The Nineteenth Byte

caird coinheringaahing

9:00 PM

@Mr.Xcoder Then you're not computing it

Mr. Xcoder

What's p-adic?

Zacharý

The only numerical things I know are square root approximation and Euler's method

Pavel

@LeakyNun tio.run/##y00syUjNTSzJTE78/98v2jXOWEfB0MDAIPb///@6BUWZeSUA

caird coinheringaahing

@LeakyNun I probably shouldn't go down this rabbit hole but what does p-adic mean?

ಠ_ಠ Goddamn ninja's

Leaky Nun

I got two people interested :D

AdmBorkBork

9:00 PM

Just not your day, @caird

Riker

@LeakyNun phd paper detected /s

Leaky Nun

now let's work with 10-adic integers first, although in the proper setting it should be a prime

so work with the integers in base 10

and then pretend that they can go infinitely to the left

so you have integers like ...9999

and then ....9999 + 1 = 0, because the "1" that is carried gets lost to infinity

Pavel

So ...99999 == -1

Riker

> get email saying "event is cancelled"
> person emails me asking if I saw email
> I respond "yes I did but I couldn't go anyway"
> they ask if I am going to event

wut

Leaky Nun

@Pavel precisely

totallyhuman

9:03 PM

...what

Riker

@LeakyNun is this the same thing about that like 1 + 2 + 3 ... = -12

AdmBorkBork

@Riker Cancelled is the new not-cancelled.

totallyhuman

ok...

@Riker *-1/12

Zacharý

@Riker WRONG

-1/12

Leaky Nun

@Riker that still doesn't converge in p-adic

totallyhuman

9:04 PM

Ninja'd

Riker

@Zacharý close enough

@LeakyNun not entirely certain how those work /shrug

totallyhuman

I'd like to report a bug in chat.SE >_>

Leaky Nun

@Riker I don't want to formally define them now

Riker

only off by 12 & 1/12

caird coinheringaahing

@totallyhuman Go ahead, they don't care anyway

Riker

9:04 PM

@LeakyNun np I can read wikipedia

Leaky Nun

basically it doesn't converge because the unit digit keeps changing

Riker

@cairdcoinheringaahing they fixed my plurals bug iirc

Leaky Nun

(a sum to infinity is just the limit of the partial sums)

Riker

10

Q: Chat rate limit pluralization bug

Found this out when trying to recite the alphabet in messages. When you hit the rate limit in chat, and there is 1 seconds left, it says the incorrect form. It shouldn't be plural. Example: Can this be fixed please? Seriously, this is slightly irritating. I am not just posting this to poin...

caird coinheringaahing

@Riker Jeff must have retired in that case

Leaky Nun

9:06 PM

so -32 is just ...99968

I feel like I lost half of the audience already

totallyhuman

only if you have 42.7bit integers

caird coinheringaahing

I'm still listening

totallyhuman

totally pulled that integer out of my

Leaky Nun

so it turns out that 1/n exists as long as n is coprime with 10

to compute 1/3:

...999999 = -1
...333333 = -1/3
...666667 = 1/3

totallyhuman

...so ...666667 == 0.333...

caird coinheringaahing

9:08 PM

Yes

Leaky Nun

now exp(x) := 1 + x + x^2/2! + x^3/3! + x^4/4! + ...

caird coinheringaahing

Now you've lost me

Leaky Nun

@totallyhuman yes

@cairdcoinheringaahing that's just the taylor expansion of exp :P

caird coinheringaahing

Taylor series are infinite sums that approach a value, correct?

