Algebra (Pure 3) · 代数(Pure 3)
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| partial fractions/ˈpɑːʃl ˈfrækʃnz/ | 部分分式 | bù fèn fēn shì |
| rational function/ˈræʃənl ˈfʌŋkʃn/ | 有理函数 | yǒu lǐ hán shù |
| factorise/ˈfæktəraɪz/ | 因式分解 | yīn shì fēn jiě |
| repeated factor/rɪˈpiːtɪd ˈfæktə/ | 重复因式 | chóng fù yīn shì |
| irreducible quadratic/ɪrɪˈdjuːsɪbl kwɒˈdrætɪk/ | 不可约二次式 | bù kě yuē èr cì shì |
| binomial expansion/baɪˈnəʊmɪəl ekˈspænʃn/ | 二项展开式 | èr xiàng zhǎn kāi shì |
Splitting the unsplitable
- You can't integrate $\dfrac{1}{x^2 - 1}$ directly. But split it into $\dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)}$ and each piece is a simple logarithm.
- Partial fractions 部分分式 turn hard rational functions 有理函数 into easy pieces — essential for integration and series expansion.
拆分不可积项
- 你无法直接积分 $\dfrac{1}{x^2 - 1}$。但将其拆分为 $\dfrac{1}{2(x-1)} - \dfrac{1}{2(x+1)}$，每一部分都是简单的对数形式。
- 部分分式 (Partial fractions) 将复杂的有理函数转化为简单的部分——这对于积分和级数展开至关重要。
Partial fractions
- A fraction with a factorised 因式分解 bottom splits into simpler fractions:
- This makes a rational function easy to integrate or expand.
Worked example. $\dfrac{x+4}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$. Multiply through: $x+4 = A(x-2) + B(x+1)$. Set $x = -1$: $3 = -3A \Rightarrow A = -1$. Set $x = 2$: $6 = 3B \Rightarrow B = 2$. Result: $\dfrac{2}{x-2} - \dfrac{1}{x+1}$.
The scalar product measures how much of b points along a: |b| cos theta
部分分式
- 分母因式分解后的分数可以拆分为更简单的分数：
- 这使得有理函数易于积分或展开。
例题。 $\dfrac{x+4}{(x+1)(x-2)} = \dfrac{A}{x+1} + \dfrac{B}{x-2}$。两边同乘：$x+4 = A(x-2) + B(x+1)$。令 $x = -1$：$3 = -3A \Rightarrow A = -1$。令 $x = 2$：$6 = 3B \Rightarrow B = 2$。结果：$\dfrac{2}{x-2} - \dfrac{1}{x+1}$。
标量积衡量 b 在 a 方向上的分量：|b| cos theta
Reciprocal curves · 倒数曲线
y = a/(x − b) + c
Partial fractions split a hard fraction into simple reciprocal · 倒数 pieces — each with a vertical and a horizontal asymptote. · 部分分式把一个困难的分数拆成简单的倒数片段——每个有一条垂直和一条水平渐近线。
Writing (x+4)/((x+1)(x−2)) = A/(x+1) + B/(x−2), substitute x = 2 to find B. · 写 (x+4)/((x+1)(x−2)) = A/(x+1) + B/(x−2),代入 x = 2 求 B。
At x = 2: 2 + 4 = B(2 + 1), so 6 = 3B, giving B = 2. · 在 x = 2:2 + 4 = B(2 + 1),所以 6 = 3B,给出 B = 2。
Partial fractions make a rational function easier to integrate. · 部分分式使一个有理函数更容易积分。
Each simple fraction integrates to a logarithm, so splitting first makes integration easy. · 每个简单的分数积分成一个对数,所以先拆开使积分容易。
Repeated and quadratic factors
- Repeated factor 重复因式 $(cx+d)^2$: needs $\dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}$.
- Irreducible quadratic 不可约二次式 $(x^2 + a^2)$: needs $\dfrac{Bx + C}{x^2 + a^2}$.
Don't forget the repeated factor term. $\dfrac{1}{x(x+1)^2} = \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$. Missing the $\dfrac{C}{(x+1)^2}$ term is a common error.
