diff --git a/articles/2024/daily.md b/articles/2024/daily.md
new file mode 100644
index 0000000..27f376c
--- /dev/null
+++ b/articles/2024/daily.md
@@ -0,0 +1,11 @@
+## 2024-7-12
+
+做渲染真的很苦，没啥可喜的地方，每个人的理解也是存在很大的差异的。
+
+## 2024-7-11
+
+《高效能人士的七个习惯》中提到一个事实，就是人类在技术发展的最近几十年里，把人格魅力排到了道德修养之前了。
+
+随着技术越来越多，越来越多人的理论知识的匮乏，联系实际能力的欠缺，逻辑思维能力的退化！
+
+人越来越容易产生分歧了，像电影《分歧者》中的场景越来越明显了，很多场景中悖论式的讨论越来越多，理智与非理智的对抗也越来越频繁。
diff --git a/cg/mesh/NURBS.md b/cg/mesh/NURBS.md
index 6d03135..66dfe0c 100644
--- a/cg/mesh/NURBS.md
+++ b/cg/mesh/NURBS.md
@@ -1,5 +1,5 @@
 # NURBS
->
+
 非均匀有理B样条, 这是一种数学表示3D几何的方法，能够准确描述从简单的2D线、圆、弧或曲线到非常复杂的3D自由曲面或实体的任何形状。如今，NURBS已成为CAD和计算机图形学的国际标准。
 
 ## [The Nurbs Book]()
@@ -7,6 +7,7 @@
 
 ## 参考
 
+- [OpenNURBS is an open-source NURBS-based geometric modeling library and toolset, with meshing and display / output functions. ](https://github.com/OpenNURBS/OpenNURBS)
 - [A small C++ library containing some NURBS-based geometric modeling algorithms and features I've been working on since 2013.](https://gitlab.com/ssv/Mobius)
 - [LNLib is a C++ NURBS Algorithms Library. These algorithms are primary referenced from The NURBS Book 2nd Edition. ](https://github.com/BIMCoderLiang/LNLib)
 - [LNLibViewer is a 3d viewer matches C++ NURBS algorithm library LNLib](https://github.com/BIMCoderLiang/LNLibViewer)
diff --git a/cg/threejs/grid.md b/cg/threejs/grid.md
index 5d1a8fb..c100b6f 100644
--- a/cg/threejs/grid.md
+++ b/cg/threejs/grid.md
@@ -12,4 +12,19 @@ const distance = camera.position.length();
 const logValue = 10 ** Math.ceil(Math.log(distance) / Math.log(10));
 (<ShaderMaterial>this.material).uniforms.uSize.value = logValue / this.scaleSize;
 // 对上面的无穷网格利用这个逻辑可以面前实现网格的变化
+```
+
+```c
+float getGrid(float size) {
+    vec2 r = worldPosition.${planeAxes} / size;
+    // 基于变量的平滑步函数smooth step function
+    // r-0.5 居中，使其范围在[0,1]
+    // fract(r - 0.5)使其范围在[-0.5, 0.5]
+    // fract(r - 0.5) - 0.5使其范围在[-1, 0]
+    // abs(fract(r - 0.5) - 0.5)使其范围在[0, 1]
+    // fwidth(r)返回屏幕空间宽度或模糊度
+    vec2 grid = abs(fract(r - 0.5) - 0.5) / fwidth(r);
+    float line = min(grid.x, grid.y);
+    return 1.0 - min(line, 1.0);
+}
 ```
\ No newline at end of file
diff --git a/cg/tools/glsl.md b/cg/tools/glsl.md
new file mode 100644
index 0000000..44f3406
--- /dev/null
+++ b/cg/tools/glsl.md
@@ -0,0 +1,11 @@
+# [GLSL]()
+
+## 内置函数
+在图形编程中，pow 函数常用于计算光照模型中的衰减、颜色混合的伽马校正、模拟物体表面的物理属性（如反射率、折射率等）等。
+```js
+// vndc = pos.xy / pos.w; from vertex
+// mouse = [2*offsetx/w-1, 1-2*offsety/h]
+// focused highlight
+float falloff = 10.0;
+float dif = pow(falloff, -clamp(length(mouse - vndc), 0.0, 1.0));
+```
\ No newline at end of file
diff --git a/dev-note/maven.md b/dev-note/maven.md
index 78206d1..334a9d3 100644
--- a/dev-note/maven.md
+++ b/dev-note/maven.md
@@ -1,6 +1,9 @@
-# Maven
+# [Maven](https://maven.apache.org/)
 
-maven 是打包工具，执行还是需要调用Java
+Apache Maven is a software project management and comprehension tool. Based on the concept of a project object model (POM), Maven can manage a project's build, reporting and documentation from a central piece of information. 