Leaky Nun

the thing is that if x is a real number and we are doing normal mathematics, then 1 + x + x^2/2! + ... approaches e^x

@cairdcoinheringaahing correct

caird coinheringaahing

9:11 PM

so exp(x) := (x^n)/n! for n -> ∞?

totallyhuman

exp(x) = sum (k = 0..∞) {x^k/k!}

Leaky Nun

@cairdcoinheringaahing sum

RosLuP

Taylor series for me it is just one loop one has to know where to stop

caird coinheringaahing

Yeah, totallyhuman did it better than me :P

totallyhuman

yeah except there's gotta be a better way to represent sigma notation in ascii text

caird coinheringaahing

9:12 PM

Just don;t start using mathjax, that's how to lose me completely :P

Leaky Nun

so it turns out that if we pretend that this extends to the p-adic numbers

exp(x) := 1 + x + x^2/2! + x^3/3! + x^4/4! + ... converges whenever x is divisible by p

totallyhuman

why are we pretending so much

Leaky Nun

@totallyhuman because we are pretentious

caird coinheringaahing

@LeakyNun Spending too much time in England? :P

totallyhuman

mathematicians are pretentious

Leaky Nun

9:14 PM

indeed

makes sense

0∊N ?

yes

Dennis' message on that is perfect.

Leaky Nun

so exp(10) = 1 + 10 + 50 + 500/3 + ...

Zacharý

9:15 PM

From a long time ago

totallyhuman

$\sum_{k=0}^\inf \frac{n^k}{k!}$

Leaky Nun

where 500/3 = ...333333500

caird coinheringaahing

3 mins ago, by caird coinheringaahing

Just don;t start using mathjax, that's how to lose me completely :P

Leaky Nun

and then it converges because eventually the unit place is zero, eventually the tens place is zero, etc

so every digit can be calculated using finitely many steps

totallyhuman

@cairdcoinheringaahing exactly why I did it (¬‿¬)

caird coinheringaahing

9:18 PM

So 1/3 = ...666667; 500/3 = ...33333500 so 10000/3 = ...closer to 1 than 500/3?

Leaky Nun

10000/3 = ....6666670000

caird coinheringaahing

For n/3 is it smaller or larger for a larger n?

Leaky Nun

well how do you order the numbers

AdmBorkBork

They're all infinite

caird coinheringaahing

@LeakyNun 1 < 100 < 1000 < ∞

Leaky Nun

9:20 PM

is ...112112 smaller or ...221221?

remember, they legitimately continue infinitely to the left

they don't have to be periodic at all

caird coinheringaahing

So there's no way of telling whether one is larger than any other?

Leaky Nun

well there is a linear order that you can define, but you must not be very convinced by it

append ".0" at the end, reverse the whole string, and pretend that they are real numbers and then compare them

totallyhuman

it's like trying to define parity for irrational numbers

Mr. Xcoder

Or rationals that aren’t integers

totallyhuman

or for complex numbers

Mr. Xcoder

9:24 PM

relevant

Leaky Nun

the two ways to construct the real numbers from the rationals can also be used to construct the p-adic integers from the integers, but that's too much formalism for this context

I really want p-adic to be a part in recreational maths

but the formalism always gets the better of it

caird coinheringaahing

What does p-adic actually mean?

Leaky Nun

informally, pretending that they extend infinitely to the left

in base p

caird coinheringaahing

So it's messing around with infinitely large numbers in a different base than 10?

Leaky Nun

right

caird coinheringaahing

9:27 PM

I see why its not done at high-school level :P

Leaky Nun

well is it really harder to understand than real numbers?

I mean, they continue infinitely to the right

caird coinheringaahing

It's the 'different base' part that makes it more confusing. Most people in my top set maths class can't explain what binary is

Leaky Nun

so just do it in base 10

OEIS has many 10-adic sequences

caird coinheringaahing

There is also the idea that you can cut off the end of a real number and just work with, say, 10 decimal places, rather than the whole number

AdmBorkBork

decimal or riot

caird coinheringaahing

9:29 PM

Riot?

Leaky Nun

@cairdcoinheringaahing you get the same thing in 10-adic integers

let's say you're trying to find solutions to x^2=x

to say 10 digits

5^2 = 25

25^2 = 625

caird coinheringaahing

Isn't it 1 or 0?

