网硕互联帮助中心接下来，在采用公式
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J\\left( \\mathbf{\\theta} \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}{(y_{pi} – y_{i})}^{2}
J(θ)=2m1∑i=0m−1(ypi−yi)2这样的优化函数的情况下，一起来求解模型的参数。
在极值点必有导数值为0。导数值为0则表明函数随未知变量的变化率为0。下面先对
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θ0求偏导数：
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\\frac{\\partial J\\left( \\mathbf{\\theta} \\right)}{\\partial\\theta_{0}} = \\frac{\\partial}{\\partial\\theta_{0}}\\left( \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)^{2} \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}{\\frac{\\partial}{\\partial\\theta_{0}}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)^{2}} = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)\\frac{\\partial}{\\partial\\theta_{0}}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right) \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)\\frac{\\partial}{\\partial\\theta_{0}}\\left( \\theta_{1}x_{i} + \\theta_{0} \\right) \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right) \\right) = \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)
∂θ0∂J(θ)=∂θ0∂(2m1i=0∑m−1((θ1xi+θ0)−yi)2)=2m1i=0∑m−1∂θ0∂((θ1xi+θ0)−yi)2=2m1i=0∑m−1(2((θ1xi+θ0)−yi)∂θ0∂((θ1xi+θ0)−yi))=2m1i=0∑m−1(2((θ1xi+θ0)−yi)∂θ0∂(θ1xi+θ0))=2m1i=0∑m−1(2((θ1xi+θ0)−yi))=m1i=0∑m−1((θ1xi+θ0)−yi)
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再对
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\\frac{\\partial J(\\mathbf{\\theta})}{\\partial\\theta_{1}} = \\frac{\\partial}{\\partial\\theta_{1}}\\left( \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)^{2} \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}{\\frac{\\partial}{\\partial\\theta_{1}}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)^{2}} = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)\\frac{\\partial}{\\partial\\theta_{1}}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right) \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)\\frac{\\partial}{\\partial\\theta_{1}}\\left( \\theta_{1}x_{i} + \\theta_{0} \\right) \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)\\frac{\\partial}{\\partial\\theta_{1}}(\\theta_{1}x_{i}) \\right) = \\frac{1}{2m}\\sum_{i = 0}^{m – 1}\\left( 2\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)x_{i} \\right) = \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right)x_{i} \\right) = \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right)x_{i} – y_{i}x_{i} \\right)
∂θ1∂J(θ)=∂θ1∂(2m1i=0∑m−1((θ1xi+θ0)−yi)2)=2m1i=0∑m−1∂θ1∂((θ1xi+θ0)−yi)2=2m1i=0∑m−1(2((θ1xi+θ0)−yi)∂θ1∂((θ1xi+θ0)−yi))=2m1i=0∑m−1(2((θ1xi+θ0)−yi)∂θ1∂(θ1xi+θ0))=2m1i=0∑m−1(2((θ1xi+θ0)−yi)∂θ1∂(θ1xi))=2m1i=0∑m−1(2((θ1xi+θ0)−yi)xi)=m1i=0∑m−1(((θ1xi+θ0)−yi)xi)=m1i=0∑m−1((θ1xi+θ0)xi−yixi)
由此，可得到方程组：
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\\left\\{ \\begin{matrix} \\frac{\\partial J(\\mathbf{\\theta})}{\\partial\\theta_{0}} = \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right) = 0 \\\\ \\frac{\\partial J(\\mathbf{\\theta})}{\\partial\\theta_{1}} = \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right)x_{i} – y_{i}x_{i} \\right) = 0 \\\\ \\end{matrix} \\right.\\
{∂θ0∂J(θ)=m1∑i=0m−1((θ1xi+θ0)−yi)=0∂θ1∂J(θ)=m1∑i=0m−1((θ1xi+θ0)xi−yixi)=0
（方程组1）
