(烂尾中, 更新遥遥无期的那种)【转】一口气搞定泰勒公式（泰勒展开式）的 本质 和 展开原则
虽然很欣喜有人阅读这些文字, 但是很遗憾因为个人原因没有更完.
虽然时间的原因是一方面, 但感觉主要原因是我对这方面内容仍不够透彻.
因此, 为了避免误人子弟,
我不建议您继续阅读本文. 我只将本文的原始markdown文本展示出来留作纪念, 仍坚持需要阅读原文可自行复制于md编辑器查看.
备注
- 文中对维基百科的引用可能 仅仅会加深读者对数学史的了解.
- 但是,如果您需要应对考试, 您不应看这篇文章.
2024年8月3日10:25:30
原始文本的markdown原文
# 一口气搞定泰勒公式（泰勒展开式）的**本质**和*展开原则*
Get The **Essence** and The *Expansion Principle* of Taylor formula (Taylor expansion formula) in One Sitting
> 来自 [bilibili-不看后悔 一口气搞定泰勒公式的本质及展开原则 ](https://www.bilibili.com/video/BV1WX4y1g7bx)
>本文将视频尽量以通俗语言转写为图文模式，以期便于快速阅读和学习。本文内容仅供学习参考。
[toc]
**开篇语**：*你*是谁呢？
**PREFACE**: Who are *you*?
| ![image][jpg1] | ![image][jpg2] |
|:-:|:-:|
| 洛神赋图 | 大展宏图 |
| *Someone who only learned how to fucking expand things by L 'Hopital's rule* | *Someone who only learned how to fucking expand things using Taylor's expansions* |
---
# 1. 泰勒展开式的**本质**
The **Essence** of Taylor expansion formula
> **严谨的名词解释**：*（你可以快速阅读这个备注）*
>
>在数学中，**泰勒级数**用无限项连加式——级数, 来表示一个函数，这些相加的项由函数在某一点的导数求得。泰勒级数是以于1715年发表了泰勒公式的英国数学家*布鲁克·泰勒*（Sir Brook Taylor）来命名的。通过函数在自变量零点的导数求得的泰勒级数又叫做**麦克劳林级数**，以苏格兰数学家*科林·麦克劳林*的名字命名。
>
>拉格朗日在1797年之前，最先提出带有余项的现在形式的泰勒定理。实际应用中，泰勒级数**需要截断**，只取有限项，可以用泰勒定理估算这种近似的误差。一个函数的有限项的泰勒级数叫做泰勒多项式。一个函数的泰勒级数是其泰勒多项式的极限（如果存在极限）。即使泰勒级数在每点都收敛，函数与其泰勒级数也可能不相等。在开区间（或复平面上的开区间）上，与自身泰勒级数相等的函数称为解析函数。**(来自[维基百科](https://zh.wikipedia.org/wiki/%E6%B3%B0%E5%8B%92%E7%BA%A7%E6%95%B0))**
>**A STRICT INTERPRETATION OF THE NOUN**: *(You can read this note quickly)*
>
> In mathematics, the **Taylor series of a function** is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point. Taylor series are named after *Brook Taylor*, who introduced them in 1715. A Taylor series is also called a **Maclaurin series**, when 0 is the point where the derivatives are considered, after *Colin Maclaurin*, who made extensive use of this special case of Taylor series in the mid-18th century.
