Taylor Series Revision Guide for IGCSE OCR Maths | IGCSE OCR 数学：泰勒级数考点精讲

📚 Taylor Series Revision Guide for IGCSE OCR Maths | IGCSE OCR 数学：泰勒级数考点精讲

Taylor series allow us to represent complicated functions as infinite sums of polynomial terms built from the function’s derivatives at a single point. This powerful tool not only makes difficult functions easier to handle in calculus, but also underpins many approximations used in physics and engineering. In IGCSE OCR Maths, you are expected to understand the principle behind Taylor expansions, construct series for common functions like eˣ, sin x, cos x and ln(1+x), and use them to estimate function values to a specified accuracy.

泰勒级数使我们能够将复杂的函数表示为一个由该函数在某一点的各阶导数构成的无穷多项式之和。这个强大的工具不仅能让我们在微积分中更轻松地处理复杂函数，而且还是物理学和工程学中许多近似计算的基础。在IGCSE OCR数学考试中，你需要理解泰勒展开的原理，能够构造出常用函数（如 eˣ、sin x、cos x 和 ln(1+x)）的级数，并能利用它们来估算达到指定精度的函数值。

1. What Is a Taylor Series? | 什么是泰勒级数？

Imagine you want to evaluate a function f(x) near a point a where you know its value f(a) and the values of all its derivatives f'(a), f”(a), f”'(a), etc. The Taylor series builds a polynomial approximation using this local information: the more terms you include, the closer the polynomial gets to the original function, at least near x = a.

假设你想在点 a 附近计算函数 f(x) 的值，并且你已经知道该点的函数值 f(a) 以及所有各阶导数值 f'(a)、f”(a)、f”'(a) 等。泰勒级数就是利用这些局部信息构造出一个多项式逼近：你包含的项数越多，多项式就越接近原函数，至少在 x = a 的附近如此。

The infinite Taylor series of f(x) about x = a is written:

函数 f(x) 在 x = a 处的无穷泰勒级数写作：

f(x) = f(a) + f'(a)(x − a) + f”(a)(x − a)²/2! + f”'(a)(x − a)³/3! + …

In more compact sigma notation this becomes:

用更简洁的求和符号写出来就是：

f(x) = Σₙ₌₀∞ [ f⁽ⁿ⁾(a) / n! ] (x − a)ⁿ

Here f⁽ⁿ⁾(a) means the n-th derivative of f evaluated at x = a, and n! is n factorial. The term for n = 0 is defined as f(a).

在这里，f⁽ⁿ⁾(a) 表示 f 在 x = a 处的 n 阶导数，n! 是 n 的阶乘。n = 0 时的项定义为 f(a)。

2. The General Formula and Key Ingredients | 通项公式与关键要素

To work with Taylor series successfully, you must be comfortable finding higher-order derivatives quickly and correctly. The formula itself is simple, but careless differentiation is the most common source of mistakes.

要熟练运用泰勒级数，你必须能够快速而准确地求高阶导数。公式本身很简单，但马虎的求导是导致错误的最常见原因。

Step 1: Choose the centre a (often a = 0 for Maclaurin series).
第一步：选择中心 a（如果 a = 0 就是麦克劳林级数）。
Step 2: Calculate f(a), f'(a), f”(a), f”'(a), …
第二步：计算 f(a)、f'(a)、f”(a)、f”'(a)……
Step 3: Substitute into the Taylor formula and simplify the factorial coefficients.
第三步：代入泰勒公式并化简阶乘系数。
Step 4: If asked, write the series up to a given number of terms or write the general term.
第四步：如果题目要求，就写出指定项数的级数或写出通项公式。

Note that the series may be infinite; in practice we often truncate it after a few terms to create an approximation.

请注意，级数可能是无穷的；在实践中我们通常会在几项之后截断，以得到一个近似值。

3. Maclaurin Series: Taylor Series at a = 0 | 麦克劳林级数：中心在 0 的泰勒级数

When the expansion point is chosen as a = 0, the Taylor series gets its own name: the Maclaurin series. The formula simplifies nicely because all the (x − a) terms become just x raised to a power.

当展开点选为 a = 0 时，泰勒级数便有了一个专门的名字：麦克劳林级数。公式变得很简洁，因为所有的 (x − a) 项都变成了 x 的幂次。

f(x) = f(0) + f'(0)x + f”(0)x²/2! + f”'(0)x³/3! + …

This is the version you will see most often in exam questions because it is easier to compute and still very powerful. Almost every standard series you memorise – for eˣ, sin x, cos x, ln(1+x) – is a Maclaurin series.

