11D Maclaurin Series | 11D 麦克劳林级数

📚 11D Maclaurin Series | 11D 麦克劳林级数

The Maclaurin series expresses a function as an infinite sum of terms calculated from the values of its derivatives at zero. It is a fundamental topic in IB Mathematics: Analysis and Approaches (HL), linking differentiation, sequences, and series. Mastery of Maclaurin expansions allows students to approximate functions, evaluate limits, and solve complex problems with elegance.

麦克劳林级数将函数表示为根据其在零点处的导数值计算出的无穷多项之和。这是 IB 数学分析与方法（高水平）中的基础主题，将微分、数列与级数连接起来。掌握麦克劳林展开式能让学生用简洁的方式逼近函数、计算极限以及解决复杂问题。

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1. Definition and General Form | 定义与一般形式

e at x = 0, its Maclaurin series is the power series centred at 0. The general formula is:

若函数 f(x) 在 x = 0 处无穷次可微，其麦克劳林级数就是中心在 0 的幂级数。一般公式为：

f(x) = ∑_n=0^∞ (f⁽ⁿ⁾(0) / n!) xⁿ

This expands to f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... . The coefficients depend only on the derivatives evaluated at zero. The series may converge to f(x) within a certain interval called the radius of convergence.

该级数展开为 f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... 。系数完全由零点处的导数决定。该级数在某个称为收敛半径的区间内可能收敛到 f(x)。

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2. Deriving the Maclaurin Series for e^x | 推导 e^x 的麦克劳林级数

All derivatives of e^x are again e^x. At x = 0, we have f⁽ⁿ⁾(0) = 1 for every n ≥ 0. Substituting into the Maclaurin formula gives:

e^x 的各阶导数仍为 e^x。在 x = 0 处，对任意 n ≥ 0 都有 f⁽ⁿ⁾(0) = 1。代入麦克劳林公式得到：

e^x = ∑_n=0^∞ xⁿ / n! = 1 + x + x²/2! + x³/3! + x⁴/4! + ...

The series converges for all real numbers x. This is one of the most useful expansions in calculus and will be heavily used for substitutions and limits.

该级数对所有实数 x 都收敛。这是微积分中最有用的展开式之一，将频繁用于代入法和极限计算。

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3. Maclaurin Series for sin x | sin x 的麦克劳林级数

y four steps: sin x → cos x → −sin x → −cos x → sin x. Evaluating at x = 0 gives f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=−1, f⁽⁴⁾(0)=0, and so on. Only odd powers survive with alternating signs.

sin x 的导数每四步循环一次：sin x → cos x → −sin x → −cos x → sin x。在 x = 0 处求值可得 f(0)=0，f'(0)=1，f''(0)=0，f'''(0)=−1，f⁽⁴⁾(0)=0，以此类推。只有奇次幂保留，且符号交替。

sin x = ∑_n=0^∞ (-1)ⁿ x²ⁿ⁺¹ / (2n+1)! = x − x³/3! + x⁵/5! − x⁷/7! + ...

for all real x and is the foundation for small-angle approximations such as sin x ≈ x − x³/6.

该展开式对所有实数 x 收敛，并是小角度近似（如 sin x ≈ x − x³/6）的基础。

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4. Maclaurin Series for cos x | cos x 的麦克劳林级数

ating the sin x series term by term. Alternatively, evaluating derivatives at 0: f(0)=1, f'(0)=0, f''(0)=−1, f'''(0)=0, producing only even powers.

对 sin x 求导得 cos x，因此可通过逐项求导 sin x 的级数得到 cos x 的级数。或者直接在 0 处求导数：f(0)=1，f'(0)=0，f''(0)=−1，f'''(0)=0，只产生偶次幂。

cos x = ∑_n=0^∞ (-1)ⁿ x²ⁿ / (2n)! = 1 − x²/2! + x⁴/4! − x⁶/6! + ...

sin x, the cosine series converges for all real x. The leading approximation cos x ≈ 1 − x²/2 is extremely common in physics and engineering.

与 sin x 类似，余弦级数对所有实数 x 均收敛。其主导近似 cos x ≈ 1 − x²/2 在物理和工程中极为常见。

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5. Maclaurin Series for ln(1 + x) | ln(1 + x) 的麦克劳林级数

Compute the derivatives of f(x) = ln(1 + x). We obtain f'(x) = (1+x)⁻¹, f''(x) = −(1+x)⁻², f'''(x) = 2(1+x)⁻³, and in general f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹(n−1)! for n ≥ 1. The series starts from n=1 because f(0)=0.

