Timeline for Arctangents and the golden ratio
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 10, 2014 at 22:47 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced tag 'tag-removed'
|
| Nov 10, 2014 at 20:41 | history | edited | Michael Hardy | CC BY-SA 3.0 |
added 9 characters in body
|
| Jul 30, 2010 at 13:03 | comment | added | Ken Fan | Right...I should have been more careful to restrict to $x > 0$ and perhaps give the identity with the change of sign for $x < 0$... | |
| Jul 29, 2010 at 19:07 | answer | added | Pietro Majer | timeline score: 2 | |
| Jul 29, 2010 at 18:57 | comment | added | T.. | 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. | |
| Jul 29, 2010 at 16:26 | comment | added | Michael Hardy | @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....) | |
| Jul 29, 2010 at 8:36 | answer | added | T.. | timeline score: 11 | |
| Jul 29, 2010 at 7:57 | comment | added | Andrew Stacey | 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. | |
| Jul 29, 2010 at 6:51 | history | edited | Charles Matthews | CC BY-SA 2.5 |
upper case title
|
| Jul 29, 2010 at 5:28 | comment | added | Ken Fan | 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 $x^2 - x - 1=0$. | |
| Jul 29, 2010 at 4:12 | comment | added | shreevatsa | 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. | |
| Jul 29, 2010 at 4:01 | history | asked | Michael Hardy | CC BY-SA 2.5 |