📚 Taylor Series and Approximations | 泰勒级数与近似
A Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is one of the most powerful tools in calculus, allowing complicated functions to be approximated by polynomials that are easy to compute and analyse. In IB Mathematics, Taylor and Maclaurin series appear in the Analysis & Approaches HL and Applications & Interpretation HL courses, where students learn to expand functions, estimate errors, and apply series to limits, integrals, and real-world approximations.
泰勒级数是将函数表示为无穷多项之和的一种方式，这些项的系数由函数在单点的各阶导数值确定。它是微积分中最强大的工具之一，能把复杂的函数近似为易于计算和分析的多项式。在 IB 数学中，泰勒级数与麦克劳林级数出现在分析与方法 HL 以及应用与解释 HL 课程里，学生需要学会展开函数、估计误差，并将级数应用于极限、积分和实际近似问题。
1. What Is a Taylor Series? | 什么是泰勒级数？
y differentiable at a point x = a, its Taylor series expansion about a is: f(x) = f(a) + f'(a)(x−a) + f''(a)(x−a)²/2! + f'''(a)(x−a)³/3! + … + f⁽ⁿ⁾(a)(x−a)ⁿ/n! + … pansion converges to f(x) on some interval around a, provided the remainder tends to zero.
如果函数 f(x) 在点 x = a 处无穷次可微，则它在 a 处的泰勒级数展开为：f(x) = f(a) + f'(a)(x−a) + f''(a)(x−a)²/2! + f'''(a)(x−a)³/3! + … + f⁽ⁿ⁾(a)(x−a)ⁿ/n! + … 。在 a 附近的某个区间上，只要余项趋于零，这个展开式就收敛于 f(x)。
2. Taylor Polynomials | 泰勒多项式
the n-th term, we obtain the Taylor polynomial of degree n: Pₙ(x) = Σₖ₌₀ⁿ f⁽ᵏ⁾(a)(x−a)ᵏ/k! gree n near x = a, because it matches the function value and the first n derivatives at a.
当我们在第 n 项之后截断无穷级数时，就得到了 n 次泰勒多项式：Pₙ(x) = Σₖ₌₀ⁿ f⁽ᵏ⁾(a)(x−a)ᵏ/k!。这个多项式是在 x = a 附近、次数为 n 的最佳多项式近似，因为它在 a 点与函数的值以及前 n 阶导数都完全吻合。
3. Maclaurin Series: A Special Case | 麦克劳林级数：一个特例
sion point is a = 0, the Taylor series simplifies to the Maclaurin series: f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + … and they form the foundation for both theoretical work and calculator-free approximations.
若展开点 a = 0，泰勒级数便简化为麦克劳林级数：f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + … 。IB 课程中学习的许多标准级数都是麦克劳林级数，它们是理论推导和无计算器近似计算的基础。
4. Common Maclaurin Series to Memorise | 必须记住的常用麦克劳林级数
The following expansions are essential for IB examinations. They must be known along with their intervals of convergence.
以下展开式对于 IB 考试至关重要，必须连同收敛区间一起熟记。
| Function f(x) | Maclaurin Series | Convergence Interval |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + … | (−∞, ∞) |
| sin x | x − x³/3! + x⁵/5! − x⁷/7! + … | (−∞, ∞) |
| cos x | 1 − x²/2! + x⁴/4! − x⁶/6! + … | (−∞, ∞) |
| ln(1 + x) | x − x²/2 + x³/3 − x⁴/4 + … | (−1, 1] |
| 1/(1−x) | 1 + x + x² + x³ + … | (−1, 1) |
| (1 + x)ᵏ | 1 + kx + k(k−1)x²/2! + … | (−1, 1) for general k |
5. Radius and Interval of Convergence | 收敛半径与收敛区间
The convergence of a power series Σ cₙ(x−a)ⁿ is determined by the radius of convergence R, often found using the ratio test: limn→∞ |cₙ₊₁/cₙ|·|x−a| < 1 gives |x−a| < R. At the endpoints, each series must be tested separately for conditional convergence.
