This is a beta course, so its structure, chapters, and examples may continue to change.
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
Symbol Type Pronunciation/Explanation Meaning in This Article ∑ \sum ∑ Greek letter Sigma Summation symbol, representing series ∞ \infty ∞ Mathematical symbol Infinity Represents infinite series, infinite number of terms n ! n! n ! Mathematical symbol Factorial n factorial, n ! = n × ( n − 1 ) × ⋯ × 1 n! = n \times (n-1) \times \cdots \times 1 n ! = n × ( n − 1 ) × ⋯ × 1
Convergence Sine Series Convergence
For any real number x x x
∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 = sin x \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = \sin x ∑ n = 0 ∞ ( 2 n + 1 )! ( − 1 ) n x 2 n + 1 = sin x
Using the ratio test:
a n + 1 a n = ( − 1 ) n + 1 ( 2 ( n + 1 ) + 1 ) ! x 2 ( n + 1 ) + 1 ⋅ ( 2 n + 1 ) ! ( − 1 ) n x 2 n + 1 = − x 2 ( 2 n + 2 ) ( 2 n + 3 ) \frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1}}{(2(n+1)+1)!} x^{2(n+1)+1} \cdot \frac{(2n+1)!}{(-1)^n x^{2n+1}} = -\frac{x^2}{(2n+2)(2n+3)} a n a n + 1 = ( 2 ( n + 1 ) + 1 )! ( − 1 ) n + 1 x 2 ( n + 1 ) + 1 ⋅ ( − 1 ) n x 2 n + 1 ( 2 n + 1 )! = − ( 2 n + 2 ) ( 2 n + 3 ) x 2
lim n → ∞ ∣ a n + 1 a n ∣ = lim n → ∞ x 2 ( 2 n + 2 ) ( 2 n + 3 ) = 0 < 1 \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{x^2}{(2n+2)(2n+3)} = 0 < 1 lim n → ∞ a n a n + 1 = lim n → ∞ ( 2 n + 2 ) ( 2 n + 3 ) x 2 = 0 < 1
Therefore, for any real number x x x
Examples Example 1
Find the sum of the series ∑ 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 :
This is the sine series with x = π 2 x = \frac{\pi}{2} x = 2 π
Therefore, the sum is: sin ( π 2 ) = 1 \sin\left(\frac{\pi}{2}\right) = 1 sin ( 2 π ) = 1
Example 2
Find the sum of the series ∑ 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 :
This is the sine series with x = π x = \pi x = π
Therefore, the sum is: sin ( π ) = 0 \sin(\pi) = 0 sin ( π ) = 0
Exercises Exercise 1
Find the sum of the series ∑ 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
Reference Answer sine series series summation
Approach :
This is the sine series, need to determine the value of x x x
Steps :
Identify series type: ∑ 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
Determine parameter: x = π 6 x = \frac{\pi}{6} x = 6 π
Calculate sum: S = sin ( π 6 ) = 1 2 S = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} S = sin ( 6 π ) = 2 1
Answer :
The sum is 1 2 \frac{1}{2} 2 1
Exercise 2
Find the sum of the series ∑ 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
Reference Answer sine series series summation
Approach :
This is the sine series, need to determine the value of x x x
Steps :
Identify series type: ∑ 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
Determine parameter: x = π 3 x = \frac{\pi}{3} x = 3 π
Calculate sum: S = sin ( π 3 ) = 3 2 S = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} S = sin ( 3 π ) = 2 3
Answer :
The sum is 3 2 \frac{\sqrt{3}}{2} 2 3
Exercise 3
Find the sum of the series ∑ 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
Reference Answer sine series series summation
Approach :
This is the sine series, need to determine the value of x x x
Steps :
Identify series type: ∑ 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
Determine parameter: x = π 4 x = \frac{\pi}{4} x = 4 π
Calculate sum: S = sin ( π 4 ) = 2 2 S = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} S = sin ( 4 π ) = 2 2
Answer :
The sum is 2 2 \frac{\sqrt{2}}{2} 2 2
Summary Symbols Used in This Article
Symbol Type Pronunciation/Explanation Meaning in This Article x x x Mathematical symbol Variable Variable in the sine series π \pi π Greek letter Pi Pi, approximately 3.14159 sin \sin sin Mathematical symbol Sine Sine function lim \lim lim Mathematical symbol Limit Represents the limit of a sequence or function
