Taylor Series at Other Points — IB数学：其他点处的泰勒级数

Taylor Series at Other Points: Expanding Functions About x = a | IB数学：其他点处的泰勒级数

泰勒级数是 IB DP 数学分析与方法（AA）HL 课程中最重要的工具之一。大多数同学都熟悉麦克劳林级数，也就是在 x = 0 处展开的级数，例如 e^x = 1 + x + x^2/2! + x^3/3! + …。但考试真正拉开差距的地方，往往在于”在其他点处展开”。本文将系统讲解泰勒级数的一般公式、展开步骤、典型例题、收敛区间以及 IB 考试中的高频陷阱。

A Taylor series is one of the most important tools in the IB DP Mathematics: Analysis and Approaches (AA) HL course. Most students are comfortable with Maclaurin series, which expand a function at x = 0, such as e^x = 1 + x + x^2/2! + x^3/3! + … . However, the questions that truly separate top scorers in exams are often those asking for expansions “about points other than zero”. This article explains the general Taylor formula, the step-by-step expansion procedure, worked examples, the interval of convergence, and the common traps that appear in IB examinations.

1. The General Formula of a Taylor Series: Derivatives at x = a | 泰勒级数的一般公式：在 x = a 处的各阶导数

对于在点 x = a 附近可以无限次求导的函数 f(x)，泰勒级数把它表示成关于 (x – a) 的幂的无穷和：f(x) = f(a) + f'(a)(x – a) + f”(a)/2! * (x – a)^2 + f”'(a)/3! * (x – a)^3 + … 。写成紧凑的求和形式：f(x) = 求和符号 n=0 到无穷 f^(n)(a)/n! * (x – a)^n，其中 f^(n)(a) 表示 f 在点 a 处的 n 阶导数，并约定 f^(0)(a) = f(a)，0! = 1。

For a function f(x) that can be differentiated infinitely many times near a point x = a, the Taylor series expresses it as an infinite sum of powers of (x – a): f(x) = f(a) + f'(a)(x – a) + f”(a)/2! * (x – a)^2 + f”'(a)/3! * (x – a)^3 + … . In compact summation notation: f(x) = sum from n = 0 to infinity of f^(n)(a)/n! * (x – a)^n, where f^(n)(a) denotes the n-th derivative of f evaluated at a, with the conventions f^(0)(a) = f(a) and 0! = 1.

理解这个公式的关键在于两点。第一，每一项都包含因子 (x – a)^n，这意味着级数是围绕点 a 构造的，而不是围绕原点；当 x 离 a 越近，(x – a)^n 衰减得越快，级数的前几项就越能精确地代表函数值。第二，系数 f^(n)(a)/n! 完全由函数在单一点 a 处的信息决定，这体现了泰勒级数”用局部信息描述整体行为”的深刻思想。

There are two keys to understanding this formula. First, every term contains the factor (x – a)^n, which means the series is built around the point a rather than the origin; the closer x is to a, the faster (x – a)^n decays, and the more accurately the leading terms represent the function value. Second, the coefficients f^(n)(a)/n! are determined entirely by information at the single point a, which reflects the profound idea that a Taylor series uses local information to describe global behaviour.

2. Maclaurin vs Taylor: The Expansion Point Makes the Difference | 麦克劳林级数与泰勒级数的区别：展开点不同

麦克劳林级数只是泰勒级数在 a = 0 时的特殊情况。学生熟记的那些标准展开式，e^x = 1 + x + x^2/2! + x^3/3! + …、sin x = x – x^3/3! + x^5/5! – …、cos x = 1 – x^2/2! + x^4/4! – …、ln(1 + x) = x – x^2/2 + x^3/3 – …，全部是展开中心位于 0 的结果。在 IB 考试中，题目一旦写明”about x = 1″、”about x = pi/4″或”at the point a”，就必须使用一般泰勒公式，而不是直接套用麦克劳林展开式。

A Maclaurin series is simply the special case of a Taylor series with a = 0. The standard expansions that students memorise, e^x = 1 + x + x^2/2! + x^3/3! + … , sin x = x – x^3/3! + x^5/5! – … , cos x = 1 – x^2/2! + x^4/4! – … , and ln(1 + x) = x – x^2/2 + x^3/3 – … , are all results centred at 0. In IB examinations, once a question states “about x = 1”, “about x = pi/4”, or “at the point a”, you must apply the general Taylor formula rather than quoting a Maclaurin expansion directly.

