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January 9

The last digits of pi

This question arises from my comment on a blogposting on the butterfly effect here. So, classical mechanics is uncomputable, and one can even give an explicit construction of a computer whose clock cycle will shrink exponentially and its memory will expand exponentially, so that you can execute an infinite number of computations in a finite amount of time. Now, while classical mechanics does not apply (in an exact sense) to our universe, it's nevertheless mathematically well defined if you assume the mathematical existence of infinite quantities.

The question is then if this makes the last digits of irrational numbers such as pi well defined mathematically. If the exponentially accelerating computer is a well defined mathematical construct, we can consider it running an algorithm to compute the digits of pi. If each clock cycle takes half the time of the previous clock cycle, then after twice the time it takes to execute the first clock cycle, it will have computed an infinite number of digits of pi. It should thus be possible for the device to output the last few digits it has computed at that time. Count Iblis (talk) 04:10, 9 January 2020 (UTC)[reply]

Iblis, given your history, I sort of suspect that this question ... may not be entirely sincere. But I'll take a stab at it anyway. The answer is that there are no "last few digits [...] computed at that time". If (just for definiteness) the computation ends at noon, then at 11:59, there are still infinitely many digits to be computed. At 11:59:59, there are still infinitely many digits to be computed. At 11:59:59.9, there are still infinitely many digits to be computed. And so on.

Yes, at noon, it's all over and done with. But there aren't any "last" digits, just a wild rush of digits pouring out faster and faster the closer you get to noon. --Trovatore (talk) 07:10, 9 January 2020 (UTC)[reply]

The memory on which it writes the output would become larger and larger by shrinking the size of the physical bits, but the last few bits can be written on a separate memory where the physical bits stay the same size. It would then erase the previously written bits written there. So, when the computer stops, the last few bits would be visible, while the infinite number of bits would have merged into a continuum in the main memory. Count Iblis (talk) 13:34, 9 January 2020 (UTC)[reply]

Any bits that are written into your fixed-size FIFO buffer, at some time before noon, will be overwritten by later bits written at a later time (but still before noon). So the buffer does not give you the last few bits of π. What it does give you is not well-specified by the statement of the problem.

This issue is a slight variation on Thomson's lamp. --Trovatore (talk) 17:31, 9 January 2020 (UTC)[reply]

It most certainly would be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. EvergreenFir (talk) 07:13, 9 January 2020 (UTC)[reply]

@Count Iblis: You're trying to invent a Zeno machine, aren't you? --CiaPan (talk) 13:41, 9 January 2020 (UTC)[reply]

A similar problem that I tried to tackle once when I was a kid is the square of 0.3333... If you notice the pattern: 33^2 = 1089, 333^2 = 110889, 3333^2 = 11108889 etc. If you try to square the limit of 0.3, 0.33, 0.333 etc., will your Zenoistic calculator switch from 1s to 8s on the midway of the output? Somewhat related to the idea of the Gibbs phenomenon. 89.172.38.145 (talk) 07:30, 10 January 2020 (UTC)[reply]

Zeroth

I'm not sure this belongs here or the lang reference desk. How many syllables in Zeroth. I know Wiktionary says two, and that there are only two vowels, but most of the people that I know pronounce it as if it was "Zero-eth", for three syllables.Naraht (talk) 14:01, 9 January 2020 (UTC)[reply]

A few other dictionaries that I checked also only give two syllables, which is the only way that I've personally ever heard it said as well. I don't have access to the OED unfortunately, which tends to be more comprehensive. I'd be curious to see if it listed the other. –Deacon Vorbis (carbon • videos) 14:05, 9 January 2020 (UTC)[reply]

Only two syllables in the OED. No listing for the mock-archaic zeroeth. Dbfirs 14:41, 9 January 2020 (UTC)[reply]

Zeroeth in Wikitionary. DroneB (talk) 11:01, 12 January 2020 (UTC)[reply]

This strikes me as more a discussion for the language refdesk, but oh well. I'm curious about both where this -eth version comes from, and about the perception that it's "mock-archaic". I am not aware of any archaic "foureth", for example.

There's an archaic -eth for the third-person singular, so I suppose you could come up with something like "yon reckoning engine zeroeth his memory", but that seems unrelated. --Trovatore (talk) 19:56, 12 January 2020 (UTC) [reply]

What is a pickled beet?

I'm pretty sure they didn't start out as perfect spheres, and when they were cut they weren't cut exactly in half, and then exactly in half, and then exactly in half again. But if they were, what is one eighth of a sphere?

I know a hemisphere is half a sphere but what is it when you cut it in half to make each half a quarter of the original sphere? For the Earth that would be the part north of the Equator, west of the Prime Meridian and east of the International Date Line. And then to make it an eighth you would have just what is 90 degrees or less.— Vchimpanzee • talk • contributions • 21:09, 9 January 2020 (UTC)[reply]

Hemispherical wedge. --jpgordon^{𝄢𝄆 𝄐𝄇} 23:13, 9 January 2020 (UTC)[reply]

Hemidemisemisphere, cf splits of a quaver>2A00:23C6:AA05:F900:F8C7:BEA9:47FE:489C (talk) 16:23, 10 January 2020 (UTC)[reply]

The hemispherical wedge would be cut in half to produce what I see when I eat a pickled beet. The quaver is a musical note. Perhaps there should be some change made such as a disambiguation page.— Vchimpanzee • talk • contributions • 17:05, 10 January 2020 (UTC)[reply]

Why? A wedge can be any angle; if you remove a wedge from a sphere, the remainder of the sphere is also a wedge. --jpgordon^{𝄢𝄆 𝄐𝄇} 19:35, 10 January 2020 (UTC)[reply]

Okay, but I think there may be a different name for it. As for quaver, I meant an above response linked to an article about a musical note with no hatnote for the shape.— Vchimpanzee • talk • contributions • 19:51, 10 January 2020 (UTC)[reply]

The 1/8 wedge could be described as an octant of the sphere, and people would probably understand what you mean. --{{u|Mark viking}} {Talk} 20:12, 10 January 2020 (UTC)[reply]

Sounds good. I still have to figure out what was meant by a quaver.— Vchimpanzee • talk • contributions • 21:20, 10 January 2020 (UTC)[reply]

The "quaver" reference was a connection to how musical note lengths, in the English tradition, are named "quaver" (1/8), "semiquaver" (1/16), "demisemiquaver" (1/32), etc., so maybe a half of a half of a hemisphere could be a semidemihemisphere. Just a slightly joking reference. --jpgordon^{𝄢𝄆 𝄐𝄇} 21:28, 10 January 2020 (UTC)[reply]

Okay, I understand. I've seem those terms for notes before.— Vchimpanzee • talk • contributions • 21:40, 10 January 2020 (UTC)[reply]