+
+maven是项目管理工具，将项目分成了三个生命周期，clean，default，site
+ 是打包工具，执行还是需要调用Java
 
 ## 命令
 
@@ -14,11 +17,14 @@ maven 是打包工具，执行还是需要调用Java
 - mvn install
 - mvn clean install -Dmaven.test.skip=true 
 
-### springboot
+### [Spring Boot Maven Plugin](https://docs.spring.io/spring-boot/maven-plugin/index.html)
+
+- [插件源码](https://github.com/spring-projects/spring-boot/tree/main/spring-boot-project/spring-boot-tools/spring-boot-maven-plugin)
 
 Spring Boot 通过 Spring Boot Maven Plugin 在 Apache Maven 中提供了对 Spring Boot 的支持。
 
 - mvn spring-boot:run
+- mvn spring-boot:repackage
 - mvn package 打包后用java -jar来运行
 
 调用springboot打包后的文件启动服务
diff --git a/exercises/advanced.mathematics.md b/exercises/advanced.mathematics.md
index d9fc73f..9c69964 100644
--- a/exercises/advanced.mathematics.md
+++ b/exercises/advanced.mathematics.md
@@ -1,22 +1,52 @@
 # 高等数学
 > 
 
-## 例题1 2024-5-17
+## 2024-7-10
+
+<details>
+<summary>例题2</summary>
+
+$$
+a_{1}=\sqrt(\frac{1}{2}), a_{n}=\sqrt{\frac{1+a_{n-1}}{2}}, \text{求} \lim\limits_{n \to \infty}a_{1}a_{2}a_{3}\cdots a_{n}
+$$
+
+求解：
+
+$$
+\text{由}a_{1}=\sqrt{\frac{1}{2}}, \text{不防设}a_{1}=cos{\frac{\theta}{2}}, \theta = \pi. \newline
+\text{则由}a_{2}=\sqrt{\frac{1+cos{\frac{\theta}{2}}}{2}}=\sqrt{\frac{1}{2}(1+cos{\frac{\theta}{2}})}=\sqrt{\frac{cos^2{\theta}}{2^2}}=\frac{cos\theta}{2^2} \newline
+\text{则可以推得} a_{n}=\frac{cos\theta}{2^n}. \text{则有} \newline
+\lim\limits_{n \to \infty}a_{1}a_{2}a_{3} \cdots a_{n} \newline
+= \lim\limits_{n \to \infty}\frac{cos\frac{\theta}{2} cos\frac{\theta}{2^2} cos\frac{\theta}{2^3} \cdots cos\frac{\theta}{2^n}sin\frac{\theta}{2^n}}{sin\frac{\theta}{2^n}} \newline
+= \lim\limits_{n \to \infty}\frac{cos\frac{\theta}{2} cos\frac{\theta}{2^2} cos\frac{\theta}{2^3} \cdots cos\frac{\theta}{2^(n-1)} sin\frac{\theta}{2^(n-1)}}{2sin\frac{\theta}{2^n}} \newline
+= \lim\limits_{n \to \infty}\frac{sin{\theta}}{2^n sin\frac{\theta}{2^n}} \text{有基本极限式子} \lim\limits_{x \to 0}\frac{sinx}{x}=1 \newline
+= \lim\limits_{n \to \infty}\frac{\frac{1}{2^n} sin{\theta}}{sin\frac{\theta}{2^n}} \newline
+= \frac{sin{\theta}}{\theta} \newline
+\text{因此} \lim\limits_{n \to \infty}a_{1}a_{2}a_{3}\cdots a_{n} = \frac{sin\frac{\pi}{2}}{\frac{\pi}{2}}=\frac{2}{\pi}
+$$
+
+</details>
+
+## 2024-5-17
+
+<details>
+<summary>例题1</summary>
 
 > **Note来源**
 > 吴康公众号
 
 $$
