General Formula for all arithmetic operations

[jovial rock]

oh okay, what do you mean by left sided derivative and right sided derivative?

[azure crane]

well normally it doesn't matter

but for discontinuous functions it does

here lemme show you

Well, not just discontinuous functions but you can clearly see it with them

So here

When x=the point where the function is discontinuous

you can't really define the normal derivative

bc remember limits work by approaching the value getting very close

$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

[dawn finch]

RoyalBanana

[azure crane]

See if you approach 0 by plugging in very small negative values of h there, it'll give you a different result than very small positive values of h because the function is completely different on each side of the discontinuity

But the derivative can't be two different numbers at the same time ofc so it's undefined

but, basically the left sided derivative is what you get for very small negative values of h and the right sided derivative is what you get for very small positive values

[azure crane]

Ah I can fix that

If $f(x+1)=g(f(x))$

$f(x)=\sum_{n=-\infty}^{\infty} \qty{n \le x<n+1:g_n(h(x-n)),0}$

[dawn finch]

RoyalBanana

[azure crane]

My brain hurts but it works lol

[jovial rock]

Oh Yeah Right

Is it possible to maybe aproach the derivative with an other numbersystem? Like the complex numbers

is gn(x) iterated g(x) ?

[azure crane]

..?

[azure crane]

yep

[azure crane]

Oh like doing 0.00001*i+0? Sure

[azure crane]

I wonder if the continuous thing makes it so there's only one solution

because that means all of the derivatives have to match at every integer value of the function

that seems like a very strict requirement

[azure crane]

that would be nice

[jovial rock]

hmm, yeah probably ye

[jovial rock]

I thought maybe the discontinuous functions could behave diffrently in other number systems and I thought they could be used to take derivative and stuff

[azure crane]

Mostly they're just undefined bc > and < doesn't make sense in complex numbers

[jovial rock]

okay

[azure crane]

I feel like the way to figure this out is right on the tip of my tongue lol

[jovial rock]

Lol

It‘s pretty hard for my because I’m not very good at math lol, but it’s really cool to find a solution for that and it’s fun

[azure crane]

@jovial rock

I figured it out

I think

so using the Taylor series thingy

Assuming f(x) has a Taylor series, $f(x)=\sum_{n=0}^{\infty}f^{(n)}(0)\frac{x^n}{n!}$

[dawn finch]

RoyalBanana

[azure crane]

so

$f(1)=\sum_{n=0}^{\infty}f^{(n)}(0)\frac{1}{n!}$

$f(2)=\sum_{n=0}^{\infty}f^{(n)}(0)\frac{2^n}{n!}$

$f(3)=\sum_{n=0}^{\infty}f^{(n)}(0)\frac{3^n}{n!}$

etc

[dawn finch]

RoyalBanana

[azure crane]

now if we wanna solve for f^(n)(0) we can just treat them like variables here

and just solve for them

infinite things to solve for but infinite equations so it's doable I think

it's kinda like solving

$a_1x_1+a_2x_2+a_3x_3=y_1$

$b_1x_1+b_2x_2+b_3x_3=y_2$

$c_1x_1+c_2x_2+c_3x_3=y_3$

for $x_1,$ $x_2,$ and $x_3$

[dawn finch]

RoyalBanana

[azure crane]

So basically we can solve for this, and then plug those values back into the original formula to get the Taylor series of the function!

I wonder if this actually works

[jovial rock]

hmm, I see

I'm just not sure if this function is continous

And how can we take the darivative again?

[azure crane]

wdym

[jovial rock]

the ln function because it goes negative infinity and stuff

[azure crane]

we don't need to know the function there then

[jovial rock]

okay

[jovial rock]

I thought this may be cool. I wanted to expand this to the reals: x^(x-1)^(x-2)^… so F(x) = x^F(x-1), F(1) = 1. this function explodes very fast, at F(4) it is 4^9 = 262’144. so I had the idea to use the iterated ln-> lnx. And I found out that when x aproaches infinity it aproaches the constant -0.0613312…

Maybe by expanding it to the reals there could be a way to find the iterated ln expanded to the reals

[azure crane]

interesting: 🤔

[azure crane]

this kinda reminds me of factorials

like factorials but exponential

interesting

[azure crane]

btw $\ln^s(a^b)=\ln^s(a)+\ln^s(b)$?

[dawn finch]

RoyalBanana

[azure crane]

wait no

exponentiation isn't communative so that can't be right

hmm

[jovial rock]

I did more stuff with this and this is what I found. So when x gets large lnx(F(x)) is equal to -0.0613312. I was wondering if we Can change that value by exponating every number in this exponetiation tower. Something Like that: (5^x)^(4^x)^(3^x)^(2^x). And then i remembered if you do lnx(e^^x) you get 1-> ln e =1, ln ln e^e = 1 and so on. I used iterated ln to stabilize the function to a constant. So if we would find the number to expontiate every number in the exponetiation tower that aproaches the costant 1, we would have a function that behaves exactly like e^^x because both equal 1 when they get transformed by lnx

This number is about 2.403, not sure. I hope someone can understand this lol

[jovial rock]

Yeah, I was wondering if I can to the same thing but with exponents

[jovial rock]

Yup, sadly

I was wondering if it could be something with inverse operations like - or / because they aren’t commutative

[azure crane]

?

[jovial rock]

https://youtu.be/MP3pO7Ao88o?si=UJmuHiMAKbc9SnXW

I found this video pretty interesting. Maybe it could be useful for commutative things

[jovial rock]

So ln5(5^4^3^2^1) equals -0.061…, and this is the same for lnx(F(x))

When F(x) = x^F(x-1)

[azure crane]

ah

[jovial rock]

We can rewrite this as ln5((5^1)^(4^1)^(3^1)^(2^1))

And if we change the 1, we get a diffrent constant which the function aproaches

[azure crane]

oh ig sure

[jovial rock]

And what we’re doing is taking the ln x times

And we know lnx(e^^x) is 1

[jovial rock]

So we need to find a value for 1 here that aproaches 1. What means it aproaches the behavior of e^^x

Because we have a exponetiation tower

Hmm

I need to find out how to explain this

I’ll ask chat gpt

[azure crane]

bruh

[jovial rock]

Sorry

[azure crane]

?

