📚 Maclaurin Series | 麦克劳林级数
A Maclaurin series is a powerful tool in calculus that represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point, usually zero. It allows us to approximate complicated functions with polynomials, making calculations in limits, integrals, and differential equations much simpler. For IB Mathematics, mastering Maclaurin series is essential, particularly in the Analysis and Approaches HL course.
麦克劳林级数是微积分中一个强大的工具，它将一个函数表示为一个无穷级数，其各项由该函数在零点处的导数值计算得出。它使我们能用多项式来逼近复杂函数，从而大大简化极限、积分和微分方程中的计算。对于 IB 数学，尤其是分析与方法高级课程，掌握麦克劳林级数至关重要。
1. Definition of Maclaurin Series | 麦克劳林级数的定义
A Maclaurin series is a Taylor series expansion of a function about 0. If a function f(x) has derivatives of all orders at x = 0, its Maclaurin series is given by the infinite sum:
麦克劳林级数是函数在 0 点处的泰勒级数展开。如果函数 f(x) 在 x = 0 处具有所有阶的导数，则其麦克劳林级数由以下无穷和式给出：
f(x) = ∑n=0∞ (f⁽ⁿ⁾(0)/n!) xⁿ
Here, f⁽ⁿ⁾(0) denotes the n-th derivative of f evaluated at 0, n! is n factorial, and xⁿ is x raised to the power n. The first few terms are often written as f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + … . This representation is exact only if the series converges to f(x).
这里，f⁽ⁿ⁾(0) 表示 f 在 0 处的 n 阶导数，n! 是 n 的阶乘，xⁿ 是 x 的 n 次方。通常前几项写为 f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + …。只有当级数收敛到 f(x) 时，这个表示才是精确的。
2. Derivation from Taylor Series | 从泰勒级数推导
The Maclaurin series is simply the special case of the Taylor series centred at a = 0. Recall the Taylor series of f(x) about x = a: f(x) = ∑n=0∞ (f⁽ⁿ⁾(a)/n!) (x − a)ⁿ. Setting a = 0 directly yields the Maclaurin formula. This means that all the function’s information is encoded in its derivatives at zero.
麦克劳林级数只是泰勒级数在中心 a = 0 时的特例。回想一下 f(x) 关于 x = a 的泰勒级数：f(x) = Σ (f⁽ⁿ⁾(a)/n!)(x − a)ⁿ。令 a = 0 就直接得到了麦克劳林公式。这意味着函数的所有信息都编码在它在零点的各阶导数中。
The derivation assumes that the function is infinitely differentiable and that the remainder term tends to zero as the number of terms increases. This condition is crucial for the series to converge to the function.
这一推导假设函数是无穷可微的，并且随着项数增加，余项趋近于零。这个条件对于级数收敛到函数本身至关重要。
3. Standard Series for eˣ | eˣ 的标准麦克劳林展开
The exponential function eˣ is uniquely simple because all its derivatives at 0 are 1. Thus f⁽ⁿ⁾(0) = 1 for all n, giving:
指数函数 eˣ 极其简单，因为它所有阶导数在 0 处的值都是 1。因此对所有 n 都有 f⁽ⁿ⁾(0) = 1，从而得到：
eˣ = 1 + x + x²/2! + x³/3! + … + xⁿ/n! + …
This series converges for all real x (radius of convergence R = ∞). It is often used to define eˣ in more advanced analysis. You can quickly verify that differentiating the series term-by-term reproduces itself.
该级数对所有实数 x 都收敛（收敛半径 R = ∞）。在更高层次的分析学中，它常被用来定义 eˣ。通过逐项求导，你很快就能验证它重复得到自身。
4. Standard Series for sin x and cos x | sin x 和 cos x 的标准展开
The sine and cosine functions have repeating patterns of derivatives at 0. For sin x, derivatives cycle 0, 1, 0, −1, …; for cos x, they cycle 1, 0, −1, 0, … . Their Maclaurin series are:
正弦和余弦函数在 0 处的导数呈现循环模式。对于 sin x，导数循环为 0, 1, 0, −1, …；对于 cos x，导数循环为 1, 0, −1, 0, …。它们的麦克劳林级数如下：
sin x = x − x³/3! + x⁵/5! − x⁷/7! + …
cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …
Both series converge for all real x. Notice that sin x contains only odd powers, and cos x contains only even powers, which matches their parity. These expansions form the basis for many small-angle approximations.
