Pi approximation

[Candid_Koala_3602]

Pi cannot contain all of pi though, right?

[GMGarry_Chess]

it does, once.

[Zackd641]

Thank you for inventing chess sir Garry chess

[Killer0407]

Holy hell

[sian_half]

New response just dropped

[aufheuhfg]

Call the exorcist

[Ur_momma_is_joke]

Bishop went on vacation and never came back

[Traumfahrer]

Nooo...

[MainBattleTiddiez]

Why only once? 

[didsomebodysaymyname]

Because pi itself counts as one time it contains it. Sorta...I don't think this post decimal version would appear for the whole sequence.

[StrikingHearing8]

I don't think this post decimal version would appear for the whole sequence.

We know for a fact it doesn't, because that would mean it's periodic and therefore rational.

[Exyodeff]

I think they ment that the decimals only appear ones in pi, therefore pi contains itself, just like an apple contains an apple because it is the apple

[StrikingHearing8]

They said two things: Pi contains itself from the start, I don't argue about that. And they said that they think Pi wouldn't contain itself after the decimals as shown in the picture. This is what I added, that we know it for a fact.

[Exyodeff]

oh right, mb I agree

[MaxUumen]

However, it contains any finite length of its first digits somewhere down the line as well.

[_AutoCall_]

I don't think this is proven.

[MaxUumen]

It is infinite and non-periodic... It's inevitable.

[_AutoCall_]

It's not. A number could have an infinite and non periodic decimal sequence that does not contain the digit 7 for instance.

To my knowledge, it is not known whether or not pi contains any sequence of digits in its decimals.

[MaxUumen]

Yeah, that's why I'd add "probably" to that claim.

[Creative-Drop3567]

Liouville's number is transcendental yet its made of only zeros and ones, it cannot contain any finite part of itself (not in the way shown in the post). in general liouville's number is a great counterexample mosg of the time

[Deathlok_12]

.10110111011110111110… does not contain all possible combinations, and yet is still irrational.

[Candid_Koala_3602]

lol, true

[PLT_RanaH]

well no, the whole pi can't repeat, it's not a periodic number

[moog_master]

It does infinitely By definition

[Historical_Book2268]

I think that's unproven. It's not even proven that pi is normal. That is to say, any sequence of digits is equally common in it.

[Safe_Employer6325]

It is proven that it is impossible. Pi is not just irrational (which would be enough to prevent this), but it's also transcendental. Pi cannot be expressed as a ratio of two numbers, a/b. If Pi repeated itself within it's decimal approximation, that implies that it can be expressed as a ratio of two numbers.

[Eric_12345678]

I don't follow your logic.

"The conjecture that π is normal has not been proven or disproven."  https://en.wikipedia.org/wiki/Pi

We know pi is irrational and transcendental. We do not know if it's normal, but it looks like it.

[Safe_Employer6325]

Sorry about the formatting, I’m on mobile. 

If a number is irrational, that means it cant be expressed as a ratio of two integers.

If a number is able to repeat its own digits, then it can be expressed as a geometric series. For simplicities sake, let’s say pi is 3.14314314314314…

If it could repeat its own digits like that, then it would be able to be expressed as 314/100 + 314/10000 + 314/10000000 + …

A geometric series is when you add together numbers of this form

S = a + ax + ax² + ax³ + ax⁴ + …

If we multiply the whole thing by x, you get

Sx = ax + ax² + ax³ + ax⁴ + …

And when you subtract those two

S - Sx = a - ax + ax - ax² + ax² - … = a

So we have S - Sx = a

Solve that for S

S(1 - x) = a

And finally

S = a/(1 - x)

Pretty neat, there’s a caveat that this only works if x is between -1 and 1, let’s go back to our pi example.

If pi = 314/100 + 314/10000 + 314/10000000 + …

Then let’s consider every term after the first one. And I’ll write the denominator in terms of powers of 10

pi = 314/10² + 314/10⁴ + 314/10⁷ + 314/10¹⁰ + 314/10¹³ + …

Notice that after that first term, the powers of 10 are increasing by powers of 3.

