Series · 级数
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| converges/kənˈvɜːdʒɪz/ | 收敛 | shōu liǎn |
| geometric series/ˌdʒiːəʊˈmetrɪk ˈsɪəriːz/ | 等比级数 | děng bǐ jí shù |
| binomial expansion/baɪˈnəʊmɪəl ekˈspænʃn/ | 二项展开式 | èr xiàng zhǎn kāi shì |
| binomial coefficient/baɪˈnəʊmɪəl ˌkəʊɪˈfɪʃənt/ | 二项式系数 | èr xiàng shì xì shù |
| arithmetic progression/əˈrɪθmətɪk prəˈɡreʃn/ | 等差数列 | děng chā shù liè |
| common difference/ˈkɒmən ˈdɪfrəns/ | 公差 | gōng chāi |
| geometric progression/ˌdʒiːəʊˈmetrɪk prəˈɡreʃn/ | 等比数列 | děng bǐ shù liè |
| common ratio/ˈkɒmən ˈreɪʃɪəʊ/ | 公比 | gōng bǐ |
| sum to infinity/sʌm tʊ ɪnˈfɪnɪti/ | 无穷和 | wú qióng hé |
The infinite sum that converges 收敛
- Add $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ forever. The total? Exactly $2$.
- This is a geometric series 等比级数 with ratio $\dfrac{1}{2}$. Not all infinite sums behave this nicely — but the ones that do are powerful tools in finance, physics, and computing.
收敛的无穷和
- 永远地加 $1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$。总和?恰好是 $2$。
- 这是一个比率为 $\dfrac{1}{2}$ 的几何级数(geometric series)。不是所有的无穷和都表现得这么好——但表现好的那些是金融、物理和计算中的强大工具。
The binomial expansion 二项展开式
- For a positive integer $n$:
- where $\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ is a binomial coefficient 二项式系数.
Worked example. $(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$. The coefficients $1, 4, 6, 4, 1$ are the 4th row of Pascal's triangle.
二项展开
- 对一个正整数 $n$:
- 其中 $\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ 是一个二项系数(binomial coefficient)。
算例。 $(1+x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$。系数 $1, 4, 6, 4, 1$ 是帕斯卡三角形的第 4 行。
What is the binomial coefficient C(5, 2) = 5! / (2! 3!)? · 二项系数 C(5, 2) = 5! / (2! 3!) 是多少?
C(5,2) = 120/(2×6) = 120/12 = 10. · C(5,2) = 120/(2×6) = 120/12 = 10。
Arithmetic progressions 等差数列 (AP)
- An AP adds a constant common difference 公差 $d$ each step: $a, a+d, a+2d, \ldots$
- $n$th term: $u_n = a + (n-1)d$.
- Sum of first $n$ terms: $S_n = \dfrac{n}{2}\big(2a + (n-1)d\big)$.
等差数列(AP)
- 一个 AP(arithmetic progression,等差数列)每步加一个常数的公差(common difference)$d$:$a, a+d, a+2d, \ldots$
- 第 $n$ 项:$u_n = a + (n-1)d$。
- 前 $n$ 项的和:$S_n = \dfrac{n}{2}\big(2a + (n-1)d\big)$。
An AP has first term a = 3 and common difference d = 4. What is the 5th term? · 一个等差数列首项 a = 3、公差 d = 4。第 5 项是多少?
u₅ = a + (n−1)d = 3 + 4×4 = 3 + 16 = 19. · u₅ = a + (n−1)d = 3 + 4×4 = 3 + 16 = 19。
An AP has a = 2 and d = 3. What is the sum of the first 5 terms? · 一个等差数列 a = 2、d = 3。前 5 项的和是多少?
S₅ = (5/2)(2×2 + 4×3) = (5/2)(4 + 12) = (5/2)(16) = 40. · S₅ = (5/2)(2×2 + 4×3) = (5/2)(4 + 12) = (5/2)(16) = 40。
Geometric progressions 等比数列 (GP)
- A GP multiplies by a constant common ratio 公比 $r$ each step: $a, ar, ar^2, \ldots$
- $n$th term: $u_n = ar^{n-1}$.
- Sum of first $n$ terms: $S_n = \dfrac{a(1-r^n)}{1-r}$ (for $r \neq 1$).