[jovial rock]

For Not being able to explain it

Not sure if this helps

[azure crane]

so

whats the proof that it matches for k=2.403..

[jovial rock]

It‘s not very acurate, I only used F(5)

This is just ln5(F(5))

So ln ln ln ln ln 5^4^3^2

[azure crane]

ok but.. how do you prove it

[jovial rock]

And a Bit simplified because the Numbers would get too big

I guess solve for k when x aproaches infinity here: Definition for function L: L(x,k) = (x^k)^L(x-1,k), solve for k: lnx(L(x,k)) = 1

[azure crane]

oh you just meant as x goes to infinity? ok

wait what's the point of this again?

[jovial rock]

The bigger x is, the more accurate the constant is. If the constant is 1, it means the function has the same growth like e^^x. So we would have a Connection between the descending power tower and tetration with e

[azure crane]

what's the point though?

it's only as x gets large

[jovial rock]

Wait, I need to think

I thought we could find some interval where x is large and work it all the way back, but am not sure if it works

[azure crane]

?

[jovial rock]

It’s not very hard to find non integer values for L(x) = lnx(F(x)), when F(x) = x^F(x-1) because it aproaches a constant value very fast, what means, L(x+1) ≈ L(x) when x is bigger than 6. That means L(x+0.5) ≈ L(x). So we can use this to find an interval for e^^x when x is big, and than work our way back

Like in the factorial vid or harmonic numbers vid

But I’m not sure if it works

[azure crane]

if you want to get a converging value out of tetration just do 1/e^^x

?

[jovial rock]

I’m just not sure if it works with this, because it converges to 0 very quickly

It may work

Idk, I just had some idea I was playing around with in my head with the descending power tower

[azure crane]

It doesn't matter how fast it does

as long as it doesn't converge within a finite range

[jovial rock]

Yeah right

I think the only big problem is finding the recrusive formula for exponentiation

[azure crane]

f(x+1)=e^f(x)?

[jovial rock]

Yeah, but Write it in discrete form

[azure crane]

wdym?

[jovial rock]

with the capital sigma

[azure crane]

$\sum_{n=?}^{?}?$

like this?

[dawn finch]

RoyalBanana

[azure crane]

woah

[jovial rock]

yup

hmmm

maybe we need iterated sums or something like that

[azure crane]

I assume we can write it with $\prod_{n=?}^{?}?$

[dawn finch]

RoyalBanana

[azure crane]

then again maybe we need some sort of operation like that for exponentiation, idk

[jovial rock]

btw I tried to make iterated products and i got higher degree factorials which i found pretty cool. They are based on the pascal's triangle. 5! * 4! * 3! * 2! * 1! = 1^5 * 2^4 * 3^3 * 4^2 * 5^1. this is the same as f(x) = product(x,n=1)(product(n,k=1)(k)) for f(5), f(x) = product(x,n=1)(product(n,k=1)(product(k, m=1)(m))) for f(5) is 1^35 * 2^20 * 3^10 * 4^4 * 5^1

[jovial rock]

yeah, but exponetiation is not commutative

[azure crane]

so?

[jovial rock]

I just was messing arount with iterated products

[azure crane]

we'll just define which order you go in

after all

you can do that

that's what the sum and product ones do

anyway

[jovial rock]

I tried doing it, but I dont get how to combine two of them

hmm

[azure crane]

Wdym

[jovial rock]

maybe it could work

[azure crane]

what is there to not work ?

$\bigwedge_{n=?}^{?}?$ is meant for something else but it works for this

[dawn finch]

RoyalBanana

[azure crane]

it looks like the exponent symbol ^

[jovial rock]

yeah right

maybe I'll try to solve it again

[azure crane]

So do we define for example $\bigwedge_{n=1}^{3}n$=1^{2^3}$ or $3^{2^1}$

[jovial rock]

3^2^1

[dawn finch]

RoyalBanana
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[jovial rock]

I'll say

[azure crane]

I feel like it's kinda weird to go from top to bottom

[jovial rock]

because it's exponetiation

1^2^3 = 1

0^1^2^3 = 0?

[azure crane]

ye

[jovial rock]

3^2^1 is a bit more safe because 3^2^1 = 3^2^1^0^-1^-2^...

[azure crane]

..?

[jovial rock]

thats my thought

[azure crane]

um

[jovial rock]

idk

[azure crane]

$\bigwedge_{n=1}^{3}{(3-n)}$

[jovial rock]

we can do 2 versions

[dawn finch]

RoyalBanana

[azure crane]

why?

[jovial rock]

for 1^2^3 and 3^2^1

[azure crane]

This works though

[jovial rock]

ah I see

[azure crane]

well, typo, it should be 4-n, but still

[jovial rock]

yeah you're right

so you like 1^2^3 more?

[azure crane]

hm

I feel like I always imagined $\prod$ and $\sum$ putting in each term left to right

[dawn finch]

RoyalBanana

[jovial rock]

yeah right

[azure crane]

well, it doesn't matter that much bc you can just use this if you want the other way around

[jovial rock]

maybe you're right. lets do it like that for know

yeah right

[azure crane]

If n is whole, ${}^nx=\bigwedge_{k=1}^{n}x$

[dawn finch]

RoyalBanana

[jovial rock]

yup

so if we find a way to cancel out the k we have tetration

and then n can be a non integer

maybe

[azure crane]

sure

[jovial rock]

we could use that ln(x+n) = ln(n) for huge x

but it's harder with exponentiation. ln(exponentiation(x)) is not product(ln(x)) i think

[azure crane]

Here's that exponential version of the factorial btw $\bigwedge_{k=0}^{x}(x-k)$

[dawn finch]

RoyalBanana

[jovial rock]

okay, yeah that's nice

[azure crane]

hm

I had an idea

$\ln(\prod_{n=1}^{k}f(n))=\sum_{n=1}^{k}\ln(f(n))$

[dawn finch]

RoyalBanana

[azure crane]

$\ln^s(\bigwedge_{n=1}^{k}f(n))=...$

[dawn finch]

RoyalBanana

[jovial rock]

Oh yeah

I was thinking about that too

Not sure if it works

[azure crane]

Well, for sure $\ln(\bigwedge_{n=1}^{k}f(n))=\ln(f(1))\bigwedge_{n=2}^{k}f(n)$

[dawn finch]

RoyalBanana

[azure crane]

hmm

[jovial rock]

lnk

Hm

There is a problem

ln(5^x) = x * ln(5), ln(x * ln(5)) = ln(x) + ln(ln(5)), ln(ln(x) + ln(ln(5))) = ?

