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little optimize.

fetches/scruel/master
scruel 7 years ago
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9fc58f733c
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      README.md
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      week1.md

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README.md View File

@@ -59,7 +59,9 @@ This work is licensed under a [Creative Commons Attribution-NonCommercial 4.0 In

Copyright © Scruel. All Rights Reserved.

**注:转载请注明出处,禁止用于任何商业用途(包括广告形式)**


本人所作的所有笔记内容**禁止任何商业形式(包括广告形式)的转载**,非商业形式的转载**请先联系我授权**,在承诺不会用于商业用途后方可使用,且需声明转载地址(GitHub)。

## Donate



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week1.md View File

@@ -178,7 +178,7 @@ $h_\theta(x)=\theta_0+\theta_1x$,为解决房价问题的一种可行表达式
>
> $\left(x, y\right)$: 训练集中的实例
>
> $\left(x^\left(i\right),y^\left(i\right)\right)$: 训练集中的第 $i$ 个样本实例
> $\left(x^{\left(i\right)},y^{\left(i\right)}\right)$: 训练集中的第 $i$ 个样本实例

![](images/20180105_224648.png)

@@ -200,7 +200,7 @@ $$J(\theta_0,\theta_1)=\dfrac{1}{2m}\displaystyle\sum_{i=1}^m\left(\hat{y}_{i}-y

- 假设函数(Hypothesis): $h_\theta(x)=\theta_0+\theta_1x$
- 参数(Parameters): $\theta_0, \theta_1$
- 代价函数(Cost Function): $ J\left( \theta_0, \theta_1 \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}}} $
- 代价函数(Cost Function): $J\left( \theta_0, \theta_1 \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}}}$
- 目标(Goal): $\underset{\theta_0, \theta_1}{\text{minimize}} J \left(\theta_0, \theta_1 \right)$

为了直观理解代价函数到底是在做什么,先假设 $\theta_1 = 0$,并假设训练集有三个数据,分别为$\left(1, 1\right), \left(2, 2\right), \left(3, 3\right)$,这样在平面坐标系中绘制出 $h_\theta\left(x\right)$ ,并分析 $J\left(\theta_0, \theta_1\right)​$ 的变化。
@@ -304,7 +304,7 @@ $\begin{align*} & \text{repeat until convergence:} \; \lbrace \newline \; &{{\th
线性回归模型

- $h_\theta(x)=\theta_0+\theta_1x$
- $ J\left( \theta_0, \theta_1 \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}}} $
- $J\left( \theta_0, \theta_1 \right)=\frac{1}{2m}\sum\limits_{i=1}^{m}{{{\left( {{h}_{\theta }}\left( {{x}^{(i)}} \right)-{{y}^{(i)}} \right)}^{2}}}$

梯度下降算法
- $\begin{align*} & \text{repeat until convergence:} \; \lbrace \newline \; &{{\theta }_{j}}:={{\theta }_{j}}-\alpha \frac{\partial }{\partial {{\theta }_{j}}}J\left( {\theta_{0}},{\theta_{1}} \right) \newline \rbrace \end{align*}$


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