把展开中心从 0 移到任意点 a，这一改动看起来微小，却能解决两类关键问题。第一类：函数在 x = 0 处根本没有定义，例如 ln x 和 1/x，它们不可能有麦克劳林级数，但可以在 x = 1 等有定义的点处展开。第二类：我们希望近似在某个特定点附近最精确，例如物理问题中在平衡位置附近展开势能函数，或需要快速估算某个数值（如 e^0.5 或 sin 1）时，选择离目标最近的展开点能显著提高精度。

Moving the expansion centre from 0 to an arbitrary point a looks like a minor change, but it solves two crucial classes of problems. First, some functions are not even defined at x = 0, such as ln x and 1/x; they cannot have a Maclaurin series, yet they can be expanded at defined points like x = 1. Second, we often want the approximation to be most accurate near a specific point, for example expanding a potential energy function about an equilibrium position in physics, or estimating a value such as e^0.5 or sin 1 quickly; choosing the expansion point closest to the target dramatically improves the accuracy.

3. Step-by-Step Method: Five Steps to Expand at x = a | 分步方法：在 x = a 处展开的五步流程

在 x = a 处展开泰勒级数的过程可以归纳为五步。第一步，选定展开点 a 并计算 f(a)。第二步，依次求各阶导数 f'(x)、f”(x)、f”'(x)、…，直到所需的阶数。第三步，把 x = a 代入每一阶导数，得到数值 f'(a)、f”(a)、f”'(a)、…。第四步，把得到的数值代入泰勒公式，按 (x – a) 的升幂写出前几项。第五步，观察系数规律，写成求和符号形式（若题目要求），并注明级数有效的范围。

The process of expanding a Taylor series at x = a can be organised into five steps. Step one: choose the expansion point a and evaluate f(a). Step two: compute successive derivatives f'(x), f”(x), f”'(x), … up to the required order. Step three: substitute x = a into each derivative to obtain the numbers f'(a), f”(a), f”'(a), … . Step four: substitute these numbers into the Taylor formula and write out the leading terms in ascending powers of (x – a). Step five: recognise the pattern and write the result in summation notation if requested, then state the range of validity.

为加快计算速度，下列常用导数必须熟练掌握，它们是展开的基础：

To speed up the computation, you must be fluent with the following standard derivatives, which form the foundation of every expansion:

函数 f(x)	导数 f'(x)	n 阶导数 f^(n)(x)
e^x	e^x	e^x（所有阶导数相同）
sin x	cos x	以 sin、cos、-sin、-cos 四阶循环
cos x	-sin x	以 cos、-sin、-cos、sin 四阶循环
ln x	1/x	(-1)^(n-1) * (n-1)! / x^n
1/(1 + x)	-1/(1 + x)^2	(-1)^n * n! / (1 + x)^(n+1)

表中的规律值得记忆：指数函数求导不变，所以展开最省力；正弦和余弦的导数每四阶循环一次，因此系数呈现 +、- 交替的规律；ln x 和 1/(1 + x) 的 n 阶导数带有 (n – 1)! 或 n! 因子，代入泰勒公式后恰好与分母 n! 相消，产生形式简洁的级数。

The patterns in the table are worth memorising: the exponential function is its own derivative, so it is the easiest to expand; the derivatives of sine and cosine cycle every four orders, producing alternating signs in the coefficients; the n-th derivatives of ln x and 1/(1 + x) carry factors of (n – 1)! or n!, which cancel exactly with the n! in the denominator of the Taylor formula, yielding neatly structured series.

