General Formula for all arithmetic operations

[jovial rock]

I tried to find a general formula for the arithmetic operations +,,^,^^,^^^,^^^^, and I realized that 2#2 is for all of them 4.,^,^^ and the others are created by iterating operations, the smallest one we know is addition. So we can say that there is more and more iteration complexity being added the higher n is in the Hyperoperation function H(n,a,b). So when n is going below the addition operation there should be something like even less iteration complexity. I just don’t get how this iteration complexity into a general operation formula. With a general formula we could make operations between multiplication and addition and we could understand why we are losing this commutative concept when crossing the exponentiation operation. For values of n smaller than 1 or even negative we could find out how operations could get more abstact and maybe we could find mathematical objects more abstract than numbers. We also could input complex numbers for n to get some other weird operations. So, how to make a formula to generalize the arithmetic operations?

[azure crane]

Some recursive form for H probably then solve for H using that

Is addition H(1,a,b) or H(0,a,b)?

Ah, you mean this one specifically
https://en.m.wikipedia.org/wiki/Hyperoperation

So H(1,a,b) is addition

[jovial rock]

Yeah right

I couldn‘t figure out how to use non integervalues for n

And n=0 is kinda boring because it‘s not really a new operation

It‘s just b+1

a gets ignored

I feel like n=0 is just the assigment Operation but i‘m not sure

I think just recursion wouldn’t solve the problem, because we can’t do non integer. We don’t even really know how to use tetrarion for non integer values for b when it’s a^^b

[azure crane]

It's a start

H(n,a,b)=H(n-1,a,H(n-1,a,...H(n-1,a,a)))
b-1 Hs being applied on the right side
eg. 5^^6=H(4,5,6)=H(3,5,H(3,5,H(3,5,H(3,5,H(3,5,5)))))=5^5^5^5^5^5

recursive functions are a mess jeeze

[jovial rock]

Oh I See, I just don‘t really get the thing with b-1. pretty complicated

I found this website one time: http://myweb.astate.edu/wpaulsen/tetcalc/tetcalc.html
it expands tetration to non integer values and complex numbers. I don‘t really know how it works but it‘s cool

[jovial rock]

Not sure if tetration has this rule

[azure crane]

Oh I see they only have that rule bc multiplication and addition are commutative

exponentiation isn't ofc

[jovial rock]

Yes, exponetiation has some rules because multiplication is commutative, but it doesn‘t work with tetration because exponetiation isn‘t commutative

[azure crane]

$H(n,a,b)=H(n-1,H(n,a,b-1),1)$

[dawn finch]

RoyalBanana

[azure crane]

wait no hmm how do I notate this

[jovial rock]

Hmm, not sure. There should be a new Type of notation for recursion that allowes to notate this stuff easier

[azure crane]

$H(n,a,b)=H(n-1,a,H(n,a,b-1))$

[dawn finch]

RoyalBanana

[azure crane]

based on:
a×b=a+a(b-1)

a^b=a×a^(b-1)

a^^b=a^a^^(b-1)

etc

by definition kinda

[jovial rock]

So it‘s a(^n-2)b = a(^n-3)a(^n-2)(b-1)

Maybe

(^) = exponentiation

[azure crane]

from this you can get $H(1,a,b-a+1)=H(0,a,b)

[jovial rock]

(^-1) = multiplication

[azure crane]

so H(0,a,b)=a+b-a+1=b+1

[jovial rock]

I understand

[azure crane]

now

lets plug in n=0 hehe

H(-1,a,b)=b+1

ah

[jovial rock]

Huh

[azure crane]

$H(0,a,b)=H(-1,a,H(0,a,b-1))$

[dawn finch]

RoyalBanana

[azure crane]

$b+1=H(-1,a,H(0,a,b-1))$

[dawn finch]

RoyalBanana

[azure crane]

$b+1=H(-1,a,b-1+1)$

[dawn finch]

RoyalBanana

[azure crane]

$b+1=H(-1,a,b))$

[dawn finch]

RoyalBanana

[azure crane]

I suppose the same applies for H(-2,a,b)

[jovial rock]

That doesn‘t make sense. If we would repeat b+1 we would get addition?

[azure crane]

hm

weird..

the formula is right tho i think

[jovial rock]

It‘s a similar problem like root(x) = -1

[azure crane]

well you can't say it doesn't hold for n<0 really, there isn't a definition yet we're making one

[jovial rock]

I think it‘s only defined for positive integers

We a more complex logic to solve this i think

[azure crane]

possibly

[azure crane]

I wonder what the inverse of tetration is btw

[jovial rock]

Supersquare Root?

It‘s e^W(ln(x))

[azure crane]

$x^{1/2}=\sqrt{x}$

[dawn finch]

RoyalBanana

[azure crane]

maybe there's something like that here

[jovial rock]

Yeah, tetration gets complex very quickly, we would Need some operation that behaves like exponentiation but is commutative

I Wonder what operations with n equal to a complex number would Look like

Imagine the inverse of the imaginary operation with the imaginary base of the imaginary number

[azure crane]

lol I dont even know what that means

[jovial rock]

Like H(i,i,i) but the inverse operation of that

[azure crane]

lol

what would H(4,x,i) even mean lol

[jovial rock]

Ah right

Huh

Hyperoperation algebra…

[azure crane]

I mean this is basically equivalent to saying $H(0,a,b)=H(0,a,b-1)+1$ if you use that formula above, so it makes sense

[dawn finch]

RoyalBanana

[jovial rock]

Yeah probably

Imagine the will be questions like: what is x when H(x,5,5) = 15

[azure crane]

interesting

[jovial rock]

x is probably something between 2 and 3

No

1 and 2

[azure crane]

$H^{-1}(n,a,b)$

[dawn finch]

RoyalBanana

[jovial rock]

huh

[azure crane]

usual inverse notation

[jovial rock]

But aren‘t there 2 inverses?

Root and log

[azure crane]

uhh

I guess it depends if you mean by inverse

[jovial rock]

Only for non commutative operations

[azure crane]

wait there's three inverses

bc there's three inputs

[jovial rock]

How?

Oh, for this function yeah

Intresting

If it‘s commutative only 2 because H(1,2,x) = H(1,x,2) i think

[azure crane]

true

$H(3,x,2)=5,$ so $x^2=5,$ so $x=\sqrt{5}$

$H(3,2,x)=5,$ so $2^x=5,$ so $x=log_{2}(5)$

$H(x,2,3)=5,$ so ???