[azure crane]

nothing useful there

[jovial rock]

Take the ln k times so there are only products

[azure crane]

Doesn't work

[jovial rock]

I know

[azure crane]

hm

[jovial rock]

It would be Nice if we could do some transformation that brings it back into multiplication without taking away the ln

[azure crane]

why don't we just define an operation that does that

[jovial rock]

Yeah, maybe

[azure crane]

Although it would be undefined for 1 and 0 lol

bc 1^anything=1

same with 0 (mostly)

[jovial rock]

Space for new number systems lol

For example taking the log of 0 multiple times and still continue calculating with it Like with the Square Root of -1

[azure crane]

actually this can't work

that function

5^3=5^(3^((whatever)^0))

also 9^1 and 3^2 would give different outputs

9 and 6

Hm

why does ln work then?

Oh I see

ln(3*6)=ln(3)+ln(6)

they're in the function still

ln

hm

So.. L(9^1)=L(9)*L(1)?

(I'm calling it L idk)

[jovial rock]

This guy did that thing with inventing new number systems that work when you take the ln too many times

[jovial rock]

Uh. So that would mean * is not commutative?

[azure crane]

no that just means L(1^9)=L(9^1) ig

hm not useful

ok

maybe

we do something else hmm

L(9^1)=9*L(1)?

L(3^2)=3*L(2)

Wait

L(1^x)=1L(x)
L(1)=L(x)
dangit

ok..

maybe

L(9^1)=L(9)*1

yes

this works I think

L(1^x)=L(1)x so we just say L(1)=0

kinda like ln

[azure crane]

ln(ab)=ln(a)+ln(b), so ln(1)+ln(b)=ln(b), so ln(1) must be 0 that kinda thing

L(x^1)=L(x)*1 yeah that's true

hmm

do you think this is good?

[jovial rock]

Hmm

[azure crane]

Wait

that's just ln..

[jovial rock]

But wouldn‘t that be just ln?

Ye

[azure crane]

yeah lmao

agh

[jovial rock]

L(9^2) = 9-2 or 9/2 ?

Maybe

[azure crane]

you got a point I think it needs to be some non commutative operation bc ^ is non commutative

[jovial rock]

Ye

[jovial rock]

hmm

a constant which is based on a and b

[azure crane]

ooh nice symbols lol

ok so first of all plug in some important values

a=1

T(1^x)=T(1)+AT(x)

T(1)=T(1)+AT(x)

either T(1) is infinite, or, AT(x)=0

[jovial rock]

hmm

[azure crane]

by the way, couldn't you just say.. A(a,b)?

[jovial rock]

yeah probably

[azure crane]

$T(a^b)=T(a)+A(a^b)T(b)$

[dawn finch]

RoyalBanana

[azure crane]

how's that

?

hmm plugging in a=1

$T(1)=T(1)+A(1)T(b)$

[dawn finch]

RoyalBanana

[azure crane]

so either, T(1) is infinite, or A(1)=0

interesting

plugging in b=1,

$T(a)=T(a)+A(a)T(1)$

[dawn finch]

RoyalBanana

[azure crane]

possibly unless a=1, because T(1) may be infinite, $0=A(a)T(1)$

[dawn finch]

RoyalBanana

[azure crane]

hm

so basically T(1) has to be 0

[azure crane]

since T(1) isn't infinite, A(1)=0

[jovial rock]

hmm, this is hard

[azure crane]

gotta love process of elimination lol

$T(a^b)=T(a)+A(a^b)T(b)$

$T(1)=0, A(1)=0$

[dawn finch]

RoyalBanana

[azure crane]

ok lemme try b=0 now

$0=T(a)$

oh

[dawn finch]

RoyalBanana

[azure crane]

uhh

so this doesn't work..

:/

unless

wait

unless

T(0) is undefined or infinite

bc then A(1)*T(0) isn't nessicarily 0

ok..

$T(a^b)=T(a)+A(a^b)T(b)$

$T(1)=0, A(1)=0, T(0)$ is undefined/infinite

[dawn finch]

RoyalBanana

[azure crane]

that's fine, after all ln(0) is -infinity

hmm ok lemme try.. a=0

oh wait nvm

that just outputs an undefined thing

since T(0) is

hmm

[jovial rock]

right?

wait

[azure crane]

I don't like writing the A like that bc you can't see it's a function

[jovial rock]

yeah true, you don't need to. I'm just trying something

[azure crane]

that's wrong though

see you got misled by writing it as a constant, bc there the inputs for A are different

its not A^2

[jovial rock]

yeah right

[azure crane]

hm

lemme try b=log_a(x)

[jovial rock]

Probably something like A(A(5,4),2)

[azure crane]

$T(x)=T(a)+A(x)T(log_a(x))$

[dawn finch]

RoyalBanana

[azure crane]

hmm

so this works for any a

and x

well, log is undefined for a=0 and 1 so ignore that and x<=0..

ok so anyway

I think we can work with this..

hmm

let me set a to x-1

$T(x)=T(x-1)+A(x)T(log_{x-1}(x))$

[dawn finch]

RoyalBanana

[azure crane]

this is recursive

hm

I'm gonna start graphing this

[jovial rock]

idk if this is useful

[azure crane]

h

[jovial rock]

when T(x) = ln(x)

[azure crane]

btw ln(a)/ln(b)=log_b(a)

[jovial rock]

Yeah

[azure crane]

hey btw I think we can define a starting value for T, like used in recursive formulas
T(2)=..?