 \text{设}F(x) = sinx \cdot sin2x \cdots sin(2022x), \text{求}F^{2024}(0). \newline
-\text{首先，把公式一般化，就是}F(x) = \prod _{\substack{1 \le k \le n}} {sin(kx)}, n \in N^+. \text{求 }F^{n+2}(0)
+\text{首先，把公式一般化，就是}F(x) = \displaystyle \prod _{\substack{1 \le k \le n}} {sin(kx)}, n \in N^+. \text{求 }F^{n+2}(0)
 $$
 
 解：
 
 $$
 \text{引入定理，证略} \newline
-sinx = x - \frac{x^3}{6} + o(x^3), x \to 0 \text{带入可以得到} F(x) = \prod _{\substack{1 \le k \le n}} { \[ kx - \frac{kx^3}{6} + o(x^3) \]}, x \to 0 \newline
+sinx = x - \frac{x^3}{6} + o(x^3), x \to 0 \text{带入可以得到} F(x) = \displaystyle \prod _{\substack{1 \le k \le n}} { \[ kx - \frac{kx^3}{6} + o(x^3) \]}, x \to 0 \newline
 F(x) \text{的展开式中}x^{n+2}\text{项为} \newline
-\sum_{1 \le k \le n}{\[\prod_{1 \le k \le n}\]}(kx)^{-1} \cdot (-6^{-1})(kx)^3=-6^{-1}\lgroup{\sum_{1 \le k \le n}{k^2}}\rgroup n!x^{n+2} = -36^{-1}n(2n+1)(n+1)!(n+2)!
+ {\displaystyle \sum_{1 \le k \le n}}{\displaystyle \prod_{1 \le k \le n}}(kx)^{-1} \cdot (-6^{-1})(kx)^3=-6^{-1}\lgroup{\displaystyle \sum_{1 \le k \le n}{k^2}}\rgroup n!x^{n+2} = -36^{-1}n(2n+1)(n+1)!(n+2)!
 $$
 
+</details>
\ No newline at end of file
diff --git a/exercises/index.md b/exercises/index.md
index a6cbcd4..67f177e 100644
--- a/exercises/index.md
+++ b/exercises/index.md
@@ -37,4 +37,8 @@
 
 ### 感受
 
-- [2024](/exercises/2024.md)
\ No newline at end of file
+- [2024](/exercises/2024.md)
+
+<details>
+<summary>xxx</summary>
+</details>
diff --git a/exercises/linear.algebra.md b/exercises/linear.algebra.md
index 821ad3d..a2b16f0 100644
--- a/exercises/linear.algebra.md
+++ b/exercises/linear.algebra.md
@@ -69,4 +69,16 @@ AA^{T}=UDU^{T} \newline
 D\text{为特征值的对角矩阵，}V,U\text{为正交矩阵}
 $$
 
+[利用Weyl's不等式证明 Eckhart-Young定理](https://mp.weixin.qq.com/s/ErHlXWNvZC6vdPUZs2ycGQ)
+
+在现代机器学习中，降维处理数据中，很重要的工具也是SVD，即
+
+$$
+A_{m \times n}=U_{m \times m} \sum_{m \times n} V_{n \times n}^{T} = \begin{bmatrix} u_{1} & \cdots & u_{m} \end{bmatrix} \begin{bmatrix} \sigma_{1} & & & & \newline & \cdots & & & \newline & & \sigma_{r} & & \newline & & & \cdots & \newline & & & & 0 \end{bmatrix} \begin{bmatrix} v_{1}^T \newline \vdots \newline v_{n}^T \end{bmatrix} = \textstyle \sum_{i=1}^{r} \sigma^{i}u_{i}v_{i}^T
+$$
+
+可以看到右边式子中的项数是r，与矩阵的秩相同，实际应用中，r很大，一个想法就是降低其值，用一个秩较低的矩阵来近似。SVD分解就具有这样的特性：
+
+任何秩k小于r的截断分解都能以最佳方式近似A，这是Eckhart-Young 定理的内容：当B等于A的截断SVD 时，Frobenius 范数（衡量原始矩阵A与任何低秩矩阵B之间的差异）将最小化
+
 </details>