[dawn finch]

RoyalBanana

[jovial rock]

Hmm, we would Need a new notation

But I think the Solution would be 1

[azure crane]

we should probably finish extending H(x,a,b) to non integer numbers before we tackle the inverses lol

[jovial rock]

This feels like creating a function like x! and trying to find the non integer values

[jovial rock]

Yeah true

[azure crane]

https://www.youtube.com/watch?v=v_HeaeUUOnc&t=170s&pp=ygUZZXh0ZW5kaW5nIHRoZSBmYWN0b3JpYWxzIA%3D%3D
like this?

[jovial rock]

Yeah, the gama function

Gamma*

[azure crane]

there isn't really a "correct" definition, you just want to invent a "simple" solution/ intuitive one

[jovial rock]

Yes, but how?

[azure crane]

well, you gotta use what you already know ofc

what are some facts about this function

obvious or not

[jovial rock]

It works with recrusive structure and I have no idea how this is possible

[jovial rock]

Hmm

Maybe graph some points for diffrent values of a and b and try to find the non integer values

[jovial rock]

H(x,2,2) is always 4 I think, exept when it‘s 0

uh

I mean x can be any positive integer number and the function would output 4

[azure crane]

true

I think it's good to work on a simpler version of the problem first

and see if we can solve that

so how about just extending tetration

H(4,a,b)

what's the notation for tetration?

I wonder if this looks good ${}^{b}a$

[dawn finch]

RoyalBanana

[azure crane]

noice

that's the notation

[jovial rock]

Yeah, looks good

[azure crane]

alr

[jovial rock]

What do you mean with extending tetration?

[azure crane]

make ${}^{b}a$ make sense with non integer values of b

[dawn finch]

RoyalBanana

[azure crane]

or rather, make a formula of some sorts for it

like how we can do x^2.4

thats an extension

[jovial rock]

Yeah I know, I used to work on it a bit. But I don‘t know how to do it with out the commutativity of exponetiation

[azure crane]

well, maybe there's something useful to get out of properties of tetration

[jovial rock]

I was thinking about 2^^1.5 could be 2^(e^LambertW(ln(2))) But it’s not

[azure crane]

let's see.. ${}^{b}a=a^{{}^{b-1}a}$

[jovial rock]

That tetration calc says that 2^^1.5 is 2.7487616545898224907

[dawn finch]

RoyalBanana

[jovial rock]

Hm

[azure crane]

so ${}^{b-1}a=log_a({}^{b}a)$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah true

[azure crane]

well hey we already got an extension for negative numbers that's good!

[jovial rock]

Looks like it, but I think it‘s not very stable

[azure crane]

well, it works at least

[jovial rock]

Isn‘t log(0) infinity or something?

[azure crane]

yeah

:/

[jovial rock]

Kinda sad, but at least something

[azure crane]

so for $a \ne 0, {}^{0}a=1$

[dawn finch]

RoyalBanana

[azure crane]

makes sense

[jovial rock]

What would 0^^-1 be?

0?

[azure crane]

so for $a \ne 0, {}^{-1}a=0$

[dawn finch]

RoyalBanana

[azure crane]

and then beyond that undefined

[jovial rock]

Yep

[azure crane]

I guess it makes sense after all tetration is definitely not stable in the positives either it grows extremely fast

[jovial rock]

I guess exponentiation goes from 0 to infinity with real numbers and tetration enters the infinity zone at b=-2

And -3,-4 wouldn‘t make sense

Sadly we can‘t find pentation with the height of negative numbers because we don‘t have formula for superlog

[azure crane]

${}^{b}1=1$

[dawn finch]

RoyalBanana

[azure crane]

i think that's fair to say

[jovial rock]

True

[azure crane]

https://www.desmos.com/3d/suc3rb3rkn

[jovial rock]

Nice

I think (-1)^^n = -1 when n≠0

[azure crane]

If n is natural

Same with 0 too

[jovial rock]

And we know that n^^0.5 is e^W(ln(n)), so we can find the supersquareroot of -1

[jovial rock]

True

[azure crane]

?how do you know that

[jovial rock]

But 0^^2 would be 0^0?

[azure crane]

Wait

yeah

oops

[jovial rock]

Solve x^x = n for x

[azure crane]

Ah W

[jovial rock]

xe^x I think

[azure crane]

Where are you getting .5 from?

[jovial rock]

Because it’s supersquare root

And then it should be the half I think

[azure crane]

the reason this is true is bc $x^a*x^b=x^{a+b}$

[dawn finch]

RoyalBanana

[azure crane]

so $x^\frac{1}{2}*x^\frac{1}{2}=x$

[dawn finch]

RoyalBanana

[azure crane]

which is the definition of sqrt x

[jovial rock]

Oh

Okay

I understand

[azure crane]

I assume there's something similar for tetration tho..🤔

[jovial rock]

Yeah am wrong

It’s not .5

This sucks

[azure crane]

${}^{{}^{b}a}a=?$

[dawn finch]

RoyalBanana

[jovial rock]

Huh

I’m confused

[azure crane]

${}^{({}^{b}a)}a=?$

[dawn finch]

RoyalBanana

[azure crane]

wait

uhh..

hmm

what would the tetration root be a solution to

[jovial rock]

I don’t know

[azure crane]

Like how the square root is a solution to ?^2=x

[jovial rock]

I’m really confused right now haha

[azure crane]

Like, what would the super square root mean

the normal square root is the number that when squared is x

[jovial rock]

I thought it’s supersquareroot(2)^ supersquareroot(2)=2

[azure crane]

${}^2?=x$

[dawn finch]

RoyalBanana

[azure crane]

?

[azure crane]

Oh yeah that's another way of writing this

True

idk what symbol we should use for it

[jovial rock]

On wiki they are using something like that

[azure crane]

alr

[jovial rock]

The supersquareroot was my only hope reaching the non integer values but it turns out it doesn’t work…

[azure crane]

I'm pretty sure ?^?=x has a solution in terms of W lambart function

[jovial rock]

Hmm

But wouldn‘t that be ? = e^LambertW(ln(x))

Or what do you mean?

[jovial rock]

How would the -1th superroot of 2 look like?

Wouldn‘t that be x^^-1 = 2?

So thats undefined…

And the -1th superroot of 0 would equal everything

At least we can take squareroots. They are approximately ^^0.602

Would be cool if they are some value that is based on e

Maybe we need a new type of e. e is a constant for exponentiation, maybe we need a constant for tetration

[azure crane]

you mean like, $\lim_{N \to \infty} {}^N(1+\frac{1}{N})$?

