Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T04:43:57Z http://mathoverflow.net/feeds/question/40334 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40334/logarithm-of-am-gm-ratio-sqrt-logxy-2-sqrtxy Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ S. Sra 2010-09-28T16:02:00Z 2010-09-29T21:14:56Z <p>Recently, while playing around with infinite-divisibility, i arrived at the following metric:</p> <p>$$d(x,y) := \sqrt{\log\left(\frac{x+y}{2\sqrt{xy}}\right)},$$</p> <p>defined for positive reals $x$ and $y$. Proving that $d$ is a metric is trivial, except for the triangle-inequality. However, we can bypass a direct proof by appealing to Schoenberg's theorem (I. J. Schoenberg. <em>Metric spaces and positive-definite functions</em>, TAMS, 1938), from which the metricity follows easily because $-\log(x+y)$ is a conditionally positive-definite kernel.</p> <p>However, i have been searching for following:</p> <blockquote> <ol> <li>Applications / situations where this metric shows up? </li> <li>An elementary proof of $d(x,y)$ being a metric.</li> </ol> </blockquote> <p><strong>Remarks</strong></p> <p>a. A google search on "ratio arithmetic geometric mean" yields some applications of the <em>ratio</em> alone;</p> <p>b. An elementary proof should exist, but my initial attempts have not been that successful, especially as i stubbornly did not want to use differential calculus.</p> <p>c. Notice that while proving $$d(x,y) \le d(x,z) + d(y,z),$$ we may assume wlog $x &lt; 1$ and $y > 1$ and $z=1$, as proving the other cases ranges from very-trivial to trivial.</p> http://mathoverflow.net/questions/40334/logarithm-of-am-gm-ratio-sqrt-logxy-2-sqrtxy/40502#40502 Answer by Willie Wong for Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ Willie Wong 2010-09-29T17:40:24Z 2010-09-29T17:40:24Z <p>Take the coordinate transformation from $\mathbb{R}$ to $\mathbb{R}_+$ by the exponential map. Then $(x+y) / \sqrt{xy} = e^{a-b} + e^{b-a}$ where $x = e^{2a}$ and $y = e^{2b}$. So we re-write</p> <p>$$d(e^{2a}, e^{2b}) = \sqrt{\log \cosh (a-b) }$$</p> <p>So it suffices to consider the function $q(x) = \sqrt{ \log \cosh x}$, which is an even function with a unique minimum at $x = 0$, where $q(0) = 0$. Triangle inequality reduces then to checking that $q$ is sub-additive. If you were willing to use calculus, then this follows from the fact that $x q'(x)$ is negative except at $0$, where the derivative is not well defined. </p> <p>In any case, a simple lemma for situations like this is</p> <p><strong>Lemma</strong> Let $f(x)$ be an increasing function on the positive real line with $f(0) = 0$. Then $f$ is sub-additive (i.e. $f(s+t) \leq f(s) + f(t)$) if $f(x)/x$ is decreasing.</p> <p>Thus the triangle inequality boils down to checking $\log \cosh x / x^2$ is a decreasing function of (positive) $x$. At this point, I am lost as to how to check this fact without using any calculus at all. (Somewhere you will need to delve into properties of the natural logarithm or hyperbolic cosine, and calculus is the most natural tool there.)</p> <p>As to where this metric comes up: the function $q(x)^2$ behaves like $x^2/2$ near the origin and $|x| - \log 2$ for large values. This makes it a Huber function*, which has found many uses in the literature of applied mathematics, especially with image processing (related to fMRI and things like that) and in robust statistics. (My knowledge here is only superficial, so you are probably better off asking a local expert in those fields about the $\log \cosh$ function.)</p> <p>*Huber, P. J., 1973, Robust regression: Asymptotics, conjectures, and Monte Carlo: Ann. Statist., 1, 799-821</p> http://mathoverflow.net/questions/40334/logarithm-of-am-gm-ratio-sqrt-logxy-2-sqrtxy/40543#40543 Answer by fedja for Logarithm of AM/GM ratio: $\sqrt{\log((x+y)/(2\sqrt{xy}))}$ fedja 2010-09-29T21:14:56Z 2010-09-29T21:14:56Z <p>This is just the $L^2(\frac{dt}{t})$ distance between $e^{-xt}$ and $e^{-yt}$ (Frullani integrals) up to some positive factor. I'm not sure whether you'll call this "elementary" though. </p>