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Rational sums of inverse powers of fixed points of e
This post doesn't have much if anything to do with tetration, but I discovered it while analyzing logarithms at the fixed points. I had mentioned this previously:

The fixed points of e are irrational. However, the sums of the inverse n-th powers are rational.

For example, the sum of the inverse third powers of all the fixed points of e is -1/2:

Note: Here, is the fixed point in the kth branch of the natural logarithm, i.e., . If there were a single fixed point in the 0th branch of the natural logarithm, we could use and run our sum from negative infinity to positive. Alas, with two fixed points in the 0th branch, I thought it clearer to use complex conjugates, but perhaps we could still run the sum from negative to positive infinity if we had a way to account for both points in the 0th branch.

Continuing with a couple more examples, the sum of the inverse fifth powers is 3/8. The sum of the inverse eighth powers is -29/630.

Well, as it turns out, the denominators roughly increase as factorials. For example, the denominator of the 7th sum (i.e., the sum of the inverse 8th powers) is 630. This is 7!/8. The denominator of the 10th sum is 10!/3.

So, here's a list of the numerators for the first 26 sums (inverse 2nd to 27th powers), when using the factorials as the denominators (starting with 1!, not 0!):


I've currently got a program running to calculate the first 100,000 fixed points to 5120 bits of precision, to allow me to calculate the first few hundred sums. I'll then use continued fraction expansion, and look for the first "very large" number, which indicates that the previous fraction would most likely be exact.

Using only the first 5,000 fixed points, I had typically seen situations where the subsequent convergent fraction would have a denominator with three to four times as many digits, e.g., 14 digits versus 50. I can't provide a specific example at the moment because SAGE is still crunching numbers (takes a good fraction of a second to calculate a fixed point to the desired precision, so 100,000 of them will take a good fraction of a day [Edit: It took 47200 seconds!]). With 100,000 digits, I'm hoping to make the difference something like 14 digits versus 60 or 70.

Anyway, in the meantime, I wanted to try to figure out how to calculate the numerators. It seems to me that they deserve their own Sloane sequence, considering how surprising it was (to me) to get rational sums.
~ Jay Daniel Fox

Messages In This Thread
Rational sums of inverse powers of fixed points of e - by jaydfox - 11/20/2007, 07:55 PM

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