这将是你在考题中最常看到的版本，因为它更容易计算，同时功能也非常强大。你该记住的几乎每一个标准级数——eˣ、sin x、cos x、ln(1+x)——都是麦克劳林级数。

4. Building the Maclaurin Series for eˣ | 构建 eˣ 的麦克劳林级数

The exponential function f(x) = eˣ is special because all its derivatives are also eˣ. At x = 0, e⁰ = 1, so every derivative evaluates to 1.

指数函数 f(x) = eˣ 很特别，因为它所有的导数依然都是 eˣ。在 x = 0 处，e⁰ = 1，因此每一阶导数代入后都等于 1。

Thus:

因此有：

f(0) = 1, f'(0) = 1, f”(0) = 1, f”'(0) = 1, …

Substituting into the Maclaurin formula gives the well-known series:

代入麦克劳林公式就得到熟悉的级数：

eˣ = 1 + x + x²/2! + x³/3! + x⁴/4! + …

You can use this expansion to approximate e to a high accuracy by setting x = 1. For example, the sum of the first five terms (up to x⁴/4!) gives e ≈ 2.70833, already close to the true value 2.71828.

你可以通过令 x = 1，用这个展开式来高精度地估算 e。例如，前五项的和（算到 x⁴/4!）得到 e ≈ 2.70833，已经很接近真实值 2.71828 了。

5. Maclaurin Series for sin x and cos x | sin x 和 cos x 的麦克劳林级数

These two trigonometric functions produce a repeating cycle of derivatives that is easy to master.

这两个三角函数的导数呈现出容易掌握的循环规律。

For f(x) = sin x:

对于 f(x) = sin x：

f(0) = 0, f'(0) = cos 0 = 1, f”(0) = −sin 0 = 0, f”'(0) = −cos 0 = −1, and then the pattern repeats every four derivatives.
f(0) = 0，f'(0) = cos 0 = 1，f”(0) = −sin 0 = 0，f”'(0) = −cos 0 = −1，之后每四次求导循环一次。

Insert these values into Maclaurin’s formula:

将这些值代入麦克劳林公式：

sin x = x − x³/3! + x⁵/5! − x⁷/7! + …

For f(x) = cos x:

对于 f(x) = cos x：

f(0) = 1, f'(0) = −sin 0 = 0, f”(0) = −cos 0 = −1, f”'(0) = sin 0 = 0, f⁽⁴⁾(0) = 1, etc.
f(0) = 1，f'(0) = −sin 0 = 0，f”(0) = −cos 0 = −1，f”'(0) = sin 0 = 0，f⁽⁴⁾(0) = 1，依此类推。

This yields:

由此得出：

cos x = 1 − x²/2! + x⁴/4! − x⁶/6! + …

Notice that sin x only contains odd powers of x, while cos x only contains even powers. This is a quick check for correctness in the exam.

注意 sin x 只含有 x 的奇数次幂，而 cos x 只含有偶数次幂。这是在考试中快速检验正误的好方法。

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

The natural logarithm function f(x) = ln(1 + x) is defined for x > −1 and has a Maclaurin series that is only valid for −1 < x ≤ 1.

自然对数函数 f(x) = ln(1 + x) 的定义域是 x > −1，其麦克劳林级数仅在 −1 < x ≤ 1 时收敛。

Calculate the derivatives:

计算各阶导数：

f'(x) = (1+x)⁻¹, so f'(0) = 1
f”(x) = −(1+x)⁻², so f”(0) = −1
f”'(x) = 2(1+x)⁻³, so f”'(0) = 2
f⁽⁴⁾(x) = −6(1+x)⁻⁴, so f⁽⁴⁾(0) = −6

The pattern gives f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)! for n ≥ 1. Substituting into the Maclaurin formula with f(0)=0 leads to:

由此可以得出模式：对于 n ≥ 1，f⁽ⁿ⁾(0) = (−1)ⁿ⁻¹ (n−1)!。代入麦克劳林公式，且 f(0)=0，得到：

ln(1 + x) = x − x²/2 + x³/3 − x⁴/4 + …

This alternating harmonic series is a favourite in exam questions because it converges slowly and thus illustrates the need for many terms to achieve good accuracy near x = 1.