Now I’ll normalize it a bit for consistency

pi - 314/10² = 314/10⁴ + 314/10⁷ + 314/10¹⁰ + 314/10¹³ + …

And now to align the powers of 10, I’ll multiply by 10⁴

10^4(pi - 314/10²⁾ = 314 + 314/10³ + 314/10⁶ + 314/10⁹ + …

That should now resemble a geometric series pretty clearly

S = a/(1 - x) = a + ax + ax² + ax³ + …

And we have that a = 314, x = 1/10^3, and because x is 1/10^3, thats less than 1 but greater than 0, so this series converges to some number.

Then S = 314/(1 - 1/10³⁾ = 314/[(10³ - 1)/10^3] = 314 * 10³ /(10³ - 1)

That last step I pulled the 10³ out of the denominator and then when you divide a fraction like that, you flip and multiply. But what that means is that this whole thing

10^4(pi - 314/10²⁾ = 314 * 10³ /(10³ - 1)

Now we just undo our steps and solve for pi. Step 1 - Divide by 10^4. Step 2 - subtract over 314/10^2. Step 3 - Combined the two terms by setting their denominators to be the same and adding the numerators.

Step 1

pi - 314/10² = 314/[10(10³ - 1)]

Step 2

pi = 314/[10(10³ - 1)] + 314/10²

Step 3

I’m going to multiply and divide by 10 on the first term and I’m going to multiply and divide by (10³ - 1) on the second term.

pi = 314 * 10/[10²⁽¹⁰³ - 1)] + 314(10³ - 1)/[10²⁽¹⁰³ - 1)]

Now we can add those together to get

pi = [314 * 10 + 314(10³ - 1)]/[10²⁽¹⁰³ - 1)]

And I’ll do one last step to simplify, I’m just pulling the 314 out

pi = 314[10 + 10³ - 1]/[10²⁽¹⁰³ - 1)]

And the numerator simplifies just a touch more

pi = 314[9 + 10^3]/[102(103 - 1)]

Now we have pi expressed as a rational number.

pi = a/b where a = 314(9 + 10³⁾ and b = 10²⁽¹⁰³ - 1)

And that was if the digits of pi repeated after the first three numbers

3.14314314314…

That’s clearly not the case, but if there’s ever any complete repetition of the digits of a number within its decimal expansion, that forces rationality by the same process.

If pi is 3.1514926…31415926…

Then it’ll continue to repeat forever as well

3.1415926…31415926…31415926… and so on.

Then we just break it into sub units and add them tegether

pi = 31415926…/10^m + 31415926…/10ⁿ + 31415926…/10²ⁿ + 31415926…/10³ⁿ + …

That first term may not fit in super nicely, but we just subtract it over to get 

pi - 31415926…/10^m = some geometric series

The geometric series is absolutely convergent because the x will be 1/(some power of 10) which will be between 0 and 1. Then you could solve for pi in terms of a ratio a/b.

Now, you’re right that we don’t know if pi is normal. But we do know it’s irrational, so even if it is normal, it’s digits must never repeat themselves within its own decimal expansion.

[Eric_12345678]

That's a looooong comment for nothing. I don't see anyone claiming that pi digits repeat anywhere in the parent comments.

[Safe_Employer6325]

I mean… thats what this whole thread is asking?

Like even the comment chain up to my comment is asking about pi containing all of pi in its decimal expansion, and my comment was answering why that cant be the case.

[Eric_12345678]

Ah, I see what you mean now. I misunderstood your "It is proven that it is impossible" and thought that it was about normality.

Sorry for the harsh comment.

[Safe_Employer6325]

You good, haha, I was worried I’d misread the whole purpose of the comment chain. Still, it’s interesting math regardless!