Leaky Nun

@cairdcoinheringaahing there are 4 solutions

625^2 = 390625

so the number goes like ...625 (it's in OEIS!)

and you can work with only 10 decimal places instead of the entire number

...625^2 = ...625

mudkip201

10 digits? ...

lol

I think you meant 3

Leaky Nun

well I don't have enough time to deal with 10 digits so I just showed you 3 digits

mudkip201

9:35 PM

:P

ah

fair enough

Leaky Nun

and also you can just look it up OEIS

caird coinheringaahing

@LeakyNun what degree are you doing at uni?

Leaky Nun

the basic one

caird coinheringaahing

Scratch that. What exactly is 'Pure mathematics'?

Leaky Nun

mathematics without perservatives

Thomas Ward

9:38 PM

@cairdcoinheringaahing 1 + 2 + 3 = 6

:P

Leaky Nun

log(1+2+3) = log(1)+log(2)+log(3)

mudkip201

... true

derp. duh

caird coinheringaahing

This user's top tags are brillinat when taken out of context :P

Leaky Nun

have I lost all my audience lol

caird coinheringaahing

I kinda don't see the point of it, and you've explained what p-adic means, so I've kinda lost interest :/

Leaky Nun

9:48 PM

ok

precision = 1000
n = 47

m = 1
res = 0
for _ in range(precision):
	while res*n%(m*10) != 1:
		res += m
	m *= 10
print(res)
print(res*n)

tio.run/…

compute 1/n where n is coprime to 10

@cairdcoinheringaahing does it interest you more?

ATaco

I'm assuming you mean without the usage of /

Leaky Nun

@ATaco ah, you missed the context

55 mins ago, by Leaky Nun

now let's work with 10-adic integers first, although in the proper setting it should be a prime

read from here, it isn't really long, and there are quite a lot of irrelevant messages

ATaco

Sorry, Trains make bad work on my internet.

Leaky Nun

shouldn't take you more than 2 minutes to read the whole thing

ATaco

Nissed a bit.

Leaky Nun

9:56 PM

or I can just repeat the whole thing

ATaco

Nah, alllll good

Adám

@LeakyNun What do you say, which should precede which in a total array order (where 100 precedes 'a') 0 1⍴0 or 0 0⍴''?

Leaky Nun

heh...

Luis Mendo

@flawr Done

The syntax is cumbersome, though

Adám

@LuisMendo Oush?

Luis Mendo

10:12 PM

Aw. I can't edit the link anymore

Zacharý

An abbreviation for Out sh?

Or for Aw, hush!

Pedants.

:-)

No, I think it is the Opposite push.

@Mr.Xcoder Nice edit.

Luis Mendo

10:13 PM

^^ And is that pop, or pull?

Adám

@LuisMendo Oush.

@LuisMendo Yank?

Luis Mendo

Funny thing, I only learned that verb, yank, because of the V Language :-)

Mr. Xcoder

s/because of/thanks to/ /s

Adám

@LuisMendo I use it as mnemonic for APL's dyadic function ↑ (officially called "Take") but is located on the Y key.

DJMcMayhem

@LuisMendo :)

Luis Mendo

10:17 PM

Heh. Does mentioning V summon you?

DJMcMayhem

Hehe

Yes

Zacharý

The letter V eVen ?

DJMcMayhem

I should change my username to DJMcVhem

flawr

@LuisMendo cool, I mean you'd probably only need it for testing/debugging

Luis Mendo

@flawr Or as a form or source coding (compression). What's the golfiest way to produce the array [0.1612309217453 0.793517112731934 0.292634516954422 0.803473114967346]? :-P

@DJMcMayhem Sure, you haven't changed your name enough times

Leaky Nun

10:39 PM

ideone.com/8Hrtda

1000 digits of e^3 in Z_3 (3-adic integer)

flawr

@LuisMendo Go make a kolmogorov-complexity out of that XD

Luis Mendo

:)

HyperNeutrino

DJMcMayhem
DJMcJunehem
DJMcJulyhem...

11

DJMcMayhem

@HyperNeutrino Hahahaha

Best thing about that is that since a new username sticks around for 30 days, I could totally do that (although I'd have to be DJMcFebruaryhem for the first 2 days of DJMcMarchhem)

HyperNeutrino

or for the last 2 days of January