Sine Series
Definition
The series ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 is called the sine series.
∑ \sum ∑ (sigma) : summation symbol.
∞ \infty ∞ (infinity) : infinitely many terms.
n ! n! n ! : factorial, n ! = n × ( n − 1 ) × ⋯ × 1 n! = n \times (n-1) \times \cdots \times 1 n ! = n × ( n − 1 ) × ⋯ × 1 .
Convergence
Proof sketch
Ratio test:
a n + 1 a n = − x 2 ( 2 n + 2 ) ( 2 n + 3 ) , lim n → ∞ ∣ a n + 1 a n ∣ = 0 < 1 , \frac{a_{n+1}}{a_n} = -\frac{x^2}{(2n+2)(2n+3)}, \quad \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = 0 < 1, a n a n + 1 = − ( 2 n + 2 ) ( 2 n + 3 ) x 2 , lim n → ∞ a n a n + 1 = 0 < 1 ,
so it converges for all real x x x .
Examples
Example 1
Sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ( π 2 ) 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{2}\right)^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n ( 2 π ) 2 n + 1 .
Solution : x = π 2 x = \frac{\pi}{2} x = 2 π , so sin ( π 2 ) = 1 \sin\left(\frac{\pi}{2}\right) = 1 sin ( 2 π ) = 1 .
Example 2
Sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! π 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \pi^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n π 2 n + 1 .
Solution : x = π x = \pi x = π , so sin ( π ) = 0 \sin(\pi) = 0 sin ( π ) = 0 .
练习题
练习 1
Sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ( π 6 ) 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{6}\right)^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n ( 6 π ) 2 n + 1 .
参考答案
思路 :Sine series with x = π 6 x = \frac{\pi}{6} x = 6 π .
答案 :sin ( π 6 ) = 1 2 \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} sin ( 6 π ) = 2 1 。
练习 2
Sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ( π 3 ) 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{3}\right)^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n ( 3 π ) 2 n + 1 .
参考答案
思路 :x = π 3 x = \frac{\pi}{3} x = 3 π .
答案 :sin ( π 3 ) = 3 2 \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} sin ( 3 π ) = 2 3 。
练习 3
Sum ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ( π 4 ) 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left(\frac{\pi}{4}\right)^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n ( 4 π ) 2 n + 1 .
参考答案
思路 :x = π 4 x = \frac{\pi}{4} x = 4 π .
答案 :sin ( π 4 ) = 2 2 \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} sin ( 4 π ) = 2 2 。
总结
本文出现的符号
符号 类型 读音/说明 在本文中的含义 ∑ \sum ∑ 希腊字母 Sigma(西格玛) 求和符号,表示级数 ∞ \infty ∞ 数学符号 无穷大 表示无穷级数,项数无限 n ! n! n ! 数学符号 阶乘 n n n 的阶乘,n ! = n × ( n − 1 ) × ⋯ × 1 n! = n \times (n-1) \times \cdots \times 1 n ! = n × ( n − 1 ) × ⋯ × 1 x x x 数学符号 变量 正弦级数中的变量 π \pi π 希腊字母 Pi(派) 圆周率,约等于 3.14159 sin \sin sin 数学符号 正弦 正弦函数 lim \lim lim 数学符号 极限 表示数列或函数的极限
中英对照
中文术语 英文术语 音标 说明 正弦级数 sine series /saɪn ˈsɪəriːz/ 形如 ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 的级数 收敛 convergence /kənˈvɜːdʒəns/ 级数部分和序列有有限极限 比值判别法 ratio test /ˈreɪʃiəʊ test/ 通过相邻项比值判断收敛性的方法
1 Exploring Functions in Advanced Mathematics
先修课程
Functions are a core idea of advanced mathematics. This course walks through foundational concepts, key properties, and classic constants so you can read, reason, and compute with confidence.
前往课程
2 Sequences bridge discrete thinking and calculus. This track covers core definitions, limits, convergence, and classic models.
前往课程
3 The World of Limits in Advanced Mathematics
先修课程
Limits are the foundation of calculus and one of the most important ideas in advanced mathematics.
前往课程
4 Explore convergence tests, summation, power-series expansions, and applications.
前往课程