怎么求解这个方程组呢？可根据方程组的第1个方程，得到：
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\\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right) – y_{i} \\right) = 0 \\Longrightarrow \\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\theta_{1}x_{i} + \\theta_{0} – y_{i} \\right) = 0
m1i=0∑m−1((θ1xi+θ0)−yi)=0⟹m1i=0∑m−1(θ1xi+θ0−yi)=0
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\\Longrightarrow \\sum_{i = 0}^{m – 1}\\theta_{0} = \\sum_{i = 0}^{m – 1}\\left( y_{i} – \\theta_{1}x_{i} \\right) \\Longrightarrow m\\theta_{0} = \\sum_{i = 0}^{m – 1}\\left( y_{i} – \\theta_{1}x_{i} \\right)
⟹i=0∑m−1θ0=i=0∑m−1(yi−θ1xi)⟹mθ0=i=0∑m−1(yi−θ1xi)
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\\Longrightarrow m\\theta_{0} = \\sum_{i = 0}^{m – 1}y_{i} – \\sum_{i = 0}^{m – 1}\\left( \\theta_{1}x_{i} \\right) \\Longrightarrow \\text{mθ}_{0} = \\sum_{i = 0}^{m – 1}y_{i} – \\theta_{1}\\sum_{i = 0}^{m – 1}x_{i}
⟹mθ0=i=0∑m−1yi−i=0∑m−1(θ1xi)⟹mθ0=i=0∑m−1yi−θ1i=0∑m−1xi
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\\Longrightarrow \\theta_{0} = \\frac{1}{m}\\sum_{i = 0}^{m – 1}y_{i} – {\\frac{1}{m}\\theta}_{1}\\sum_{i = 0}^{m – 1}x_{i}
⟹θ0=m1i=0∑m−1yi−m1θ1i=0∑m−1xi
根据方程组4-5的第2个方程，可得：
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\\frac{1}{m}\\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right)x_{i} – y_{i}x_{i} \\right) = 0 \\Longrightarrow \\sum_{i = 0}^{m – 1}\\left( \\left( \\theta_{1}x_{i} + \\theta_{0} \\right)x_{i} – y_{i}x_{i} \\right) = 0
m1i=0∑m−1((θ1xi+θ0)xi−yixi)=0⟹i=0∑m−1((θ1xi+θ0)xi−yixi)=0
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\\Longrightarrow \\sum_{i = 0}^{m – 1}\\left( \\theta_{1}{x_{i}}^{2} + \\theta_{0}x_{i} – y_{i}x_{i} \\right) = 0 \\Longrightarrow \\sum_{i = 0}^{m – 1}\\left( \\theta_{0}x_{i} \\right) = \\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} – \\theta_{1}{x_{i}}^{2} \\right)
⟹i=0∑m−1(θ1xi2+θ0xi−yixi)=0⟹i=0∑m−1(θ0xi)=i=0∑m−1(yixi−θ1xi2)
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\\Longrightarrow \\theta_{0}\\sum_{i = 0}^{m – 1}x_{i} = \\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2}
⟹θ0i=0∑m−1xi=i=0∑m−1(yixi)−θ1i=0∑m−1xi2
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\\Longrightarrow \\theta_{0} = \\frac{\\left( \\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2} \\right)}{\\sum_{i = 0}^{m – 1}x_{i}}
⟹θ0=∑i=0m−1xi(∑i=0m−1(yixi)−θ1∑i=0m−1xi2)
因此，可得：
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\\frac{1}{m}\\sum_{i = 0}^{m – 1}y_{i} – {\\frac{1}{m}\\theta}_{1}\\sum_{i = 0}^{m – 1}x_{i} = \\frac{\\left( \\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2} \\right)}{\\sum_{i = 0}^{m – 1}x_{i}}
m1i=0∑m−1yi−m1θ1i=0∑m−1xi=∑i=0m−1xi(∑i=0m−1(yixi)−θ1∑i=0m−1xi2)
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\\Longrightarrow \\frac{1}{m}\\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i} – \\frac{1}{m}\\theta_{1}\\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} = \\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2}
⟹m1i=0∑m−1yii=0∑m−1xi−m1θ1i=0∑m−1xii=0∑m−1xi=i=0∑m−1(yixi)−θ1i=0∑m−1xi2
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\\Longrightarrow \\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i} – \\theta_{1}\\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} = m\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – m\\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2}