这个正负交替的调和级数是考试中的常见考点，因为它收敛很慢，能很好地说明在 x = 1 附近要想获得足够的精度需要很多项。

7. Taylor Series for Functions Not Centred at Zero | 中心不在零的泰勒级数

Sometimes an exam question will ask you to expand a function about a point other than 0, for example a = 1 or a = π/4. The procedure is identical: compute f(a), f'(a), f”(a), etc., and plug into the general Taylor formula with (x − a).

有时考题会要求你围绕某个非零点展开函数，比如 a = 1 或 a = π/4。步骤完全一样：计算 f(a)、f'(a)、f”(a) 等，然后代入含有 (x − a) 的通式。

Example: Expand f(x) = √x about a = 4 up to the (x−4)² term.

例：将 f(x) = √x 在 a = 4 处展开到 (x−4)² 项。

f(4) = 2
f'(x) = ½ x⁻½, so f'(4) = ¼
f”(x) = −¼ x⁻³/², so f”(4) = −1/32

Thus:

因此：

√x ≈ 2 + ¼ (x−4) − (1/64)(x−4)²

Such local approximations are extremely useful when you only need to estimate values near a known point.

当你只需要估算一个已知点附近的函数值时，这样的局部近似非常有用。

8. Using Taylor Polynomials to Approximate Function Values | 用泰勒多项式估算函数值

In IGCSE OCR Maths, you will often be asked to use the first few terms of a Taylor series to find an approximate value of a function and to compare it with the true value or to bound the error.

在IGCSE OCR数学中，你经常会被要求用泰勒级数的前几项来求一个函数的近似值，并将其与真实值比较，或者给出误差范围。

Approximation steps:

近似步骤：

Identify the function and the point a around which to expand.
确定函数以及所要展开的中心点 a。
Write down the Taylor polynomial of the required degree.
写出题目要求的泰勒多项式次数。
Substitute the given x-value and compute the result.
代入给定的 x 值并计算结果。
If the exact value is known, calculate the absolute error. If not, use the next term to estimate the maximum error (Lagrange remainder).
如果已知真实值，计算绝对误差；如果未知，则用下一项来估算最大误差（拉格朗日余项）。

For instance, using the Maclaurin polynomial of degree 3 for sin x: sin x ≈ x − x³/6. To estimate sin 0.2, substitute x = 0.2 to get 0.2 − (0.008)/6 ≈ 0.198667. The true value is about 0.198669, showing excellent agreement.

例如，用 sin x 的三次麦克劳林多项式：sin x ≈ x − x³/6。要估算 sin 0.2，代入 x = 0.2 得到 0.2 − (0.008)/6 ≈ 0.198667。真实值约为 0.198669，两者吻合得非常好。

9. Error Bound and the Lagrange Remainder | 误差界与拉格朗日余项

When we truncate a Taylor series after the term of degree n, the error introduced is given by the Lagrange remainder:

当我们在 n 次项之后截断泰勒级数时，引入的误差由拉格朗日余项给出：

Rₙ(x) = f⁽ⁿ⁺¹⁾(c) (x − a)ⁿ⁺¹ / (n+1)!

where c is some number between a and x. Although you are not required to compute c exactly, you can find an upper bound for the error by maximizing |f⁽ⁿ⁺¹⁾(c)| on the interval.

其中 c 是介于 a 和 x 之间的某个数。虽然你不必精确求出 c，但你可以通过求区间内 |f⁽ⁿ⁺¹⁾(c)| 的最大值来找到误差的上界。

Example: For eˣ expanded around 0 to degree 3, the remainder for approximating e⁰·⁵ is bounded by (e⁰·⁵ × (0.5)⁴)/4!. Since e⁰·⁵ < 1.65, the error is less than 0.0043. Such reasoning shows how many terms are needed to guarantee a desired precision.