这两个级数对所有实数 x 都收敛。注意 sin x 只含奇次幂，cos x 只含偶次幂，这与它们的奇偶性一致。这些展开是许多小角度近似的基础。
5. Series for ln(1+x) | ln(1+x) 的级数
The natural logarithm ln(1+x) has derivatives at 0 that follow a factorial-type pattern: f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)! for n ≥ 1. This gives the alternating harmonic-like series:
自然对数 ln(1+x) 在 0 处的导数遵循阶乘型规律：对 n ≥ 1，f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)!。由此得到交错调和型级数：
ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + …
This series converges for −1 < x ≤ 1. At x = 1, it gives the alternating harmonic series converging to ln 2; at x = −1, it diverges. Always state the interval of convergence when using this series.
该级数在 −1 < x ≤ 1 时收敛。在 x = 1 处，它给出收敛于 ln 2 的交错调和级数；在 x = −1 处，它发散。使用此级数时，务必说明收敛区间。
6. Binomial Series (1+x)ᵏ | 二项式级数 (1+x)ᵏ
For any real exponent k, the function (1+x)ᵏ can be expanded as a Maclaurin series. The derivatives at 0 yield the general binomial coefficients:
对于任意实数指数 k，函数 (1+x)ᵏ 可以展开为麦克劳林级数。0 点处的各阶导数产生推广的二项式系数：
(1+x)ᵏ = 1 + kx + k(k−1)x²/2! + k(k−1)(k−2)x³/3! + …
This series converges for |x| < 1, unless k is a non‑negative integer, in which case the series terminates and becomes a polynomial. For IB, the case where k is a positive integer gives the familiar binomial theorem, but Maclaurin series extends this to rational and negative exponents.
该级数在 |x| < 1 时收敛，除非 k 是非负整数——此时级数终止成为多项式。对 IB 而言，k 为正整数时对应于熟悉的二项式定理，但麦克劳林级数将其推广到有理数和负数指数。
7. Series for arctan x | arctan x 的级数
The derivative of arctan x is 1/(1+x²), which can be expanded as a geometric series 1 − x² + x⁴ − x⁶ + … . Integrating term‑by‑term from 0 to x gives the Maclaurin series:
arctan x 的导数是 1/(1+x²)，它可以展开为几何级数 1 − x² + x⁴ − x⁶ + …。从 0 到 x 逐项积分即得到麦克劳林级数：
arctan x = x − x³/3 + x⁵/5 − x⁷/7 + …
This series converges for |x| ≤ 1, with conditional convergence at the endpoints. It is historically significant as it leads to Leibniz’s formula for π when substituting x = 1.
该级数在 |x| ≤ 1 时收敛，在端点处条件收敛。它在历史上意义重大，因为代入 x = 1 就得出莱布尼茨关于 π 的公式。
8. Convergence and Interval of Convergence | 收敛性与收敛区间
A Maclaurin series is useful only within its interval of convergence, the set of x-values for which the infinite sum approaches a finite limit equal to the function. Use the ratio test or root test to find the radius of convergence R. For example, eˣ, sin x, cos x have R = ∞; ln(1+x) has R = 1, and the binomial series (1+x)ᵏ (k not a non‑negative integer) has R = 1.
麦克劳林级数仅在其收敛区间内才有用，即无穷和趋近于一个等于该函数的有限极限的 x 值集合。运用比值判别法或根值判别法可求出收敛半径 R。例如 eˣ、sin x、cos x 的 R = ∞；ln(1+x) 的 R = 1；二项式级数 (1+x)ᵏ（k 不是非负整数）的 R = 1。
Endpoints must be checked separately. For ln(1+x), x = 1 yields convergence, but x = −1 yields divergence. Always write the final answer as an inequality such as −1 < x ≤ 1.
端点必须单独检查。对 ln(1+x)，x = 1 收敛，但 x = −1 发散。最终答案务必写为不等式，例如 −1 < x ≤ 1。
| Function | Maclaurin Series (first few terms) | Interval of Convergence |
|---|---|---|
| eˣ | 1 + x + x²/2! + x³/3! + … | (−∞, ∞) |
| sin x | x − x³/3! + x⁵/5! − … | (−∞, ∞) |
| cos x | 1 − x²/2! + x⁴/4! − … | (−∞, ∞) |
| ln(1+x) | x − x²/2 + x³/3 − x⁴/4 + … | (−1, 1] |
| (1+x)ᵏ | 1 + kx + k(k−1)x²/2! + … | |x| < 1 (unless k ∈ ℕ₀) |
| arctan x | x − x³/3 + x⁵/5 − … | [−1, 1] |
9. Differentiation and Integration of Series | 级数的逐项微积分
Within the open interval of convergence, a Maclaurin series can be differentiated or integrated term‑by‑term. This property is extremely useful for finding series of functions whose derivatives are easier to expand. For instance, the series for −ln(1−x) can be obtained by integrating 1/(1−x) = 1 + x + x² + … .
