From 21d8aec01eae21e9b108ff9f25b842f3011fe992 Mon Sep 17 00:00:00 2001 From: wizardforcel <562826179@qq.com> Date: Sun, 4 Mar 2018 19:59:38 +0800 Subject: [PATCH] tex --- 15.md | 6 +++--- 17.md | 8 ++++---- 4.md | 4 ++-- 6.md | 4 ++-- img/tex-15-1.gif | Bin 0 -> 864 bytes img/tex-15-2.gif | Bin 0 -> 864 bytes img/tex-15-3.gif | Bin 0 -> 1123 bytes img/tex-17-1.gif | Bin 0 -> 975 bytes img/tex-17-2.gif | Bin 0 -> 4792 bytes img/tex-17-3.gif | Bin 0 -> 4867 bytes img/tex-17-4.gif | Bin 0 -> 881 bytes img/tex-4-1.gif | Bin 0 -> 1153 bytes img/tex-4-2.gif | Bin 0 -> 2495 bytes img/tex-6-1.gif | Bin 0 -> 921 bytes img/tex-6-2.gif | Bin 0 -> 876 bytes 15 files changed, 11 insertions(+), 11 deletions(-) create mode 100644 img/tex-15-1.gif create mode 100644 img/tex-15-2.gif create mode 100644 img/tex-15-3.gif create mode 100644 img/tex-17-1.gif create mode 100644 img/tex-17-2.gif create mode 100644 img/tex-17-3.gif create mode 100644 img/tex-17-4.gif create mode 100644 img/tex-4-1.gif create mode 100644 img/tex-4-2.gif create mode 100644 img/tex-6-1.gif create mode 100644 img/tex-6-2.gif diff --git a/15.md b/15.md index 0e372ed..3954ace 100644 --- a/15.md +++ b/15.md @@ -384,7 +384,7 @@ array([ 0.59610766, -0.19065363]) 我们如何实现呢? 在二维空间中,这非常简单。 如果我们在坐标`(x0, y0)`处有一个点,而在`(x1, y1)`处有另一个点,则它们之间的距离是: -![](http://latex.codecogs.com/gif.latex?D%20%3D%20%5Csqrt%7B%28x_0-x_1%29%5E2%20+%20%28y_0-y_1%29%5E2%7D) +![](img/tex-15-1.gif) (这是从哪里来的?它来自勾股定理:我们有一个直角三角形,边长为`x0 - x1`和`y0 - y1`,我们想要求出斜边的长度。) @@ -635,11 +635,11 @@ ax.scatter(banknotes.column('WaveletSkew'), 我们知道如何在二维空间中计算距离。 如果我们在坐标`(x0, y0)`处有一个点,而在`(x1, y1)`处有另一个点,则它们之间的距离是: -![](http://latex.codecogs.com/gif.latex?D%20%3D%20%5Csqrt%7B%28x_0-x_1%29%5E2%20+%20%28y_0-y_1%29%5E2%7D) +![](img/tex-15-2.gif) 在三维空间中,点是`(x0, y0, z0)`和`(x1, y1, z1)`,它们之间的距离公式为: -![](http://latex.codecogs.com/gif.latex?D%20%3D%20%5Csqrt%7B%28x_0-x_1%29%5E2%20+%20%28y_0-y_1%29%5E2%20+%20%28z_0-z_1%29%5E2%7D) +![](img/tex-15-3.gif) 在 N 维空间中,东西有点难以可视化,但我想你可以看到公式是如何推广的:我们总结每个独立坐标差的平方,然后取平方根。 diff --git a/17.md b/17.md index 4c1de68..fa645c8 100644 --- a/17.md +++ b/17.md @@ -129,13 +129,13 @@ students.pivot('Major', 'Year') 后验概率。这些是考虑专业声明状态的信息后,二年级的概率。我们计算了其中的一个: -假设学生已经声明,学生是三年级的后验概率表示为 ![](http://latex.codecogs.com/gif.latex?P%28%5Cmbox%7BThird%20Year%7D%20%7E%5Cbig%7B%7C%7D%7E%20%5Cmbox%7BDeclared%7D%29),计算如下。 +假设学生已经声明,学生是三年级的后验概率表示为 ![](img/tex-17-1.gif),计算如下。 -![](http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7D%20P%28%5Cmbox%7BThird%20Year%7D%20%7E%5Cbig%7B%7C%7D%7E%20%5Cmbox%7BDeclared%7D%29%20%7E%20%26%3D%7E%20%5Cfrac%7B%200.4%20%5Ctimes%200.8%7D%7B0.6%20%5Ctimes%200.5%20%7E+%7E%200.4%20%5Ctimes%200.8%7D%20%5C%5C%20%5C%5C%20%26%3D%7E%20%5Cfrac%7B%5Cmbox%7B%28prior%20probability%20of%20Third%20Year%29%7D%20%5Ctimes%20%5Cmbox%7B%28likelihood%20of%20Declared%20given%20Third%20Year%29%7D%7D%20%7B%5Cmbox%7Btotal%20probability%20of%20Declared%7D%7D%20%5Cend%7Balign*%7D) +![](img/tex-17-2.gif) 另一个后验概率是: -![](http://latex.codecogs.com/gif.latex?