[jovial rock]

yeah right

am not sure which one

[azure crane]

we can just choose it later ig

[jovial rock]

ln(6^5^4^3^2) = ln(2) * (ln(6^5^4^3)/ln(2)) + ln(3) * (ln(6^5^4^2)/ln(3)) + ln(4) * (ln(6^5^3)/ln(4)) + ln(5) * (ln(6^4)/ln(5)) + ln(6)

[azure crane]

?

[jovial rock]

ln(7^6^5^4^3^2) = ln(2) * (ln(7^6^5^4^3)/ln(2)) + ln(3) * (ln(7^6^5^4^2)/ln(3)) + ln(4) * (ln(7^6^5^3)/ln(4)) + ln(5) * (ln(7^6^4)/ln(5)) + ln(6) * (ln(7^5)/ln(6)) + ln(7)

[azure crane]

um

[azure crane]

wdym

[jovial rock]

I tried to put a exponential factorial into a sum

[jovial rock]

with this

Wait leme find a general formula

[azure crane]

dude

[jovial rock]

ye?

[azure crane]

howd you get this..?

[jovial rock]

I used the same technique like for this one: ln(2) * 12 + ln(3) * (ln(16)/ln(3)) + ln(4) = ln(4^3^2)

[azure crane]

show the steps

[azure crane]

for this

[jovial rock]

okay give me a sec

[azure crane]

also you can just cancel out the fractions lol that's just overcomplicating it

[azure crane]

ok so like, if this is true, how's it useful?

[jovial rock]

ln(6^5^4^3^2) = ln(2) * (((2-1) * ln(6^5^4^3))/ln(2)) + ln(6^5^4^3) = ln(2) * (ln(6^5^4^3))/ln(2)) + ln(6^5^4^3) = ln(2) * ((ln(6^5^4^3))/ln(2)) + ln(3) * ((3-1) * ln(6^5^4)/ln(3)) = ln(2) * ((ln(6^5^4^3))/ln(2)) + ln(3) * (ln(6^5^4^2)/ln(3)) + ln(6^5^4)

I hope ln(6^5^4) is clear how it continues to ln(4) * (ln(6^5^3)/ln(4)) + ln(5) * (ln(6^4)/ln(5)) + ln(6)

[azure crane]

^

[jovial rock]

You could use a sum notation for this

and replace all exponation factorials with sums too

[azure crane]

ok.. but how's it useful?

[jovial rock]

use them to make this discrete capital sigma thing, shuffle around some stuff so the capital sigma has big N over it and then aproximate

I'm using the techique from the factorial vid all the time, and it helps me

[azure crane]

ok..

[jovial rock]

hard to explain

[azure crane]

no i get it

that just seems really impractical

[jovial rock]

ah

wait

nvm

I think I have a stupid idea rn

[azure crane]

nice

[jovial rock]

L(x) = ln(2^^x)

[azure crane]

Just cancel out the ln(2)s-

[jovial rock]

oh, yeah

uh

[azure crane]

btw you know ln(2^^x)=2^^(x-1)ln(2) right

[jovial rock]

I guess yea

[azure crane]

..sorry but, this is in a form a+b*c^x

not ln(2^^x)

[jovial rock]

I was just trying to find a general formula for this: ln(2^^4) = ln(2) * (ln(2^^3)/ln(2)) +ln(2) * (ln(2^^2)/ln(2)) + ln(2) * (ln(2)/ln(2))
so: L(x) = ln(2^^x). L(x) = Sum(x,k=1)(ln(2) * (ln(2^^(x-k)/ln(2))

[jovial rock]

uhh

[azure crane]

What's the point of the ln(2)s just cancel them out

[jovial rock]

yeah, I only saw that right now

hmm

could there be a better Form for T(a^b) ?

maybe T(a) + T(b) + A(a,b)

[azure crane]

Uh, this is incorrect..

[azure crane]

At that point just do A(a,b)

[jovial rock]

I'll try to find the formula for it using ln.

[azure crane]

..?

formula for what..?

[jovial rock]

for A(a,b)

it's just x= (b−1)⋅ln(a)−ln(b)

ln a + ln b + (b−1)⋅ln(a)−ln(b) = ln a^b

[azure crane]

that's not a function you can use though

What's T(64)?

[jovial rock]

uhh, I don't understand

wait

[azure crane]

What is T(64)?

[jovial rock]

with the log thing?

[jovial rock]

this one?

[azure crane]

no

[azure crane]

this

[jovial rock]

ah

I'll use ln for T(x). It would be ln(8) + ln(2) + ln(4)

I used 8^2

plug in 8 for a and 2 for b in (b−1)⋅ln(a)−ln(b) to get ln(4)

[jovial rock]

ln(2^^x) = x* ln(2) + Sum(x,k=1)(2^^(k-2))

idk

It would be nice to have a function that completly eliminates the tetration

[azure crane]

64=4^3 too

and 2^6

[azure crane]

you mean lns(x)?

[jovial rock]

yeah but we can't compute it

[azure crane]

lns(e^e)=2 wdym?

[jovial rock]

lns(2^3) ?

we can only compute it for some numbers but not all

[azure crane]

hey btw

I just remembered something

remember that one graph of like S_k(x)?

how it was linear as x->infinity or something

we can use the same method used in the extending factorial video

ye?

i forgot to try that lol

so if S_k(x)~linear for big x

then S_k(N+x)~S_k(N)+m(k)x (m(k) is the slope)

so.. say S_k(N+1/2)~S_k(N)+m(k)/2

so then just do the backtracking thing from there

I think wasn't $S_k(x)$ like $\sum_{n=0}^{x}ln_n(k)$

[dawn finch]

RoyalBanana

[azure crane]

yeah

I think?

here it was

ohhh

it doesn't approach a line for big x it's for big k

ok so

S_k(x)~linear for big k

then S_{N+k}(x)~S_{N}(x)+m(x)k (m(x) is the slope it approaches, which depends on x)

so.. say S_{N+1/2}(x)~S_N(x)+m(x)/2

so then just do the backtracking thing from there

ah..

