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Arctangent Series

Definition

Definition of Arctangent Series

The series ∑n=0∞(−1)n2n+1x2n+1\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} is called the arctangent series.

符号说明
SymbolTypePronunciation/ExplanationMeaning in This Article
∑\sumGreek letterSigmaSummation symbol, representing series
∞\inftyMathematical symbolInfinityRepresents infinite series, infinite number of terms

Convergence

Convergence of Arctangent Series
  • Interval of convergence: [−1,1][-1, 1]
  • The sum is:

∑n=0∞(−1)n2n+1x2n+1=arctan⁡x\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} x^{2n+1} = \arctan x

Applications

The arctangent series is particularly useful for calculating the value of π\pi. When x=1x = 1:

arctan⁡1=π4=∑n=0∞(−1)n2n+1=1−13+15−17+⋯\arctan 1 = \frac{\pi}{4} = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots

This is the famous Leibniz series.

Examples

Example 1

Use the arctangent series to approximate arctan⁡12\arctan \frac{1}{2} (using the first 5 terms).

Solution: arctan⁡12≈12−13⋅23+15⋅25−17⋅27+19⋅29\arctan \frac{1}{2} \approx \frac{1}{2} - \frac{1}{3 \cdot 2^3} + \frac{1}{5 \cdot 2^5} - \frac{1}{7 \cdot 2^7} + \frac{1}{9 \cdot 2^9}

Calculation yields: arctan⁡12≈0.4636\arctan \frac{1}{2} \approx 0.4636

Exercises

Exercise 1

Use the arctangent series to approximate arctan⁡13\arctan \frac{1}{3} (using the first 4 terms).

Reference Answer (2 个标签)
arctangent series series summation

Problem-solving approach: Use the arctangent series formula and substitute x=13x = \frac{1}{3}.

Detailed steps:

  1. Identify the series type: This is the arctangent series
  2. Determine the parameter: x=13x = \frac{1}{3}
  3. Calculate the first 4 terms:
    • Term 1: 13\frac{1}{3}
    • Term 2: −13⋅33=−181-\frac{1}{3 \cdot 3^3} = -\frac{1}{81}
    • Term 3: 15⋅35=11215\frac{1}{5 \cdot 3^5} = \frac{1}{1215}
    • Term 4: −17⋅37=−115309-\frac{1}{7 \cdot 3^7} = -\frac{1}{15309}
  4. Summation: arctan⁡13≈13−181+11215−115309≈0.3218\arctan \frac{1}{3} \approx \frac{1}{3} - \frac{1}{81} + \frac{1}{1215} - \frac{1}{15309} \approx 0.3218

Answer: arctan⁡13≈0.3218\arctan \frac{1}{3} \approx 0.3218 (approximation using the first 4 terms).

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
xxMathematical symbolVariableVariable in the arctangent series
π\piGreek letterPiCircular constant, approximately 3.14159
arctan⁡\arctanMathematical symbolArctangentArctangent function

Chinese-English Glossary

Chinese TermEnglish TermIPA PronunciationExplanation
反正切级数arctangent series/ɑːkˈtændʒənt ˈsɪəriːz/Series expansion of the arctangent function
收敛区间interval of convergence/ˈɪntəvəl əv kənˈvɜːdʒəns/Interval where the series converges
收敛convergence/kənˈvɜːdʒəns/Sequence of partial sums has a finite limit
莱布尼茨级数Leibniz series/ˈlaɪbnɪts ˈsɪəriːz/Series used to calculate π\pi
Previous Tangent Series Learn the definition, convergence, and applications of tangent series.

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