Holiday figure
重复因子与二次因子
- 重复因子 (Repeated factor) $(cx+d)^2$：需要 $\dfrac{B}{cx+d} + \dfrac{C}{(cx+d)^2}$。
- 不可约二次式 (Irreducible quadratic) $(x^2 + a^2)$：需要 $\dfrac{Bx + C}{x^2 + a^2}$。
不要忘记重复因子项。 $\dfrac{1}{x(x+1)^2} = \dfrac{A}{x} + \dfrac{B}{x+1} + \dfrac{C}{(x+1)^2}$。遗漏 $\dfrac{C}{(x+1)^2}$ 项是常见的错误。
假期插图
For 1/(x(x+1)²), the partial fraction form includes: · 对于 1/(x(x+1)²),部分分式形式包括:
A repeated factor (x+1)² needs both B/(x+1) and C/(x+1)² terms. · 一个重复因式 (x+1)² 需要 B/(x+1) 和 C/(x+1)² 两项。
Binomial expansion 二项展开式 for rational powers
- Works for fractional/negative $n$ when $|x| < 1$:
- e.g. $(1+x)^{1/2} = 1 + \tfrac12 x - \tfrac18 x^2 + \cdots$
有理指数的二项展开式
- 当 $n$ 为分数/负数且 $|x| < 1$ 时适用：
- e.g. $(1+x)^{1/2} = 1 + \tfrac12 x - \tfrac18 x^2 + \cdots$
The binomial expansion (1+x)ⁿ for a fractional or negative n is valid only when: · 对分数或负数的 n,二项展开 (1+x)ⁿ 只在以下时有效:
For non-positive-integer powers the series converges only for |x| < 1. · 对非正整数的幂,级数只对 |x| < 1 收敛。
In the expansion of (1+x)^(1/2), the coefficient of x² is n(n-1)/2! where n=1/2. Find it. · 在 (1+x)^(1/2) 的展开中,x² 的系数是 n(n-1)/2!,其中 n=1/2。求它。
(1/2)(1/2 - 1)/2 = (1/2)(-1/2)/2 = -1/8 = -0.125. · (1/2)(1/2 - 1)/2 = (1/2)(-1/2)/2 = -1/8 = -0.125。
Worked example — integrating with partial fractions
- $\displaystyle\int \frac{x+4}{(x+1)(x-2)}\,dx = \int\left(\frac{2}{x-2} - \frac{1}{x+1}\right)dx = 2\ln|x-2| - \ln|x+1| + C$.
allow_no_figure: Algebraic technique (partial fractions decomposition) — the visual content is the algebraic manipulation itself, not a diagram.
- Use the factor theorem and remainder theorem: dividing a polynomial gives a quotient and a remainder.
例题——使用部分分式进行积分
- $\displaystyle\int \frac{x+4}{(x+1)(x-2)}\,dx = \int\left(\frac{2}{x-2} - \frac{1}{x+1}\right)dx = 2\ln|x-2| - \ln|x+1| + C$.
allow_no_figure: 代数技巧（部分分式分解）——视觉内容本身就是代数运算，而非图表。
- 使用因式定理和余数定理：多项式除法会产生商式和余式。
∫ 2/(x-2) dx = 2 ln|x-2| + C. At x = 3, what is 2 ln|x-2| (to 2 dp)? · ∫ 2/(x-2) dx = 2 ln|x-2| + C。在 x = 3,2 ln|x-2| 是多少(到 2 位小数)?
2 ln|3-2| = 2 ln(1) = 2 × 0 = 0. · 2 ln|3-2| = 2 ln(1) = 2 × 0 = 0。
You've got it
- partial fractions split a rational function for integrating/expanding
- match the form to the bottom (repeated factor → an extra $\dfrac{C}{(cx+d)^2}$ term)
- the binomial works for rational $n$ when $|x| < 1$
你掌握了
- 部分分式将有理函数拆分以便积分/展开
- 根据分母匹配形式（重复因子 → 增加一个 $\dfrac{C}{(cx+d)^2}$ 项）
- 二项式在 $n$ 为有理数且 $|x| < 1$ 时适用