[azure crane]

I just realized that would mean extending it for all k instead, but that's not useful at all we already know what it is for all k..

agh

wait

that

that is S_2(x)

ohhhhhhhhhhhhhhhhhhhhhhhhh i made a typo i switched x and k

wait

so

this might work

bc then we are extending x

the thing we don't know

okkkkk

I wonder what slope S_k(2) approaches

it seems to be about 0.318

oh wow this actually converges very fast to a constant

0.318131505205...

i wonder what this number is..

My first thought was it was 1-ln2 but nope..

[jovial rock]

lnk(2^^n) = En * ln(2) + ln(ln(2)/2)

I think

[jovial rock]

maybe

[jovial rock]

ye sadly

[jovial rock]

I have no idea why I did that, but I think it should work kinda

I mean there is a iterated log in there but what ever

[azure crane]

Nono that was because of a typo, the one that would be extended is the one we don't know

[jovial rock]

oh really?

[azure crane]

Yeah!

[jovial rock]

nicee

[azure crane]

So.. if we figure out how to extend $S_k(2)$ then we can just use $S_{k+1}(2)-S_k(2)=ln_k(2)$!

[dawn finch]

RoyalBanana

[azure crane]

lemme resend the formula lol

[jovial rock]

ah I remember

It does converge extremly fast but I think at some value we're adding undefined numbers

[azure crane]

nope!

lemme show the graph

Real(s_k(2))

[jovial rock]

Huh

I thought you get undefined at ln3(2)

[azure crane]

I figured out, nope

that only happens if x is a whole superpower of e

like e^^3

or e^^2

Oh wait did you know how you can take ln of negative numbers?

[azure crane]

see it's only undefined eventually if a whole number of ln's makes it 0, bc then another ln would make it ln(0) which is undefined

like, lemme take 2 as an example..

ln(2)=0.69..

ln(ln(2))=-0.33..

ln(ln(ln(2)))=-1.003..+pi*i

you see

[azure crane]

It just skips right past 0

[azure crane]

Oh I figured out the slope

so that's only the real part of it I was using real(s_k(2))

the whole part is 0.3181315...+1.3372357...i

but you see

ln(that number)=that number

in fact

it's the same number for s_k(3)

And s_k(4)

and actually I think every input

they all approach this number

slope

(except for numbers that are exact superpowers of e, aka e^^-1, e^^0, e^^1, e^^2..., then the slope is just undefined)

this is cooooool

yep

woah we might actually be able to get an extension for lnk

Hell yeah

hold on so..

[azure crane]

Ok I looked up solutions to x=lnx

and a number that came up is W(-1)

W(-1)=-0.3181315...+1.3372357...i

so it seems the constant that they all approach is -Re(W(-1))+Im(W(-1))

weird

I'm too lazy to write that I'm just gonna call this constant $\omega$

[dawn finch]

RoyalBanana

[azure crane]

I'll just replace that with the number later

alrighty

so now that we got the slope it approaches we can do the stepping back thing

hmmm

so..

$S_k(x)=\sum_{n=1}^{k}\ln_n(x)$

[dawn finch]

RoyalBanana

[azure crane]

we need that super recursive formula like in the extending harmonic numbers video

[azure crane]

$S_{k+1}(x)=S_{k}(x)+ln_{k+1}(x)$

[dawn finch]

RoyalBanana

[azure crane]

$S_{k+2}(x)=S_{k}(x)+ln_{k+1}(x)+ln_{k+2}(x)$

[dawn finch]

RoyalBanana

[azure crane]

hmm I can see the pattern..

$S_{k+m}(x)=S_{k}(x)+\sum_{n=1}^{m}ln_{k+n}(x)$

[dawn finch]

RoyalBanana

[azure crane]

so it approaching a line means

For large N, $S_{N+k}(x)$ gets closer to $S_N(x)+k*$slope

yeah

[dawn finch]

RoyalBanana

[azure crane]

so as we just said before the slope it approaches is that constant that I'm calling w for now

hmm

so $\lim_{N \to \infty}S_{N+k}(x)-S_{N}(x)=\omega k$

[dawn finch]

RoyalBanana

[azure crane]

It's workingg

so $\lim_{N \to \infty}S_{k+N}(x)-S_{N}(x)=\omega k$

[dawn finch]

RoyalBanana

[azure crane]

ok so we have this super recursive formula

So.. $S_{k+N}(x)=S_{k}(x)+\sum_{n=1}^{N}ln_{k+n}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Now let's replace S_{k+N}(x) with that in the limit thingy

$\lim_{N \to \infty}S_{k}(x)+\sum_{n=1}^{N}ln_{k+n}(x)-S_{N}(x)=\omega k$

[dawn finch]

RoyalBanana

[azure crane]

replace S_N(x) with the definition of S..

[azure crane]

$S_k(x)=\sum_{n=1}^{k}\ln_n(x)$ I mean

[dawn finch]

RoyalBanana

[azure crane]

$\lim_{N \to \infty}S_{k}(x)+\sum_{n=1}^{N}ln_{k+n}(x)-\sum_{n=1}^{N}\ln_n(x)=\omega k$

[dawn finch]

RoyalBanana

[azure crane]

$S_k(x)=\omega k+\lim_{N \to \infty}\sum_{n=1}^{N}(ln_n(x)-ln_{k+n}(x))$

[dawn finch]

RoyalBanana

[azure crane]

oh

wait

oh

man

you can't use this at all for non integer k

bc of the ln_{k+n}(x)

you need to know what that is for non integer k to use this..

[azure crane]

:/

[jovial rock]

Yeah right

We Need a function that helps us solve this to input non integers

But Pretty cool

[azure crane]

well, at least we know that $ln_k(x)$ approaches $\omega$ for very large k, unless x is a superpower of e like ${}^3e$

[dawn finch]

RoyalBanana

[azure crane]

that makes sense, ln(w)=w

[jovial rock]

I‘ll try to find a number System which works for all negative ln and can be translated into complex numbers

[jovial rock]

Nvm

It‘s too Hard

[azure crane]

Wdym?

