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Taylor Series About Other Points | 关于其他点的泰勒级数

2026-08-03 01:59 · 本文部分内容由AI辅助创作

📚 Taylor Series About Other Points | 关于其他点的泰勒级数

situations, we need to expand a function about a different point to obtain a series that converges more rapidly or to approximate values far from the origin. This article explores Taylor series about other points, their formulas, error analysis, and applications, tailored for the IB Mathematics Analysis and Approaches HL curriculum.

泰勒级数是将函数表示为无穷级数的有力工具，由函数在某一点的导数值计算各项系数。学生最先接触的通常是麦克劳林级数（在 x = 0 处展开）。然而，在许多实际问题中，我们需要在其他点上展开函数，以便获得收敛更快的级数，或近似计算远离原点的函数值。本文探讨在其他点展开的泰勒级数，涵盖公式、误差分析与应用，紧扣 IB 数学分析与方法 HL 大纲。

1. Review of Taylor Polynomials and Series | 泰勒多项式与级数回顾

nomial of degree n for a function f(x) about x = a is constructed by matching the function's value and its first n derivatives at x = a. The formula is Pn(x) = f(a) + f'(a)(x − a) + (f''(a)/2!)(x − a)2 + … + (f(n)(a)/n!)(x − a)n infinitely differentiable and the remainder tends to zero, the infinite Taylor series converges to the function.

函数 f(x) 在 x = a 处的 n 次泰勒多项式通过匹配函数值及其在 a 点的前 n 阶导数值来构造。公式为 Pn(x) = f(a) + f'(a)(x − a) + (f''(a)/2!)(x − a)2 + … + (f(n)(a)/n!)(x − a)n。随着 n 增大，多项式在 a 附近更好地逼近 f(x)。如果 f 无限可导且余项趋于零，则无穷泰勒级数收敛到原函数。

ift is essential when the function or its derivatives are undefined at 0, or when we need high accuracy around a non-zero value.

在 IB 数学中，通常通过重复求导来求麦克劳林级数，但同样的原理适用于任意中心 x = a。这种转移对于函数或其导数在 0 处无定义，或需要在非零点附近获取高精度近似时十分关键。

2. General Formula: Taylor Series About x = a | 一般公式：在 x = a 处的泰勒级数

unction f(x) about x = a (also called the expansion about a) is given by the infinite sum:

函数 f(x) 在 x = a 处的一般泰勒级数（也称关于 a 点的展开）由以下无穷级数给出：

f(x) = Σn=0∞ (f(n)(a) / n!) (x − a)n

Here f(n)(a) denotes the n-th derivative evaluated at x = a, with the convention that f(0)(a) = f(a) and 0! ing the terms involve powers of (x − a). When a = 0, this reduces to the Maclaurin series.

这里 f(n)(a) 表示在 x = a 处的 n 阶导数值，约定 f(0)(a) = f(a) 且 0! = 1。该级数以 a 为中心，各项包含 (x − a) 的幂。当 a = 0 时，即为麦克劳林级数。

To write the series correctly, we must compute the derivatives of f at the chosen point a. The series representation is valid only for x within the interval of convergence, which is centred at a.

要正确写出级数，我们必须计算函数在选定点 a 处的各阶导数。级数表示仅在以 a 为中心的收敛区间内有效。

3. Why Expand About Other Points? | 为何要在其他点展开？

There are several compelling reasons to use a Taylor series about a point other than zero:

有以下几个重要理由需要在非零点展开泰勒级数：

By shifting the centre a, we effectively translate the power series to better fit the local behaviour of the function. This is analogous to zooming in on a different part of the graph.

通过平移中心 a，我们可以有效地将幂级数转移，以更好地匹配函数的局部行为。这就像将图像放大到图形的另一部分。

4. Change of Variable: Reducing to Maclaurin Form | 变量替换：化归为麦克劳林形式

Then x = t + a, and we can find the Maclaurin series in t. That is, g(t) = f(t + a) = Σ (g(n)(0)/n!) tn. After expanding, substitute back t = x − a to obtain the series in terms of x.

