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## 2024-7-12 | ||
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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。 | ||
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## 2024-7-11 | ||
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《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。 | ||
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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! | ||
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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。 |
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# [GLSL]() | ||
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## 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)); | ||
``` |
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# 高等数学 | ||
> | ||
## 例题1 2024-5-17 | ||
## 2024-7-10 | ||
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<details> | ||
<summary>例题2</summary> | ||
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$$ | ||
a_{1}=\sqrt(\frac{1}{2}), a_{n}=\sqrt{\frac{1+a_{n-1}}{2}}, \text{求} \lim\limits_{n \to \infty}a_{1}a_{2}a_{3}\cdots a_{n} | ||
$$ | ||
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求解: | ||
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$$ | ||
\text{由}a_{1}=\sqrt{\frac{1}{2}}, \text{不防设}a_{1}=cos{\frac{\theta}{2}}, \theta = \pi. \newline | ||
\text{则由}a_{2}=\sqrt{\frac{1+cos{\frac{\theta}{2}}}{2}}=\sqrt{\frac{1}{2}(1+cos{\frac{\theta}{2}})}=\sqrt{\frac{cos^2{\theta}}{2^2}}=\frac{cos\theta}{2^2} \newline | ||
\text{则可以推得} a_{n}=\frac{cos\theta}{2^n}. \text{则有} \newline | ||
\lim\limits_{n \to \infty}a_{1}a_{2}a_{3} \cdots a_{n} \newline | ||
= \lim\limits_{n \to \infty}\frac{cos\frac{\theta}{2} cos\frac{\theta}{2^2} cos\frac{\theta}{2^3} \cdots cos\frac{\theta}{2^n}sin\frac{\theta}{2^n}}{sin\frac{\theta}{2^n}} \newline | ||
= \lim\limits_{n \to \infty}\frac{cos\frac{\theta}{2} cos\frac{\theta}{2^2} cos\frac{\theta}{2^3} \cdots cos\frac{\theta}{2^(n-1)} sin\frac{\theta}{2^(n-1)}}{2sin\frac{\theta}{2^n}} \newline | ||
= \lim\limits_{n \to \infty}\frac{sin{\theta}}{2^n sin\frac{\theta}{2^n}} \text{有基本极限式子} \lim\limits_{x \to 0}\frac{sinx}{x}=1 \newline | ||
= \lim\limits_{n \to \infty}\frac{\frac{1}{2^n} sin{\theta}}{sin\frac{\theta}{2^n}} \newline | ||
= \frac{sin{\theta}}{\theta} \newline | ||
\text{因此} \lim\limits_{n \to \infty}a_{1}a_{2}a_{3}\cdots a_{n} = \frac{sin\frac{\pi}{2}}{\frac{\pi}{2}}=\frac{2}{\pi} | ||
$$ | ||
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</details> | ||
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## 2024-5-17 | ||
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<details> | ||
<summary>例题1</summary> | ||
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> **Note来源** | ||
> 吴康公众号 | ||
$$ | ||
\text{设}F(x) = sinx \cdot sin2x \cdots sin(2022x), \text{求}F^{2024}(0). \newline | ||
\text{首先,把公式一般化,就是}F(x) = \prod _{\substack{1 \le k \le n}} {sin(kx)}, n \in N^+. \text{求 }F^{n+2}(0) | ||
\text{首先,把公式一般化,就是}F(x) = \displaystyle \prod _{\substack{1 \le k \le n}} {sin(kx)}, n \in N^+. \text{求 }F^{n+2}(0) | ||
$$ | ||
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解: | ||
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$$ | ||
\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)! | ||
$$ | ||
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</details> |
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### 感受 | ||
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- [2024](/exercises/2024.md) | ||
- [2024](/exercises/2024.md) | ||
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<details> | ||
<summary>xxx</summary> | ||
</details> |
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