[jovial rock]

Like e^C = 0, when C is some constant

Hm, but that’s just ln 0

[azure crane]

ye

$e^{-\infty}=0$

[dawn finch]

RoyalBanana

[jovial rock]

What if we aproach ln 0 from the negative side?

Like Limit x -> infinity ln(-1/x)

I think it would be pi * i - infinity

[azure crane]

ye

anything finite times 0 is 0

$e^{ai-\infty}=0$

[dawn finch]

RoyalBanana

[jovial rock]

Ye

• infinity * - infinity = - infinity?

ln(0) * ln(0) = ln(0+0) = ln(0)

[azure crane]

uhh ye

btw

[jovial rock]

Hmm

[azure crane]

$e^{2\pi i}=1$ btw

[dawn finch]

RoyalBanana

[azure crane]

idk if you knew this btw

but that means $\ln(1)=2\pi i$ also

[jovial rock]

Yeah it‘s just e^pi I Squared

[dawn finch]

RoyalBanana

[jovial rock]

I guess ye

[azure crane]

you know about that thing?

All the different solutions

[jovial rock]

2 * ln(-1) = ln((-1)^2) = ln(1) ?

[azure crane]

you can add a $+2k\pi i$ on there where k is any integer

[dawn finch]

RoyalBanana

[jovial rock]

For what?

[azure crane]

for ln

[jovial rock]

Oh okay

[azure crane]

$e^{2k\pi i}=1$ bc of this

[dawn finch]

RoyalBanana

[jovial rock]

Oh, that’s cool

[azure crane]

yep

[jovial rock]

2 * ln(-2) = ln(4). So ln(-2) = ln(4)/2 ?

[azure crane]

ye

uh

[jovial rock]

Idk

Doesn‘t Look True

[azure crane]

$\ln(-x)=\ln(x)+\ln(-1)=\ln(x)+\pi i$

[azure crane]

It is

[jovial rock]

The pi*i is missing

[dawn finch]

RoyalBanana

[azure crane]

Nope

You're just looking at the wrong ln(4)

[jovial rock]

Ah

Now I See

[azure crane]

remember ln(4) is also the normal ln(4)+2pi i

[jovial rock]

Yeah Right I guess

[azure crane]

you gotta be careful which you use lol

[jovial rock]

Yeah I see

It‘s Bit complicated

Hmm

Maybe ln ln x = ln x + C

Or ln ln x = ln x * C

And then use that rule for iterated log

Idk

[azure crane]

Hmm

I wonder if we can see a pattern in the Taylor series of ln x, ln_2(x) ln_3(x) etc

[jovial rock]

I would like to have a function like this: we have a function L(x), so that L(2^^x) = x * L(2)

This would be very practical

[azure crane]

Just replace x with slog_2(x)

$L(x)=\log^s_2(x)L(2)$

[dawn finch]

RoyalBanana

[azure crane]

I assume you meant in general tho lol

I suggest $L(a^x)=f(a,L(x))$

[dawn finch]

RoyalBanana

[azure crane]

f is some function

but it has to have some properties

to work for any a and x

I think

[jovial rock]

I think I like L(x^^x) = L(x)^x more

[jovial rock]

Yeah i guess

Pretty cool that logs apears

[azure crane]

would x^^x be useful though?

(that's x^^^2 btw)

[jovial rock]

I saw some cool propeties with that one: L(x^x) = L(x) * L(x)

[azure crane]

ohh I see the point

[jovial rock]

I don‘t know. Maybe 2^^x

[azure crane]

It's like how ln(a^b)=ln(a)*b?

[jovial rock]

Yeah kinda, just that if a ≠ b, L(a) * L(b) ≠ L(a^b)

There would be an other structure for that probably

[azure crane]

L(1)=1 btw

[jovial rock]

How?

[azure crane]

bc L(1)=L(0^^0)=L(0)^0=1

[jovial rock]

Ohh

Nice

[azure crane]

well, unless L(0) is undefined or something

[jovial rock]

Ye

But I think that should work

Wait, isn‘t 0^^0 = 0 ?

I saw that on wiki

[azure crane]

Oh wait

hold on

what was the rule

$log_{a}({}^ba)={}^{(b-1)}a$

[dawn finch]

RoyalBanana

[azure crane]

you can't take log 0 but we can just take the limit basically just ignore that

anyway

$log_{a}({}^1a)={}^{(0)}a$

[dawn finch]

RoyalBanana

[azure crane]

$log_{a}(a)={}^{(0)}a$

[dawn finch]

RoyalBanana

[azure crane]

$\lim_{a \to 0} log_{a}(a)={}^{0}0$

[dawn finch]

RoyalBanana

[azure crane]

ahh log_x(x)=1 soo

$\lim_{a \to 0}1={}^{0}0$

[dawn finch]

RoyalBanana

[azure crane]

$1={}^{0}0$

[dawn finch]

RoyalBanana

[azure crane]

Huh? It says in the bottom right it equals 1

[jovial rock]

Btw L(x) = L(a)^log(a)s(x)

[azure crane]

what's s(x)?

[jovial rock]

(a) is the base

[azure crane]

oh do you mean $log^s_a(x)$?

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

[jovial rock]

Yeah maybe

Not sure

[azure crane]

In the same way 0^0=1

[jovial rock]

Yeah I guess. It‘s confusing. Sometimes 0^0 = 1 sometimes 0

[azure crane]

tbh I've never really seen it be 0

[jovial rock]

Btw logas(x) = ln L(x)/ln L(a)

Pretty cool

I think L(x) has pretty much good propeties for now

Should we give it a name?

[jovial rock]

Same, but I heard it is true sometimes

[azure crane]

Wait huh?