[dawn finch]

RoyalBanana

[azure crane]

hmm

I'm pretty sure that just equals 1

[jovial rock]

Yeah probably, something else i guess

[unkempt tinsel]

addition:
a+b
multiplication:
exp(log(a) + log(b))
the next one:
exp(exp(log(log(a)) + log(log(b))))
with this you get a commutative operation for every integer

i've written "the next one" in a nonsimplified way so the pattern is clear, it's actually a^log(b) = b^log(a)

[azure crane]

I mean, idk how that's useful though

[jovial rock]

I think I have something, not sure if this is useful. I was trying out new things in my Head and messing around with the supersquareroot. I thought what would happen if I take the supersquareroot of e, and I got this number e^W(1). This is also a solution to a problem I had in the past where I wanted to take the derivative of y^x and get y^x ln(y)^x. I wanted to see what is the optimal value for y when x approaches infinity that the gradient of the function stays at 0 so it’s a straight line. That value of y is actually e^W(1), the supersquareroot of e. What if we try to take the deriva again and find the optimal value? Could we found higher superroots of e?

[unkempt tinsel]

me when people mention tetration

this gives a hyperoperation that accepts non integer values of b, at least

[jovial rock]

Looks intresting, I’ll check it out when I’m at home

Do you maybe have python code for that?

[unkempt tinsel]

just loop the exp function and the log function

[jovial rock]

👍

[jovial rock]

So I took the derivative again and got ln(y^y)^x * ln(ln(y^y))^x. I calculated the optimal value an I got e^W(e^W(1)), it‘s very similar to the e^W(1), the optimal value of the next derivative would probably be e^W(e^W(e^W(1)))

So it‘s a recursive function i guess

After I tetrated that value by 2 I got something like 5.8312001

I don‘t know if this helps but i‘ll just try running it through many functions

I researched a bit and found that e has a diffrent height for it‘s supersquareroot than 2, 2 had approximately a height of ^^0.602 and e has approximately a height of ^^0.572. So there must be a value bigger than e, that has ^^0.5 as its height for it’s super square root

The equation would kinda look like that I think: x^^0.5 =e^W(ln(x))

Probably impossible to solve

But we would get a number that could be stable

With 10 I could find 10^^0.5045=e^LambertW(ln(10)), so even bigger than 10. Hopefully not infinity

[jovial rock]

I finally aproximated this value

it's about 12.3823

if anyone wants to keep aproximating it, here is the file

[azure crane]

What if you take the half derivative

[jovial rock]

is it a new helpful constant? can we express it with lambert functions? i dont know

[jovial rock]

not sure, I didn't really take the derivate, I took the nth derivative

[jovial rock]

and replaced the n with x

[azure crane]

So if the a certain derivative gives you the a certain tetration then you can use fractional derivatives to get fractional tetrations?

[jovial rock]

yeah probably

I would probably need to take the n^0.5th derivative

[jovial rock]

Maybe I could do this in 0.5 steps through all the numbers, graph all the points and find a function that goes through all the points

[jovial rock]

but how do you take the n^0.5th derivative?

I think it's (ln y)^x^0.5 * y^x

[azure crane]

https://en.m.wikipedia.org/wiki/Caputo_fractional_derivative

[azure crane]

Wait so if I'm understanding right you're taking the nth derivative and inputting a number at that derivative and it gives you a nice tetration
Sounds like you could make a Maclaurin series type thingy function out of it 🤔

[jovial rock]

I got e^(W(2)/2)

[jovial rock]

yeah kinda, but I couldn't take the nth derivative of ln(y^y)^x * ln(ln(y^y))^x for somereason. I tried it with 2 but I couldn't generalize it more: ln(4)^x * ln(ln(4))^x

[jovial rock]

I don't really understand this type of stuff, I just used ln(y)^x^0.5 * y^x

and I solved it for y

with out the x, because it cancels out

[jovial rock]

I hoped for a second this is the supersquareroot of 2, but it saldy isn't. This is about 1.53.., and the supersquareroot of 2 is 1.55..

[jovial rock]

My goal with the recursive derivative was that I could take the x^x th derivative of the function and than get a new function with a nice optimum constant but I just can‘t get past ln(y)^x * ln(ln(x))^x

[jovial rock]

I found the third nth derivative: (ln(ln(y^y)*ln(ln(y^y))))^x * (ln(y^y)*ln(ln(y^y)))^x

It‘s something like this y = e^W(y * ln(y))

I can‘t solve it, I need help

Wait

That doesn‘t make sense, this would equal every number, because the equation says the supersquareroot of x^^2 is x, and it cancels out, so it‘s x=x. Huh

[unkempt tinsel]

supersquareroot(x) = y when y^y = x?

[jovial rock]

I just need to now what the solution to this equation is: (ln(ln(y^y)*ln(ln(y^y)))) * (ln(y^y)*ln(ln(y^y))) = 1

[jovial rock]

Yes

[unkempt tinsel]

so what are you doing

[jovial rock]

Trying to solve this for y

[unkempt tinsel]

why

[jovial rock]

To get the value where the gradient is 0

[unkempt tinsel]

y^y = x
e^yln(y) = x
x(1 + ln(y))y' = 1
y' = 0 => no solution

= 0?

i lost a y' somewhere

[jovial rock]

Wait

[unkempt tinsel]

actually, when 1 + ln(y) ~ infinity, i.e. y ~ 0, i.e. (1,0), the curve is not differentiable, but it's close enough to y' = 0

or when x ~ infinity

[jovial rock]

Hmm

It‘s somewhere around 5.9

It could be e^W(W(1)) tetrated by 2

e(W(e^W(1)))*

I think it actually is, but I‘m not sure

[unkempt tinsel]

what are you doing?

[unkempt tinsel]

do you disagree with this

[jovial rock]

I mean I tried to aproach a value and I got this

And I think it‘s right

Pretty cool, at first I had 1, then e^W(1), then e^W(e^W(1)) and now e^W(e^W(1))^^2

What do you think?

[jovial rock]

Did you try to solve this?

[unkempt tinsel]

this is just y = y

[jovial rock]

I know, I did a mistske there

It‘s the incorrect way of solving that

[jovial rock]

This is the correct way

I tried to approach it in geogebra and found that it’s about 5.8

[jovial rock]

I remembered that I had this number before here:

[unkempt tinsel]

i just copied and pasted your equation

i think you must have made an error

[jovial rock]

(ln(ln(y^y)*ln(ln(y^y)))) * (ln(y^y)*ln(ln(y^y))) = 1?

[unkempt tinsel]

this has a unique solution of about y ~ 2.4

which is not a critical point of the implicit function of y^y = x

[jovial rock]

Ah

I think the = 1 is the problem

Hm

Can you show me the equation you pasted in?