%5Cbegin%7Balign*%7D%20P%28%5Cmbox%7BSecond%20Year%7D%20%7E%5Cbig%7B%7C%7D%7E%20%5Cmbox%7BDeclared%7D%29%20%7E%20%26%3D%7E%20%5Cfrac%7B%200.6%20%5Ctimes%200.5%7D%7B0.6%20%5Ctimes%200.5%20%7E+%7E%200.4%20%5Ctimes%200.8%7D%20%5C%5C%20%5C%5C%20%26%3D%7E%20%5Cfrac%7B%5Cmbox%7B%28prior%20probability%20of%20Second%20Year%29%7D%20%5Ctimes%20%5Cmbox%7B%28likelihood%20of%20Declared%20given%20Second%20Year%29%7D%7D%20%7B%5Cmbox%7Btotal%20probability%20of%20Declared%7D%7D%20%5Cend%7Balign*%7D) +![](img/tex-17-3.gif) ```py (0.6 * 0.5)/(0.6 * 0.5 + 0.4 * 0.8) @@ -148,7 +148,7 @@ students.pivot('Major', 'Year') 正因为如此,贝叶斯方法有时被归纳为比例陈述: -![](http://latex.codecogs.com/gif.latex?%5Cmbox%7Bposterior%7D%20%7E%20%5Cpropto%20%7E%20%5Cmbox%7Bprior%7D%20%5Ctimes%20%5Cmbox%7Blikelihood%7D) +![](img/tex-17-4.gif) 公式非常便于高效地描述计算。 但是在我们的学生示例这样的情况中,不用公式来思考更简单。 我们仅仅使用树形图。 diff --git a/4.md b/4.md index 809c359..edad78f 100644 --- a/4.md +++ b/4.md @@ -321,13 +321,13 @@ array([ 1.5, 1. , 0.5, 0. , -0.5, -1. , -1.5]) 伟大的德国数学家和哲学家戈特弗里德·威廉·莱布尼茨(Gottfried Wilhelm Leibniz,1646 ~ 1716年)发现了一个简单分数的无穷和。 公式是: -![](http://latex.codecogs.com/gif.latex?%5Cpi%20%3D%202%20%5Ccdot%20%5Cleft%28%20%5Cfrac%7B2%7D%7B1%7D%5Ccdot%5Cfrac%7B2%7D%7B3%7D%5Ccdot%5Cfrac%7B4%7D%7B3%7D%5Ccdot%5Cfrac%7B4%7D%7B5%7D%5Ccdot%5Cfrac%7B6%7D%7B5%7D%5Ccdot%5Cfrac%7B6%7D%7B7%7D%5Cdots%20%5Cright%29) +![](img/tex-4-1.gif) 虽然需要一些数学来确定它,但我们可以用数组来说服我们自己,公式是有效的。 让我们计算莱布尼茨的无穷和的前 5000 个项,看它是否接近 π。 我们将计算这个有限的总和,首先加上所有的正项,然后减去所有负项的和 [1]: -![](http://latex.codecogs.com/gif.latex?%5Cpi%20%5Capprox%202%20%5Ccdot%20%5Cleft%28%20%5Cfrac%7B2%7D%7B1%7D%20%5Ccdot%20%5Cfrac%7B4%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B6%7D%7B5%7D%20%5Ccdots%20%5Cfrac%7B1%2C000%2C000%7D%7B999999%7D%20%5Cright%29%20%5Ccdot%20%5Cleft%28%20%5Cfrac%7B2%7D%7B3%7D%20%5Ccdot%20%5Cfrac%7B4%7D%7B5%7D%20%5Ccdot%20%5Cfrac%7B6%7D%7B7%7D%20%5Ccdots%20%5Cfrac%7B1%2C000%2C000%7D%7B1%2C000%2C001%7D%20%5Cright%29) +![](img/tex-4-2.gif) > [1] 令人惊讶的是,当我们将无限多个分数相加时,顺序可能很重要。但是我们对 π 的近似只使用了大量的数量有限的分数,所以可以按照任何方便的顺序,将这些项相加。 diff --git a/6.md b/6.md index 2745751..09889a6 100644 --- a/6.md +++ b/6.md @@ -555,9 +555,9 @@ heights 如果我们只查看表格的第一行,计算就会变得清晰。 -请记住,数据集中有 200 部电影。这个`[300,400)`的桶包含 81 部电影。这是所有电影的 40.5%:![](http://latex.codecogs.com/gif.latex?%5Cmbox%7BPercent%7D%20%3D%20%5Cfrac%7B81%7D%7B200%7D%20%5Ccdot%20100%20%3D%2040.5)。 +请记住,数据集中有 200 部电影。这个`[300,400)`的桶包含 81 部电影。这是所有电影的 40.5%:![](img/tex-6-1.gif)。 -`[300, 400)`桶的宽度是`400-300 = 100`。所以 ![](http://latex.codecogs.com/gif.latex?%5Cmbox%7BHeight%7D%20%3D%20%5Cfrac%7B40.5%7D%7B100%7D%20%3D%200.405)。 +`[300, 400)`桶的宽度是`400-300 = 100`。所以 ![](img/tex-6-2.gif)。 用于计算高度的代码使用了总共​​有 200 个电影,以及每个箱的宽度是 100 的事实。 diff --git a/img/tex-15-1.gif b/img/tex-15-1.gif new file mode 100644 index 0000000000000000000000000000000000000000..53b620bb44a779a6325eaed2763108b3643df96f GIT binary patch literal 864 zcmV-m1E2gyNk%w1Vd?-E0J8u9|Ns90004G&b|NAoW@ctYL`1r}x>QtD?(XiGn3&w$ 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