AP grows linearly (add $d$); GP grows exponentially (multiply by $r$). For $r > 1$, the GP eventually dwarfs the AP.
等比数列(GP)
- 一个 GP(geometric progression,等比数列)每步乘以一个常数的公比(common ratio)$r$:$a, ar, ar^2, \ldots$
- 第 $n$ 项:$u_n = ar^{n-1}$。
- 前 $n$ 项的和:$S_n = \dfrac{a(1-r^n)}{1-r}$(对 $r \neq 1$)。
AP 线性增长(加 $d$);GP 指数增长(乘以 $r$)。对 $r > 1$,GP 最终使 AP 相形见绌。
Build a sequence · 构建一个数列
uₙ = a · rⁿ⁻¹
Each term multiplies by r. When |r| < 1 the terms shrink and the sum to infinity converges. · 每一项乘以 r。当 |r| < 1 时项缩小,无穷和收敛。
A GP has a = 2 and r = 3. What is the 4th term? · 一个等比数列 a = 2、r = 3。第 4 项是多少?
u₄ = ar³ = 2 × 3³ = 2 × 27 = 54. · u₄ = ar³ = 2 × 3³ = 2 × 27 = 54。
Sum to infinity 无穷和
- A GP converges when $|r| < 1$ — the terms get smaller and smaller.
- Sum to infinity: $S_\infty = \dfrac{a}{1-r}$.
$|r| < 1$ is essential. If $|r| \geq 1$, the terms don't shrink and the infinite sum diverges (grows without bound). The formula $S_\infty = \dfrac{a}{1-r}$ only applies when the series converges.
- An arithmetic progression has a common difference; a geometric progression a common ratio. A convergent geometric progression ($|r|<1$) shows convergence to a sum to infinity.
无穷和
- 一个 GP 在 $|r| < 1$ 时收敛(converges)——项变得越来越小。
- 无穷和(sum to infinity):$S_\infty = \dfrac{a}{1-r}$。
$|r| < 1$ 是必不可少的。 如果 $|r| \geq 1$,项不缩小,无穷和发散(无界增长)。公式 $S_\infty = \dfrac{a}{1-r}$ 只在级数收敛时适用。
- 等差数列(arithmetic progression)有公差;等比数列(geometric progression)有公比。收敛的等比数列(convergent geometric progression,$|r|<1$)体现收敛(convergence),有无穷和。
A GP has a = 8 and r = 0.5. What is the sum to infinity? · 一个等比数列 a = 8、r = 0.5。无穷和是多少?
S∞ = a/(1−r) = 8/(1−0.5) = 8/0.5 = 16. · S∞ = a/(1−r) = 8/(1−0.5) = 8/0.5 = 16。
A geometric series with r = 1.5 has a finite sum to infinity. · 一个 r = 1.5 的几何级数有一个有限的无穷和。
The sum to infinity formula only applies when |r| < 1. With r = 1.5, the series diverges. · 无穷和公式只在 |r| < 1 时适用。当 r = 1.5 时,级数发散。
Match each formula to what it computes. · 把每个公式和它计算的量配对。
d marks the arithmetic family, r the geometric one — recognising the formula is half the exam question. · 有 d 的是等差家族,有 r 的是等比家族——认出公式,考题就成功了一半。
You've got it
- binomial expansion: $(a+b)^n = \sum \binom{n}{r} a^{n-r} b^r$
- AP: $u_n = a + (n-1)d$, $S_n = \dfrac{n}{2}(2a+(n-1)d)$
- GP: $u_n = ar^{n-1}$, $S_n = \dfrac{a(1-r^n)}{1-r}$
- if $|r|<1$, sum to infinity $S_\infty = \dfrac{a}{1-r}$
你掌握了
- 二项展开:$(a+b)^n = \sum \binom{n}{r} a^{n-r} b^r$
- AP:$u_n = a + (n-1)d$,$S_n = \dfrac{n}{2}(2a+(n-1)d)$
- GP:$u_n = ar^{n-1}$,$S_n = \dfrac{a(1-r^n)}{1-r}$
- 如果 $|r|<1$,无穷和 $S_\infty = \dfrac{a}{1-r}$