4. Worked Example 1: Expanding e^x About x = 1 | 例题一：把 e^x 在 x = 1 处展开

题目：求 f(x) = e^x 在 x = 1 处的泰勒级数，写出前四项。由于 e^x 的所有阶导数都是 e^x，代入 x = 1 后得到 f^(n)(1) = e 对一切 n 成立。于是泰勒公式给出 e^x = e + e(x – 1) + e/2! * (x – 1)^2 + e/3! * (x – 1)^3 + … 。把 e 提出来，就得到 e^x = e * [1 + (x – 1) + (x – 1)^2/2 + (x – 1)^3/6 + …]。

Question: find the Taylor series of f(x) = e^x about x = 1, up to the fourth term. Since every derivative of e^x is e^x itself, substituting x = 1 gives f^(n)(1) = e for all n. The Taylor formula then yields e^x = e + e(x – 1) + e/2! * (x – 1)^2 + e/3! * (x – 1)^3 + … . Factoring out e gives e^x = e * [1 + (x – 1) + (x – 1)^2/2 + (x – 1)^3/6 + …].

这个结果可以和麦克劳林级数对照理解。麦克劳林展开 e^x = 1 + x + x^2/2 + x^3/6 + … 的系数是 1、1、1/2、1/6，而这里每项多乘了一个常数 e，并且幂的底从 x 变成了 (x – 1)。数值验证：取 x = 0.5，它距离展开点 1 有 0.5 个单位。前三项给出 e^0.5 约等于 e * (1 – 0.5 + 0.125) = 2.71828 * 0.625 = 1.6989，真实值 1.6487；加上第四项 e * (-0.0208) 后约为 1.6424，与真实值的误差缩小到 0.4% 以内。这说明在展开点附近，级数收敛得很快。

This result can be understood in parallel with the Maclaurin series. The Maclaurin expansion e^x = 1 + x + x^2/2 + x^3/6 + … has coefficients 1, 1, 1/2, 1/6, whereas here every term carries an extra factor of e and the power base changes from x to (x – 1). Numerical check: take x = 0.5, which is 0.5 units away from the expansion point 1. The first three terms give e^0.5 approximately equal to e * (1 – 0.5 + 0.125) = 2.71828 * 0.625 = 1.6989, while the true value is 1.6487; adding the fourth term e * (-0.0208) gives about 1.6424, bringing the error below 0.4%. This shows how quickly the series converges near the expansion point.

5. Worked Example 2: Expanding sin x About x = pi/4 | 例题二：把 sin x 在 x = pi/4 处展开

题目：求 sin x 在 x = pi/4 处的泰勒级数，展开到 (x – pi/4)^4 项。先计算各阶导数在 pi/4 处的值：f(pi/4) = sin(pi/4) = 根号2/2，f'(pi/4) = cos(pi/4) = 根号2/2，f”(pi/4) = -sin(pi/4) = -根号2/2，f”'(pi/4) = -cos(pi/4) = -根号2/2，f””(pi/4) = sin(pi/4) = 根号2/2。注意导数值是正负交替的。

Question: find the Taylor series of sin x about x = pi/4, up to the term in (x – pi/4)^4. First evaluate the derivatives at pi/4: f(pi/4) = sin(pi/4) = sqrt(2)/2, f'(pi/4) = cos(pi/4) = sqrt(2)/2, f”(pi/4) = -sin(pi/4) = -sqrt(2)/2, f”'(pi/4) = -cos(pi/4) = -sqrt(2)/2, and f””(pi/4) = sin(pi/4) = sqrt(2)/2. Note that the derivative values alternate in sign.