I’m really confused right know

[unkempt tinsel]

just look at a graph of y^y = x

you will see that the only things that kind of pretend to be critical points are (1,0) and (infinity, infinity)

[jovial rock]

Where do you see (1,0) and (infinity,infinity), sorry for my question but I’m just really confused

For 2.46 I got a complex number

[unkempt tinsel]

i think you have made at least 2 errors in the process of your calculations

[jovial rock]

Which ones?

Maybe you’re right

It needs to be 1 not 0

So I approximated it for 1

[jovial rock]

So I looked over the stuff again and I found the problem in my calculations

I used y, instead of y^y

So yeah, I did something wrong

[jovial rock]

This value is like I predicted e^W(e^W(e^W(1)))

[jovial rock]

.

The next value for the nth derivative would be e^W(e^W(e^W(e^W(1))))

I think this Value e^W(e^W(e^....W(e^W(1)))....)) is aproaching e

[azure crane]

Well, the cool thing about infinite function things like that, is that:
let's say s=e^W(e^W(e^W(.....)))))...
e^W(s)=s
you can just solve for the second equation which is easier

e^W(s)=s
W(s)=ln(s)
s=ln(s)e^ln(s)
s=ln(s)s
s=0, or ln(s)=1
s=0, or s=e
s=0 is clearly an incorrect solution to the original problem, so it must be e!

[jovial rock]

Yesss, that’s very nice

So at some point e would just be the last value that would stabilize the nth derivative.

[azure crane]

neat

[jovial rock]

Can we maybe find a function for the nth super root? We know that the second one is e^W(ln(x)

I don’t know if the a iterated W helps a lot

[azure crane]

so $({e^{W(ln(x))}})^{(e^{W(ln(x))})}=x?$

[dawn finch]

RoyalBanana

[azure crane]

Oooh

I see

That's cool

how it cancels so nicely

[azure crane]

do you still need to solve this btw

[jovial rock]

Yeah, that’s nice

[jovial rock]

I solved it thanks

It’s this e^W(… value

[azure crane]

I'm seeing if I can find a similar one for the third root

Agh
$W(a)e^a=x$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm, what would this one do?

[azure crane]

I did some substitutions so if you solve this you solve the third superroot

Or just $ae^a-a=x$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

For the 3rd root?

Superroot*

[jovial rock]

I found this on wiki

I don’t exactly understand how it works

[jovial rock]

I think this is the solution for n—> infinity

[azure crane]

it's really quite simple actually

[azure crane]

It's basically like making an infinite function thing out of the equation instead of the other way around

the x_n is just a shorthand so they don't have to write so many terms

[jovial rock]

Okay, understandable

It‘s just a bit complicated

Btw I didn‘t got further with tetration so I jumped to pentation. I think a^^^-1 = 0

But I don’t know what sloga(0) is for a^^^-2

Probably infinity too

Wait, I think it’s -1, because a^-1 = 0

And wiki said that 0^^0 is 1

That means the pentation squareroot of 1 would be 0 too

The pentation squareroot of 0 would be -1

The pentation is a bit more stable than tetration with negative values

[jovial rock]

I analyzed that derivative thing I was working on and I found a recrusive structure but I just can't figure out how to describe it with n because it's pretty complex:
1: 2^x
2: 2^x * ln(2)^x
3: 2^x * ln(2)^x * ln(2 * ln(2))^x
4: 2^x * ln(2)^x * ln(2 * ln(2))^x * ln(2 * ln(2) * ln(2 * ln(2)))^x
5: 2^x * ln(2)^x * ln(2 * ln(2))^x * ln(2 * ln(2) * ln(2 * ln(2)))^x * ln(2 * ln(2) * ln(2 * ln(2)) * ln(2 * ln(2) * ln(2 * ln(2))))^x
6: 2^x * ln(2)^x * ln(2 * ln(2))^x * ln(2 * ln(2) * ln(2 * ln(2)))^x * ln(2 * ln(2) * ln(2 * ln(2)) * ln(2 * ln(2) * ln(2 * ln(2))))^x * ln(2^x * ln(2) * ln(2 * ln(2)) * ln(2 * ln(2) * ln(2 * ln(2))) * ln(2 * ln(2) * ln(2 * ln(2)) * ln(2 * ln(2) * ln(2 * ln(2)))))^x

It just multiplies the natrual log of the whole previous function with the previous function to get the new function

how can this be generalized with n?

[azure crane]

What's sloga?

[jovial rock]

super log with the base a

[azure crane]

f(x,n)=f(x,n-1)ln(xthroot(f(x,n-1)))^x

you can group all the ^xs together

$a^x*b^x=(ab)^x$ ofc

[jovial rock]

you could replace 2 with y to get the equation to get the e^W(... sequence. n=1 is 1, n=2 is e^W(1), n=3 is e^W(e^W(1))... . This would allow us to get non integer things for e^W(... sequence

[dawn finch]

RoyalBanana

[jovial rock]

yeah, but this would only work for integers

[azure crane]

Don't forget $ln(ab)=ln(a)+ln(b)$ btw

[dawn finch]

RoyalBanana

[jovial rock]

I wonder what is this function for n=0 or n=-1

[jovial rock]

I know, but wouldn't the function be more complicated with + ?

[azure crane]

not really

One sec

hold on wait

this is just tetration

[jovial rock]

n=0 should be y/ln(y) = 1

[azure crane]

It's ln(ln(... Over and over that's just reverse tetration

[jovial rock]

maybe, I'm not sure

[azure crane]

2
ln(2)
ln(2)+ln(ln(2))
ln(2)+ln(ln(2))+ln(ln(ln(2)))
the terms written out

[jovial rock]

I'm confused

[azure crane]

$f(x,n)=2^x\prod_{m=1}^{n-1}(\sum_{k=1}^{m}ln_k(2))^x$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k(x)$ is ln repeated k times
For example $ln_3(x)=ln(ln(ln(x)))$

[dawn finch]

RoyalBanana

[azure crane]

#1330832852819906661 message

[azure crane]

Aka
$f(x,n)=(2\prod_{m=1}^{n-1}\sum_{k=1}^{m}ln_k(2))^x$

[dawn finch]

RoyalBanana

[jovial rock]

Oh, yeah you are right. With the base e

[azure crane]

remember $\log_a(x)=\frac{\ln(x)}{\ln(a)}$

[dawn finch]

RoyalBanana

[azure crane]

This just needs solved then it's writable in terms of tetration

${}^{x}e=2$

[dawn finch]

RoyalBanana

[jovial rock]

So lnx(2) = e ?