[jovial rock]

L(x) = L(a)^log(a)s(x),ln L(x) = ln L(a) *log(a)s(x), log(a)s(x) = ln L(x) / ln L(a)

I just solved for log(a)s(x)

This reminds me of ln x/ ln a = loga(x)

I think L(x) has a close relationship to the logarithmic functions

So ln L(2^^x) = x * ln L(2)

[azure crane]

How did you get $L(x)=L(a)^{\log^s_a(x)}$

[dawn finch]

RoyalBanana

[azure crane]

Oh I see

L(x)=L(a^^logs(a)(x))=that

wait

no

that doesn't work

[jovial rock]

I think I did something like L(a^^x) = L(a) ^x and replaced the x with the superlogs so a^^x canlcels to x

[azure crane]

You said L(x^^x)=L(x)^x

only for x

[jovial rock]

Ye

[azure crane]

^

[jovial rock]

I thought this is only for multiplication

Wait

[azure crane]

..?

it's your function?

[jovial rock]

Ah no it works for L(x^^a) But Not for L(x^a)

Because it will get mutliplied a times

And that’s Not the Same as just multipling a and x together

[jovial rock]

So this is only for the case when there is a x to the power of x

[azure crane]

..?

[jovial rock]

If you rewite it to L(x^^2) it works

[azure crane]

Um, what are you talking about

What is the definition of your function that everything else is based off of

just write that

The recursive formula or something

[jovial rock]

L(x^x) = L(x) * L(x) = L(x)^2
L(x^^x) = L(x)^x
So L(a^^x) = L(a)^x
L(a^^logas(x)) = L(a)^logas(x) = L(x)
logas(x) = ln L(x) / ln L(a)

It‘s the Same logic like ln, just one operation higher

[azure crane]

Which one of the top two is the starting one

[jovial rock]

L(a^^x) = L(a)^x

[azure crane]

ok

do you know there's no contradictions with that

[jovial rock]

I can‘t find one

So I guess there shouldn‘t be one

I wonder what it‘s inverse would be

A new Type of exponential

[azure crane]

Hm

Lemme plug in x=0

L(1) = 1

Lemme plug in a=0

unless x<=0, L(0) = L(0)^x

Hmm

so L(0)=0

Or 1

[jovial rock]

Hmm

ln L(4) / ln L(2) = 2

ln L(16) / ln L(2) = 3

L(1) should be 0 i think

Or

No

Wait

ln L(2) / ln L(1) = infinity i think

[azure crane]

^

[azure crane]

Undefined not infinity

1/0 is not infinity

[jovial rock]

Because log1s(2)

[jovial rock]

Yeah Right

[azure crane]

Lemme continue checking for contradictions

[jovial rock]

Okay

[azure crane]

Lemme plug in x=1

L(a)=L(a)

alr

Lemme try a=1

L(1)=L(1)^x

that makes sense, L(1)=1

[jovial rock]

1/L(a) = 0 ?

If you plug in -1 for x

And a^^-1 = 0

Thats weird

[azure crane]

L(a^^x) = L(a)^x

uhh

[jovial rock]

L(a)^-1 = 1/L(a)

[azure crane]

wait so x^^0=1

Yeah x^^-1=0

[azure crane]

Plugging -1 in for x
L(0)=1/L(a)

yep

That's what I'm talking about

[jovial rock]

Okay

[azure crane]

There's the contradiction

you gotta check

:/

[jovial rock]

It would have been Pretty cool

[azure crane]

you could say this is for x>=0

[jovial rock]

Yeah

Or just a new number system lol

[azure crane]

?

idk why you keep saying that lol

[jovial rock]

ln(-1) = undefinded, but we used complex numbers to get pi * i

[azure crane]

ye

[azure crane]

so what's that have to do with this

[jovial rock]

L(a) could be some number bigger than infinity, so 1/L(a) = 0

Idk, I think I‘ll just keep L(x) as a trick to solve or simplify. Maybe L(a) is some complex number, who knows

For example logL(a)(L(x)) is just logas(x)

Because of ln x / ln a, and replace x with L(x) and a with L(a)

It‘s a transformation from normal log to super log

L(x) numbers could still be treated as normal numbers but their numerical approximation would be a bit hard

[azure crane]

Umm.. what?

a is a variable

??

[jovial rock]

Yeah but all numbers the Numbers L(1), L(2), L(3), L(4),…

[azure crane]

..?

[jovial rock]

Wait

Uhh

I need to rethink this

Something what we could do is Delfine this function for a single number for x

Or a

For example e

I think I get it

I think L(x), has a base like the normal log

I‘ll come back to this tommorow and rework the definitions with bases

[azure crane]

..?

oh

Ok

Ig

[jovial rock]

Or L(0) could just be undefined

[azure crane]

Then the rest of L would be too

[jovial rock]

How?

[azure crane]

do the log bases thing

[azure crane]

we just said L(0)=1/L(a)

If L(0) is undefined then 1/L(a) is too so L(a) is

Also

[azure crane]

If you plug in a=0 here you get that L(0)=1/L(0)

so L(0)=1 or -1

so actually L(0) is not undefined

[jovial rock]

Yeah, I‘ll see if I can do something with the logs or add something so it doesn‘t make that contradiction

[azure crane]

you just have to make a different function

this one doesn't work

so by bases..

I assume you meant something like $L_n(x)$?

[dawn finch]

RoyalBanana

[azure crane]

We could say $L_a({}^xa) = L_a(a)^x$
or $L_x({}^xa) = L_x(a)^x$

[dawn finch]

RoyalBanana

[azure crane]

Hmm

[jovial rock]

L_a(a^^x) = L_a(a)^x would make sense here

For 1/L_a(a)

Hmm

[jovial rock]

I think it wouldn‘t

I was thinking of L(a^b) = L(a)^loga(a) * L(a)^loga(b)

And if this would work and a = b, then it would be L(a^^2) = L(a)^2 which would lead to contradiction

[azure crane]

how would that lead to a contradiction

[azure crane]

loga(a) is just 1 btw

[jovial rock]

Because L(x)^-1

Or I’ll just say x always has to be e

[azure crane]

But how do you plug that in to L(a^^2)=L(a)^2

that contradiction only happens if you allow L(a^^b)=L(a)^b in general

not just 2

so you're good

no contradiction

[azure crane]

What's the point of L again?

