Where does your birthday land on pi?


#1

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#2

A mathematical question for someone versed in this kind of stuff:

So pi’s decimal representation is infinite and never-repeating - Does it follow that it contains all numerical combinations?

(i.e. is there somewhere deep in the googols of numerals a stretch of pi with a thousand 7s in a row? Million 7s? Can it be proven or disproven? If not, what precludes certain patterns from forming?)


#3

It depends on whether or not it’s a normal number. AFAIK, they’re not yet sure that pi is normal, but they think it is.


#4

Mine lands on Pi “a la mode”.


#5

Yay!

Happy Pi Day!


#6

A more difficult problem is to find the smallest decimal number that never appears in the digits of pi.


#7

Today’s Pi/pie themed Something Positive made me LOL:


#8

Reminds me of this clip from “Person of Interest”–


#9

Pah, automatically reduced the year to two digits. I want to find my birthday in ISO 8601 format, the only true date format.


#10

Amen 


#11

My ISO 8601 birthdate is much earlier in pi than my MMDDYY birthdate.


#12

So what is the likely answer?


#13

That’s essentially a subset of the problem I stated. If pi is truly infinite with no hard limit on possible combinations, it should contain all numbers. Unless we’re getting into some Cantor stuff about various dimensions of infinity…


#14

I noticed that, too. Isn’t truncating to YY what got us into the Y2K mess? Good lord! Does Pi have a Y2K problem we haven’t remediated?!?!?!?

OK. I’ve calmed down again now.


#15

I haven’t done the math, but I think the problem is that, on average, you have to go 100 times further into pi to find a given 8-digit number than a given 6-digit number.

So the probable reason that they’re not doing 4-digit years is that that it would require that much more computing power.


#16

The likely answer is that pi is normal, on the grounds that most real numbers appear to be normal. But there’s no proof as yet.

Should? I don’t know why that would follow. A sense of duty?


#17

Because otherwise it would settle into a loop. Since it doesn’t loop (as has been proven), it “needs to explore”* the phase space of combinations further and further. Which on the plane of infinity means it either: i) explores all combinations; or ii) skips some and keeps going further. (Which is the core of my question. Does it skip? If so, why and what combinations?)

Or so my logic tells me. Admittedly though, I’m not a mathematician. Which is why I asked for an answer from a mathematician.

*Please excuse the agency-flavored vocabulary.


#18

I meant: Based on the assumption that pi is most likely normal, what is the answer to my original question?


#19

Many numbers don’t settle into a loop, and yet don’t contain many possible strings. For example the Thue-Morse sequence is non-periodic but contains no triples (three-in-a-row) of any item.


#20

I can point you to the wikipedia snippet which relates the most closely to your expression above, which says:

Even though there will be sequences such as 10, 100, or more consecutive tails (binary) or fives (base 6) or even 10, 100, or more repetitions of a sequence such as tail-head (two consecutive coin flips) or 6-1 (two consecutive rolls of a die), there will also be equally many of any other sequence of equal length. No digit or sequence is "favored".

But I’m not convinced myself that any arbitrary sequence of digits will eventually turn up in any given normal number (such as pi, if it is normal).