>
>The partial sum formed by the first n + 1 terms of a Taylor series is a polynomial of degree n that is called *the nth Taylor polynomial of the function*. Taylor polynomials are approximations of a function, which become generally better as n increases. Taylor's theorem gives quantitative estimates on the error introduced by the use of such approximations. If the Taylor series of a function is convergent, its sum is the limit of the infinite sequence of the Taylor polynomials. A function may differ from the sum of its Taylor series, even if its Taylor series is convergent. A function is analytic at a point x if it is equal to the sum of its Taylor series in some open interval (or open disk in the complex plane) containing x. This implies that the function is analytic at every point of the interval (or disk). **(FROM [wikipedia](https://en.wikipedia.org/wiki/Taylor_series))**
---
>**关于名词使用的备注：** 本文中，利用**泰勒级数**的展开求解函数极限*的过程中*，泰勒级数的具体展开形式也被写作了 **“泰勒公式” “泰勒展开式”** 以及 **“泰勒展开”**。为了使得本文文字兼容更广泛的说法，本文在尽量较少歧义的情况下**混用**了这些名词，即使他们确实存在微小的意义区别。这一点请读者留意。
>**A NOTE ON THE USE OF NOUNS:** In this short article, *in the process of* solving the function limit by using the expansion of **Taylor series**, the specific expansion form of Taylor series is also written as **"Taylor formula", "Taylor expansion formula"** or **"Taylor expansion"**. In ORDER TO MAKE THE TEXT COMPATIBLE WITH a WIDER RANGE OF terms, the words are **MIXED** with as little ambiguity as possible, even if they do have minor differences in meaning. Please take note of this.
## 1.1 泰勒展开式
Taylor expansion formula
我们回忆一下什么是**泰勒展开式**：
Let's remember what **Taylor's expansion** is:
设 $ f(x) $ 在 $ x_0 $ 处，有 $ n $ 阶导数，则有公式：
Let $ f(x) $ have the derivative of $ n $ at $ x_0 $ , then we have the formula:
$$
f(x) = f(x_0)+{
f'(x_0) \over 1!
} (x - x_0) + {
f''(x_0) \over 2!
} (x - x_0)^2 + \dots + {
f^{(n)}(x_0) \over n!
}(x - x_0)^n + o((x-x_0)^n )
$$
**只**需要记住他是一个非常长的式子。
**Just** remember that it's a very long formula.
它利用**幂函数**的相加，来近似任意一个函数。就类似于：
It approximates any function by adding **powers**. Something like this:
$$ \pi = 3.14159 \dots $$
可以写作：
Can be written as:
$$ \pi = 3 + 0.1 + 0.04 + 0.001 + 0.0005 +\dots $$
## 1.2 麦克劳林展开式
McLaurin's expansion
设 $ f(x) $ 在 $ 0 $ 处，有 $ n $ 阶导数，则有公式：
Let $ f(x) $ have the derivative of $ n $ at $ 0 $ , then we have the formula:
$$
f(x) = f(0)+{
f'(0) \over 1!
} x + {
f''(0) \over 2!
} x^2 + \dots + {
f^{(n)}(0) \over n!
}x^n + o(x^n)
$$
## 1.3 一些常见的，算好的公式
Some formulas provided in advance, they are commonly used.
（方块即x）
$$
\begin{align*}
e^\square &= 1 + \square + {\square^2 \over 2!} + {\square^3 \over 3!}+ {\square^4 \over 4!} + o(\square^4)\\
\sin \square &= \square-{\square^3 \over 3!}+{\square^5 \over 5!}-{\square^7 \over 7!} + o(\square^8)\\
\cos \square &= 1-{\square^2 \over 2!}+{\square^4 \over 4!}-{\square^6 \over 6!} + o(\square^7)\\
（几何级数）：{1\over 1-\square} &= 1+\square+\square^2+\square^3+\square^4+ o(\square^4)\\
{1\over 1+\square}&=1-\square+\square^2-\square^3+\square^4+ \dots\\
由于 (\ln(1+\square))' &= {1\over 1+\square}\\
\ln(1+\square) &= \square - {\square^2\over 2} + {\square^3\over 3} - {\square^4\over 4} + \dots\\
{1 \over 1+\square^2} &= 1-\square^2+\square^4-\square^6+\square^8+\dots\\
由于 (\arctan \square)' &= {1 \over 1+\square^2} \\
\arctan \square &=\square-{\square^3 \over 3}+{\square^5 \over 5}-{\square^7 \over 7} + {\square^9 \over 9} + \dots\\
（二项式级数）：(1+\square)^\alpha &= 1+\alpha \square + {\alpha(\alpha-1)\over 2!} \cdot \square^2 + {\alpha(\alpha-1)(\alpha-2)\over 3!} \cdot \square^3 + \dots\\
\arcsin \square &= \square + {\square^3 \over 6}+o(\square^3) \\
\tan \square &= \square + {\square^3 \over 3}+o(\square^3)
\end{align*}
$$
## 1.4 你需要知道的三点总结
Three things you need to know
> 1. 等价无穷小的**本质**是*特殊的泰勒公式*（这不准确，但是现阶段你可以这样去近似的理解）
> 2. 泰勒公式计算的**本质**是*近似*
> 3. 洛必达计算的**本质**是*降阶*
> 1. The **essence** of the equivalent infinitesimal is a *special Taylor formula* (this is not exact, but you can approximate it this way at this stage).
