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A formula for the nth digit of πœ‹ and πœ‹^n

arxiv.org

71–80 of 143 posts

Re: A formula for the nth digit of πœ‹ and πœ‹^n

#74

It is all curious, but it seems that procedure that calculates the n-th digit using some other functions that require O(n) calculations (i.e. Bernoulli numbers) is not that exciting, as it's just a speedup comparing to a naive calculation (maybe a big one but still). Although because pi is a transcendental number maybe it is impossible to have an algorithm to return the n-th digit in O(1) operations? Does anyone know…

Even rational numbers which have repeating digits at the end are not O(1) because in order to calculate the n-th digit you need to calculate n modulo k where k is the length until the number repeats its digits, which needs to read the entire input number n (thus runs in O(log(n)). The only exception are rational numbers with k being a power of two. Take 0.1010101010…, you can get the n-th digit by reading off the smallest digit of n and checking whether it is a 0 or a 1.

There are transcendental numbers where the n-th digit can be computed in O(log n) like Champerowne constant though.

Re: A formula for the nth digit of πœ‹ and πœ‹^n

#75

Earlier quoted context omitted.

Spoiler: it's zero, recurring.

Can you prove it's not nine, recurring?

Yes... proving that would require an infinite-capacity rounding mechanism, which cannot exist, because when trying to build it, you always run out of universe. Thusly, "impossible" is equivalent to "false", by default.

Re: A formula for the nth digit of πœ‹ and πœ‹^n

#76
post #33

Earlier quoted context omitted.

The article title says "powers of πœ‹", so presumably an arbitrary digit.

I’m not sure that’s right. The abstract and text say the nth digit of \pi^n, and the worked example on page 2 uses the same n=1000 to find the thousandth digit of \pi^{1000}.

IOW, the arXiv title might have lost the "nth" because of some formatting thing. The title inside the PDF says "A formula for nth digit of \pi and \pi^n"

Re: A formula for the nth digit of πœ‹ and πœ‹^n

#79

Earlier quoted context omitted.

I don't think that would be true seeing 1/3 = 0.3333.. , or am I reading this wrong?

3 recurring represents a particular quantity continuing forever. 0 recurring represents the end of a quantity, and the absence of any further quantity, forever. Eg: 0.012500000000000... The significant portion is 0.0125 - the recurring zeroes serve a mathematical role akin to that of a full-stop in a sentence. Hence zero being (jokingly, but in a sense truthfully) always the "last digit".

This is just an artifact of representing it in base-10. In base-3 0.0125 has the same value but would have a non-terminating representation of 0.00010001...

I will grant that in base-Ο€, Ο€ is 10, however.

Re: A formula for the nth digit of πœ‹ and πœ‹^n

#80
post #4

Earlier quoted context omitted.

Ο€fs: Never worry about data again! - https://github.com/philipl/pifs

It's actually not known if pi contains every combination of digits (in any base 2 or greater) or not. It feels likely but really all we know is it's transcendental and seems pretty random from the parts we've generated. The library of babel could be a good "useless" backing though https://libraryofbabel.info/bookmark.cgi?hnexample

You just now convinced me how that site wasn't a sham. Thank you.

It takes the input, converts it into N, then calculates position using the N.

Thank you. I argued for a long time simply by being ignorant.

good bless ya sir.

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