求 f(x) = ln(1 + x) 的导数。可得 f'(x) = (1+x)⁻¹，f''(x) = −(1+x)⁻²，f'''(x) = 2(1+x)⁻³，一般地，对于 n ≥ 1 有 f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹(n−1)!。因 f(0)=0，级数从 n=1 开始。

ln(1 + x) = ∑_n=1^∞ (-1)ⁿ⁻¹ xⁿ / n = x − x²/2 + x³/3 − x⁴/4 + ...

x ≤ 1. At x=1, it gives the alternating harmonic series ln 2 = 1 − 1/2 + 1/3 − 1/4 + ... .

该展开式在 −1 < x ≤ 1 内有效。在 x=1 处，它给出交错调和级数 ln 2 = 1 − 1/2 + 1/3 − 1/4 + ... 。

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6. Binomial Series: (1 + x)^k | 二项级数：(1 + x)^k

For any real exponent k, the Maclaurin expansion of (1 + x)^k generalises the ordinary binomial theorem. The derivatives at 0 produce coefficients involving factorial-like products.

对任意实数指数 k，(1 + x)^k 的麦克劳林展开式是对普通二项式定理的推广。其在 0 处的导数产生的系数包含阶乘形式的乘积。

(1 + x)^k = 1 + kx + [k(k−1)/2!] x² + [k(k−1)(k−2)/3!] x³ + ... = ∑_n=0^∞ C(k, n) xⁿ

where C(k, n) = k(k−1)...(k−n+1)/n! is the generalised binomial coefficient. Convergence holds for |x| 1; if k is a non-negative integer, the series terminates and becomes a finite polynomial.

其中 C(k, n) = k(k−1)...(k−n+1)/n! 为广义二项式系数。当 |x| < 1 时收敛；若 k 为非负整数，级数在有限项终止而成为一个多项式。

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7. Manipulating Series: Differentiation and Integration | 级数运算：逐项求导与积分

Within the interval of convergence, Maclaurin series can be differentiated and integrated term by term. This provides a powerful strategy to derive new series from known ones.

在收敛区间内，麦克劳林级数可以逐项求导和积分。这为由已知级数推导新级数提供了强有力的策略。

ting sin x gives cos x immediately. Integrating the geometric series 1/(1 + x) = 1 − x + x² − x³ + ... (|x|<1) yields the series for ln(1 + x). Similarly, the series for arctan x is found by integrating 1/(1 + x²):

例如，对 sin x 求导立即得 cos x。对几何级数 1/(1 + x) = 1 − x + x² − x³ + ... (|x|<1) 积分可得 ln(1 + x) 的级数。类似地，arctan x 的级数可由积分 1/(1 + x²) 求得：

arctan x = ∑_n=0^∞ (-1)ⁿ x²ⁿ⁺¹ / (2n+1) = x − x³/3 + x⁵/5 − ... , |x| ≤ 1

You can also multiply series or combine them to obtain expansions for functions like e^x sin x, though this is less common in IB exams.

你还可以将级数相乘或组合，以获得如 e^x sin x 等函数的展开式，不过在 IB 考试中较少见到。

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8. Substitution to Form New Series | 代入法构造新级数

If you know a standard Maclaurin series, you can replace x with a function of x (as long as the composition respects the convergence interval). This quickly builds series for related functions without recalculating derivatives.

若已知一个标准的麦克劳林级数，你可以将 x 替换成一个关于 x 的函数（只要该复合符合收敛区间要求）。这就无需重新计算导数即可快速构造出相关函数的级数。

For instance, to expand e^−x², substitute −x² into the e^x series: e^−x² = 1 − x² + x⁴/2! − x⁶/3! + ... . Expanding sin(3x) uses the sin x series with x replaced by 3x, producing factors 3²ⁿ⁺¹ in each term. Similarly, (1 + 2x)^k uses the binomial series with x replaced by 2x. Always check the resulting convergence condition: for (1 + 2x)^k, we need |2x| < 1, i.e. |x| < 1/2.

例如，要将 e^−x² 展开，可将 −

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