Taylor Series at Other Points | 其他点处的泰勒级数

📚 Taylor Series at Other Points | 其他点处的泰勒级数

In IB Mathematics, students first meet Taylor series through the Maclaurin expansion, where a function is represented as an infinite sum of powers of \(x\) around \(x=0\). However, many functions are better approximated or analysed around a different point, such as \(x=1\), \(x=\pi/4\), or any other value in their domain. This article explores how to construct and apply Taylor series centred at arbitrary points \(x=a\), a key skill for IB Higher Level paper 3 questions and beyond.

在IB数学中，学生首先通过麦克劳林展开认识泰勒级数，即在 \(x=0\) 附近把函数表示为 \(x\) 的幂的无穷和。然而，许多函数在域内其他点（如 \(x=1\)、\(x=\pi/4\) 等）附近能得到更好的近似或分析。本文探讨如何构造并应用以任意点 \(x=a\) 为中心的泰勒级数，这是IB高级别试卷三及其他高阶内容中的关键技能。

1. General Taylor Series Definition | 一般泰勒级数定义

For a function \(f(x)\) that is infinitely differentiable at \(x=a\), the Taylor series centred at \(a\) is given by:

\(f(x) = f(a) + f'(a)(x-a) + \frac{f”(a)}{2!}(x-a)^2 + \frac{f”'(a)}{3!}(x-a)^3 + \cdots\)

Equivalently, in sigma notation:

\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)

Here \(f^{(n)}(a)\) denotes the \(n\)-th derivative of \(f\) evaluated at \(x=a\), and \(n!\) is the factorial of \(n\). The point \(a\) is called the centre or expansion point.

对于在 \(x=a\) 处无限可微的函数 \(f(x)\)，以 \(a\) 为中心的泰勒级数为：

\(f(x) = f(a) + f'(a)(x-a) + \frac{f”(a)}{2!}(x-a)^2 + \frac{f”'(a)}{3!}(x-a)^3 + \cdots\)

等价地，用求和符号表示：

\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n\)

其中 \(f^{(n)}(a)\) 表示 \(f\) 在 \(x=a\) 处的 \(n\) 阶导数，\(n!\) 是 \(n\) 的阶乘。点 \(a\) 称为展开的中心点。

2. Difference from Maclaurin Series | 与麦克劳林级数的区别

The Maclaurin series is a special case of the Taylor series with \(a=0\). When \(a=0\), the formula simplifies to:

\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\)

This is sometimes called the Taylor series at the origin. In IB, you must recognise that any Maclaurin series is just a Taylor series centred at \(0\), but when asked for a Taylor expansion about \(x=a\), you must use powers of \((x-a)\), not \(x\).

麦克劳林级数是泰勒级数在 \(a=0\) 时的特例。当 \(a=0\) 时，公式简化为：

\(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\)

这有时也称作在原点的泰勒展开。在IB中，你需要知道麦克劳林级数只是以 \(0\) 为中心的泰勒级数，但若题目要求展开关于 \(x=a\)，则必须使用 \((x-a)\) 的幂，而不是 \(x\) 的幂。

3. Computing Coefficients | 计算系数

To construct a Taylor series at \(x=a\), the key steps are:

Find \(f(a)\): evaluate the function at the centre.
Find \(f'(x)\), \(f”(x)\), \(f”'(x)\), etc., and evaluate each at \(x=a\).
Substitute into the general formula and simplify each term.
Write the first few non-zero terms, and state the general term if possible.

构造在 \(x=a\) 处的泰勒级数，关键步骤如下：

求 \(f(a)\)：计算函数在中心点的值。
求 \(f'(x)\)、\(f”(x)\)、\(f”'(x)\) 等，并分别代入 \(x=a\) 求值。
代入一般公式并化简每一项。
写出前几个非零项，并尽可能写出通项公式。

Remember that the factor \(n!\) in the denominator must be included. For example, the coefficient of \((x-a)^2\) is \(f”(a)/2\), not simply \(f”(a)\).