代入泰勒公式并提取公因子 根号2/2，得到 sin x = (根号2/2) * [1 + (x – pi/4) – (x – pi/4)^2/2 – (x – pi/4)^3/6 + (x – pi/4)^4/24 + …]。数值验证：估算 sin 1。1 弧度与 pi/4 约 0.7854 相差约 0.2146。代入前三项：1 + 0.2146 – 0.0230 = 1.1916，乘以 0.7071 得到 0.8426；加上第四项 -0.00165 的影响后约为 0.8414，而 sin 1 的真实值为 0.84147。仅用四项就精确到了小数点后三位。

Substituting into the Taylor formula and factoring out sqrt(2)/2 gives sin x = (sqrt(2)/2) * [1 + (x – pi/4) – (x – pi/4)^2/2 – (x – pi/4)^3/6 + (x – pi/4)^4/24 + …]. Numerical check: estimate sin 1. One radian differs from pi/4, about 0.7854, by roughly 0.2146. Substituting the first three terms gives 1 + 0.2146 – 0.0230 = 1.1916, which multiplied by 0.7071 gives 0.8426; including the fourth term, about -0.00165, yields 0.8414, while the true value of sin 1 is 0.84147. Four terms alone give accuracy to three decimal places.

这道题还揭示了一个实用技巧：如果要在 x = 1 附近估算正弦值，与其在原点展开（x = 1 距离原点较远，需要很多项），不如选择最近的有特殊角的点 pi/4 展开，这样 (x – a) 的幂很小，收敛更快。考试中遇到”估算 sin 1″这类问题，先找一个离目标最近的特殊角作为展开点，是拿高分的常见策略。

This example also reveals a practical technique: to estimate the sine near x = 1, rather than expanding about the origin (where x = 1 is far away and many terms are needed), choose the nearest special angle, pi/4, as the expansion point so that the powers of (x – a) are small and convergence is faster. In an exam, when asked to “estimate sin 1”, locating the nearest special angle as the expansion point is a common strategy for scoring full marks.

6. Worked Example 3: Expanding ln x About x = 1 | 例题三：把 ln x 在 x = 1 处展开

题目：求 f(x) = ln x 在 x = 1 处的泰勒级数。这是”函数在 0 处无定义”的典型例子：ln 0 没有意义，因此麦克劳林级数根本不存在。在 x = 1 处，f(1) = ln 1 = 0，f'(x) = 1/x，所以 f'(1) = 1；f”(x) = -1/x^2，所以 f”(1) = -1；f”'(x) = 2/x^3，所以 f”'(1) = 2；f””(x) = -6/x^4，所以 f””(1) = -6。

Question: find the Taylor series of f(x) = ln x about x = 1. This is the classic example of a function that is undefined at 0: ln 0 has no meaning, so no Maclaurin series exists. At x = 1 we have f(1) = ln 1 = 0, f'(x) = 1/x so f'(1) = 1, f”(x) = -1/x^2 so f”(1) = -1, f”'(x) = 2/x^3 so f”'(1) = 2, and f””(x) = -6/x^4 so f””(1) = -6.

代入泰勒公式并利用 (n – 1)!/n! = 1/n 化简，得到 ln x = (x – 1) – (x – 1)^2/2 + (x – 1)^3/3 – (x – 1)^4/4 + …，即 ln x = 求和符号 n=1 到无穷 (-1)^(n+1) * (x – 1)^n / n。数值验证：估算 ln 1.1。令 x = 1.1，则 x – 1 = 0.1，前三项给出 0.1 – 0.005 + 0.000333 = 0.095333，而 ln 1.1 的真实值是 0.095310。这个级数在 0 到 2 之间收敛，即 |x – 1| 小于 1。

Substituting into the Taylor formula and simplifying with (n – 1)!/n! = 1/n gives ln x = (x – 1) – (x – 1)^2/2 + (x – 1)^3/3 – (x – 1)^4/4 + … , that is, ln x = sum from n = 1 to infinity of (-1)^(n+1) * (x – 1)^n / n. Numerical check: estimate ln 1.1. Let x = 1.1, so x – 1 = 0.1; the first three terms give 0.1 – 0.005 + 0.000333 = 0.095333, while the true value of ln 1.1 is 0.095310. This series converges for 0 to 2, that is, for |x – 1| less than 1.

