4
$\begingroup$

Why is the golden ratio lurking in $(d/dx)\arctan\left( x + \frac{1}{x} \right)$ $$ = \frac{\left(\frac{1+\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1+\sqrt{5}}{2}\right)^2} + \frac{\left(\frac{1-\sqrt{5}}{2}\right)}{x^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2}\text{ ?} $$ Is this merely an instance of its (unbeknownst to me) lurking everywhere, or is something special about this particular arctangent of a sum?

(An arctangent of a sum seems like a bit of a freak, though.)

(This was inspired by a related question that someone posted to http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics.)

Improve this question
$\endgroup$
6
  • 3
    $\begingroup$ I don't think there's anything very special going on. Wolfram Alpha gives that the derivative is $(1-1/x^2)/((x+1/x)^2+1)$ (which is also $((x-1)(x+1))/(x^4+3 x^2+1)$. Most probably, expressing this with partial fractions etc. gives the golden ratio. $\endgroup$
    – shreevatsa
    Commented Jul 29, 2010 at 4:12
  • 14
    $\begingroup$ The question might be seeming to suggest that there's something going on in the derivative process that causes the golden ratio to appear. But, the golden ratio already appears before the derivative, as can be seen by the identity: $\arctan(x + \frac{1}{x}) = \frac{\pi}{2} + \arctan(\frac{x}{\alpha}) + \arctan(\frac{x}{\beta})$, where $\alpha$ and $\beta$ are the roots of $x^2 - x - 1=0$. $\endgroup$
    – Ken Fan
    Commented Jul 29, 2010 at 5:28
  • 2
    $\begingroup$ To follow up on Ken's comment, the golden ratio appears on the left-hand side in slightly disguised form: whenever I see things like "x + 1/x" then I expect the golden ratio to appear because those expressions are closely related to the defining equation of the golden ratio. It's more usual to see "x - 1/x", but nonetheless, I don't find it surprising. $\endgroup$
    – Andrew Stacey
    Commented Jul 29, 2010 at 7:57
  • $\begingroup$ @Ken: Very nice. A sum of arctangents is a natural thing; an arctangent of a sum doesn't seem so, but you've shown how to view it in the way that seems natural. @Andrew: By hindsight I wonder why I didn't think of that. The two roots of $x + 1/x = 1$ are the golden ratio and its conjugate. (Although the $=1$ part is not explicitly there....) $\endgroup$
    – Michael Hardy
    Commented Jul 29, 2010 at 16:26
  • 1
    $\begingroup$ Ken, that formula is multivalued and holds only when you restrict the motion of $x$, e.g., to positive values. (Compare the values at small positive and negative $x$ to see a discrepancy of $\pi$). Here the role of differentation is not to overcomplicate a simpler identity, but to make that identity single-valued. Golden ratio appears because $\log (x^2 - x - 1) = \log(x-\alpha) + \log(x-\beta)$ but for any other polynomial there would be a similar formula with arctan as a sum over the roots. $\endgroup$
    – T..
    Commented Jul 29, 2010 at 18:57

2 Answers 2

Reset to default
11
$\begingroup$

"Welcome to $K_1( \mathbb{C}(t))$!"

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).

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.

(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.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ Very interesting answer. I'm not done digesting it yet, but it appears that by $K_1$ you don't mean the same thing as what that notation means at en.wikipedia.org/wiki/Brauer%E2%80%93Siegel_theorem . Can you define that? $\endgroup$
    – Michael Hardy
    Commented Jul 29, 2010 at 16:36
  • 1
    $\begingroup$ $K_1$ as in K -theory of the field of rational functions C(t). Formulas of this type come up in topology and number theory when calculating regulators on K groups. For metabolizing the answer above, here's a philosophy-free rephrasing of the recipe: given real f(x) with all real roots, compute imaginary part of dlogf(ix) in two ways -- (1) find imaginary/real parts of f(ix) and take arctan; and (2) as sum of imaginary parts of $d \log$ of the linear factors of f(ix)$. $\endgroup$
    – T..
    Commented Jul 29, 2010 at 19:47
2
$\begingroup$

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 $\frac{x^2-1}{x^2+(x^2+1)^2}$ &c).

Improve this answer
$\endgroup$
5
  • $\begingroup$ I hope you are aware that this is a matter of points of view. One could as well argue that arctan and x+1/x are not essential at all: everything that produces a fraction with poles at a number $\alpha$ will show it in its partial fracion decomposition. And the golden number is quite common to appear as root of small polynomials. An explanation doesn't need to explain everything: what counts (to me) is the ratio "explained matter" over "length of explanation". Under this respect, a more satisfactory explanation than the one I gave is to me the plain computation. And yours, of course. $\endgroup$
    – Pietro Majer
    Commented Jul 29, 2010 at 19:54
  • $\begingroup$ Hey T. where's your comment? ;-) $\endgroup$
    – Pietro Majer
    Commented Jul 29, 2010 at 19:55
  • $\begingroup$ Given a rational function $f(t)$, it's certainly possible to express $d arctan(f(x))$ as a sum of partial fractions of $f'(x) / (1 + f(x)^2)$. However, the answer that running this somewhat arbitrary calculation on the somewhat arbitrary polynomial $x+1/x$ happened to hit the golden ratio by chance, is unsatisfactory, because there is clearly a deeper and generalizable pattern behind the formula, as seen from the other answers. $\endgroup$
    – T..
    Commented Jul 29, 2010 at 20:05
  • $\begingroup$ re: "where" -- I replaced with a more specific remark. $\endgroup$
    – T..
    Commented Jul 29, 2010 at 20:05
  • $\begingroup$ (written before I saw your latest one, for which I don't disagree with anything, except to add that one should distinguish "explanation" from "verification" when possible. Sometimes only a verification is available, that's life. But here we have better luck.) $\endgroup$
    – T..
    Commented Jul 29, 2010 at 20:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .