Stanford ML - Lecture 4 - Neural Networks: Representation

最新推荐文章于 2020-02-19 13:59:44 发布

Quebradawill

最新推荐文章于 2020-02-19 13:59:44 发布

阅读量892

点赞数

CC 4.0 BY-SA版权

分类专栏： Machine Learning ML-Stanford-Andrew Ng

本文链接：https://blog.youkuaiyun.com/qiudw/article/details/8670649

Machine Learning 同时被 2 个专栏收录

19 篇文章

订阅专栏

ML-Stanford-Andrew Ng

12 篇文章

订阅专栏

本文介绍了引入非线性假设的原因，探讨了神经网络的发展历程及其在高维数据处理中的作用。文章还详细解释了神经元模型及激活函数，并通过实例展示了神经网络如何实现AND、OR及XNOR等逻辑函数，最后讨论了多分类问题的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1. Non-linear hypotheses

why introduce non-linear hypotheses?
- high dimensional data
- non-linear hypotheses

2. Neurons and the brain

Neural networks
- origins: algorithms that try to mimic the brain
- was very widely used in 80s and early 90s, popularity diminished in late 90s
- recent resurgence: state-of-the-art technique for many applications

3. Model representation I

neuron model: logistic unit
sigmoid (logistic) activation function

$a_i^{(j)} = \textrm{"activation" of unit} \ i \ \textrm{in layer} \ j$

$\begin{array}{ll} \theta^{(j)} = & \textrm{matrix of weights controlling function mapping from} \\ & \textrm{layer} \ j \ \textrm{to layer} \ j+1 \\ \end{array}$

$a_1^{(2)} = g \left( \theta_{10}^{(1)} x_0 + \theta_{11}^{(1)} x_1 + \theta_{12}^{(1)} x_2 + \theta_{13}^{(1)} x_3 \right )$

$a_2^{(2)} = g \left( \theta_{20}^{(1)} x_0 + \theta_{21}^{(1)} x_1 + \theta_{22}^{(1)} x_2 + \theta_{23}^{(1)} x_3 \right )$

$a_3^{(2)} = g \left( \theta_{30}^{(1)} x_0 + \theta_{31}^{(1)} x_1 + \theta_{32}^{(1)} x_2 + \theta_{33}^{(1)} x_3 \right )$

$h_{\theta}(x) = a_1^{(3)} = g \left( \theta_{10}^{(2)} a_0 + \theta_{11}^{(2)} a_1 + \theta_{12}^{(2)} a_2 + \theta_{13}^{(2)} a_3 \right )$

$\begin{array}{ll} \textrm{if network has} \ s_j \ \textrm{units in layer} \ j, s_{j+1} \textrm{units in layer} \\ j+1, \textrm{then} \theta^{(j)} \textrm{will be of dimension} \ s_{j+1} \times (s_j + 1) \end{array}$

4. Model representation II

forward propagation: vectorized implementation

$z^{(2)} = \theta^{(1)} x$

$a^{(2)} = g(z^{(2)})$

$\textrm{add} \ a_0^{(2)} = 1$

$z^{(3)} = \theta^{(2)} a^{(2)}$

$h_{\theta} (x) = a^{(3)} = g(z^{(3)})$