> 2. The **essence** of Taylor's calculation is *approximation*
> 3. The **essence** of L 'Hopital computation is to *reduce the power*
# 2. 泰勒公式的使用原则
The principle of Taylor's formula
> 1. 乘除位置，使用**等价无穷小**
> 2. 加减位置，展开到上下**次方数相同**
> 3. 没有分母提示时，展开到前后**不能抵消的**最低次幂
> 1. Multiplication and division positions, using the **equivalent infinitesimal**
> 2. The positions of addition and subtraction are expanded to the **same power** in the numerator and denominator
> 3. Without a hint from the denominator, expand to the lowest power where the **front and back do not cancel**
> **备注：** 我们还记得，在学习等价无穷小的时候，被告知**加减不能用等价无穷小**，如果强行使用可能会招致计算错误。这是由于加减时，计算精度可能不够导致的。
>
> 令人振奋的是，我们正在学习如何避免这一问题。而泰勒公式可以*增加精度*，避免了这一问题。
>
> 因此，在**同时**出现加减法和乘除法时，应当对可展开的项**先使用泰勒展开**。即使这样做可能会显得麻烦一些，但这样做是正确的。
>
> 例如: $ \ln(1+x) - xe^{-{1 \over 2}x} $，你应该将它化为 $ (x-{x^2 \over 2}) - x\left(1 + ({(-{1 \over 2}x) \over 1}) + ({(-{1 \over 2}x)^2 \over 2}) \right)$, 而不是 $ (x-{x^2 \over 2}) - x\left(1+(-{1 \over 2}x)\right)$
> **NOTE:** We also remember that when we learned the equivalent infinitesimal, we were told that **addition and subtraction could not be used as the equivalent infinitesimal**, and that it might lead to calculation errors if forced to use it. This is due to the addition and subtraction, the calculation accuracy may not be enough.
>
> Encouragingly, we are learning how to avoid this problem. Taylor's formula can *increase the accuracy* and avoid this problem.
>
> Therefore, when **both** addition and subtraction and multiplication and division occur, **Taylor's expansion should be used first** for the expansible term. Even if it may seem cumbersome, it is the right thing to do.
>
> For example, for $ \ln(1+x) - xe^{-{1 \over 2}x} $ , you SHOULD convert it to $ (x-{x^2 \over 2}) - x\left(1 + ({(-{1 \over 2}x) \over 1}) + ({(-{1 \over 2}x)^2 \over 2}) \right)$, NOT $ (x-{x^2 \over 2}) - x\left(1+(-{1 \over 2}x)\right)$
Example 1:
$$
\lim\limits_{x \to 0}{
\frac{
\arcsin x \cdot \left(
\ln(1+x) -
xe^{
-{1 \over 2}x
}
\right)
} {
(1 - \cos \sqrt{x})
\tan x^3
}
}
$$
它告诉了我们泰勒展开和等价无穷小 的在使用上的*微小区别*。
It tells us the *slight difference* in use between the Taylor expansion and the equivalent infinitesimal.
未完待续 2022年10月31日11:04:29
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[jpg1]: https://img2022.cnblogs.com/blog/1912999/202210/1912999-20221031110523766-1909138815.jpg
[jpg2]: https://img2022.cnblogs.com/blog/1912999/202210/1912999-20221031110523805-91445474.jpg
