spiral of Theodorus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T02:33:21Z http://mathoverflow.net/feeds/question/3440 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3440/spiral-of-theodorus spiral of Theodorus Jason S 2009-10-30T13:25:15Z 2010-01-14T21:34:23Z <p>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 <a href="http://en.wikipedia.org/wiki/Spiral%5Fof%5FTheodorus" rel="nofollow">Spiral of Theodorus</a>.)</p> <p><img src="http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Spiral%5Fof%5FTheodorus.svg/400px-Spiral%5Fof%5FTheodorus.svg.png" alt="alt text" /></p> <p>I spent a long time trying to prove that the series of points approximated a spiral R = K&theta; + &phi;, by trying to show the limit of the difference &phi; = sqrt(N+1) - K*sum(atan(1/sqrt(N)) 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!)</p> <p>Is this a known problem, and is there a closed-form solution to K and &phi; ? </p> http://mathoverflow.net/questions/3440/spiral-of-theodorus/3444#3444 Answer by Michael Lugo for spiral of Theodorus Michael Lugo 2009-10-30T14:52:29Z 2009-10-30T14:52:29Z <p>Here's a sketch of a proof that the constant you want exists, and how to find it.</p> <p>Let f(n) = arctan(1) + arctan(1/sqrt(2)) + arctan(1/sqrt(3)) + ... + 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.) </p> <p>Then take the power series expansion of arctan(1/sqrt(k)); this is</p> <p>(*) k^(-1/2) - 1/3 k^(-3/2) + 1/5 k^(-5/2) + ...</p> <p>So summing over 1 to n, we should get</p> <p>f(n) = (1^(-1/2) + 2^(-1/2) + ... + n^(-1/2)) - 1/3 (1^(-3/2) + 2^(-3/2) + ... + n^(-3/2)) + 1/5 (1^(-5/2) + 2^(-5/2) + ... + n^(-5/2)) - ...</p> <p>Now, 1^(-1/2) + 2^(-1/2) + ... + n^(-1/2) has the asymptotic form</p> <p>2 sqrt(n) + Zeta(1/2) + O(n^{-1/2}) </p> <p>where I cheated a bit and asked Maple. Zeta is the Riemann zeta function. And 1^(-j/2) + 2^(-j/2) + ... + n^(-j/2) has the asymptotic form</p> <p>Zeta(j/2) - O(n^{-j/2 + 1})</p> <p>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,</p> <p>f(n) = 2 sqrt(n) + (Zeta(1/2) - 1/3 Zeta(3/2) + 1/5 Zeta(5/2) - ...) + o(1).</p> <p>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 &phi;, 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.</p>