注意分母中的 \(n!\) 不能遗漏。例如 \((x-a)^2\) 的系数是 \(f”(a)/2\)，而不是 \(f”(a)\)。

4. Convergence and Validity | 收敛性与有效性

A Taylor series centred at \(a\) converges to \(f(x)\) only for \(x\) within the radius of convergence \(R\), meaning \(|x-a|

以 \(a\) 为中心的泰勒级数仅在收敛半径 \(R\) 内（即 \(|x-a|

For example, the Taylor series for \(e^x\) centred at \(0\) has \(R=\infty\), but the series for \(\ln x\) centred at \(1\) has radius \(R=1\), so it only converges for \(0

例如，\(e^x\) 在 \(0\) 处的泰勒级数 \(R=\infty\)，但 \(\ln x\) 在 \(1\) 处的级数半径 \(R=1\)，因此仅在 \(0

In IB problems, you may be asked to state the interval of validity. Check whether the series is a standard geometric-like series or use the ratio test with the general term.

在IB题目中，可能会要求你写出有效区间。判断方法是看级数是否为类似几何级数的形式，或对通项使用比值审敛法。

5. Example: \(e^x\) at \(x=1\) | 实例：\(e^x\) 在 \(x=1\) 处展开

Let \(f(x)=e^x\). Since \(f^{(n)}(x)=e^x\) for every positive integer \(n\), we have \(f^{(n)}(1)=e\), a constant. Thus the Taylor series about \(x=1\) is:

\(e^x = e + e(x-1) + \frac{e}{2!}(x-1)^2 + \frac{e}{3!}(x-1)^3 + \cdots\)

This simplifies to \(e^x = e \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!}\). Note that this series is valid for all real \(x\).

设 \(f(x)=e^x\)。因为对所有正整数 \(n\)，\(f^{(n)}(x)=e^x\)，所以 \(f^{(n)}(1)=e\)，是常数。因此在 \(x=1\) 处的泰勒级数为：

\(e^x = e + e(x-1) + \frac{e}{2!}(x-1)^2 + \frac{e}{3!}(x-1)^3 + \cdots\)

可简化为 \(e^x = e \sum_{n=0}^{\infty} \frac{(x-1)^n}{n!}\)。注意该级数对所有实数 \(x\) 都有效。

This example shows that even when the centre changes, the exponential function retains an infinite radius of convergence because its derivatives are always well-behaved.

这个例子说明，即使中心改变，指数函数仍具有无穷收敛半径，因为其各阶导数始终表现良好。

6. Example: \(\sin x\) at \(x=\pi/4\) | 实例：\(\sin x\) 在 \(x=\pi/4\) 处展开

Let \(f(x)=\sin x\). We need derivatives at \(a=\pi/4\):

\(f(\pi/4) = \sin(\pi/4) = \sqrt{2}/2\)
\(f'(\pi/4) = \cos(\pi/4) = \sqrt{2}/2\)
\(f”(\pi/4) = -\sin(\pi/4) = -\sqrt{2}/2\)
\(f”'(\pi/4) = -\cos(\pi/4) = -\sqrt{2}/2\)
\(f^{(4)}(\pi/4) = \sin(\pi/4) = \sqrt{2}/2\)

Thus the Taylor series begins:

\(\sin x = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) – \frac{\sqrt{2}}{2 \cdot 2!}(x-\frac{\pi}{4})^2 – \frac{\sqrt{2}}{2 \cdot 3!}(x-\frac{\pi}{4})^3 + \cdots\)

设 \(f(x)=\sin x\)。我们需要在 \(a=\pi/4\) 处的各阶导数：

\(f(\pi/4) = \sin(\pi/4) = \sqrt{2}/2\)
\(f'(\pi/4) = \cos(\pi/4) = \sqrt{2}/2\)
\(f”(\pi/4) = -\sin(\pi/4) = -\sqrt{2}/2\)
\(f”'(\pi/4) = -\cos(\pi/4) = -\sqrt{2}/2\)
\(f^{(4)}(\pi/4) = \sin(\pi/4) = \sqrt{2}/2\)

因此泰勒级数开头为：

\(\sin x = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x-\frac{\pi}{4}) – \frac{\sqrt{2}}{2 \cdot 2!}(x-\frac{\pi}{4})^2 – \frac{\sqrt{2}}{2 \cdot 3!}(x-\frac{\pi}{4})^3 + \cdots\)

Notice the alternating pattern of signs. This expansion is particularly useful when approximating \(\sin x\) near \(x=\pi/4\), such as in numerical integration or solving differential equations.