7. Why Expand at Other Points? Three Practical Reasons | 为什么要在其他点展开？三个实用理由

理由一：函数在 x = 0 处没有定义或不可导。ln x、1/x、根号 x 在原点都存在定义或可导性问题，无法使用麦克劳林级数，但可以在 x = 1 或其它正数点展开。这类题目在 IB AA HL 试卷中几乎每年出现，目的就是考察学生能否跳出”展开点必须是 0″的思维定式。

Reason one: the function is undefined or non-differentiable at x = 0. Functions such as ln x, 1/x, and sqrt(x) have definition or differentiability problems at the origin, so a Maclaurin series is impossible; however, they can be expanded at x = 1 or another positive point. Questions of this kind appear in IB AA HL papers almost every year, precisely to test whether students can break free of the mindset that “the expansion point must be 0”.

理由二：在目标值附近展开收敛更快。级数的第 n 项含有 (x – a)^n，|x – a| 越小，高阶项贡献越小。例如估算 e^0.5 时，在 x = 1 处展开比在原点展开需要的项数少得多；估算 sin 1 时，在 pi/4 处展开几乎只需三项就达到足够精度。在物理与工程中，这一思想被广泛用于在小扰动条件下线性化复杂的非线性系统。

Reason two: expanding near the target value converges faster. The n-th term of the series contains (x – a)^n, and the smaller |x – a| is, the smaller the contribution of high-order terms. For example, estimating e^0.5 requires far fewer terms when expanding at x = 1 than at the origin, and estimating sin 1 needs only about three terms at pi/4 for sufficient accuracy. In physics and engineering, this idea is widely used to linearise complicated nonlinear systems under small perturbations.

理由三：展开点的选择可以匹配问题的物理背景。在许多应用题中，展开点是问题本身给定的，例如在势能函数的平衡点、温度变化的时间起点或利率为 0 的基准点处展开。此时必须严格围绕题目指定的点 a 展开，而不是自作主张地使用麦克劳林级数，否则即使计算正确，也会因为与题目要求不符而失分。

Reason three: the choice of expansion point can match the physical context of the problem. In many application questions the expansion point is given by the problem itself, such as the equilibrium position of a potential energy function, the starting time of a temperature change, or a baseline interest rate of zero. In such cases you must expand strictly about the specified point a rather than using a Maclaurin series of your own choice; otherwise, even a correct computation loses marks for not matching the question.

8. Interval of Convergence: How Far Does the Approximation Stay Valid? | 收敛区间：近似值在多大范围内有效

IB AA HL 考试通常不要求使用比值检验求收敛半径，但要求考生能说明标准级数的适用范围，并理解截断误差会随 |x – a| 增大而迅速变大。常见结果如下：e^x 在 x = a 处的泰勒级数对一切实数 x 都收敛；sin x 和 cos x 的级数同样对一切实数收敛；ln x 在 x = 1 处的级数仅当 0 小于 x 小于 2（即 |x – 1| 小于 1）时收敛；1/(1 + x) 在 x = 1 处的级数仅当 |x – 1| 小于 2 时收敛。

The IB AA HL examination usually does not require the ratio test to find the radius of convergence, but it does expect candidates to state the range of validity of standard series and to understand that the truncation error grows quickly as |x – a| increases. The common results are: the Taylor series of e^x about x = a converges for all real x; the series for sin x and cos x also converge for all real x; the series of ln x about x = 1 converges only for 0 less than x less than 2, that is, |x – 1| less than 1; and the series of 1/(1 + x) about x = 1 converges only for |x – 1| less than 2.

