5. Manipulation of Taylor Series | 泰勒级数的变换与应用

📚 5. Manipulation of Taylor Series | 泰勒级数的变换与应用

In IB Mathematics, Taylor and Maclaurin series allow us to represent functions as infinite polynomials. While evaluating coefficients directly using derivatives can be tedious, we can manipulate a handful of standard series to obtain expansions for many related functions. This approach saves time and reduces errors, but requires careful attention to algebra and the radius of convergence.

在IB数学中，泰勒级数和麦克劳林级数可以将函数表示为无限多项式。虽然直接用导数求系数可能很繁琐，但我们可以通过对少数几个标准级数进行变换，得出许多相关函数的展开式。这种方法省时且减少错误，但需要注意代数运算和收敛半径。

1. Standard Maclaurin Series to Memorise | 必记的标准麦克劳林级数

t important Maclaurin series (centred at x = 0) along with their intervals of convergence.

在变换级数之前，你必须对这些基本展开式了如指掌。下表列出了最重要的麦克劳林级数（以 x = 0 为中心）及其收敛区间。

These five expansions form the backbone of series manipulation. You should be able to reproduce them instantly and state the radius or interval of convergence for each.

这五个展开式构成级数变换的基础。你应该能够瞬间写出它们，并说明每个级数的收敛半径或区间。

2. Substitution: Creating New Series from Old | 代入法：从已知级数生成新级数

One of the simplest manipulations is direct substitution. Replace every x in a standard series with a new expression, and simplify the terms. The new series is valid wherever the substituted expression lies within the original interval of convergence.

最简单的变换之一就是直接代入。把标准级数中的每一个 x 替换成一个新的表达式，然后化简各项。只要替换后的表达式落在原收敛区间内，新级数就成立。

For example, to expand e^2x, substitute 2x for x in the series for e^x:

例如，要展开 e^2x，将 e^x 级数中的 x 替换为 2x：

e^2x = 1 + (2x) + (2x)²/2! + (2x)³/3! + ... = 1 + 2x + 4x²/2! + 8x³/3! + ...

The convergence remains all real x because 2x is real for all x. Similarly, e^x² gives 1 + x² + x⁴/2! + x⁶/3! + ... . For sin(3x), simply write 3x − (3x)³/3! + (3x)⁵/5! − ... .

收敛域仍然是全体实数，因为 2x 对所有 x 都是实数。类似地，e^x² 可得 1 + x² + x⁴/2! + x⁶/3! + ...。对于 sin(3x)，直接写出 3x − (3x)³/3! + (3x)⁵/5! − ...。

Be cautious with the interval of convergence. If you substitute x² for x in the series for 1/(1−x), which converges for |x| < 1, you get 1/(1−x²) converging for |x²| < 1, i.e. |x| < 1. Substituting √x into ln(1+x) would require x ≥ 0 and −1 < √x ≤ 1, so 0 ≤ x ≤ 1.

要注意收敛区间。如果把 x² 代入 1/(1−x) 的级数（收敛域 |x| < 1），得到 1/(1−x²)，其收敛条件为 |x²| < 1，即 |x| < 1。把 √x 代入 ln(1+x) 时则需 x ≥ 0 且 −1 < √x ≤ 1，即 0 ≤ x ≤ 1。

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3. Multiplication by a Power or Constant | 乘以幂函数或常数

Multiplying a known series by x^m or by a constant shifts powers or scales coefficients. This is straightforward but extremely useful.

用一个幂函数 x^m 或常数去乘已知级数，可以平移幂次或缩放系数。这非常简单，但极为有用。

Example: to find the Maclaurin series for x e^x, multiply the series for e^x by x:

示例：要得到 x e^x 的麦克劳林级数，把 e^x 的级数乘以 x：

x e^x = x(1 + x + x²/2! + x³/3! + ...) = x + x² + x³/2! + x⁴/3! + ...

All terms are simply shifted up by one degree. Similarly, 2 cos x = 2 − 2x²/2! + 2x⁴/4! nvergence interval remains unchanged because multiplication by a non-zero constant or by x^m does not affect the radius of convergence (except possibly at endpoints).

每一项只是升高了一次幂。类似地，2 cos x = 2 − 2x²/2! + 2x⁴/4! − ...。收敛区间保持不变，因为乘以非零常数或 x^m 不影响收敛半径（端点可能除外）。

You will often use this technique to obtain series for functions like x² sin x, 3 ln(1+x), or √x cos x (though be careful with domains). Such products appear frequently in integration problems and differential equations.

你经常会用这种方法来得到 x² sin x、3 ln(1+x) 或 √x cos x 等函数的级数（注意定义域）。这类乘积在积分问题和微分方程中经常出现。

4. Addition and Subtraction of Series | 级数的加减运算

Adding or subtracting two series term by term is legitimate within the common interval of convergence. This is especially helpful for building hyperbolic functions or simplifying combinations.

在共同的收敛区间内，对两个级数逐项做加减是合理的。这一点在构造双曲函数或化简组合时尤其有用。

Recall that sinh x = (e^x − e^−x)/2. Using the series for e^x and substituting −x for e^−x, we get:

回忆 sinh x = (e^x − e^−x)/2。利用 e^x 的级数并对 e^−x 代入 −x，可得：

e^−x = 1 − x + x²/2! − x³/3! + ...

sinh x = [ (1 + x + x²/2! + x³/3! + ...) − (1 − x + x²/2! − x³/3! + ...) ] / 2

= (2x + 2x³/3! + 2x⁵/5! + ...)/2 = x + x³/3! + x⁵/5! + ...

ed for an odd function. The convergence is all real x. The same approach gives cosh x = (e^x + e^−x)/2 = 1 + x²/2! + x⁴/4! + ... .

只有奇次幂保留下来，正如奇函数的预期。收敛域是所有实数。同样的方法可得 cosh x = (e^x + e^−x)/2 = 1 + x²/2! + x⁴/4! + ...。

Addition and subtraction also let you combine series for polynomials with those for transcendental functions, which is common when handling expressions like (1 + x + x²)e^x or sin x + cos x.

加减运算还能让你把多项式的级数与超越函数的级数组合起来，这在处理如 (1 + x + x²)e^x 或 sin x + cos x 时很常见。

5. Multiplying Two Series Together | 两个级数的乘法

To multiply two infinite series, we use the Cauchy product: collect all pairs of terms whose indices sum to the same power. In practice, we only need terms up to a specified order.

要乘两个无穷级数，我们使用柯西乘积：收集所有下标之和等于同一幂次的项对。实际上，我们只需要到指定阶的项。

Suppose we want the Maclaurin series for e^x sin x up to the x⁴ term. Write the series:

假设我们要求 e^x sin x 的麦克劳林级数，直到 x⁴ 项。写出级数：

e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ...

sin x = x − x³/3! + ... (we can stop at x³ because higher terms will exceed x⁴ when multiplied)

e^x sin x = (1)(x) + (x)(x) + (1)(−x³/6) + (x²/2)(x) + (x³/6)(x) + ...

x²; x³ term: 1·(−x³/6) + (x²/2)·x = −x³/6 + x³/2 = x³/3; x⁴ term: (x³/6)·x + (x²/2)·(x²/2)?? Wait, let' te step by step. It is often clearer to list terms up to x³ for sin x: sin x = x − x³/6. e^x ltiply and discard powers > 4. x⁴ contributions: from 1·(none from sin x), from x·(x³ term? sin has x³, product x⁴), from x²/2·(sin x² term? sin has no x² term), from x³/6·(x) = x⁴/6, from x⁴/24·(none). So x⁴ coefficient: (x

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