Length of a general Archimedean spiral

The Endeavour 2024-04-19

This post ties together the previous three posts.

In this post, I said that an Archimedean spiral has the polar equation

r = b θ^1/n

and applied this here to rolls of carpet.

When n = 1, the length of the spiral for θ running from 0 to T is approximately

½ bT²

with the approximation becoming more accurate [1] as T increases.

In this post we want to look at the general case where n might not be 1. In that case the arc length is given by a hypergeometric function, and finding the asymptotic behavior for large T requires evaluating a hypergeometric function at a large argument.

Arc length

According to MathWorld, the length of the spiral with n > 1 and θ running from 0 to T is given by

$L = b T^{1/n} \,_2F_1\left(-\frac{1}{2}, \frac{1}{2n}; 1 + \frac{1}{2n}; -n^2 T^2 \right)$

As I discussed here, the power series defining a hypergeometric function ₂F₁ diverges for arguments outside the unit disk, but the function can be extended by analytic continuation using the identity

$\begin{align*} F(a, b; c; z) &= \frac{\Gamma(c) \Gamma(b-a)}{\Gamma(b) \Gamma(c-a)} \,(-z)^{-a\phantom{b}} F\left(a, 1-c+a; 1-b+a; \frac{1}{z}\right) \\ &+ \frac{\Gamma(c) \Gamma(a-b)}{\Gamma(a) \Gamma(c-b)} \,(-z)^{-b\phantom{a}} F\left(\,b, 1-c+b; 1-a+b; \,\frac{1}{z}\right) \\ \end{align*}$

Asymptotics

Most of the terms above will drop out in the limit as z = −n²T² gets large. The 1/z terms go to zero, and hypergeometric functions equal 1 when z = 0, and so the F terms on the right hand side above go to 1 as T increases.

In our application the hypergeometric parameters are a = −1/2, b = 1/2n, and c = 1 + 1/2n. The term

(−z)^−a = (−z)^1/2

matters, but the term

(−z)^−b = (−z)^−1/2n

goes to zero and can be ignored.

We find that

L ≈ k T^{1 + 1/n}

where the constant k is given by

k = bn Γ(1 + 1/2n) Γ(1/2 + 1/2n) / Γ(1/2n) Γ(3/2 + 1/2n)

When n = 1, this reduces to L ≈ ½ bT² as before.

[1] The relative approximation error decreases approximately quadratically in T. But the absolute error grows algorithmically.

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