📚 Maclaurin Expansion | 麦克劳林展开考点精讲
h simple polynomials near x = 0. Understanding the derivation and standard expansions of eˣ s x, ln(1+x), and (1+x)ⁿ is essential for both calculation and proof-based exam questions.
麦克劳林展开是 IGCSE 附加数学中一项强有力的工具，它让我们可以用简单的多项式在 x=0 附近逼近复杂的函数。掌握 eˣ、sin x、cos x、ln(1+x) 和 (1+x)ⁿ 等标准展开式的推导与应用，对计算题和证明题都至关重要。
1. What is Maclaurin Expansion? | 什么是麦克劳林展开？
entiable function f(x) as an infinite sum of terms built from the values of its derivatives at zero. This means we can replace a difficult function with a polynomial that is much easier to evaluate, as long as x is near zero.
麦克劳林级数是中心在 x=0 处的泰勒级数。它将可导函数 f(x) 表示成无穷多项的和，各项由该函数在零点处的导数值构造而成。这意味着，只要 x 靠近零，我们就可以用更容易计算的多项式来替代复杂的函数。
The core idea is that smooth functions can be approximated by polynomials whose coefficients are determined by f(0), f'(0), f''(0), and so on. The more terms we include, the closer the approximation is to the true function, provided we stay within the radius of convergence.
核心思想是，光滑函数可以用多项式来近似，多项式的系数由 f(0)、f'(0)、f''(0) 等决定。只要在收敛半径内，包含的项越多，近似值就越接近真实函数。
2. General Formula | 一般公式
The Maclaurin series for a function f(x) is written as:
f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... + f⁽ⁿ⁾(0)xⁿ/n! + ...
函数 f(x) 的麦克劳林级数一般形式为：
f(x) = f(0) + f'(0)x + f''(0)x²/2! + f'''(0)x³/3! + ... + f⁽ⁿ⁾(0)xⁿ/n! + ...
Here, f⁽ⁿ⁾(0) denotes the n-th derivative of f evaluated at zero, and n! is the factorial of n. By convention, 0! = 1, so the constant term f(0) is simply f⁽⁰⁾(0)x⁰/0!.
这里 f⁽ⁿ⁾(0) 表示 f 在 0 点的 n 阶导数，n! 是 n 的阶乘。规定 0! = 1，因此常数项 f(0) 正是 f⁽⁰⁾(0)x⁰/0!。
If f is a polynomial of degree m, then all derivatives of order higher than m are zero, so the series terminates and becomes an exact finite expression rather than an infinite sum.
如果 f 是 m 次多项式，则所有高于 m 阶的导数都为零，级数将终止，成为一个精确的有限表达式，而不是无穷级数。
3. Expansion of eˣ | eˣ 的展开
For f(x) = eˣ, every derivative is eˣ itself, so f⁽ⁿ⁾(x) = eˣ for all n. Evaluating at zero gives f⁽ⁿ⁾(0) = e⁰ = 1. Therefore, the Maclaurin series is:
对 f(x) = eˣ，它的所有阶导数还是 eˣ，因此 f⁽ⁿ⁾(x) = eˣ 对所有 n 成立。代入零得 f⁽ⁿ⁾(0) = e⁰ = 1。于是麦克劳林级数为：
eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + ... + xⁿ/n! + ...
exceptionally useful. For a quick approximation, we can truncate after a few terms, e.g. eˣ ≈ 1 + x + x²/2 + x³/6. This polynomial can estimate e⁰·² accurately with only a small error.
这个展开式对所有实数 x 都成立，因此非常有用。若要快速近似，可截取前几项，比如 eˣ ≈ 1 + x + x²/2 + x³/6。用此多项式估算 e⁰·² 只会产生很小的误差。
4. Expansion of sin x | sin x 的展开
(x) = sin x, the derivatives cycle through sin x, cos x, -sin x, -cos x. At x = 0 we have:
对 f(x) = sin x，导数依次循环：sin x, cos x, -sin x, -cos x。在 x=0 处有：
f(0) = sin 0 = 0 / f(0) = sin0 = 0
f'(0) = cos 0 = 1 / f'(0) = cos0 = 1
f''(0) = -sin 0 = 0 / f''(0) = -sin0 = 0
f'''(0) = -cos 0 = -1 / f'''(0) = -cos0 = -1
f''''(0) = sin 0 = 0, and the pattern repeats. / f''''(0) = sin0 = 0，依此重复。
Since all even-order derivatives vanish at zero, the series contains only odd powers of x, with alternating signs:
因为所有偶数阶导数在零点均为零，级数中只含 x 的奇次幂，且符号交替：
sin x = x - x³/3! + x⁵/5! - x⁷/7! + ...
e approximation sin x ≈ x is its first term and is widely used in physics and engineering.
