| 【转】一口气搞定泰勒公式（泰勒展开式）的 本质 和 展开原则 | 您所在的位置：网站首页 › √x泰勒展开式 › 【转】一口气搞定泰勒公式（泰勒展开式）的 本质 和 展开原则 |
【转】一口气搞定泰勒公式（泰勒展开式）的 本质 和 展开原则
|
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 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。 目录一口气搞定泰勒公式（泰勒展开式）的本质和展开原则1. 泰勒展开式的本质1.1 泰勒展开式1.2 麦克劳林展开式1.3 一些常见的，算好的公式1.4 你需要知道的三点总结2. 泰勒公式的使用原则开篇语：你是谁呢？ PREFACE: Who are you? 
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) 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) 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: \[f(x) = f(x_0)+{ f'(x_0) \over 1! } (x - x_0) + { f''(x_0) \over 1! } (x - x_0)^2 + \dots + { f^{(n)}(x_0) \over n! }(x - x_0)^n + o((x-x_0)^n ) \]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: \[f(x) = f(0)+{ f'(0) \over 1! } x + { f''(0) \over 1! } x^2 + \dots + { f^{(n)}(0) \over n! }x^n + o(x^n) \]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 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） 泰勒公式计算的本质是近似 洛必达计算的本质是降阶 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). The essence of Taylor's calculation is approximation 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 乘除位置，使用等价无穷小 加减位置，展开到上下次方数相同 没有分母提示时，展开到前后不能抵消的最低次幂 Multiplication and division positions, using the equivalent infinitesimal The positions of addition and subtraction are expanded to the same power in the numerator and denominator 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 |
【本文地址】