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A long time ago when I was in college I read about making a spiral out of right triangles with sides 1 and $\sqrt{N}$. (A google search seems to indicate that this is called the Spiral of Theodorus.)

Spiral of Theodorus built out of a sequence of right triangles

I spent a long time trying to prove that the series of points approximated a spiral $R = K\theta + \varphi$, by trying to show the limit of the difference $\varphi = \sqrt{N+1} - K \sum_{i=1}^N \arctan(1/\sqrt{i})$ existed for some $K$. I think I managed to do it but it was confusing and can't find my papers. (and I'm still an amateur mathematician!)

Is this a known problem, and is there a closed-form solution to $K$ and $\varphi$?

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I'm glad this got asked and answered. I too "invented" the Theodorus spiral as a youngster. –  Q.Q.J. Mar 1 '10 at 13:00
    
I better point out this book: amazon.com/dp/1568810105 ; there's a nice article there by Walter Gautschi –  J. M. Aug 14 '10 at 6:50

1 Answer 1

up vote 15 down vote accepted

Here's a sketch of a proof that the constant you want exists, and how to find it.

Let $$ f(n) = \arctan(1) + \arctan(1/\sqrt{2}) + \arctan(1/\sqrt{3}) + \ldots + \arctan(1/\sqrt{n}). $$ You want to show that $f(n) = \sqrt{n} + C + o(1)$ for some constant $C$. (If you're not familiar with the $o$-notation, think of $o(1)$ as representing some function which goes to $0$ as $n$ goes to infinity.)

Then take the power series expansion of $\arctan(1/\sqrt{k})$; this is

$$ (*) ~~~~~~~k^{-1/2} - \frac{1}{3} k^{-3/2} + \frac{1}{5} k^{-5/2} + \ldots $$

So summing over $1$ to $n$, we should get \begin{align*} f(n) = & (1^{-1/2} + 2^{-1/2} + ... + n^{-1/2}) \\\ - \, \frac{1}{3} &(1^{-3/2} + 2^{-3/2} + ... + n^{-3/2}) \\\ + \, \frac{1}{5}& (1^{-5/2} + 2^{-5/2} + ... + n^{-5/2}) - \ldots \end{align*} Now, $1^{-1/2} + 2^{-1/2} + \ldots + n^{-1/2}$ has the asymptotic form $$ 2 \sqrt{n} + \zeta(1/2) + O(n^{-1/2}) $$ where I cheated a bit and asked Maple, and $\zeta$ is the Riemann zeta function. And $1^{-j/2} + 2^{-j/2} + \ldots + n^{-j/2}$ has the asymptotic form $$ \zeta(j/2) - O(n^{-j/2 + 1}) $$ where, if you're not familiar with the $O$-notation, $O(n^{-j/2+1})$ should be thought of as a function that goes to zero at least as fast as $n^{-j/2 + 1}$ as n goes to infinity. So, assuming that we can rearrange series however we like, $$ f(n) = 2 \sqrt{n} + (\zeta(1/2) - \frac{1}{3} \zeta(3/2) + \frac{1}{5} \zeta(5/2) - \ldots) + o(1). $$ Since $\zeta(s)$ is very close to $1$ when $s$ is a large real number, that alternating series should converge. Again cheating and using Maple, I claim it converges to about $−2.157782997$. This is the constant you call $\varphi$, and what you called $K$ is equal to $2$. (An easier way to see that your $K$ is $2$ is to note that $\arctan(1/\sqrt{n})$ is about $1/\sqrt{n}$, and approximate the sum by an integral.

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2  
See also the unfinished paper "Analytic Continuation of the Theodorus Spiral" by Jörg Waldvogel (math.ethz.ch/~waldvoge/Papers/theopaper.html), which gets that some constant by what seems to be a different route. (I have not read it carefully.) –  Michael Lugo Oct 30 '09 at 15:02
    
thanks ! –  Jason S Oct 30 '09 at 17:19
    
uni-graz.at/~gronau/monthly230-237.pdf is a newer paper discussing the analytic continuation of the Theodorus spiral. –  J. M. Dec 16 '11 at 14:32

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