幂级数 Σ cₙ(x−a)ⁿ 的收敛性由收敛半径 R 决定，通常使用比值审敛法求得：limn→∞ |cₙ₊₁/cₙ|·|x−a| < 1 给出 |x−a| < R。在端点处，需要单独检验级数是否条件收敛。
verges conditionally at x = 1. Knowing these intervals helps avoid invalid approximations.
sin x 和 cos x 的麦克劳林级数对所有实数 x 都收敛，而 ln(1+x) 的级数在 x = −1 处发散，在 x = 1 处条件收敛。掌握这些区间有助于避免无效的近似。
6. Lagrange Error Bound | 拉格朗日误差界
x) when approximating a function by its n-th degree Taylor polynomial is given by Lagrange's remainder: Rₙ(x) = f⁽ⁿ⁺¹⁾(c) (x−a)ⁿ⁺¹/(n+1)! for some c between a and x. If there exists an upper bound M for |f⁽ⁿ⁺¹⁾(t)| on the interval, then |Rₙ(x)| ≤ M|x−a|ⁿ⁺¹/(n+1)!.
用 n 次泰勒多项式逼近函数时，误差 Rₙ(x) 由拉格朗日余项给出：Rₙ(x) = f⁽ⁿ⁺¹⁾(c) (x−a)ⁿ⁺¹/(n+1)!，其中 c 介于 a 和 x 之间。如果 |f⁽ⁿ⁺¹⁾(t)| 在该区间上有上界 M，则 |Rₙ(x)| ≤ M|x−a|ⁿ⁺¹/(n+1)!。
This bound is crucial in IB questions that ask "how many terms are needed to achieve an error less than 0.001". Students must estimate M, often noticing that derivatives of sin and cos are bounded by 1.
该误差界在 IB 考题中非常关键，例如要求“确保误差小于 0.001 所需的最少项数”。学生必须估算 M，通常会注意到 sin 和 cos 的导数值受限于 1。
7. Alternating Series Error Bound | 交错级数误差界
When the Taylor series is alternating and decreasing in magnitude after a certain term, the error after truncating at the n-th term is at most the absolute value of the (n+1)-th term. This makes error estimation extremely quick for sin x, cos x, and ln(1+x) within their intervals.
当泰勒级数为交错级数且项的绝对值在某项后递减时，在第 n 项截断后的误差至多为第 (n+1) 项的绝对值。这使得对 sin x、cos x 和 ln(1+x) 在其区间内的误差估计变得极其迅速。
8. Approximating Function Values | 近似计算函数值
To approximate e², we can use the Maclaurin polynomial for eˣ at x = 2: e² ≈ 1 + 2 + 2²/2! + 2³/3! + … r strategy is to use e² = (e¹)² and approximate e, or to expand about a value closer to 2.
要近似计算 e²，可以使用 eˣ 的麦克劳林多项式在 x = 2 处展开：e² ≈ 1 + 2 + 2²/2! + 2³/3! + … 。虽然收敛是保证的，但因为 x 不接近 0，所以需要很多项。更好的策略是利用 e² = (e¹)² 先近似 e，或者在更接近 2 的点处展开。
e is about 0.198669. The error is tiny because the first omitted term is about 2.6 × 10⁻⁶.
对于 sin(0.2)，只需几项就能得到高精度：sin(0.2) ≈ 0.2 − 0.2³/6 = 0.2 − 0.008/6 ≈ 0.198667，而真实值约为 0.198669。误差很小，因为第一个被忽略的项大约为 2.6 × 10⁻⁶。
9. Using Taylor Series to Evaluate Limits | 利用泰勒级数求极限
Taylor expansions are excellent for resolving indeterminate forms like 0/0. For example, consider limx→0 (sin x − x)/x³. Substitute sin x = x − x³/6 + x⁵/120 − … to obtain ( −x³/6 + x⁵/120 − … )/x³ = −1/6 + x²/120 − … . As x → 0, the limit is −1/6. This method avoids multiple applications of L'Hôpital's rule.
