Arctangents and the golden ratio - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:42:38Z http://mathoverflow.net/feeds/question/33753 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33753/arctangents-and-the-golden-ratio Arctangents and the golden ratio Michael Hardy 2010-07-29T04:01:40Z 2010-09-24T03:58:57Z <p>Why is the golden ratio lurking in $(d/dx)\arctan\left( x + \frac{1}{x} \right)$ $$= \frac{\left(\frac{1+\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1+\sqrt{5}}{2}\right)^2} + \frac{\left(\frac{1-\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2}$$ Is this merely an instance of its (unbeknownst to me) lurking everywhere, or is something special about this particular arctangent of a sum?</p> <p>(An arctangent of a sum seems like a bit of a freak, though.)</p> <p>(This was inspired by a related question that someone posted to <a href="http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics" rel="nofollow">http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics</a> .)</p> http://mathoverflow.net/questions/33753/arctangents-and-the-golden-ratio/33768#33768 Answer by T. for Arctangents and the golden ratio T. 2010-07-29T08:36:43Z 2010-07-29T09:17:55Z <p>"Welcome to $K_1( \mathbb{C}(t))$!"</p> <p>The identity instantiates the fact that if $f(z)$ is a rational function, the complex, multivalued $\log(f(z))$ is a sum of logarithms of the linear factors of $f(z)$. This fact can be made single-valued (by differentiating the identity) and real (by taking the real or, in this case, imaginary part of the formula, i.e., symmetrizing under Gal(C/R) which replaces logarithm with arctangent).</p> <p>Specializing the fact to $f(x)=g(ix)$, where $g(x) = x^2 - x - 1$ and $x$ is real, produces the identity with the golden ratio. The imaginary part of $\log(f)$ is $(1/2i)\log(g(ix)/g(-ix))$, which can be expanded as a sum over the roots and differentiated. </p> <p>(Remember also that $\arctan t = \arg (q+itq)$ for real $t$ and $q$, so that $\log f(x)$ can be evaluated without factorization, by computing real and imaginary parts of $g(ix)$. Equating the two expressions for the imaginary part of ($d\log(f)$) gives the formula in the question.)</p> http://mathoverflow.net/questions/33753/arctangents-and-the-golden-ratio/33830#33830 Answer by Pietro Majer for Arctangents and the golden ratio Pietro Majer 2010-07-29T19:07:24Z 2010-07-29T19:28:34Z <p>At a first glance, the golden ratio lurking there may seem indeed a curious phenomenon. But everything looks more regular observing that the RHS of the equality is a partial fraction decomposition, and that the golden ratio is a root of very simple polynomials, not unlikely to appear as denominators (here the fraction is $\frac{x^2-1}{x^2+(x^2+1)^2}$ &amp;c). </p>