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修改week8几个错别字
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fengdu78 committed Jun 22, 2018
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Expand Up @@ -327,7 +327,7 @@ $$z^{(i)}=U^{T}_{reduce}*x^{(i)}$$
![](../images/a4477d787f876ae4e72cb416a2cb0b8a.jpg)

其中的$S$是一个$n×n$的矩阵,只有对角线上有值,而其它单元都是0,我们可以使用这个矩阵来计算平均均方误差与训练集方差的比例:
$$\dfrac {\dfrac {1}{m}\sum^{m}_{i=1}\left\| x^{\left( i\right) }-x^{\left( i\right) }_{approx}\right\| ^{2}}{\dfrac {1}{m}\sum^{m}_{i=1}\left\| x^{(i)}\right\| ^{2}}=1-\dfrac {\Sigma^{m}_{i=1}S_{ii}}{\Sigma^{k}_{i=1}S_{ii}}\leq 1\%$$
$$\dfrac {\dfrac {1}{m}\sum^{m}_{i=1}\left\| x^{\left( i\right) }-x^{\left( i\right) }_{approx}\right\| ^{2}}{\dfrac {1}{m}\sum^{m}_{i=1}\left\| x^{(i)}\right\| ^{2}}=1-\dfrac {\Sigma^{k}_{i=1}S_{ii}}{\Sigma^{m}_{i=1}S_{ii}}\leq 1\%$$

也就是:$$\frac {\Sigma^{k}_{i=1}s_{ii}}{\Sigma^{n}_{i=1}s_{ii}}\geq0.99$$

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