注意符号交替出现的规律。该展开在 \(x=\pi/4\) 附近近似 \(\sin x\) 时尤其有用，例如在数值积分或求解微分方程时。

7. Example: \(\ln x\) at \(x=2\) | 实例：\(\ln x\) 在 \(x=2\) 处展开

For \(f(x)=\ln x\), derivatives are:

\(f'(x)=1/x\), so \(f'(2)=1/2\)
\(f”(x)=-1/x^2\), so \(f”(2)=-1/4\)
\(f”'(x)=2/x^3\), so \(f”'(2)=2/8 = 1/4\)
\(f^{(4)}(x)=-6/x^4\), so \(f^{(4)}(2)=-6/16 = -3/8\)

Also \(f(2)=\ln 2\). The Taylor series about \(x=2\) is:

\(\ln x = \ln 2 + \frac{1}{2}(x-2) – \frac{1}{4 \cdot 2!}(x-2)^2 + \frac{1}{4 \cdot 3!}(x-2)^3 – \frac{3}{8 \cdot 4!}(x-2)^4 + \cdots\)

对于 \(f(x)=\ln x\)，各阶导数为：

\(f'(x)=1/x\)，所以 \(f'(2)=1/2\)
\(f”(x)=-1/x^2\)，所以 \(f”(2)=-1/4\)
\(f”'(x)=2/x^3\)，所以 \(f”'(2)=2/8 = 1/4\)
\(f^{(4)}(x)=-6/x^4\)，所以 \(f^{(4)}(2)=-6/16 = -3/8\)

同时 \(f(2)=\ln 2\)。在 \(x=2\) 处的泰勒级数为：

\(\ln x = \ln 2 + \frac{1}{2}(x-2) – \frac{1}{4 \cdot 2!}(x-2)^2 + \frac{1}{4 \cdot 3!}(x-2)^3 – \frac{3}{8 \cdot 4!}(x-2)^4 + \cdots\)

The radius of convergence here is \(R=2\), so the series is valid for \(0

此处收敛半径为 \(R=2\)，因此级数在 \(0

8. Application: Approximating Values | 应用：近似计算

Taylor series at other points allow us to approximate function values near the centre with high accuracy, especially when direct calculation is difficult. For instance, to estimate \(\ln(2.2)\) using the series centred at \(x=2\), set \(x-2 = 0.2\). The first few terms give:

\(\ln(2.2) \approx \ln 2 + 0.5(0.2) – 0.25(0.2)^2/2 + 0.25(0.2)^3/6 – \cdots\)

其他点处的泰勒级数可以在中心附近高精度近似函数值，尤其当直接计算困难时。例如，用 \(x=2\) 处的级数估计 \(\ln(2.2)\)，令 \(x-2 = 0.2\)。前几项给出：

\(\ln(2.2) \approx \ln 2 + 0.5(0.2) – 0.25(0.2)^2/2 + 0.25(0.2)^3/6 – \cdots\)

Calculating this yields \(\ln(2.2) \approx 0.7885\), which is close to the actual value \(0.788457…\). The more terms you include, the better the approximation, provided the point lies within the radius of convergence.

计算得 \(\ln(2.2) \approx 0.7885\)，接近实际值 \(0.788457…\)。只要点在收敛半径内，项数越多近似越精确。

9. Common Mistakes | 常见错误

When working with Taylor series at other points, students often make these errors:

Using \(x\) instead of \((x-a)\) in the terms. The powers must be \((x-a)^n\).
Forgetting the factorial in the denominator.
Evaluating derivatives at \(0\) instead of at the centre \(a\).
Ignoring the radius of convergence when using the series for approximation.
Confusing \(n!\) and \(n\) in the general term.

在处理其他点处的泰勒级数时，学生常犯以下错误：

在各幂次中使用 \(x\) 而不是 \((x-a)\)。幂必须是 \((x-a)^n\)。
忘记分母中的阶乘。
在 \(0\) 处求导数值，而不是在中心 \(a\) 处求。
用级数做近似时忽略收敛半径。
混淆通项中的 \(n!\) 与 \(n\)。

To avoid these, write the general formula every time and carefully substitute \(a\) before simplifying.