得到在 x = a 处泰勒级数的一个便捷技巧是定义新变量 t = x − a。那么 x = t + a，我们就可以求关于 t 的麦克劳林级数。即 g(t) = f(t + a) = Σ (g(n)(0)/n!) tn。展开后，将 t = x − a 代回，得到关于 x 的级数。

ly, as the derivatives are evaluated at 0 in the t-domain. For example, to expand f(x) = ex about x = 1, set t = x − 1, then g(t) = et+1 = e·et. The Maclaurin series for et is known, so the desired series is e·ex−1 = Σ e·(x−1)n/n!.

这种方法常常能极大简化代数运算，因为在 t 域中导数在 0 处计算。例如，要展开 f(x) = ex 关于 x = 1，令 t = x − 1，则 g(t) = et+1 = e·et。已知 et 的麦克劳林级数，于是所需级数为 e·ex−1 = Σ e·(x−1)n/n!。

5. Computing Coefficients by Direct Differentiation | 通过直接求导计算系数

uate at x = a, we can compute the Taylor coefficients directly. The coefficient of (x − a)n is an = f(n)(a) / n! The steps are: differentiate repeatedly, substitute x = a, and divide by n factorial. Let's illustrate with f(x) = ln(x) about a = 2. We have f'(x) = 1/x, f''(x) = −1/x2, f'''(x) = 2/x3, f(4)(x) = −6/x4, and in general f(n)(x) = (−1)n−1(n−1)! / xn for n ≥ 1. At a = 2: f(n)(2) = (−1)n−1(n−1)! / 2n. Then the coefficient an = (−1)n−1(n−1)! / (n! 2n) = (−1)n−1 / (n 2n). The series is ln(2) + Σn=1∞ (−1)n−1/(n 2n) (x − 2)n.

当函数及其导数在 x = a 处容易求值时，可以直接计算泰勒系数。 (x − a)n 的系数为 an = f(n)(a)/n!。步骤为：重复求导，代入 x = a，再除以 n 的阶乘。以 f(x) = ln(x) 在 a = 2 处展开为例。f'(x) = 1/x，f''(x) = −1/x2，f'''(x) = 2/x3，f(4)(x) = −6/x4，一般地当 n ≥ 1 时 f(n)(x) = (−1)n−1(n−1)!/xn。在 a = 2 处：f(n)(2) = (−1)n−1(n−1)!/2n。于是系数 an = (−1)n−1(n−1)!/(n! 2n) = (−1)n−1/(n 2n)。级数为 ln(2) + Σn=1∞ (−1)n−1/(n 2n) (x − 2)n。

This direct method is systematic and works for any infinitely differentiable function. For trigonometric functions, be mindful of the cyclic patterns of derivatives when evaluating at non-zero angles like π/6 or π/4.

这种直接方法系统且适用于任何无限可导函数。对于三角函数，在非零角如 π/6 或 π/4 处求导时，需注意导数的循环规律。

6. Worked Examples: Common Functions | 典型例题：常见函数展开

Example 1: f(x) = sin(x) about a = π/3. Derivatives cycle: f'(x)=cos(x), f''(x)=−sin(x), f'''(x)=−cos(x), f(4)(x)=sin(x). At π/3: sin(π/3)=√3/2, cos(π/3)=1/2, −sin(π/3)=−√3/2, −cos(π/3)=−1/2. The Taylor series is: √3/2 + (1/2)(x−π/3) − (√3/4)(x−π/3)2 − (1/12)(x−π/3)3 + ...

例1：f(x) = sin(x) 在 a = π/3 处展开。 导数循环：f'(x)=cos(x)，f''(x)=−sin(x)，f'''(x)=−cos(x)，f(4)(x)=sin(x)。在 π/3 处：sin(π/3)=√3/2，cos(π/3)=1/2，−sin(π/3)=−√3/2，−cos(π/3)=−1/2。泰勒级数为：√3/2 + (1/2)(x−π/3) − (√3/4)(x−π/3)2 − (1/12)(x−π/3)3 + ...

Example 2: f(x) = 1/x expanded about a = 1. Derivatives: f'(x)=−1/x2, f''(x)=2/x3, f'''(x)=−6/x4, f(n)(x) = (−1)n n! / xn+1. At a=1, f(n)(1) = (−1)n n!. The coefficient an = (−1)n. Hence the series is 1 − (x−1) + (x−1)2 − (x−1)3 + ... = Σn=0∞ (−1)n (x−1)n.

