Skip to content

Fix KaTeX parse errors in notebook math blocks #79

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Open
KaynXu wants to merge 1 commit into
base: main
Choose a base branch
from
Open

Fix KaTeX parse errors in notebook math blocks #79

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
20 changes: 15 additions & 5 deletions LA_01_01_01.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -17,9 +17,13 @@
"metadata": {},
"source": [
"本代码首先导入了用于数值计算的 NumPy 库，目的是利用其高效的数组操作功能。数学上，我们可以把数据看作一个向量，这里定义的 Python 列表 \n",
"$$a\\_list = [1,\\,2,\\,3,\\,4,\\,5,\\,6,\\,7,\\,8,\\,9,\\,10]$$ \n",
"$$\n",
"a\\_list = [1,\\,2,\\,3,\\,4,\\,5,\\,6,\\,7,\\,8,\\,9,\\,10]\n",
"$$\n",
"包含 $10$ 个元素。接下来，利用 NumPy 的 array 函数将这个列表转换为一个一维数组，我们将其视为一个行向量，记作 \n",
"$$a = [1,\\; 2,\\; 3,\\; 4,\\; 5,\\; 6,\\; 7,\\; 8,\\; 9,\\; 10].$$ \n",
"$$\n",
"a = [1,\\; 2,\\; 3,\\; 4,\\; 5,\\; 6,\\; 7,\\; 8,\\; 9,\\; 10].\n",
"$$\n",
"\n",
"数组的维数由属性 $a.ndim$ 得到，此处 $dim = 1$，表明这是一个一维向量。数组的形状由 $a.shape$ 给出，返回值为 $(10,)$，说明行向量在其唯一的维度上共有 $10$ 个元素。\n",
"\n",
Expand All @@ -30,11 +34,17 @@
"- $a[-2]$ 返回倒数第二个元素（对应 $a_9$）。\n",
"\n",
"切片操作则用于提取向量的子部分。切片表达式采用半开区间的形式，如 $a[0:2]$ 表示从索引 $0$ 开始，直到索引 $2$（不含 $2$）的所有元素，即得到 \n",
"$$a_{\\text{slice}} = [1,\\; 2].$$ \n",
"$$\n",
"a_{\\text{slice}} = [1,\\; 2].\n",
"$$\n",
"同样，$a[:2]$ 是简写形式，效果相同。表达式 $a[1:]$ 则提取从索引 $1$（即数学中的 $a_2$）开始到最后的所有元素，得到 \n",
"$$a_{\\text{slice}} = [2,\\; 3,\\; 4,\\; 5,\\; 6,\\; 7,\\; 8,\\; 9,\\; 10].$$ \n",
"$$\n",
"a_{\\text{slice}} = [2,\\; 3,\\; 4,\\; 5,\\; 6,\\; 7,\\; 8,\\; 9,\\; 10].\n",
"$$\n",
"此外，$a[-3:]$ 和 $a[7:]$ 均提取了最后 $3$ 个元素，因为索引 $7$ 对应的是第 $8$ 个元素，于是得到 \n",
"$$a_{\\text{slice}} = [8,\\; 9,\\; 10].$$\n",
"$$\n",
"a_{\\text{slice}} = [8,\\; 9,\\; 10].\n",
"$$\n",
"\n",
"总体来说，这段代码展示了如何从数学上定义的行向量转化为 NumPy 数组，并通过索引与切片操作实现对向量元素的访问与提取，同时强调了 Python 中零索引与数学中常见的一索引之间的差别。"
]
Expand Down
8 changes: 6 additions & 2 deletions LA_01_07_03.ipynb
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -25,9 +25,13 @@
"已知一个二维向量 $\\mathbf{a} = \\begin{bmatrix} 3 \\\\ 4 \\end{bmatrix}$，我们希望将它分解为两个正交方向上的分量：\n",
"\n",
"- 方向一：与单位向量 $\\mathbf{v}_1$ 方向相同，定义为 \n",
" $$\\mathbf{v}_1 = \\begin{bmatrix} \\cos\\theta \\\\ \\sin\\theta \\end{bmatrix}$$\n",
" $$\n",
" \\mathbf{v}_1 = \\begin{bmatrix} \\cos\\theta \\\\ \\sin\\theta \\end{bmatrix}\n",
" $$\n",
"- 方向二：与单位向量 $\\mathbf{v}_2$ 方向相同，垂直于 $\\mathbf{v}_1$，定义为 \n",
" $$\\mathbf{v}_2 = \\begin{bmatrix} -\\sin\\theta \\\\ \\cos\\theta \\end{bmatrix}$$\n",
" $$\n",
" \\mathbf{v}_2 = \\begin{bmatrix} -\\sin\\theta \\\\ \\cos\\theta \\end{bmatrix}\n",
" $$\n",
"\n",
"$\\mathbf{v}_1$ 和 $\\mathbf{v}_2$ 构成一个二维空间中的**标准正交基底（orthonormal basis）**。\n",
"\n",
Expand Down