该展开对任意实数 x 收敛。小角近似 sin x ≈ x 正是其第一项，在物理和工程中应用广泛。
5. Expansion of cos x | cos x 的展开
For f(x) = cos x, derivatives cycle as -sin x, -cos x, sin x, cos x. At x = 0:
对 f(x) = cos x，导数依次为 -sin x, -cos x, sin x, cos x。在 x=0 处：
f(0) = 1 / f(0)=1
f'(0) = 0 / f'(0)=0
f''(0) = -1 / f''(0)=-1
f'''(0) = 0 / f'''(0)=0
f''''(0) = 1, and repeats. / f''''(0)=1，此后循环。
Every odd derivative at zero is zero, so the Maclaurin series has only even powers of x, also with alternating signs:
所有奇数次导数在零点均为零，因此麦克劳林级数只含 x 的偶次幂，符号也交替：
cos x = 1 - x²/2! + x⁴/4! - x⁶/6! + ...
1 - x²/2 is extremely common when x is small, for example in simple pendulum calculations.
该级数也对所有实数 x 成立。两项近似 cos x ≈ 1 - x²/2 在处理小量 x 时极为常见，例如单摆小角度近似。
6. Expansion of ln(1 + x) | ln(1 + x) 的展开
For f(x) = ln(1 + x), we begin with f(0) = ln 1 = 0. The first few derivatives are:
对 f(x) = ln(1 + x)，先有 f(0) = ln1 = 0。前几阶导数为：
f'(x) = (1 + x)⁻¹ → f'(0) = 1
f''(x) = -(1 + x)⁻² → f''(0) = -1
f'''(x) = 2(1 + x)⁻³ → f'''(0) = 2
f⁽⁴⁾(x) = -6(1 + x)⁻⁴ → f⁽⁴⁾(0) = -6
The resulting Maclaurin series is:
因此麦克劳林级数为：
ln(1 + x) = x - x²/2 + x³/3 - x⁴/4 + ... + (-1)ⁿ⁻¹ xⁿ/n + ...
4 + ... At x = -1, the series diverges. Always state the domain when using this expansion.
该展开只在 -1 < x ≤ 1 时有效。x = 1 时，得到著名的交错调和级数 ln2 = 1 - 1/2 + 1/3 - 1/4 + ...；x = -1 时级数发散。使用时务必指明定义域。
7. Binomial (1 + x)ⁿ as Maclaurin | (1 + x)ⁿ 的二项展开作为麦克劳林展开
ation of Maclaurin series. Let f(x) = (1 + x)ⁿ. Then f(0) = 1, and the derivatives follow:
对于任意实数 n，二项展开式 (1 + x)ⁿ 是麦克劳林级数的直接应用。令 f(x) = (1 + x)ⁿ，有 f(0)=1，各阶导数如下：
f'(x) = n(1 + x)ⁿ⁻¹ → f'(0) = n
f''(x) = n(n-1)(1 + x)ⁿ⁻² → f''(0) = n(n-1)
f'''(x) = n(n-1)(n-2)(1 + x)ⁿ⁻³ → f'''(0) = n(n-1)(n-2)
Thus the series is:
因此级数为：
(1 + x)ⁿ = 1 + nx + [n(n-1)/2!] x² + [n(n-1)(n-2)/3!] x³ + ...
When n is a positive integer, the series stops after the xⁿ term and is a finite polynomial valid for all x. When n is fractional or negative, the series is infinite and converges only for |x| < 1. This is a crucial distinction in exam validations.
当 n 为正整数时，级数在 xⁿ 项后停止，成为对所有 x 都成立的有限多项式。当 n 为分数或负数时，级数为无穷级数，仅在 |x| < 1 时收敛。这是考试中验证有效性的关键区别。
8. Finding a Specific Coefficient | 求特定项的系数
A common exam question is: "Find the coefficient of x³ in the Maclaurin expansion of f(x)". The direct method is to differentiate f three times, evaluate at zero, and divide by 3!. However, when f is a product or composition of known functions, multiplying the known series can save a great deal of time.
常见的考题是：“求 f(x) 的麦克劳林展开中 x³ 的系数”。直接方法是求导三次后代入零并除以 3!。然而，当 f 是已知函数的乘积或复合时，将已知级数相乘可以节省大量时间。
For example, to obtain the cubic term in eˣ sin x, write:
例如，要求 eˣ sin x 中的三次项，可写出：
eˣ ≈ 1 + x + x²/2 + x³/6, sin x ≈ x - x³/6.
Multiply and collect terms up to x³:
将二者相乘并收集到 x³ 项：
²/2 + x³/6)(x - x³/6) = x + x² + (1/2)x³ - (1/6)x³ + higher terms = x + x² + (1/3)x³ + ...
Thus the coefficient of x³ is 1/3. This is far quicker than computing the third derivative of eˣ sin x, which involves product rule three times.
因此 x³ 的系数为 1/3。这比用乘法法则对 eˣ sin x 求三次导快得多。
Similarly, for compositions like esin x, substitute the sin x series into the eˣ series and expand. Always be
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