[jovial rock]

Or maybe lnx(e) = 2

[jovial rock]

It‘s this one. I approximated the solution with the tetration calculator, it‘s about 0.71

We would need a function for this

This function would be called the super log with the base e, or in other words the super natrual log or tetration natrual log. This would allow us to take non integer natural logs and I could use it to solve my problem with generalizing this ln(2)^x function

[azure crane]

${}^{0.71}e \approx 2$?

[dawn finch]

RoyalBanana

[azure crane]

${}^{b-1}a=log_a({}^{b}a)$

${}^{b-1}e=ln({}^{b}e)$

${}^{~0.71-1}e=ln({}^{~0.71}e)$

${}^{~0.71-1}e=ln(2)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{~0.71-1-1}e=ln(ln(2))$

[dawn finch]

RoyalBanana

[azure crane]

etc

so ${}^{~0.71-k}e=ln_k(2)$

[dawn finch]

RoyalBanana

[azure crane]

I think that makes sense

[jovial rock]

Yeah looks right

can we find it for lnk(e)?

[azure crane]

${}^{1-k}e=ln_k(e)$

the number in the top left is just the supernaturallog of the number being ln-ed

[dawn finch]

RoyalBanana

[jovial rock]

Ah Right

[azure crane]

Btw can you come up with some notation for the tetration natural log?

[jovial rock]

Hm

[azure crane]

Or just any log

you'd need to have space for the log base

[jovial rock]

We could maybe do the same thing like with the supersquareroot and put a "s" after the log(x). Like here:

[azure crane]

interesting

[jovial rock]

Or we can just add a s, like slog, or sln

[azure crane]

I think this works sure

[jovial rock]

Okay 👍

Sounds good

[azure crane]

${}^{ln(x)_s-k}e=ln_k(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Uh, wait. How does this work with the -k?

[azure crane]

maybe we put the s above it's kinda confusing for the log

[jovial rock]

yeah true

[azure crane]

${}^{(ln^s(x)-k)}e=ln_k(x)$

[dawn finch]

RoyalBanana

[azure crane]

negative tetration is like taking the log over and over, which makes sense bc log is like the inverse of exponents

[jovial rock]

Oh Yeah, that makes sense, that’s why it doesn‘t work with the root

So if k=0, it‘s equal to x?

[azure crane]

yep

like taking the natural log of x 0 times

so its just x

[jovial rock]

Ah nice, now I understand

[azure crane]

I suppose there's no reason this has to be only for ln and e

${}^{(log_a^s(x)-k)}a=log_{a,k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

btw do you think the a,k is a good way to write it

like, log a applied k times

Anyway so $f(x,n)=(2\prod_{m=1}^{n-1}\sum_{k=1}^{m}{}^{(ln^s(2)-k)}e)^x$

[dawn finch]

RoyalBanana

[jovial rock]

I think I found something: e = x^x, we know that this is e^W(1). Lets take the tetration natrual log on Both sides. lns(e) = e^W(1)* lns(e^W(1)). 1 = e^W(1) * lns(e^W(1)). lns(e^W(1)) = 1/e^W(1)

Oh

No I‘m wrong

It‘s x^x, not x^^x…

[jovial rock]

Yeah, I guess this is true

Hmm. What if 4 = x^^x, log4s(4), 1 = 2 * log4s(2), 0.5 = log4s(2)

i‘m trying to find These decimal values for tetration

[azure crane]

the ln(a^b)=bln(a) rule doesn't hold for the super log i think

[jovial rock]

I know

It‘s lns(e^x) = 1+ lns(x)

I just relized

Realised*

[azure crane]

btw, for higher logs than tetration I suppose you can use the n number
so like, for pentation $log^5_{a,k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Instead of s for higher ones

$\log^4_{a,k}(x)=\log^s_{a,k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Lol 4 inputs wow

[azure crane]

👍

[jovial rock]

Isn‘t it without the k?

[azure crane]

I mean if you wanna apply it multiple times then do it with the k

[jovial rock]

Yeah true

Nice notation

[azure crane]

$\log^n_{a,1}(x)=\log^n_{a}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Better than just ln(ln(…ln(…

[jovial rock]

Nice

[azure crane]

I can't tell if this is sarcastic or not

lol

[jovial rock]

It‘s not, I really Like that notation

[azure crane]

ooh oke

[jovial rock]

It‘s a flexible function

[azure crane]

btw this also allows you to have non integer k values

[jovial rock]

Yeah, I think this will be helpful for the general arithmetic function

[azure crane]

The heck does $\ln_{\frac{1}{2}}(x)$ mean lol

[dawn finch]

RoyalBanana

RoyalBanana

[jovial rock]

Yeah probably that

But we need to undestand the fractional Numbers in the height

[jovial rock]

Is this true?

This would mean 4^^0.5 = 2

But I‘m not sure

[azure crane]

Where are you getting 1=2*log4s(2) from?

[jovial rock]

log4s(4) = log4s(2^^2)

[azure crane]

Remember not normal log rules

[jovial rock]

Yeah but it‘s tetration

[azure crane]

how do you know that works

[jovial rock]

Oh

Huh

[azure crane]

The bases being multiples thing

[jovial rock]

Oh shit

I forgot

yeah i see the problem

[azure crane]

Can use this tho

No reason that has to apply only for ln and e

$\log^s_a(a^x)=1+log^s_a(x)$

[dawn finch]

RoyalBanana

[azure crane]

Hmmmm!

$\log^s_a(a^{log_a(b^x)})=1+log^s_a({log_a(b^x))$

[dawn finch]

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

[azure crane]

a^log_a cancels out

$\log^s_a(b^x)=1+log^s_a(log_a(b^x))$

[dawn finch]

RoyalBanana

[azure crane]

h

Could be useful

[azure crane]

There's probably some rule for superlogs like that tho

[azure crane]

How'd you get the last one?

[jovial rock]

anynumber^^-2 = undefined i think

[azure crane]

${}^{-k}a=log_{a,k}(1)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{-2}a=\log_{a,2}(1)=\log_a(0)=undefined$

[dawn finch]

RoyalBanana

[azure crane]

yup

[jovial rock]

Hmm, but could there be more rules?

[azure crane]

Hmm $\log_a^s({}^{x}(2a))=?$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

That one is hard

[azure crane]

$\log_a^s({}^{x}(2a))=?$

$\log_a^s({}^{x}(a^2))=?$

$\log_a^s({}^{x}({}^2a))=?$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm, is there a possibility to change the base of the function by transforming it from the outside?