在收敛区间的开区间内，麦克劳林级数可以逐项求导或积分。这一性质对于寻找那些导数更易展开的函数的级数极其有用。例如，−ln(1−x) 的级数可以通过积分 1/(1−x) = 1 + x + x² + … 得到。
If f(x) = ∑ aₙ xⁿ, then f'(x) = ∑ n aₙ xⁿ⁻¹ and ∫₀ˣ f(t) dt = ∑ (aₙ/(n+1)) xⁿ⁺¹. These operations do not change the radius of convergence, though endpoint behavior might change. Be careful to adjust the index and constant terms.
若 f(x) = ∑ aₙ xⁿ，则 f'(x) = ∑ n aₙ xⁿ⁻¹ 且 ∫₀ˣ f(t) dt = ∑ (aₙ/(n+1)) xⁿ⁺¹。这些运算不改变收敛半径，但端点行为可能会变。注意调整下标和常数项。
10. Applications: Approximating Values | 应用：近似计算
One of the most practical uses of Maclaurin series is to approximate function values with a few polynomial terms. For small x, truncating after the first few terms often gives high accuracy. For example, sin(0.1) ≈ 0.1 − (0.1)³/6 = 0.0998333…, compared to the true value 0.0998334…, giving an error of only about 2 × 10⁻⁷.
麦克劳林级数最实用的用途之一是用少量多项式项来近似函数值。对于很小的 x，仅取前几项通常就能给出高精度。例如 sin(0.1) ≈ 0.1 − (0.1)³/6 = 0.0998333…，与真实值 0.0998334… 比较，误差仅为约 2 × 10⁻⁷。
Such approximations are used in physics, engineering, and even in calculators. Always estimate the error using the next term (Taylor’s remainder) to ensure the required precision. In IB questions, you may be asked to find an approximation to a given degree of accuracy.
这种近似在物理、工程甚至计算器中都有应用。要始终利用下一项（泰勒余项）估计误差，以确保满足所需精度。在 IB 考题中，你可能会被要求求出特定精度的近似值。
11. Applications: Limits and Integrals | 应用：极限与积分
Maclaurin series provide an elegant way to evaluate tricky limits, especially those of the form 0/0. By expanding numerator and denominator as series, dominant terms cancel, revealing the limit. For example, limx→0 (sin x − x)/x³ = limx→0 (−x³/6 + …)/x³ = −1/6.
麦克劳林级数为求解棘手的极限，尤其是 0/0 型极限，提供了一种优雅的方法。将分子和分母展开为级数后，主导项互相抵消，极限便显现出来。例如 limx→0 (sin x − x)/x³ = limx→0 (−x³/6 + …)/x³ = −1/6。
Similarly, integrals that lack elementary antiderivatives, such as ∫ e⁻ˣ² dx, can be handled by expanding the integrand into a Maclaurin series and integrating term‑by‑term. This yields an infinite series representation of the antiderivative.
类似地，对于那些没有初等原函数的积分，例如 ∫ e⁻ˣ² dx，可以先将被积函数展开为麦克劳林级数，再逐项积分，从而得到原函数的无穷级数表示。
12. Common Mistakes and Exam Tips | 常见错误与备考提示
Students often forget to divide by factorial coefficients, confuse the signs, or miswrite the expansion for ln(1+x) missing the alternating signs. Also, many neglect to state the interval of convergence, losing marks. Another common error is using a series expansion beyond its radius of convergence, which leads to incorrect approximations.
学生经常忘记除以阶乘系数、搞错符号，或漏写了 ln(1+x) 的交错符号。还有很多人忽略写明收敛区间而丢分。另一个常见错误是在收敛半径之外使用级数展开，导致近似结果错误。
When deriving a series, always compute the first few derivatives carefully. If a known standard series can be used by substitution, do so — for example, e²ˣ can be found by replacing x with 2x in the eˣ series. Also remember that a Maclaurin polynomial of degree n uses terms up to xⁿ. Practice past paper questions to become fluent.
推导级数时，务必仔细计算最初几阶导数。如果可以代入已知的标准级数，就尽量使用——例如 e²ˣ 只需将 eˣ 级数中的 x 替换为 2x 即可。同时记住 n 次麦克劳林多项式包含到 xⁿ 项。通过练习往年真题来达到熟练。
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