例2：f(x) = 1/x 在 a = 1 处展开。 导数：f'(x)=−1/x2，f''(x)=2/x3，f'''(x)=−6/x4，f(n)(x) = (−1)n n! / xn+1。在 a=1 处，f(n)(1) = (−1)n n!。系数 an = (−1)n。因此级数为 1 − (x−1) + (x−1)2 − (x−1)3 + ... = Σn=0∞ (−1)n (x−1)n。

Notice that this series is a geometric series with ratio −(x−1). It converges for |x−1| < 1. This highlights how the centre a determines the interval of convergence.

注意此级数为公比 −(x−1) 的几何级数。收敛于 |x−1| < 1。这显示了中心 a 如何决定收敛区间。

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7. Interval and Radius of Convergence About x = a | 关于 x = a 的收敛区间与收敛半径

series in (x − a). Its radius of convergence R can be found using the ratio test: R = limn→∞ |an/an+1| erval is (0, 4] (check endpoints separately). For 1/x about a = 1, R = 1, interval (0, 2).

在 x = a 处的泰勒级数是关于 (x − a) 的幂级数。其收敛半径 R 可用比值判别法求得：若极限存在，R = limn→∞ |an/an+1|。收敛区间为 (a − R, a + R)，根据端点处的收敛性可能包含端点。例如，ln(x) 在 a = 2 的级数半径为 2，区间为 (0, 4]（需单独判别端点）。对于 1/x 在 a = 1 处，R = 1，区间为 (0, 2)。

Shifting the centre changes the interval. Expanding about a larger a can produce a series that converges over a range that does not include 0, which is useful when the function has a singularity at 0, as seen with 1/x and ln(x). Always specify the interval of convergence in IB exam answers.

移动中心会改变收敛区间。以更大的 a 为中心展开可以得到不包含 0 的收敛范围，当函数在 0 处有奇点时十分有用，例如 1/x 和 ln(x)。IB 考试答案中务必写明收敛区间。

8. Taylor's Theorem and the Lagrange Remainder | 泰勒定理与拉格朗日余项

To use a Taylor polynomial for approximation, we need to bound the error. Taylor's theorem states that for a function f with n+1 continuous derivatives on an interval containing a and x, there exists a number c between a and x such that the remainder Rn(x) = f(x) − Pn(x) can be written as:

Rn(x) = (f(n+1)(c) / (n+1)!) (x − a)n+1

This is the Lagrange form of the remainder. It allows us to estimate the maximum possible error over an interval, which is crucial in numerical approximations like estimating sin(61°) using a third-degree polynomial about π/3.

要用泰勒多项式进行近似，需要限定误差范围。泰勒定理指出：若函数 f 在包含 a 和 x 的区间上有 n+1 阶连续导数，则存在介于 a 与 x 之间的数 c，使得余项 Rn(x) = f(x) − Pn(x) 可写为：

Rn(x) = (f(n+1)(c) / (n+1)!) (x − a)n+1

此即拉格朗日余项。它使我们能够在整个区间上估计最大可能误差，这在数值近似中至关重要，例如利用三次泰勒多项式计算 sin(61°) 时估计误差。

In IB problems, you may be asked to find the maximum error of a given Taylor polynomial on a specified interval by maximising |f(n+1)(x)|. For instance, to approximate e1.1 using the expansion of ex about a = 1 to the 2nd degree, the error term involves the 3rd derivative ec with c ∈ [1, 1.1], so the maximum error is bounded by e1.1·(0.1)3/6.

在 IB 题目中，可能会要求通过最大化 |f(n+1)(x)| 来求特定泰勒多项式在给定区间上的最大误差。例如，用 ex 在 a = 1 处的二次展开近似 e1.1，误差项涉及三阶导数 ec，c ∈ [1, 1.1]，因此最大误差受限于 e1.1·(0.1)3/6。

9. Practical Approximation and Error Bounds | 实际近似与误差界限

When using a truncated series, we compute an approximate value and an error bound. Let's approximate √4.2 using f(x) = √x expanded about a = 4. Derivatives: f'(x) = 1/(2√x), f''(x) = −1/(4x3/2). P1(x) = 2 + (1/4)(x−4). At x = 4.2, P1(4.2) = 2 + 0.05 = 2.05. For the error, the second derivative term gives R1(x) = f''(c)/2! (0.2)2 with c ∈ [4, 4.2]. |f''(c)| ≤ 1/(4·43/2) = 1/32, so |R1| ≤ (1/32)/2 × 0.04 = 0.000625. Thus √4.2 ≈ 2.05 with error less than 0.00063.