I tried aproching it with lns and x = 2 for the first one and I got: 2.8066

Aproching it with x=1 I got: 1.53

If we could factor out x, the transformation wouldn‘t be very big

I‘ll try the second one

lns with x=2: 2.99147

For x=1 it‘s 1.701

It could be that the factoring out depends on the base too

I think this is the Same problem as trying to say that x^^1.5 is x^squareroot(x)

I think it has something to do with more logarithms. Maybe lns((2e)^^x) = lns(x) * lns(2e) or something like that

[jovial rock]

lns((2e)^^2) = lns(2e) * lns(1.834)

I don‘t know

[azure crane]

I feel like lns((2a)^^x) doesn't have a useful property about it but lns((e^2)^^x) and lns((e^^2)^^x) do

[jovial rock]

Yeah true

[azure crane]

maybe it's like how the higher the type (exponentional, tetration, pentation) the higher the operation has to be too to have a useful property

ln(a+b) doesn't have a nice answer for it but ln(ab) and ln(a^b) do

[jovial rock]

You‘re right, but how could this work? I feel like it‘s something with logarithms

lns((e^2)^^2) = lns(e^2) * lns(8.297)

I don‘t know how to approx the third one

It‘s Pretty big with the tetration

maybe the e^2 is something with the normal ln

[azure crane]

hold on I'm cooking

[jovial rock]

lets see

[azure crane]

expanding out the third one and plugging in different values of x

[jovial rock]

oh cool, I'll try that too

[azure crane]

and using this

[jovial rock]

oh, I forgot about that one

[azure crane]

You see, ofc
$(e^e)^{(e^e)}=e^{e*e^e}=e^{e^{e+1}}$

[dawn finch]

RoyalBanana

[jovial rock]

Oh, I thought about that one, but wasn't sure

so thats lns(e^e^(e+1)) = 2 + lns(e+1) i guess

[azure crane]

Interesting

[jovial rock]

yeah, some (1+e)^^(n-1) structure

[azure crane]

Rewrote it so it's easier to see

[azure crane]

except the 1 isn't in the parenthesis its like separate hmm

[jovial rock]

I think it's lns((e^^2)^^n) = 2 + lns((e+1)^^(n-1))

[azure crane]

nono

it's not (1+e)^(1+e)...

it's 1+e^(1+e^(...

[jovial rock]

oh

[azure crane]

:/

yeah

[jovial rock]

hmm

What if lns((e^^3)^^n) ?

[azure crane]

lemme see

[azure crane]

The first one is 2+lns(e) so it just simplies to 3

[jovial rock]

oh okay, the addition in the lns() is hard to tackle because we can't simplify it

[azure crane]

yeah :/

[azure crane]

oops I put extra parenthesis, whatever lol

[jovial rock]

Maybe we would need something like pentation to make it multiplication in the lns

idk

[azure crane]

yeah

[jovial rock]

I tried to aproximate the x=2 value, it's somewhere between 3.7 and 4. Idk if that helps with anything

[jovial rock]

at least it's something what we got here

log2s((2^^2)^^^2)

hmmm

[azure crane]

my brain hurts

[jovial rock]

same haha

I think thats log2s((2^^2)^^(2^^2)^^(2^^2)^^(2^^2))

[azure crane]

yeah

[jovial rock]

uhh

big number

[azure crane]

2^^2=2^2

[jovial rock]

log2s(4^^4^^4^^4)

uhhhh

lets just change to log4s(). would this be 4^^4^^4?

[azure crane]

yes

[jovial rock]

hmm, it's kinda big ngl

but why can't there be something like log4s() = x^2 or x^3

because logas(a^^x) = x

wait, what would logas(a^^^x) be?

[azure crane]

a^^^{x-1}

[jovial rock]

Oh okay

We just step 1 layer higher and we go from x to a^^^(x-1)

Can we make an equation where we get logas() for an non integer value?

[azure crane]

maybe

[jovial rock]

log2s(3) = x, 3 = 2^^x, hmm

[jovial rock]

x^^logxs(0.5) = 0.5

e = (e^W(1))^(e^W(1), ssrt(e) = e^W(1), lns(ssrt(e)) =1 + lns(W(1))

lns(ssrt(x)) = 1 + lns(W(ln(x)) I think

Hmm, I there a rule for supersquareroots in tetration natrual logs too?

lns(ssrt(x)) - lns(W(ln(x)) = 1

e^^(lns(ssrt(x)) - lns(W(ln(x))) = e

[jovial rock]

Wait

I this allowed?: ssrt(x) = lns(W(ln(x))) + lns(1)

And this means: ssrt(x) = lns(W(ln(x)))

[azure crane]

where are you getting this from?

[jovial rock]

Nvm

I think it doesn‘t work

Because it would be this with this rule: lns(ssrt(x) )= lns(W(ln(x)) + lns(0)

[jovial rock]

An this should have been lns(ssrt(x)) = lns(W(ln(x)) + lns(1)

No wait

[azure crane]

lns(1)=0

[jovial rock]

I think lns(lns(ssrt(x)) - lns(W(ln(x)) )= lns(1)

Idk

[azure crane]

Yes lns(ssrt(x)) - lns(W(ln(x))=1

Btw about non integer superbases they work fine!

$x={}^{log^s_{\frac{1}{2}}(x)}(\frac{1}{2})$

[dawn finch]

RoyalBanana

[jovial rock]

ssrt(x) * W(ln(x)) = ln(x)

[jovial rock]

Yeah true

So (0.5)^^log0.5s(x) = x. i guess

[azure crane]

log0.5s(1/√2)=2

[jovial rock]

(0.5)^^log0.5(0.5^0.5) = squareroot of 0.5

[jovial rock]

Oh

Does that work?

[azure crane]

yeah

[jovial rock]

This would be awsome

[azure crane]

1/√2=(1/2)^(1/2)

it's by definition

[jovial rock]

Oh right

Ah

Now I See

Hmm

[azure crane]

hm $2^{\frac{1}{\sqrt{2}}}$

[dawn finch]

RoyalBanana

[azure crane]

weird

uhhh

$\log^s_{-1}(2)=?$

[dawn finch]

RoyalBanana

[jovial rock]

Undefined I guess

[azure crane]

Maybe it's not -1 for non integer values or something

like what's ${}^{0.5}(-1)$

[dawn finch]

RoyalBanana

[jovial rock]

Probably complex number…

Or the same quastion but in other words: log-1s(x) = 0.5

[jovial rock]

This one is similar to that one

0.5^^log0.5s(1/√2) = 1/√2

0.5^^log0.5s(2) = 2

0.5^^log0.5s(2) = log0.5(1/√2)

0.5^^0.5^^log0.5(2) = 1/√2

[azure crane]

ye

[jovial rock]

Hm

I though I could find something like this but nvm 0.5^^0.5^^log0.5s(1/√2) = 2

[jovial rock]

lns(ssrt(x)) - lns(lnk(W(ln(x))) = k+1 I think

Can we maybe turn this other Part of the equation into a k*x?

