EQUATION

Taylor Series for Pendulum Period

Expresses the exact elliptic-integral period as a power series in θ₀: T = T₀·Σ [(2n)!/(2²ⁿ(n!)²)]²·θ₀²ⁿ

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The equation

EQ.TAYLOR-EXPANSION-PERIOD
T = T_0\sum_{n=0}^{\infty}\left[\frac{(2n)!}{2^{2n}(n!)^2}\right]^2\theta_0^{2n}
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What it solves

Expresses the exact elliptic-integral period as a power series in θ₀: T = T₀·Σ [(2n)!/(2²ⁿ(n!)²)]²·θ₀²ⁿ. The first two non-trivial coefficients give the familiar 1 + θ₀²/16 + 11θ₀⁴/3072 approximation.

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When to use it

When you need explicit correction coefficients to quantify how much amplitude increases the period. Useful for error analysis: at what amplitude does the small-angle formula err by more than X%?

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When NOT to use it

High-order truncations converge slowly for θ₀ > π/2. For amplitudes above 60°, the full elliptic integral or numerical integration is more practical.

§ 05

Common mistakes

Treating the series as exact at finite order — every truncated series has a remainder. Forgetting to use radians for θ₀ when computing the coefficients. Misidentifying the n = 0 term (it equals 1, contributing T₀) as the only term needed for small angles.

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Topics that use this equation

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Problems using this equation