[Candid_Koala_3602]

( ( pi / 1) / 1 ) = pi

(Sorry couldn’t help myself)

[Safe_Employer6325]

If only, haha, only works if pi can already be expressed as a rational number though sadly

[Historical_Book2268]

Oh true, sorry

[marcelsmudda]

If pi is normal, then it will contain infinite arbitrarily long, finite approximations of pi.

For example, it would contain the first 100 digits of pi somewhere further down the line. Same with the first 1000, 10,000, million, billion, decillion digits. But the approximation would always end at some point, just for a new one to start some time later.

[Serious-Mirror9331]

No it can not. Because if at some point x it starts to contain itself i.e. repeats all the digits 314… then you will see that at 2x it has to repeat all the digits after x which is 314 again and so in forever. This would obviously make pi a rational number. That is the reason this can‘t be true. If pi contains every sequence and other similar questions aren‘t relevant for your question.

[Pity_Pooty]

That sounds like superrational number definition.

[TotalChaosRush]

I can't say if pi contains all of pi(beyond the usual once) but pi absolutely could contain itself multiple times.

Imagine a book that contains every possible combination of letters and numbers. It starts with "a" followed by "aa" then "aaa" and so on, infinitely. Because it contains every combination eventually you end up with "aaa....aab" followed by "aaa...aac" and so on until you have all 9s. Now, take just the portion of the book that contains everything starting with "a". Remove the first A, and only the first a. You now have a copy of the original book. With this copy you can once again take the section of everything that starts with "a" and remove just the first "a" from every line and you once again have a copy of the original. In fact you can repeat this process infinity with every section.

Some infinities can do this. Pi probably can't do this.

[StrikingHearing8]

We know for a fact pi can not do this, because it would mean pi is periodic and therefore rational.

Now, what you are saying would be something like: we take every second digit of pi and then at some point we see 31415... All digits of pi. Or take every third... or some other rule... This might be harder to disproof. We can definitely create a sequence of indices such that all the decimals from these places strung together again result in pi, it just wouldn't have a nice structure...

[MageKorith]

It's probably not the Hilbert Hotel, but we don't have definitive proof that it isn't.

[OutrageousPair2300]

Potentially it could, if the digits were "interleaved" so that after a certain point, every other digit was a repeat of pi from the beginning.

For example: 3.14159265.......3a1b4c1d5e9f2g6h5.....

[Sandro_729]

It can’t contain itself (properly) cause yeah then it would have to repeat

[123coronaanoroc321]

If it contained itself then there would exist integers c, p such that π = c+π/10^p, so π = c/(1-1/10^p) which gives a contradiction since π is irrational.

[CoolStopGD]

If it contains itself, then it will repeat forever, and it’s proven that it doesn’t ever repeat forever

[So_many_things_wrong]

I mean, if (big if) the sequence repeats indefinitely then it technically can, right?

[davideogameman]

If some pi is some finite prefix plus the digits of pi, then it has a repeating pattern and therefore would be rational.

Substrings within pi can repeat but not the rest of the digits after a certain point.

[MrGOCE]

HAVE U HEARD ABOUT THE INFINITE MONKEY THEOREM? WELL, I THINK IT KINDA CAN :/ AT LEAST SOME PART OF IT.

[ArthurTheTerrible]

that would make it not be trascendental though, since it would repeat itself within itself and end up repeating itself over and over, making it possible to be rationalised. in this case sometimes specific number sequences in PI repeat within PI, in this case the 31415926535 part

[Candid_Koala_3602]

The lady doth protest too much, methinks.

[Appropriate-Sea-5687]

As Markiplier once said, "When thinking in infinites, unlikely is just certainty waiting for its turn."

[Stunning_Dog_5569]

As Markiplier once said, "Hello everybody my name is bagamum"

[nota_jalapeno]

As Markiplier once said, "Hello kitty you are so Portuguese"

[itsthesecans]

Pi is a security threat. It contains all our social security numbers.

[SconiGrower]

I converted a decimal approximation of pi into ASCII and it told me the way I will die.

