Math 2.0

Special functions Example

Stirling's series

Example

Let Γ\Gamma be the gamma function, let ln⁡Γ\ln \Gamma be its principal branch on ℂ∖(−∞,0]\C \setminus (-\infty, 0], and let B2kB_{2k} be the Bernoulli numbers. Fix δ\delta with 0<δ<π0 < \delta < \pi. As z→∞z \to \infty in the sector ∣arg⁡z∣≤π−δ\abs{\arg z} \le \pi - \delta, ln⁡Γ(z)∼(z−12)ln⁡z−z+12ln⁡(2π)+∑k=1∞B2k2k(2k−1) z2k−1.\ln \Gamma(z) \sim \left(z - \tfrac{1}{2}\right) \ln z - z + \tfrac{1}{2} \ln(2\pi) + \sum_{k=1}^{\infty} \frac{B_{2k}}{2k(2k-1) \, z^{2k-1}}. This is an asymptotic expansion in the scale zln⁡z,z,ln⁡z,1,z−1,z−3,z−5,…z \ln z, z, \ln z, 1, z^{-1}, z^{-3}, z^{-5}, \ldots. With B2=16B_2 = \frac{1}{6}, B4=−130B_4 = -\frac{1}{30} and B6=142B_6 = \frac{1}{42}, the first correction terms are 112z−1360z3+11260z5.\frac{1}{12 z} - \frac{1}{360 z^3} + \frac{1}{1260 z^5}.

Remarks

The series diverges for every zz. Indeed ∣B2k∣\abs{B_{2k}} grows like 2(2k)!/(2π)2k2 (2k)! / (2\pi)^{2k}, so the terms grow like a factorial divided by a power of zz. For real z=x>0z = x > 0, the error after the term in z−(2N−1)z^{-(2N-1)} has the sign of the first omitted term and is smaller in absolute value. For x=10x = 10, the first correction 1120\frac{1}{120} already gives ln⁡Γ(10)=ln⁡(9!)\ln \Gamma(10) = \ln(9!) with an error below 1360 000<3⋅10−6\frac{1}{360\,000} < 3 \cdot 10^{-6}. Taking the exponential of the first terms gives Stirling's formula n!∼2πn (n/e)nn! \sim \sqrt{2\pi n} \, (n/e)^n.

Further reading

  1. Web https://en.wikipedia.org/wiki/Stirling%27s_approximation
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