使用截断级数时，我们计算近似值和误差界限。让我们用 f(x) = √x 在 a = 4 处展开来近似 √4.2。导数为：f'(x) = 1/(2√x)，f''(x) = −1/(4x3/2)。P1(x) = 2 + (1/4)(x−4)。在 x = 4.2 时，P1(4.2) = 2 + 0.05 = 2.05。至于误差，二阶导数项给出 R1(x) = f''(c)/2! (0.2)2，c ∈ [4, 4.2]。|f''(c)| ≤ 1/(4·43/2) = 1/32，故 |R1| ≤ (1/32)/2 × 0.04 = 0.000625。因此 √4.2 ≈ 2.05，误差小于 0.00063。

Increasing the degree and choosing a centre close to the evaluation point drastically improves accuracy. This is why Taylor series about other points are preferred for specific numerical tasks.

增加次数并选择靠近求值点的中心可大幅提高精度。这就是为何在其他点展开的泰勒级数在特定数值任务中更受青睐。

10. Operations on Series and Composition | 级数运算与复合

using substitution, multiplication, differentiation, and integration. For example, given ex about a = 1: e·ex−1 = e Σn=0∞ (x−1)n/n!, we can find e2x about a = 1 by replacing x with 2x, but careful: the series is then in powers of (2x − 1)? Actually, it is easier to expand e2x directly about 1. However, composition with linear functions: to expand g(x) = e(x+3) about a = 1, set t = x−1, g = et+4 = e4 et, so series is e4 Σ (x−1)n/n!. This flexibility is a key exam skill.

一旦有了在 x = a 处的基本泰勒级数，便可通过代换、乘法、微分和积分进行操作。例如，已知 ex 在 a = 1 处的级数：e·ex−1 = e Σn=0∞ (x−1)n/n!，我们可将 x 替换为 2x 来求 e2x 在 a = 1 附近的展开，但需谨慎：所得级数以 (2x−1) 的幂表示？实际上，直接展开 e2x 更简单。不过与线性函数复合：要展开 g(x) = e(x+3) 关于 a = 1，令 t = x−1，则 g = et+4 = e4 et，故级数为 e4 Σ (x−1)n/n!。这种灵活操作是重要的应试技巧。

ion u = x+1 transforms 1/(1+x) = 1/u about u = 0, then expand about u = 0, yielding Σ (−1)n un, then back substitute.

另一个常见技巧是逐项积分得到新的级数。例如，1/(1+x) 在 a = 0 的级数已知；将中心移到 a = −1 可得关于 x = −1 的级数，但需注意定义域。设 u = x+1，则 1/(1+x) = 1/u 关于 u = 0，展开得 Σ (−1)n un，再回代。

11. Applications: Solving Limits and Differential Equations | 应用实例：求极限与解微分方程

Taylor series about an appropriate point can evaluate limits where direct substitution yields an indeterminate form. For example, limx→1 (ln(x)) / (x−1). Expanding ln(x) about a = 1: ln(x) = (x−1) − (x−1)2 ... , and the limit is 1. Here the choice a = 1 is natural because the limit approaches 1.

在一定点展开的泰勒级数可用于求直接代入得到不定型的极限。例如 limx→1 (ln(x))/(x−1)。在 a = 1 处展开 ln(x)：ln(x) = (x−1) − (x−1)2/2 + ...。于是比值变为 1 − (x−1)/2 + ...，极限为 1。这里选择 a = 1 很自然，因为极限趋近于 1。

e a power series solution about a point (often an initial condition). For example, solve y' = y + x with y(1) = 2 using a series about x = 1. Let y = Σ an(x−1)n, then differentiate and equate coefficients. The IB syllabus includes such series solutions for simple first-order equations.

在微分方程中，若找不到闭式解，可假设关于某点（通常为初始条件）的幂级数解。例如，求解 y' = y + x，y(1) = 2，用关于 x = 1 的级数。设 y = Σ an(x−1)n，求导后比较系数。IB 大纲包含此类简单一阶方程的级数解。

12. Summary and IB Exam Tips | 总结与 IB 考试要点

Taylor series about other points generalise the Maclaurin series. Key takeaways:

在其他点展开的泰勒级数是麦克劳林级数的推广。要点如下：