Timeline for A very very good approximation to Ramanujan constant. Why? [closed]
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 9, 2009 at 21:27 | comment | added | Michael Lugo | As requested by buzzard and JSE, I've asked a related question: mathoverflow.net/questions/4775/… | |
| Nov 9, 2009 at 19:58 | comment | added | JSE | Seconding that Michael should ask this as a separate question, unless he wants to work it out himself. | |
| Nov 9, 2009 at 18:39 | comment | added | Kevin Buzzard | Either prod Anton to re-open, or post another question. I can prove that c^2 is close to an integer, where c is Ramanujan's constant, and this isn't a formal consequence of c being close to an integer (because it's not close enough, as it were). But my proof doesn't stretch to c^5. | |
| Nov 9, 2009 at 18:18 | comment | added | Michael Lugo | buzzard, I agree; I was trying to produce such a proof when this question was closed. | |
| Nov 9, 2009 at 17:31 | comment | added | Kevin Buzzard | Michael's comment is much more surprising than Mark's. But I suspect Michael's can be explained using the same sorts of ideas which go into the proof that Ramanujan's constant is almost an integer (rather than just the statement that it's almost an integer, which isn't strong enough). | |
| Nov 9, 2009 at 17:17 | answer | added | Kevin Buzzard | timeline score: 7 | |
| Nov 9, 2009 at 17:09 | history | closed | Anton Geraschenko | not a real question | |
| Nov 9, 2009 at 17:08 | comment | added | Michael Lugo | Somewhat surprisingly, exp(Pisqrt(163))^5 is within 10^(-5) of an integer. And this integer is not N^5 where N is the integer closest to exp(Pisqrt(163)). | |
| Nov 9, 2009 at 17:05 | comment | added | David E Speyer | Agreed. That number is about 90 digits long (I might have miscounted a few.) The spacing between 5th roots of 90 digit numbers is about 10^{-72}. So it's hardly a surprise that you can find some number in that range whose 5th root agrees with a given real number to 70 decimal places. | |
| Nov 9, 2009 at 16:57 | comment | added | Reid Barton | It's extremely unsurprising, because you put more than 70 digits of information into that long number. | |
| Nov 9, 2009 at 16:53 | history | asked | Mark Thomas | CC BY-SA 2.5 |