为避免这些错误，每次写出一般公式，并在化简前仔细代入 \(a\)。

10. Problem-Solving Strategy for IB | IB解题策略

For IB exam questions involving Taylor series at other points, follow a systematic approach:

Read the question carefully to identify the centre \(a\). If it says “about \(x=1\)” or “at \(x=1\)”, use \(a=1\).
Compute as many derivatives as needed (usually up to order 3 or 4).
Construct the series term by term, showing your substitution clearly.
If asked for a definite integral or approximation, use the series to integrate or evaluate term by term.
Always state the domain of validity when possible.

对于涉及其他点泰勒级数的IB考题，采用系统方法：

仔细读题，确定中心 \(a\)。若题中说“关于 \(x=1\)”或“在 \(x=1\) 处”，就用 \(a=1\)。
计算所需阶数的导数（通常到3阶或4阶）。
逐项构造级数，清晰展示代入过程。
若要求定积分或近似值，逐项积分或计算。
尽可能写出有效性域。

In IB, you may also be asked to use the Taylor series of a known function (like \(e^x\) or \(\sin x\)) to obtain another function’s series by substitution, even if the centre is not 0.

在IB中，还可能会要求你利用已知函数（如 \(e^x\) 或 \(\sin x\)）的泰勒级数，通过换元得到另一个函数的级数，即使中心不是0。

11. Practice Questions | 练习问题

Try these exercises to consolidate your understanding:

尝试以下练习以巩固理解：

Find the Taylor series for \(\cos x\) centred at \(x=\pi/3\) up to the \(x\) term in \((x-\pi/3)^4\).
Determine the Taylor series for \(1/x\) about \(x=1\) and state its radius of convergence.
Use the series for \(e^x\) at \(x=1\) to estimate \(e^{1.1}\) to 4 decimal places.
Given that the Taylor series of \(f(x)\) about \(x=1\) is \(2 + 3(x-1) – (x-1)^2 + \cdots\), find \(f(1)\), \(f'(1)\), and \(f”(1)\).

利用 \(x=\pi/3\) 为中心的泰勒级数展开 \(\cos x\)，写至 \((x-\pi/3)^4\) 项。

求 \(1/x\) 关于 \(x=1\) 的泰勒级数，并说明收敛半径。

用 \(e^x\) 在 \(x=1\) 处的级数估计 \(e^{1.1}\)，精确到4位小数。

已知 \(f(x)\) 关于 \(x=1\) 的泰勒级数为 \(2 + 3(x-1) – (x-1)^2 + \cdots\)，求 \(f(1)\)、\(f'(1)\) 和 \(f”(1)\)。

Answers: (i) \(\cos x = \frac{1}{2} – \frac{\sqrt{3}}{2}(x-\pi/3) – \frac{1}{4}(x-\pi/3)^2 + \frac{\sqrt{3}}{12}(x-\pi/3)^3 + \frac{1}{48}(x-\pi/3)^4 + \cdots\)

(ii) \(\frac{1}{x} = 1 – (x-1) + (x-1)^2 – (x-1)^3 + \cdots\), radius \(R=1\).

(iii) \(e^{1.1} \approx 3.0042\).

(iv) \(f(1)=2\), \(f'(1)=3\), \(f”(1)=-2\).

Check your answers carefully. If you make a mistake, revisit the derivative evaluations and the factorial factors.

仔细核对答案。若出错，请重新检查导数求值和阶乘因子。

12. Summary | 总结

Taylor series at other points extend the Maclaurin expansion to any centre \(a\). The key is to replace \(x\) by \((x-a)\) and to evaluate all derivatives at \(a\). This technique is essential for approximating functions near specific points, understanding local behaviour, and solving problems that involve non-zero centres in IB Mathematics.

其他点处的泰勒级数将麦克劳林展开推广到任意中心 \(a\)。关键是把 \(x\) 替换为 \((x-a)\)，并在 \(a\) 处求各阶导数值。该技巧对于在特定点附近近似函数、理解局部行为，以及解决IB数学中涉及非零中心的问题至关重要。

Remember: always check the radius of convergence, and practise with multiple examples to become fluent in constructing these series.

请记住：务必检查收敛半径，并通过多个例子练习，以熟练构造这些级数。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Taylor Series at Other Points | 其他点处的泰勒级数