This post ties together the previous three posts.

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

*r* = *b* Î¸^{1/n}

and applied this here to rolls of carpet.

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

Â½ *bT*Â²

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.

Hereâ€™s an example with *n* = 3.

Now the arms are not equally spaced but instead grow closer together.

## Arc length

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

As I discussed here, the power series defining a hypergeometric function _{2}*F*_{1} diverges for arguments outside the unit disk, but the function can be extended by analytic continuation using the identity

## Asymptotics

Most of the terms above will drop out in the limit asÂ *z* =Â âˆ’*n*Â²*T*Â² gets large. The 1/*z* terms go to zero, and hypergeometric functions equal 1 when *z* = 0, and so theÂ *F* terms on the right hand side above go to 1 as *T* increases.

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

(âˆ’*z*)^{âˆ’a}Â = (âˆ’*z*)^{1/2}

matters, but the term

(âˆ’*z*)^{âˆ’b}Â = (âˆ’*z*)^{âˆ’1/2n}

goes to zero and can be ignored.

We find that

*L* â‰ˆ *k T*^{1 + 1/n}

where the constant *k* is given by

*k* = *bn* Î“(1 + 1/2*n*) Î“(1/2 + 1/2*n*) / Î“(1/2*n*) Î“(3/2 + 1/2*n*)

When *n* = 1, this reduces to *L* â‰ˆ Â½ *bT*Â² as before.

Â

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