为什么 ln x 的级数会失效？因为当 x 接近 0 时，ln x 趋向负无穷，而任何多项式在有限区间内都是有限的，所以多项式近似不可能覆盖 x 趋向 0 的奇异性。理解这一点比死记硬背区间更有价值：收敛区间的边界通常与函数本身的奇点位置有关。当估算值超出收敛区间时，即使增加项数，误差也不会减小，这是考试中常见的失分点。

Why does the series for ln x fail? Because as x approaches 0, ln x tends to negative infinity, while any polynomial is finite on a bounded interval, so a polynomial approximation cannot capture the singularity as x tends to 0. Understanding this is more valuable than memorising the interval: the boundary of the interval of convergence is usually related to the location of the singularities of the function itself. When you estimate values outside the interval of convergence, adding more terms does not reduce the error, a common source of lost marks in exams.

9. Common Mistakes in IB Exams: Six Traps to Avoid | IB 考试中的常见错误：六个需要避开的陷阱

陷阱一：展开点使用错误。题目要求 about x = a，却把公式里的 (x – a) 写成了 x，或者把 f^(n)(a) 算成了 f^(n)(0)。建议先在草稿纸上写下 a 的值，再逐项代入。陷阱二：忘记阶乘。泰勒公式分母是 n!，漏掉阶乘会导致系数完全错误；特别要注意 2! = 2、3! = 6、4! = 24 这些小数值。

Trap one: using the wrong expansion point. The question asks for about x = a, but you write x instead of (x – a) in the formula, or you evaluate f^(n)(a) as f^(n)(0). A good habit is to write down the value of a on the rough paper first, then substitute term by term. Trap two: forgetting the factorials. The denominator of the Taylor formula is n!, and missing a factorial makes every coefficient wrong; in particular, remember the small values 2! = 2, 3! = 6, 4! = 24.

陷阱三：三角函数求导的符号错误。sin 求导得 cos，cos 求导得负 sin，符号每隔两阶交替一次。在展开 sin x 或 cos x 时，先把导数列成一行并核对符号，再代入展开点。陷阱四：把级数项数算错。题目说”up to and including the term in x^3″时，需要写到第三阶导数对应的项；”first four non-zero terms”则要求跳过系数为零的项，两者完全不同。

Trap three: sign errors when differentiating trigonometric functions. The derivative of sin is cos, the derivative of cos is negative sin, and the sign alternates every two orders. When expanding sin x or cos x, list the derivatives in a row and check the signs before substituting the expansion point. Trap four: miscounting the number of terms. “Up to and including the term in x^3” requires the term corresponding to the third derivative, while “the first four non-zero terms” requires skipping terms whose coefficients are zero; the two instructions are completely different.

陷阱五：遗漏收敛范围。写出级数后，题目若要求 state the set of values of x for which the expansion is valid，必须给出区间，例如 0 小于 x 小于 2；只写级数不写范围会丢分。陷阱六：截断误差处理不当。用有限项近似函数时，余项 R_n(x) 的数量级由下一项决定；比较两个近似的好坏时，要比较它们在目标点的实际误差，而不是比较展开式的项数多少。

Trap five: omitting the range of validity. After writing the series, if the question asks you to state the set of values of x for which the expansion is valid, you must give the interval, for example 0 less than x less than 2; writing only the series without the range loses marks. Trap six: mishandling the truncation error. When approximating a function with finitely many terms, the order of magnitude of the remainder R_n(x) is determined by the next term; when comparing two approximations, compare their actual errors at the target point rather than the number of terms in each expansion.

10. IB AA HL Exam-Style Practice Questions | IB AA HL 考试风格练习

练习一：求 f(x) = cos x 关于 x = pi/2 的泰勒级数，展开到 (x – pi/2)^4 项。提示：cos(pi/2) = 0，f'(pi/2) = -1，f”(pi/2) = 0，f”'(pi/2) = 1，f””(pi/2) = 0。答案：cos x = -(x – pi/2) + (x – pi/2)^3/6 + …（下一非零项为 -(x – pi/2)^5/120），因为奇偶项交错消失，展开式中只有奇数次幂。

Practice 1: find the Taylor series of f(x) = cos x about x = pi/2, up to the term in (x – pi/2)^4. Hint: cos(pi/2) = 0, f'(pi/2) = -1, f”(pi/2) = 0, f”'(pi/2) = 1, f””(pi/2) = 0. Answer: cos x = -(x – pi/2) + (x – pi/2)^3/6 + … (the next non-zero term is -(x – pi/2)^5/120), because even and odd terms vanish alternately, leaving only odd powers in the expansion.