例：对于 eˣ 在 0 点附近展开到三次来估算 e⁰·⁵，其余项被 (e⁰·⁵ × (0.5)⁴)/4! 所限定。因为 e⁰·⁵ < 1.65，误差小于 0.0043。这样的推理展示出为了保证所需的精度需要多少项。

10. Taylor Series and the Binomial Expansion (1 + x)ⁿ | 泰勒级数与二项式展开 (1 + x)ⁿ

The binomial theorem for integer powers is a special case of Taylor expansion. For any real exponent n, the Maclaurin series for (1 + x)ⁿ is:

整数次幂的二项式定理其实是泰勒展开的一个特例。对于任意实数指数 n，(1+x)ⁿ 的麦克劳林级数是：

(1 + x)ⁿ = 1 + n x + [n(n−1)/2!] x² + [n(n−1)(n−2)/3!] x³ + …

This series is valid for |x| < 1 when n is not a positive integer. IGCSE OCR often tests the ability to derive this series directly from Taylor’s formula by computing derivatives of (1+x)ⁿ at x=0.

当 n 不是正整数时，此级数在 |x| < 1 时成立。IGCSE OCR经常考查用泰勒公式通过计算 (1+x)ⁿ 在 x=0 处的导数来直接推导这个级数的能力。

For n = −1, this becomes the geometric series: 1 − x + x² − x³ + …

当 n = −1 时，这就变成了几何级数：1 − x + x² − x³ + …

11. Common Exam Mistakes and How to Avoid Them | 常见考试错误与避坑指南

Even well-prepared candidates lose marks on Taylor series questions because of small algebraic slips. Here are the most frequent pitfalls:

即便是准备充分的考生也会因为小小的代数失误而在泰勒级数题目上丢分。以下是最常见的陷阱：

Forgetting factorials in denominators: Always divide the n-th derivative by n!. Write a quick check: term in x³ must have 3! = 6 in the denominator.
忘记分母中的阶乘：务必用第 n 阶导数除以 n!。速检法：x³ 项的分母必须含有 3! = 6。
Mixing up signs: Differentiate carefully, especially for sin, cos, and ln. A sign error early on propagates through the whole series.
弄混正负号：求导时要格外小心，尤其是 sin、cos 和 ln 的导数。前期的符号错误会波及整个级数。
Wrong centre point: If a ≠ 0, don’t accidentally use the Maclaurin form. Remember to include (x − a) and not just x.
搞错展开中心：如果 a ≠ 0，千万别误用麦克劳林的形式。记住括号里是 (x − a) 而不只是 x。
Truncating too early: If a question specifies ‘up to the term in x⁴’, include all terms up to x⁴, i.e. do not stop at x³.
过早截断：如果题目要求“直到含 x⁴ 项”，请写出所有直到 x⁴ 的项，亦即在 x³ 处不要停。
Ignoring validity range: For series like ln(1+x) and (1+x)ⁿ, always note the interval of convergence when asked.
忽略收敛域：对于像 ln(1+x) 和 (1+x)ⁿ 这样的级数，当被问及时必须注明收敛区间。

A good habit is to quickly differentiate your series back to see if it matches the original function’s derivative pattern near the centre.

一个好的习惯是，快速将你的级数求导回去看看它是否在原函数中心附近与导数模式相吻合。

12. Summary and Key Takeaways | 总结与关键要点

Taylor series provide a bridge between polynomials and more complex functions, making them an essential topic for IGCSE OCR Mathematics. The core formula is simple, but success lies in accurate differentiation and careful algebraic simplification.

泰勒级数在多项式和更复杂的函数之间架起了一座桥梁，因此在IGCSE OCR数学中是一个必不可少的知识点。核心公式虽然简单，但成功的关键在于准确的求导和细致的代数化简。

Key points to remember:

需要记住的要点：

Maclaurin series are Taylor series with a = 0, and they cover most exam questions.
麦克劳林级数就是 a = 0 的泰勒级数，涵盖了大部分考题。
Memorise the standard expansions for eˣ, sin x, cos x and ln(1+x) – they save time.
记住 eˣ、sin x、cos x 和 ln(1+x) 的标准展开式，这能节省时间。
Always check that factorials are placed correctly in denominators.
务必检查阶乘是否正确放在分母中。
When approximating, be ready to bound the error using the next term or the Lagrange remainder.
在近似计算时，要能用下一项或拉格朗日余项来给出误差范围。
Practise deriving series from first principles – this is a frequently examined skill.
多练习从基本原理出发推导级数，这是经常考查的技能。

With consistent practice, Taylor series will become one of the most rewarding topics, offering straightforward marks for a methodical approach.

只要坚持练习，泰勒级数将成为回报最高的课题之一，用条理清晰的方法便可轻松拿到分数。

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Taylor Series Revision Guide for IGCSE OCR Maths | IGCSE OCR 数学：泰勒级数考点精讲