数。由于损失函数中包含Sigmoid函数,这个导数形式使得计算变得非常简洁:
∂∂θjJ(θ)=1m∑i=1m(hθ(x(i))−y(i))xj(i)\frac{\partial}{\partial \theta_j}J(\theta) = \frac{1}{m}\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}∂θj∂J(θ)=m1i=1∑m(hθ(x(i))−y(i))xj(i)
- 其中hθ(x)=g(θTx)h_\theta(x) = g(\theta^T x)hθ(x)=g(θTx)。如果没有这个简洁的导数形式,梯度计算会复杂得多。
- 推导损失函数对θj\theta_jθj的偏导数:
∂∂θjJ(θ)=−1m∑i=1m(yi1hθ(xi)−(1−yi)11−hθ(xi))∂∂θjhθ(xi)=−1m∑i=1m(yi1g(θTxi)−(1−yi)11−g(θTxi))g(θTxi)(1−g(θTxi))xij=−1m∑i=1m(yi(1−g(θTxi))−(1−yi)g(θTxi))xij=1m∑i=1m(hθ(xi)−yi)xij \begin{align*} \frac{\partial}{\partial \theta_j} J(\theta) &= -\frac{1}{m}\sum_{i=1}^m \left(y_i \frac{1}{h_\theta(x_i)} - (1-y_i)\frac{1}{1-h_\theta(x_i)}\right) \frac{\partial}{\partial \theta_j} h_\theta(x_i) \\ &= -\frac{1}{m}\sum_{i=1}^m \left(y_i \frac{1}{g(\theta^T x_i)} - (1-y_i)\frac{1}{1-g(\theta^T x_i)}\right) g(\theta^T x_i)(1-g(\theta^T x_i)) x_i^j \\ &= -\frac{1}{m}\sum_{i=1}^m \left(y_i(1-g(\theta^T x_i)) - (1-y_i)g(\theta^T x_i)\right) x_i^j \\ &= \frac{1}{m}\sum_{i=1}^m (h_\theta(x_i) - y_i) x_i^j \end{align*} ∂θj∂J(θ)=−m1i=1∑m(yihθ(xi)1−(1−yi)1−hθ(xi)1)∂θj∂hθ(xi)=−m1i=1∑m(yig(θTxi)1−(1−yi)1−g(θTxi)1)g(θTxi)(1−g(θTxi))xij=−m1i=1∑m(yi(1−g(θTxi))−(1−yi)g(θTxi))xij=m1i=1∑m(hθ(xi)−yi)xij
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