泰勒展开对于求解 0/0 等不定式极为有效。例如，考虑 limx→0 (sin x − x)/x³。代入 sin x = x − x³/6 + x⁵/120 − …，得到 ( −x³/6 + x⁵/120 − … )/x³ = −1/6 + x²/120 − … 。当 x → 0 时，极限为 −1/6。这种方法避免了反复使用洛必达法则。
10. Term-by-Term Differentiation and Integration | 逐项微分与逐项积分
Within the interval of convergence, a power series can be differentiated and integrated term by term. This property allows us to find series for functions like arctan x by integrating the geometric series for 1/(1+x²).
在收敛区间内，幂级数可以逐项微分和逐项积分。这一性质使我们能够通过对 1/(1+x²) 的几何级数积分，得到 arctan x 等函数的级数。
arctan x = x − x³/3 + x⁵/5 − x⁷/7 + … , |x| ≤ 1
Differentiation of series is similarly powerful: from eˣ = Σ xⁿ/n! we immediately get d/dx(eˣ) = Σ n xⁿ⁻¹/n! = Σ xⁿ⁻¹/(n−1)! = eˣ, confirming the derivative property.
级数的微分同样强大：由 eˣ = Σ xⁿ/n! 立刻得到 d/dx(eˣ) = Σ n xⁿ⁻¹/n! = Σ xⁿ⁻¹/(n−1)! = eˣ，从而验证了其导数性质。
11. Finding Series from Known Series | 利用已知级数求新级数
y substitution, multiplication, or composition of known series. For instance, to expand e⁻ˣ², substitute −x² into the series for eˣ: e⁻ˣ² = 1 − x² + x⁴/2! − x⁶/3! + … . Likewise, x² sin x = x³ − x⁵/6 + x⁷/120 − … .
IB 题目常要求学生通过代入、相乘或复合已知级数来推导新的麦克劳林级数。例如，要展开 e⁻ˣ²，只需将 −x² 代入 eˣ 的级数：e⁻ˣ² = 1 − x² + x⁴/2! − x⁶/3! + … 。类似地，x² sin x = x³ − x⁵/6 + x⁷/120 − … 。
it is sufficient to compute terms up to the required degree: (sin x)(cos x) = (x − x³/6 + …)(1 − x²/2 + x⁴/24 − …) = x − (1/2+1/6)x³ + … = x − 2x³/3 + … , which is precisely the expansion for ½ sin 2x.
级数相乘时，只需计算到指定次数的项即可：(sin x)(cos x) = (x − x³/6 + …)(1 − x²/2 + x⁴/24 − …) = x − (1/2+1/6)x³ + … = x − 2x³/3 + … ，这恰好是 ½ sin 2x 的展开式。
12. IB Exam Tips and Common Pitfalls | IB 考试技巧与常见误区
- Always specify the interval of convergence when giving a series expansion. Simply writing the polynomial without stating the validity range can cost marks.
- 给出级数展开时，务必注明收敛区间。 只写出多项式而不说明有效范围会失分。
- Use the Lagrange error bound to justify the number of terms when accuracy is requested. For alternating series with decreasing terms, the next term suffices.
- 当题目要求一定精度时，要用拉格朗日误差界来说明所需项数。 对于项递减的交错级数，下一项的绝对值即可作为误差界。
- Check the expansion point. If the question says "expand f(x) about x = 2", you must use (x−2), not x. Forgetting to shift the variable is a classic mistake.
- 注意展开点。 若题目要求“在 x = 2 处展开 f(x)”，必须使用 (x−2)，而不能直接使用 x。忘记变量平移是经典错误。
- When evaluating limits, do not truncate too early. Keep enough terms so that the lowest-degree term that does not cancel reveals the limit. A common error is to keep only the first term of sin x, missing the −x³/6 needed for the correct limit.
- 求极限时，切勿过早截断。 保留足够多的项，使得最低次未抵消项能揭示极限。常见错误是只保留 sin x 的首项，从而丢失了正确极限所必需的 −x³/6 项。
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