little problem just thought of (evil numbers)

[pzhon]

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As k->infinity, I believe the probability that a random number in [0,1] has k as a partial sum of the scf coefficients is 0. That's because the expected value of the nth coefficient is infinite, as the probability that the value is t is about c/t^2. For a more exact limiting distribution, see this exposition.

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I thought the majority of partial quotients are 1's and 2's...and the probability that a given partial quotient is n drops off rather quickly as n gets large. So intuitively, your summing a bunch of small numbers together...it would be hard to continually "miss" the sum your looking for.

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You need to revise your intuition about the frequency of large coefficients. The expected value of any coefficient is infinite. Take a single random number to a thousand digits of precision, and you'll probably find that the average of the first thousand coefficients is quite large compared with typical coefficients. Ninety-nine random numbers (99 to estimate percentiles) produced average values of the first 1000 coefficients ranging from 7.003 to 259.116.

Here are the results of a Monte Carlo analysis of the frequency of particular partial sums with 10,000 trials:

Partial sum / Percentage scfs with that partial sum
1 50.36%
2 33.15
3 29.00
4 27.40
5 26.33
6 24.29
7 23.55
8 23.24
9 22.64
10 21.61
100 13.86
666 9.87
1000 9.48
10000 6.98

This took a few processor-hours of computing time.

[imfatandugly]

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As k->infinity, I believe the probability that a random number in [0,1] has k as a partial sum of the scf coefficients is 0. That's because the expected value of the nth coefficient is infinite, as the probability that the value is t is about c/t^2. For a more exact limiting distribution, see this exposition.

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I thought the majority of partial quotients are 1's and 2's...and the probability that a given partial quotient is n drops off rather quickly as n gets large. So intuitively, your summing a bunch of small numbers together...it would be hard to continually "miss" the sum your looking for.

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You need to revise your intuition about the frequency of large coefficients. The expected value of any coefficient is infinite. Take a single random number to a thousand digits of precision, and you'll probably find that the average of the first thousand coefficients is quite large compared with typical coefficients. Ninety-nine random numbers (99 to estimate percentiles) produced average values of the first 1000 coefficients ranging from 7.003 to 259.116.

Here are the results of a Monte Carlo analysis of the frequency of particular partial sums with 10,000 trials:

Partial sum / Percentage scfs with that partial sum
1 50.36%
2 33.15
3 29.00
4 27.40
5 26.33
6 24.29
7 23.55
8 23.24
9 22.64
10 21.61
100 13.86
666 9.87
1000 9.48
10000 6.98

This took a few processor-hours of computing time.

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I hear what your saying...while small numbers occur most frequently in a "typical" continued fraction, their average value diverges , using the typical averaging function. i.e. if x=[0,a1,a2,...,an,...] then lim n->infinity (a1+....an)/n --> infinity.... since P(k)=probability of a quotient being k is approximately 1/k^2 and the averaging function will grow roughly like the harmonic series, or log(n).

Because of this it becomes more and more likely that the sum of the partial quotients of a typical c.f. will jump over a given value as we use more and more quotients in our sum.