[jovial rock]

Oh okay

[azure crane]

(it only works for that bc you can plug in b=-1 there)

[jovial rock]

I guess for simplifing and solving tetration and eponentiation problems and getting the definition of logas(x)

So I guess L(x) changes for the value you input for a

[azure crane]

but it has one input so it can't

if you want it to do that you should write it L(a,x) or something

[jovial rock]

For L(x) = L(a)^logas(x)

[jovial rock]

Yeah right

I think it’s similar to how ssrt(2) = 2^^a, ssrt(3) = 3^^b, a≠b

Tetration is just very weird

So I guess 1/L(a,a) = L(0,a)

[azure crane]

Wdym

[azure crane]

But if you're using logas in the definition of L you're probably not gonna get anything useful

also do you mean extension

[jovial rock]

The definition of L(x), is L(a^^x) = L(a)^x

Btw L(x) = L(ssrt(x))^2

[azure crane]

sure

[jovial rock]

So L(x) = L(e^W(ln(x)))^2

[azure crane]

Well no that doesn't work remember

contradiction

[jovial rock]

Yeah

[azure crane]

so you can't use that

[jovial rock]

But if we use 2 inputs it works

[azure crane]

yeah

so don't write it like it's one

[jovial rock]

Okay

[azure crane]

We could say L(a,a^^x) = L(a,x)^x

or L(a,a^^x) = L(a,a)^x

or L(a,a^^x) = L(x,a)^x

it's just up to choice really

how you define L

[jovial rock]

Btw if ln L(a,a) = F(a,a), then F(a,a^^x) = x * F(a,a)

[jovial rock]

Yeah I guess.

[jovial rock]

This would be the correct one for my definition

[azure crane]

interesting

working on F would probably be easier

[jovial rock]

I guess

[azure crane]

and we can just take e^ to convert back

[jovial rock]

F(a,x)/F(a,a) = logas(x) I think

[jovial rock]

For this Definition the first parameter can be anything

F(x,b)/F(x,a) = logas(b) I think

[azure crane]

?

[jovial rock]

.

[azure crane]

that was with the old contradictory formula remember

[jovial rock]

So it has to be this?

[azure crane]

Try refiguring it out with the new formula ig

[jovial rock]

L(a,x) = L(a,a)^log(a)s(x),ln L(a,x) = ln L(a,a) *log(a)s(x), log(a)s(x) = ln L(a,x) / ln L(a,a)

[azure crane]

btw

$L(a,a^x)=L(a,x)L(a,a)$

[dawn finch]

RoyalBanana

[azure crane]

hold on lemme remember the proof for this lol

$L(a,{}^xa) = L(a,a)^x$

[dawn finch]

RoyalBanana

[azure crane]

So $L(a,{}^{x+1}a) = L(a,a)^{x+1}$

[dawn finch]

RoyalBanana

[azure crane]

remember a^(a^^x)=a^(x+1)

Imma also rewrite the part on the right of the equation

[azure crane]

So $L(a,a^{({}^xa)}) = L(a,a)^x*L(a,a)$

[dawn finch]

RoyalBanana

[azure crane]

now new equation

$\frac{L(a,a^{({}^xa)})}{L(a,{}^xa)}=\frac{L(a,a^{({}^xa)})}{L(a,{}^xa)}$

[dawn finch]

RoyalBanana

[azure crane]

ofc

now imma replace the ones on the right with the equations we just got

$\frac{L(a,a^{({}^xa)})}{L(a,{}^xa)}=\frac{L(a,a)^x*L(a,a)}{L(a,a)^x}$

[dawn finch]

RoyalBanana

[azure crane]

ofc we can just cancel out those

$\frac{L(a,a^{({}^xa)})}{L(a,{}^xa)}=L(a,a)$

[dawn finch]

RoyalBanana

[azure crane]

now

Imma replace x with logs(a)(x)

$\frac{L(a,a^x)}{L(a,x)}=L(a,a)$

[dawn finch]

RoyalBanana

[azure crane]

multiplying to cancel out the fraction

We get $L(a,a^x)=L(a,a)L(a,x)$!

[dawn finch]

RoyalBanana

[azure crane]

btw if you take the ln of both sides this means that

$F(a,a^x)=F(a,a)+F(a,x)$

[dawn finch]

RoyalBanana

[azure crane]

Bc of it being like a base thing I think we should rewrite $L(a,x)$ as $L_a(x)$ btw

[dawn finch]

RoyalBanana

[azure crane]

$L_a(a^x)=L_a(a)L_a(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Whoa

This is cool

[jovial rock]

logxs(a^x) = La(a)

Is that true?

If Yes, then La(a) = 2 I think

[azure crane]

huh?

[jovial rock]

No I think that’s Not true

[azure crane]

Btw $L_1(a)$ and $L_a(1)=1$

[dawn finch]

RoyalBanana

[azure crane]

for any a

[jovial rock]

Oh okay. I guess this is true

[jovial rock]

Huh, can x be any number?

[azure crane]

Sorry, where'd you get this from?

[jovial rock]

This

[azure crane]

Lemme convert that to latex so I can see it easier

$L_a({}^xa) = L_a(a)^x$

$L_a(x) = L_a(a)^{log^s_a(x)}$

$\ln(L_a(x)) = \ln(L_a(a))*log^s_a(x)$

$log^s_a(x) = \frac{\ln(L_a(x))}{\ln(L_a(a))}$

[dawn finch]

RoyalBanana

[azure crane]

yup

[jovial rock]

But it doesn‘t work with this

[azure crane]

?

[jovial rock]

If x is any number

[azure crane]

Where'd you get this

[jovial rock]

You Can write this as logas form, replace x with a^x and the base a with x

[azure crane]

Wdym logas form

[jovial rock]

Ohhhh

[azure crane]

So $\frac{L_a(a^x)}{L_a(x)}=L_a(a)$

[dawn finch]

RoyalBanana

[jovial rock]

I have severe stupidity

[azure crane]

same /j