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# 2024 | ||
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## [数学手册(原书第10版)](https://book.douban.com/subject/35350415/) | ||
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有种观点,学习数学,起始是从自己开始的,并不是从书本上规划的路径学习起来的。能认识字后,找准自己的能理解的方向下去就可以了。我也不知道我是从那里看到这种观点的。 | ||
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佐证这种观点,是历史上很多数学家的传记上就是这样描述的逻辑,这样才开拓了新的数学分支。 | ||
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最近我的感受越来越支持这个观点,也是最近冒出一个人物,姜萍,读的中专,但是参加阿里巴巴全球数学竞赛得到12名的成绩,算是一个很努力的结果。 | ||
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借鉴成功的历史事件是很好的助力,但是原创性的东西必须是从第一性原理出发,自主思考才是最好也最佳的实践过程。如果还我重新来学习,我一定会遵守这个准则。 | ||
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第一性原理(First Principles)是一个哲学和科学概念,指的是基于最基本的、不可剥夺的事实或假设来推导复杂情况或理论的方法。第一性原理思维是一种强大的思考工具,它鼓励人们深入问题的本质,避免被传统思维或既定框架限制。通过这种方法,可以发现新的可能性和解决方案。然而,它也需要深入的专业知识和对基础概念的深刻理解。 |
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# 数值计算 | ||
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数值方法帮助了数学的发展,很多时候计算的人不关心是否存在根,而是用给定的方法,即算法,先计算出来看看结果才是重点,反过来又支持了算法的稳定性。 | ||
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## 夹逼法 | ||
> "夹逼法" 在数学中是一种求解数值的方法,特别是在逼近理论中。在英文中,它通常被翻译为 "squeezing theorem" 或 "sandwich theorem" | ||
>> 这个术语描述的是一种技巧,通过证明一个未知的数值位于两个已知数值之间,并且这两个已知数值可以无限逼近未知数值,从而证明未知数值的特定属性或精确值。 | ||
## 根式求解 | ||
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### 牛顿-拉夫森法 | ||
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牛顿-拉夫森法提供了一种非常有效的方法来寻找多项式的近似根,先假设一个根x,再此处画一条曲线的切线,并找出与该切线与X轴相交的点x1,这样重复下去就可以找到近似的根。用公式表示就是 | ||
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$$ | ||
\text{记}x_{n}\text{为当前根的猜测值,则可以通过计算下一个猜测值}x_{n+1} \newline | ||
x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'}(x_{n})} | ||
$$ | ||
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通常会收敛到最近的根,但也有例外,会导致得不到解 | ||
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- [牛顿拉夫森方法的意外之喜--分形图形](https://mp.weixin.qq.com/s/vjteWAtDAVHXfRwKE_DeSw) |
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# 多项式 | ||
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## 夹逼法 | ||
> "夹逼法" 在数学中是一种求解数值的方法,特别是在逼近理论中。在英文中,它通常被翻译为 "squeezing theorem" 或 "sandwich theorem" | ||
>> 这个术语描述的是一种技巧,通过证明一个未知的数值位于两个已知数值之间,并且这两个已知数值可以无限逼近未知数值,从而证明未知数值的特定属性或精确值。 | ||
## 二次方程QuadraticEquation | ||
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## 牛顿拉夫森法 | ||
### 一元二次不定方程 | ||
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牛顿-拉夫森法提供了一种非常有效的方法来寻找多项式的近似根,先假设一个根x,再此处画一条曲线的切线,并找出与该切线与X轴相交的点x1,这样重复下去就可以找到近似的根。用公式表示就是 | ||
在数学的历史中,不定方程指的是没有给出具体解法的方程,或者解的个数不是有限的,可能是无限多的解。 | ||
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$$ | ||
\text{记}x_{n}\text{为当前根的猜测值,则可以通过计算下一个猜测值}x_{n+1} \newline | ||
x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'}(x_{n})} | ||
\text{一元二次方程是指最高次项为二次的单变量(一个未知数)方程,形式通常为} \newline | ||
ax^2+bx+c=0, \text{,其中 a、b、和c是已知数,而x是未知数。当这种方程的系数a、b、和c是变量而不是具体的数值时,我们称之为不定方程。} | ||
$$ | ||
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通常会收敛到最近的根,但也有例外,会导致得不到解 | ||
<details> | ||
<summary>例题1 2024-5-16</summary> | ||
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- [牛顿拉夫森方法的意外之喜--分形图形](https://mp.weixin.qq.com/s/vjteWAtDAVHXfRwKE_DeSw) | ||
$\text{已知A、n为正整数,满足}, A = (n-7)(n+8), \text{且A为完全平方数,那么n有最几个值,n的最小值是多少,n的最大值是多少?}$ | ||
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分析:因为n在变化,可以把A缩放在某个区间来解题。使用换元法替换一个变量,这样就更容易理解了。 | ||
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解: | ||
$$ | ||
\text{设}k=n-7\text{,则} \newline | ||
A = (n-7)(n+8) \to A = k(k+15) = k^2 + 15k < k^2 + 16k < k^2 + 16k + 4 = (k+8)^2 \newline | ||
\because A=k^2 + 15k \text{是完全平方数,所以} k^2+15 \text{必然在两个完全平方数A和} (k+8)^2 \text{之间。} | ||
$$ | ||
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这样就可以一一判断两个等式之间是否成立了。 | ||
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$$ | ||
k^2+15k=(k+1)^2 \to k=\frac{1}{13} \newline | ||
k^2+15k=(k+2)^2 \to k=\frac{4}{11} \newline | ||
k^2+15k=(k+3)^2 \to k=1 \to n=8 \to A=(8-7)(8+8)=1 \cdot 16=16=4^2 \newline | ||
k^2+15k=(k+4)^2 \to k=\frac{16}{7} \newline | ||
k^2+15k=(k+5)^2 \to k=5 \to n=12 \to A=(12-7)(12+8)=5 \cdot 20=100=10^2 \newline | ||
k^2+15k=(k+6)^2 \to k=12 \to n=19 \to A=(19-7)(19+8)=12 \cdot 27 =2^2 \cdot 3 \cdot 3 \cdot 3^2=(2 \cdot 3 \cdot 3)^2 = 18^2 \newline | ||
k^2+15k=(k+7)^2 \to k=49 \to n=56 \to A=(56-7)(56+8)=49 \cdot 64 =7^2 \cdot 8^2=(7 \cdot 8)^2 = 56^2 | ||
$$ | ||
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综上所述,n有4个值,最小的是8,最大的是56. | ||
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</details> | ||
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<details> | ||
<summary>例题2 2024-5-16</summary> | ||
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$\text{已知} n^3 + 2n^2 + 8n - 5 \text{是一个正整数的立方,则正整数的n的值可能是}$ | ||
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解: | ||
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$$ | ||
n^3 < n^3 + 2n^2 + 8n - 5 < n^3 + 6n^2 + 12n + 8 = (n+2)^3 \newline | ||
\text{可推出} \\\\ | ||
n^3 + 2n^2 + 8n - 5 = (n+1)^3 \to n^2 - 5n + 6 = 0 \to \{因式分解可得n=2或3} | ||
$$ | ||
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$$ | ||
\text{把结果待人方程也可以验证} \newline | ||
\therefore 2^3 + 2 \cdot 2^2 + 8 \cdot 2 - 5 = 27 = 3^3 \newline | ||
\therefore 3^3 + 2 \cdot 3^2 + 8 \cdot 3 - 5 = 64 = 4^3 | ||
$$ | ||
</details> |
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