Timeline for Interesting result on the Euler-Maschroni constant - what is the background?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 19, 2012 at 18:34 | vote | accept | tobias | ||
| Mar 12, 2012 at 22:15 | comment | added | Gottfried Helms | @Igor Rivin: perhaps you see it easier if you subtract 1 from the value, say at i=6, and then divide by the first few bernoulli numbers, say $\small (a_6−1)/{1 \over∗6} $or $\small (a_6−1) / {1 \over∗42} $ or even by two bernoulli-numbers $\small (a_6−1)/{1 \over 42}/{1 \over 30 } $ | |
| Mar 12, 2012 at 19:26 | comment | added | user9072 | To explain the above comments: the original version was shorter. And, perhaps I should not have stressed so much (unfortunaletly I still do this) the two term version as one indeed sees already more terms. | |
| Mar 12, 2012 at 19:18 | history | edited | user9072 | CC BY-SA 3.0 |
relevant math typo corrected, sorry for the noise
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| Mar 12, 2012 at 19:10 | history | edited | user9072 | CC BY-SA 3.0 |
expanded
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| Mar 12, 2012 at 14:31 | comment | added | user9072 | @Igor Rivin: the observed errors are all very sligthly smaller 5 times 10^(-j-1) matching closely the 1/2(10^j). And the next term in the exxpansion explains why the are all slightly smaller, and by about how much. In other words, the expansion I mention explains that the error will 5 times 10^(-j-1) minus something still alot smaller. And you can also understand who much and one can also 'see' the 1/12 . And if one goes further one will 'see' the other terms very nicely at powers of 10 in base ten, or any other power in the matching digital expansion. | |
| Mar 12, 2012 at 13:40 | comment | added | Igor Rivin | How does this explain the OP's observation? | |
| Mar 12, 2012 at 12:20 | history | answered | user9072 | CC BY-SA 3.0 |