3 ms·
It wouldn't be surprising that e is not normal. After all, it's continuous fraction is regular ([2; 1, 2, 1, 1, 4, ... 1, 1, 2k, ...]), while for a true random
by Grue3 11y ago
It wouldn't be surprising that e is not normal. After all, it's continuous fraction is regular ([2; 1, 2, 1, 1, 4, ... 1, 1, 2k, ...]), while for a true random number there wouldn't be any pattern to it.
- CarolineW 11y agosqrt(2) has the CF [1;2,2,2,2,2,...] and yet it appears to be normal. Similarly sqrt(5) has the CF [2;4,4,4,4,4,...] It's generally expected that non-integer square roots are normal, and yet they always have repeating CFs. All that combines to mean that I would be surprised if e turns out not to be normal.
- irremediable 11y agoYeah, great counterexample. I was thinking, "but continued fractions... that rings a bell..." -- couldn't figure out what seemed off about it, though.