[jovial rock]

I like this equation, I hope it's true

[jovial rock]

lns(ssrt(√10)) = 1 + lns(W(ln(10)/2))

[azure crane]

Yes

lns(x)-lns(lnk(x))=k

[jovial rock]

yeah

[jovial rock]

How to solve x^^x = x^x

I guess take the xth super log on Both sides: logxs(x^^x) = logxs(x^x), x = 2

But 1 would be a solution too i guess

How to solve x^^x = x^(x-1)

Couldn‘t get farther than logxs(x* x^^x) = 2

[azure crane]

x*x^^x=x^(1+x^^{x-1})

[jovial rock]

uh

hm

is it even possible to solve x^x = x * (x-1)

if we go one step back we get: x * x = x + (x-1)

which is x^2 -2x + 1 = 0

and we can use the quadratic formula for that

so shouldn't there be a formula too for x^x = x * (x-1) and x^^x = x^(x-1)

hmmm

nvm

I don't understand

[azure crane]

Not nessicarily

there is a solution ofc

doesn't mean there's an exact way to write it with symbols and stuff

maybe if you use W

[jovial rock]

Hmm, can we use the new tetration functions to solve such equations?

It kinda doesn‘t feel like that we can find these non integer heights

The only function I know we can express with W and stuff is ssrt(x)

Finding the lnk(x) function for non integers would help a lot

[azure crane]

#1330832852819906661 message

[jovial rock]

yea, but we would need the lns for that

hmm, what if we try to balance it out that lns -k are non integer and cancel out to an integer?

and than we have an equation with e^^1, and we can try to find out lns - k = 1

[jovial rock]

something like this

ln0(e) = e, ln1(e) = 1, ln0.5(e) = ?

this would be e^^(1-0.5) = e^^0.5 aproximation: 1.6463542337511945810

kinda cool that this cancels out: ln-k(e^^(lns(x)-k)) = x

hmm, so would that mean ln2(ln3(x)) = ln5(x) ?

lns(lns(x)) = ln2e(x)

[jovial rock]

I'll try to continue with this, I don't know how to find more rules for the lns and lnk

or maybe we just need some tetration constant which we could use to compute lns or logas

[azure crane]

yeah $\ln_a(\ln_b(x))=\ln_{a+b}(x)$

[dawn finch]

RoyalBanana

[azure crane]

makes sense

[jovial rock]

we could use this for example to find maybe ln1.5 --> ln1(e^^0.5) = ln1.5(e)

[azure crane]

Oh $\ln_{\frac{1}{2}}(\ln_{\frac{1}{2}}(x))=\ln_{1}(x)=\ln(x)$

[jovial rock]

looks like a small new rule for lnk

[dawn finch]

RoyalBanana

[jovial rock]

hm

[azure crane]

It makes sense if you write out all the of the ln

[azure crane]

So you need a function that when inputting itself it outputs ln(x)
Interesting

[jovial rock]

hmmm

so for e: (e^^0.5)^^0.5 = 1

[azure crane]

What if I combine this with ${}^{(ln^s(x)-k)}e=ln_k(x)$?

[dawn finch]

RoyalBanana

[jovial rock]

yeahh, but how?

[azure crane]

Wdym

[jovial rock]

ln0.5(e) = e^^0.5 right?

[azure crane]

yeah

[jovial rock]

and when repeated again: ln0.5(e^^0.5) = 1

because ln1(e) = 1

[azure crane]

Oohh

[azure crane]

I suppose this is like the same thing I wrote in a different form

[jovial rock]

yeah

[azure crane]

interesting

[jovial rock]

waiitt. So this means a^^b^^c = a^^(b+c) ?

[azure crane]

${}^{(ln^s({}^{(ln^s(x)-b)}e)-a)}e=ln_a(ln_b(x))$

[dawn finch]

RoyalBanana

[azure crane]

I don't think so
1/2 is a special case bc 1-1/2=1/2

[jovial rock]

ah

right

I got too exited for a moment haha

[azure crane]

${}^{{}^{ln^s(x)-b-a}e=ln_{a+b}(x)$

ah nevermind nothing helpful

[dawn finch]

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

[jovial rock]

but looks kinda cool haha

[azure crane]

yeah

lol

[jovial rock]

hmm

e^^0.25^^0.25 = e^^0.5 I think

because ln0.25(ln0.25(e)) = ln0.5(e)

[azure crane]

how do you get to that from this

[jovial rock]

ohh

wit

wait

gimme a second

ln0.25(e) = e^^0.75

right?

[azure crane]

?

Oh yes

[jovial rock]

because 1-0.25

with the rule

[azure crane]

yeah

[jovial rock]

here

wait

[azure crane]

yes

[jovial rock]

e^^0.75^^0.75 = e^^0.5

that should be the right thing

[azure crane]

That is weird lol

[jovial rock]

hm

[azure crane]

So the many logs are additive and tetration is like 1-additive or something

[jovial rock]

kinda

its like subtraction inverse

because e^^1 = ln0(e)

e^^2 = ln-1(e)

[azure crane]

${}^{1-b}({}^{1-a}x)=x^{a+b}$?

[dawn finch]

RoyalBanana

[azure crane]

(doesn't have to be e)

If so that's awesome

[jovial rock]

I think we are the first people finding this

[jovial rock]

You forgot the tetration thingy I think, it‘s x^^(a+b)

Uh, I‘m confused Right now

I‘m not sure

e^^0.5^^0.5 should be e^^1 with this rule but it‘s e^^0

I think it‘s x^^1-a^^1-b = lna+b(x)

[jovial rock]

Then you can use this on this

So e^^2^^2 is e^^3

Huh

Doesn‘t work

e^^2 = ln-1(e), so this would be ln-2(e) and this would mean ln-2(e) = e^^(lns(e) + 2), and then it‘s e^^3 ???

What did I calculate wrong

Oh

Hm

e^^3 should be ln-2(e) because 1-3 =-2

e^ln-1(e) should be ln-2(e)

so it‘s e^e^^2

I don‘t get it at all

Oh

I See the problem

[jovial rock]

The lns

lns(e^e) = 2

And then it‘s e^^4

Makes sense that e^^2^^2 = e^^4 right?

