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"Is it conceivable that different irrational numbers have different "frequencies", such that high-frequency numbers exhibit their expected averages very quickly

"Is it conceivable that different irrational numbers have different "frequencies", such that high-frequency numbers exhibit their expected averages very quickly, whereas low-frequency numbers may require billions of digits to exhibit the expected averages?" Yes, and obviously so. Take a (base 10) normal irrational and divide it by 10^1000000000000. The first trillion or so digits of the result will be 0, but the result is still irrational (because duh) and (base 10) normal because the frequencies of the digits have to converge in the infinite limit and giving one digit a large finite head start won't change its overall proportion in the limit. QED.

arsenide 11y ago
Here's a (slighty) less hand-wavey proof. Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits of value a) / (number of digits of value b) ||_\infty < ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε > 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N > n this inequality holds. In the case where there are a fixed number m of leading "a"s (whatever digit a is), we have ||(m + dig_a)/(dig_b)||_\infty = ||m/dig_b + dig_a/dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N > n for some n we choose. By assumption ||dig_a / dig_b||_\infty < ε whenever n > N. Choose n' > n such that dig_b satisfies m/dig_b < ε (Note: this is possible as dig_b increases as n' grows: since m is constant and if dig_b were to be constant for n > N' then ||dig_a/dig_b||_\infty would be larger than ε for some n). Then we have ||m/dig_b||_\infty < ε and ||dig_a/dig_b||_\infty < ε so ||m/dig_b + dig_a/dig_b||_\infty < 2ε. Thus it converges -- satisfying the same definition as above chooseing n = n' and ε = 2ε.

arsenide 11y ago

Here's a (slighty) less hand-wavey proof. Note that for a base 10 normal irrational by the definition of "convergence" in this case means || (number of digits of value a) / (number of digits of value b) ||_\infty < ε (sup norm over all a, b from 0 to 9, a ≠ b) for some ε > 0 such that for some fixed n (the number of digits in the decimal expansion, halted), we have that whenever N > n this inequality holds. In the case where there are a fixed number m of leading "a"s (whatever digit a is), we have ||(m + dig_a)/(dig_b)||_\infty = ||m/dig_b + dig_a/dig_b||_\infty -- we only need this to be less than SOME fixed ε (not necessarily the same ε as above) for every N > n for some n we choose. By assumption ||dig_a / dig_b||_\infty < ε whenever n > N. Choose n' > n such that dig_b satisfies m/dig_b < ε (Note: this is possible as dig_b increases as n' grows: since m is constant and if dig_b were to be constant for n > N' then ||dig_a/dig_b||_\infty would be larger than ε for some n). Then we have ||m/dig_b||_\infty < ε and ||dig_a/dig_b||_\infty < ε so ||m/dig_b + dig_a/dig_b||_\infty < 2ε. Thus it converges -- satisfying the same definition as above chooseing n = n' and ε = 2ε.