📚 Infinite Taylor Expansions | 无穷泰勒展开
In IB Mathematics: Analysis and Approaches (HL) and Applications and Interpretation (HL), infinite Taylor expansions provide a powerful bridge between polynomials and transcendental functions. Understanding how to generate, manipulate, and apply these series is essential for tackling problems in approximation, limits, integration, and differential equations.
在IB数学：分析与方法（HL）以及应用与解释（HL）中，无穷泰勒展开在多项式和超越函数之间架起了一座强大的桥梁。理解如何生成、操作和应用这些级数，对于解决近似、极限、积分和微分方程等问题至关重要。
1. What is a Taylor Series? | 什么是泰勒级数？
y differentiable function f(x), the Taylor series about x = a is the power series f(x) = ∑n=0∞ (f(n)(a) / n!) (x – a)n. The term f(n)(a) denotes the n-th derivative evaluated at a, with f(0)(a) = f(a).
对于无限可微的函数 f(x)，关于 x=a 的泰勒级数是幂级数 f(x) = ∑n=0∞ (f(n)(a) / n!) (x – a)n。其中 f(n)(a) 表示在 a 点的 n 阶导数，f(0)(a)=f(a)。
The series captures the local behaviour of f near a: successive terms use higher derivatives to refine the approximation. A function is called analytic at a if the series converges to f(x) in a neighbourhood of a.
该级数刻画了 f 在 a 附近的局部行为：后续各项通过高阶导数逐步改进逼近精度。如果级数在 a 的某个邻域内收敛到 f(x)，则称函数在 a 点解析。
2. Maclaurin Series: Expansion about Zero | 麦克劳林级数：关于零的展开
A Maclaurin series is simply a Taylor series centred at a = 0: f(x) = ∑n=0∞ (f(n)(0) / n!) xn. Most IB expansions are Maclaurin series because they simplify computation.
麦克劳林级数就是中心在 a=0 的泰勒级数：f(x) = ∑n=0∞ (f(n)(0) / n!) xn。大多数 IB 展开都是麦克劳林级数，因为它们简化了计算。
To find a Maclaurin series, repeatedly differentiate the function, evaluate at zero, and substitute into the coefficient formula. Common examples include ex, sin x, cos x, and ln(1+x).
要求麦克劳林级数，需反复求导、在零点取值并代入系数公式。常见的例子包括 ex, sin x, cos x 和 ln(1+x)。
3. Derivation of Coefficients via Differentiation | 通过求导推导系数
Assume f(x) can be written as a power series ∑n=0∞ cn (x – a)n. Setting x = a gives c0 = f(a). Differentiating once and setting x = a yields c1 = f '(a). Repeating this process gives cn = f(n)(a) / n!.
假设 f(x) 可写成幂级数 ∑n=0∞ cn (x – a)n。令 x=a 得 c0=f(a)。求导一次后令 x=a 得到 c1=f '(a)。重复此过程可得 cn = f(n)(a) / n!。
This derivation shows why factorials appear: each differentiation brings down a factor n, so after n times, the accumulated factor n! must be divided out to isolate the derivative value.
这一推导表明了阶乘出现的原因：每次求导会带下一个因子 n，因此 n 次求导后积累的因子 n! 必须被除掉，以分离出导数值。
4. Standard Maclaurin Expansions | 标准麦克劳林展开式
ex = ∑n=0∞ xn/n! = 1 + x + x2/2! + x3/3! + …, valid for all real x.
ex = ∑n=0∞ xn/n! = 1 + x + x²/2! + x³/3! + …, 对所有实数 x 成立。
sin x = ∑n=0∞ (−1)n x2n+1/(2n+1)! = x – x3/3! + x5/5! – …, all real x.
sin x = ∑n=0∞ (−1)n x2n+1/(2n+1)! = x – x³/3! + x⁵/5! – …, 所有实数 x 成立。
cos x = ∑n=0∞ (−1)n x2n/(2n)! = 1 – x2/2! + x4/4! – …, all real x.
cos x = ∑n=0∞ (−1)n x2n/(2n)! = 1 – x²/2! + x⁴/4! – …, 对所有实数 x 成立。
1/(1 – x) = ∑n=0∞ xn = 1 + x + x2 + x3 + …, for |x| < 1.
1/(1 – x) = ∑n=0∞ xn = 1 + x + x² + x³ + …, 当 |x| < 1 时成立。
ln(1 + x) = ∑n=1∞ (−1)n−1 xn/n = x – x2/2 + x3/3 – …, for −1 < x ≤ 1.
ln(1 + x) = ∑n=1∞ (−1)n−1 xn/n = x – x²/2 + x³/3 – …, 当 −1 < x ≤ 1 时成立。
The general binomial expansion (1 + x)α = ∑n=0∞ (α choose n) xn = 1 + αx + α(α−1)x2/2! + … converges for |x| < 1.
一般二项展开式 (1 + x)α = ∑n=0∞ (α choose n) xn = 1 + αx + α(α−1)x²/2! + … 在 |x| < 1 时收敛。
5. Radius and Interval of Convergence | 收敛半径与收敛区间
A power series ∑ an (x – a)n converges inside a radius R (|x – a| < R) and diverges for |x – a| > R. The value R is the radius of convergence.