So that means e^^0.75^^0.75 = e^^(lns(0.5) -0.5)

I‘m not really sure

and for 0.5 lns is lns(1) which is 0

Uff

Idk

I think it‘s actually e^^2^^2 = e^^(lns(e)+3)

Hm

No

[jovial rock]

Thats not true

I‘ll take a Break, my Brain hurts

It should be lns(e) = 1 all the time but I don‘t really understand how e^^2^^2 works

I thought its ln-3(e)

Wait what if e^^3^^5

This should be ln-7(e) but 3^^5 is a very big number so I don‘t get it

[jovial rock]

e^^(0.5^^0.5) should be ln|log0.5(0.5)|(e) when |k|

|k| is just a notation to See what is k

(e^^0.5)^^0.5 should be logln0.5(e)|0.5|(ln0.5(e))

[azure crane]

Oh oops yeah

[azure crane]

No

${}^2({}^2e)=(e^e)^{(e^e)}=e^{(e*e^e)}=e^{(e^{e+1})}$

[dawn finch]

RoyalBanana

[azure crane]

unless you meant ${}^{({}^22)}e$

[dawn finch]

RoyalBanana

[azure crane]

Make sure to put parenthesis so it's not confusing

[jovial rock]

Yeah

but (e^^0.5)^^0.5 is 1 Right?

I thought e^^2^^2 = e^^(2^^2) ≠ (e^^2)^^2 ?

[jovial rock]

No it‘s not…

I approximated it

So (e^^0.5)^^0.5 ≠ ln0.5+0.5(e)

Why doesn‘t this work

[azure crane]

Why would that work

[azure crane]

No

you forgot about lns(x)

lns(x)-0.5 is not 0.5 when you plug in x=e^^0.5

it's 0

so really (e^^0.5)^^0=ln1/2(ln1/2(e))=ln(e)=1

Wait

I'm confused now

Ok but I know this formula is right

${}^{(ln^s(x)-1/2)}e=ln_{1/2}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Wait

ln(e) is 1

lns(e) is 1

[azure crane]

${}^{(1/2)}e=ln_{1/2}(e)$

[dawn finch]

RoyalBanana

[jovial rock]

lne(e) is lns(e) because it‘s iterated e times

not?

I‘m confused

[azure crane]

${}^{(ln^s((ln_{1/2}(e))-1/2)}e=ln_{1/2}(ln_{1/2}(e))$

[dawn finch]

RoyalBanana

[azure crane]

${}^{(ln^s({}^{1/2}e)-1/2)}e=ln(e)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{(1/2-1/2)}e=1$

[dawn finch]

RoyalBanana

[azure crane]

${}^{0}e=1$

[dawn finch]

RoyalBanana

[azure crane]

see

[azure crane]

idk what you mean here

[jovial rock]

But if we ask, what is k if lnk(e^^e) = e

[azure crane]

hmm

[jovial rock]

I think it‘s e

[azure crane]

You are confusing base and number of times the log is being used

Wait

Oh

Nevermind I thought the question was e^e

you're right

[jovial rock]

Ye

And I don‘t understand ln1(e) = lne(e)

[azure crane]

Wdym
how are they equal

[jovial rock]

lns(e) = 1

[azure crane]

Yes

[jovial rock]

And lne is lns because this

[azure crane]

No

[jovial rock]

Why not?

[azure crane]

That is a special case where they are equal

what is lne(e^^(e+1))

[jovial rock]

ye

e+1

Oh

Wait

Ah

Idk

[azure crane]

${}^{(ln^s(x)-k)}e=ln_k(x)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{(ln^s({}^{(e+1)}e)-e)}e=ln_e({}^{(e+1)}e)$

[dawn finch]

RoyalBanana

[jovial rock]

so it’s e??

[azure crane]

yep

[jovial rock]

e+1-e

Huh

[azure crane]

${}^{(ln^s({}^{(e+x)}e)-e)}e=ln_e({}^{(e+x)}e)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{x-e}e=ln_e({}^{x}e)$

[jovial rock]

Hm

I see

[dawn finch]

RoyalBanana

[azure crane]

No reason this has to be for just e btw

[jovial rock]

Wait this is kinda cool

[azure crane]

you can test this with like 2 or 3 which ofc you can actually do ln2

If it's base 2 or 3 also

[jovial rock]

2^^x-2 = ln2(2^^x)

Right?

[azure crane]

No

[jovial rock]

Or only for lne?

[azure crane]

^

[jovial rock]

Ah

[azure crane]

You can see in the step here lns(e^^(e+1))

has to be the same base to cancel out with 2^^

${}^{x-a}a=log_{a,a}({}^{x}a)$

[dawn finch]

RoyalBanana

[jovial rock]

This is actually pretty helpful

[azure crane]

Lemme test this

Actually yeah makes perfect sense

[azure crane]

Anyway

[jovial rock]

Yeah

How does that work?

[jovial rock]

I think it‘s this

[azure crane]

Wait howd you get this tho?

[azure crane]

noo lol

dont forget about how ln is base e

[jovial rock]

ln0.25(ln0.25(e))

[azure crane]

ye

How do you go from there

[jovial rock]

ln0.25(e^^0.75

I didn’t understand the rest

But it‘s probably not true

[azure crane]

$\ln_{0.25}(e)={}^{0.75}e$

[dawn finch]

RoyalBanana

[jovial rock]

Ye

What is ln0.25(e^^0.75)

[azure crane]

now what is $\ln_{0.25}({}^xe)$

[dawn finch]

RoyalBanana

RoyalBanana

[azure crane]

${}^{(x-0.25)}e=\ln_{0.25}({}^xe)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{0.5}e=\ln_{0.25}({}^{0.75}e)$

[jovial rock]

Ye

e^^0.5

[dawn finch]

RoyalBanana

[azure crane]

${}^{0.5}e=\ln_{0.5}(e)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{1-x}e=\ln_{x}(e)$

[dawn finch]

RoyalBanana

[azure crane]

ye

my brain hurts

[jovial rock]

Same

Tetration is crazy

Can we find a rule for (e^^a)^^b?

[azure crane]

Btw this means ${}^{k}a=log_{a,(log_a^s(x)-k)}(x)$

[dawn finch]

RoyalBanana

[azure crane]

${}^{k}e=ln_{(ln^s(x)-k)}(x)$

[dawn finch]

RoyalBanana

[azure crane]

for any x of your choosing

[jovial rock]

Nice

Can we put something in for x to make the equation as simple as possible?

Probably just e

So, (e^^2)^^2 would be log(e^^2)(-1)(e^^2)

I used e^^2 to cancel out the super root

[azure crane]

x=1 works

[jovial rock]

Super log*

[jovial rock]

Oh Yeah true

[azure crane]

${}^{k}e=ln_{-k}(1)$

[dawn finch]

RoyalBanana

[jovial rock]

Looks great