⟹i=0∑m−1yii=0∑m−1xi−θ1i=0∑m−1xii=0∑m−1xi=mi=0∑m−1(yixi)−mθ1i=0∑m−1xi2
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\\Longrightarrow \\ m\\theta_{1}\\sum_{i = 0}^{m – 1}{x_{i}}^{2} – \\theta_{1}\\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} = m\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i}
⟹ mθ1i=0∑m−1xi2−θ1i=0∑m−1xii=0∑m−1xi=mi=0∑m−1(yixi)−i=0∑m−1yii=0∑m−1xi
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\\Longrightarrow \\ \\theta_{1}\\left( m\\sum_{i = 0}^{m – 1}{x_{i}}^{2} – \\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} \\right) = m\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i}
⟹ θ1(mi=0∑m−1xi2−i=0∑m−1xii=0∑m−1xi)=mi=0∑m−1(yixi)−i=0∑m−1yii=0∑m−1xi
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\\Longrightarrow \\ \\theta_{1} = \\frac{\\left( m\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i} \\right)}{\\left( m\\sum_{i = 0}^{m – 1}{x_{i}}^{2} – \\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} \\right)}
⟹ θ1=(m∑i=0m−1xi2−∑i=0m−1xi∑i=0m−1xi)(m∑i=0m−1(yixi)−∑i=0m−1yi∑i=0m−1xi)
至此，已求解得直线模型的2个参数值为：
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\\left\\{ \\begin{matrix} \\theta_{1} = \\frac{\\left( m\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) – \\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i} \\right)}{\\left( m\\sum_{i = 0}^{m – 1}{x_{i}}^{2} – \\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} \\right)} \\\\ \\theta_{0} = \\frac{1}{m}\\sum_{i = 0}^{m – 1}y_{i} – {\\frac{1}{m}\\theta}_{1}\\sum_{i = 0}^{m – 1}x_{i} \\\\ \\end{matrix} \\right.\\
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⎧θ1=(m∑i=0m−1xi2−∑i=0m−1xi∑i=0m−1xi)(m∑i=0m−1(yixi)−∑i=0m−1yi∑i=0m−1xi)θ0=m1∑i=0m−1yi−m1θ1∑i=0m−1xi
（方程组2的解）
要想打好机器学习的数学基础，请参见清华大学出版社的人人可懂系列，包括《人人可懂的微积分》（已上市）、《人人可懂的线性代数》（即将上市）、《人人可懂的概率统计》（即将上市）。
提示：
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\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right)
∑i=0m−1(yixi)与
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\\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i}
∑i=0m−1yi∑i=0m−1xi不同，只要将这2个式子展开来就可以看得很清楚了：
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\\sum_{i = 0}^{m – 1}\\left( y_{i}x_{i} \\right) = y_{0}x_{0} + \\ldots + y_{m – 1}x_{m – 1}
i=0∑m−1(yixi)=y0x0+…+ym−1xm−1
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\\sum_{i = 0}^{m – 1}y_{i}\\sum_{i = 0}^{m – 1}x_{i} = (y_{0} + \\ldots + y_{m – 1})(x_{0} + \\ldots + x_{m – 1})
i=0∑m−1yii=0∑m−1xi=(y0+…+ym−1)(x0+…+xm−1)
这2个式子明显不同。再来看
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\\sum_{i = 0}^{m – 1}{x_{i}}^{2}
∑i=0m−1xi2和
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\\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i}
∑i=0m−1xi∑i=0m−1xi，将这2个式子展开来就可以看得很清晰：
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\\sum_{i = 0}^{m – 1}{x_{i}}^{2} = {x_{0}}^{2} + \\ldots + {x_{m – 1}}^{2}
i=0∑m−1xi2=x02+…+xm−12
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\\sum_{i = 0}^{m – 1}x_{i}\\sum_{i = 0}^{m – 1}x_{i} = {(x_{0} + \\ldots + x_{m – 1})}^{2}
i=0∑m−1xii=0∑m−1xi=(x0+…+xm−1)2
这2个式子看起来也明显不同。
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