练习二：求 f(x) = e^x 关于 x = 2 的泰勒级数前三项，并用它估算 e^1.8。答案：e^x = e^2 + e^2(x – 2) + e^2/2 * (x – 2)^2 + …。取 x = 1.8，则 x – 2 = -0.2，估算值为 e^2 * (1 – 0.2 + 0.02) = 7.3891 * 0.82 = 6.059，真实值 e^1.8 = 6.0496，误差约 0.16%。

Practice 2: find the first three terms of the Taylor series of f(x) = e^x about x = 2 and use them to estimate e^1.8. Answer: e^x = e^2 + e^2(x – 2) + e^2/2 * (x – 2)^2 + … . Taking x = 1.8 gives x – 2 = -0.2, so the estimate is e^2 * (1 – 0.2 + 0.02) = 7.3891 * 0.82 = 6.059, while the true value e^1.8 = 6.0496, an error of about 0.16%.

练习三：求 f(x) = 1/(1 + x) 关于 x = 1 的泰勒级数前四项。提示：f(1) = 1/2，f'(1) = -1/4，f”(1) = 1/4，f”'(1) = -3/8。答案：1/(1 + x) = 1/2 – (x – 1)/4 + (x – 1)^2/8 – (x – 1)^3/16 + …，这是公比为 -(x – 1)/2 的几何级数，收敛条件为 |x – 1| 小于 2。验证：取 x = 0，左边 1/(1 + 0) = 1，右边 1/2 + 1/4 + 1/8 + 1/16 + … = 1，两者一致。

Practice 3: find the first four terms of the Taylor series of f(x) = 1/(1 + x) about x = 1. Hint: f(1) = 1/2, f'(1) = -1/4, f”(1) = 1/4, f”'(1) = -3/8. Answer: 1/(1 + x) = 1/2 – (x – 1)/4 + (x – 1)^2/8 – (x – 1)^3/16 + … , a geometric series with common ratio -(x – 1)/2, converging for |x – 1| less than 2. Check: put x = 0; the left side 1/(1 + 0) = 1, and the right side 1/2 + 1/4 + 1/8 + 1/16 + … = 1, so they agree.

Summary | 总结

泰勒级数把展开中心从 x = 0 推广到任意点 x = a，公式为 f(x) = 求和 f^(n)(a)/n! * (x – a)^n。掌握这一知识点的关键是：第一，熟记麦克劳林级数是 a = 0 的特例，遇到 about x = a 必须使用一般公式；第二，按照”求导、代值、代入公式、写项、注明范围”的五步流程系统展开；第三，理解展开点选择的三条理由，尤其是函数在 0 处无定义和收敛速度的考虑；第四，牢记收敛区间与常见陷阱，避免符号、阶乘和项数错误。通过本文的三道例题和三道练习，你应该能够从容应对 IB AA HL 试卷中任何关于”其他点处泰勒级数”的题目。

A Taylor series generalises the expansion centre from x = 0 to an arbitrary point x = a, with the formula f(x) = sum of f^(n)(a)/n! * (x – a)^n. The keys to mastering this topic are: first, remember that a Maclaurin series is the special case a = 0, and always use the general formula when a question says about x = a; second, expand systematically using the five-step procedure of differentiating, evaluating, substituting, writing terms, and stating the range; third, understand the three reasons for choosing an expansion point, especially functions undefined at 0 and considerations of convergence speed; fourth, memorise the intervals of convergence and common traps to avoid sign, factorial, and term-counting errors. With the three worked examples and three practice questions in this article, you should be well prepared for any question about Taylor series at other points in the IB AA HL examination.

更多咨询请联系16621398022（同微信）