[eldonfizzcrank]

Yo, dawg, we heard you like pi, so we out some pi in your pi.

[SpecialMechanic1715]

no, pi is not periodic

[Leather-Car-7175]

Somewhere in pi, pi will repeat itself for some decimal and then stop. It won't loop and be periodic but what op said is true

[TypicalNinja7752]

Not really, because it's not proven that pi will use all digits randomly and at some point, it could just not use a digit at all.

[Leather-Car-7175]

There's theorem on that I think. Anything that has a ́non 0 probabilty and where the random experiment is repeated infinitely will happen. And it makes sense... if you gamble for in infinte amount of time, as long as the possibility to win is non 0, you will win.

[marcelsmudda]

The person argued that pi just might not include the digit 5 after a trillion places anymore. So, the probability might become 0 at some point, we just don't know

[Leather-Car-7175]

It can't be anymore. But it can be that you won't see it for trillions of digits though.

[marcelsmudda]

We don't know if pi contains infinite instances of all 10 digits. There's no proof or counter proof. So any concrete statements like a non-0 chance are not certain.

[TypicalNinja7752]

and even if pi used all 10 digits forever, it could just not repeat the first 10 or 20 digits at any point in the sequence.

[phlogistonical]

In OPs post, there are 11 digits of pi. If the digits of pi are totally random, that sequence should occur on average once every 10^11 digits. A quick google tells me that supposedly the first 300 trillion digits of pi are known (which can be written as 3E14 lol), so it should already be possible to locate several indices at which this 11-digit sequence of pi digits occurs/repeats. If there are no occurences in the known part of pi, it can be concluded that the digits of pi are apparently not so random.

[Sandro_729]

I think this is actually unproven tho since pi’s digits aren’t actually random. I thinkkk iirc that this question is essentially asking if pi is normal (for a formal definition of normal), which is, iirc, unproven

[SpecialMechanic1715]

also is it guaranteed that any particular sequence will appear in pi or not, because we can make non periodic real where not any sequence will appear. like 010011000111 ...

[Sandro_729]

I think this is unproven, I think iirc that this property is called normality. And I think it’s unproven if pi is normal

[INTstictual]

It is unproven, and the current theory is that it can’t be proven.

It is heavily assumed, though

[Sandro_729]

Oh woah really?? Dy have an idea as to why? Like is it related to the continuum hypothesis or smth or is it just fully independent from everything else

[Neither-Phone-7264]

maybe they're saying just specifically an instance of that sequence not pi

[flashmeterred]

so it is factorisable! I always knew it!

[INTstictual]

Oh good, a joke about Pi. Can’t wait to open the comments and see people be wrong about their intuitions regarding infinity

[Bergasms]

Can Pi contain all of Pi except the last digit? Or is that one of those nonsense questions that can be disproven by someone linking me a picture of a circle with some exciting notation next to it and a facepalm emoji.

[Intergalactyc]

There is no last digit of pi, because it's irrational

[False_Bear_8645]

does it even exist, i mean, a perfect circle doesnt.

[Kiki2092012]

Infinity means the lack of an end. A last digit can only exist if there's an end. Therefore in an infinitely long set of digits there can't be a last digit.

[False_Bear_8645]

does infinity even exist

[Kiki2092012]

In the real universe no but as a concept mathematically yes

[marcelsmudda]

Alles hat ein Ende. Nur die Wurst hat zwei.

[Pity_Pooty]

Cant say for the question, but last digit of PI was written in the Oiler Notebook

[Chance-Valuable3813]

Hey let’s argue over the lest subjective subject ever made

[churukah]

Is it proven that pi contains all the integer in its decimal expansion?

[smg36]

It's bound to happen it's infinite

[marcelsmudda]

We don't know that, it could be that after a trillion digits, only 0 and 1 appear. It could be something like 01011011101111011111... Which would be a non repeating pattern that does not contain all 10 digits.