幂级数 ∑ an (x – a)n 在半径 R 内（|x – a| < R）收敛，在 |x – a| > R 时发散。R 称为收敛半径。
To find R, use the ratio test on coefficients: R = limn→∞ |an / an+1| t exists. The endpoints x = a ± R must be tested separately for convergence or divergence.
求 R 可使用系数比值检验法：R = limn→∞ |an / an+1|，前提是极限存在。端点 x = a ± R 需单独检验收敛或发散。
For example, the series for ln(1 + x) has an = (−1)n−1/n, so R = lim |an/an+1| series converges; at x = −1, the harmonic series diverges. Hence the interval is (−1, 1].
例如，ln(1 + x) 的级数系数 an = (−1)n−1/n，因此 R = lim |an/an+1| = 1。在 x=1 时交错调和级数收敛；在 x=−1 时调和级数发散。所以收敛区间为 (−1, 1]。
6. The Ratio Test for Power Series | 幂级数的比值检验
For a series ∑ bn, the ratio test considers L = limn→∞ |bn+1/bn| the series converges absolutely; if L > 1 it diverges; if L = 1 the test is inconclusive.
对级数 ∑ bn，比值检验考察 L = limn→∞ |bn+1/bn|。若 L < 1 则级数绝对收敛；若 L > 1 则发散；若 L = 1 则检验无定论。
When applied to power series, L = |x – a| limn→∞ |an+1/an|. Setting this less than 1 gives |x – a| < R with R = 1 / (lim |an+1/an|), which leads directly to the radius of convergence.
应用于幂级数时，L = |x – a| limn→∞ |an+1/an|。令其小于 1 得 |x – a| < R，其中 R = 1 / (lim |an+1/an|)，直接导出收敛半径。
In IB problems, you are often asked to find the radius and interval of convergence for a given Maclaurin series, and to justify the interval by testing endpoints with known convergence tests.
在 IB 考题中，常要求找出给定麦克劳林级数的收敛半径和区间，并通过已知收敛检验法验证端点来给出理由。
7. Lagrange Remainder and Error Estimation | 拉格朗日余项与误差估计
The Taylor polynomial of degree n, Tn(x), approximates f(x). The error (remainder) Rn(x) = f(x) – Tn(x) can be expressed using Lagrange's form: Rn(x) = (f(n+1)(c) / (n+1)!) (x – a)n+1 for some c between a and x.
n 次泰勒多项式 Tn(x) 逼近 f(x)。误差（余项）Rn(x) = f(x) – Tn(x) 可用拉格朗日形式表达：Rn(x) = (f(n+1)(c) / (n+1)!) (x – a)n+1，其中 c 介于 a 与 x 之间。
To bound the error, find the maximum possible value of |f(n+1)(c)| on the interval between a and x. This gives an upper bound for the absolute error when truncating the series.
要估计误差界，先求出在 a 和 x 之间 |f(n+1)(c)| 的最大可能值。这样可得到截断级数时绝对误差的上界。
For alternating series, a simpler error bound applies: |Rn| is less than the absolute value of the first omitted term. This is frequently used for sin x and cos x.
对于交错级数，可使用更简单的误差界：|Rn| 小于首个省略项的绝对值。这一性质常用于 sin x 和 cos x 的逼近。
8. Using Series for Approximations | 利用级数进行近似计算
, use the Maclaurin series centred at 0. Taking the first three terms gives sin(0.2) ≈ 0.2 – (0.2)3/6 + (0.2)5/120 = 0.2 – 0.001333... + 0.00002667... ≈ 0.198693.
要逼近函数值，选择中心 a 使 f 及其导数易计算。对于 sin(0.2)，使用中心在 0 的麦克劳林级数。取前三项得 sin(0.2) ≈ 0.2 – (0.2)³/6 + (0.2)⁵/120 = 0.2 – 0.001333... + 0.00002667... ≈ 0.198693。
The Lagrange remainder or alternating-series bound can tell you how many terms are needed for a desired accuracy. For sin(0.2), the next term is −(0.2)7/5040 ≈ −5×10−7, so three terms already give 6-decimal accuracy.
拉格朗日余项或交错级数误差界可以告诉你为达到所需精度需要多少项。对于 sin(0.2)，下一项为 −(0.2)⁷/5040 ≈ −5×10⁻⁷，因此三项已可达到六位小数精度。
Always state the maximum error bound in an IB approximation question and round the answer according to the guaranteed correct digits.
在 IB 近似问题中，务必给出最大误差界，并根据保证正确的数字位数进行舍入。
9. Applications: Limits, Integrals and Differential Equations | 应用：极限、积分与微分方程
Series expansions simplify indeterminate limits. For example, limx→0 (sin x – x)/x3: substitute sin x = x – x3/3! + x5/5! – … to obtain (−x3/6 + higher order)/x3 → −1/6.
级数展开可简化不定式极限。例如，limx→0 (sin x – x)/x³：代入 sin x = x – x³/3! + x⁵/5! – … 得 (−x³/6 + 高阶项)/x³ → −1/6。
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