General Formula for all arithmetic operations

[jovial rock]

I forgot about the ln…

[azure crane]

ye :/

also youd also be replacing the other a's too

[jovial rock]

With this we could have gotten all numbers for function L

[jovial rock]

Yeah, I guess

[azure crane]

I mean

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

[dawn finch]

RoyalBanana

[jovial rock]

Ye

[azure crane]

btw

there's nothing special about the base e

$log_a(b^c)=clog_a(b)$

[dawn finch]

RoyalBanana

[azure crane]

so

[jovial rock]

Ye sure

[azure crane]

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

[dawn finch]

RoyalBanana

[jovial rock]

The base for log can be any number

[azure crane]

ye

[jovial rock]

Hmm

[azure crane]

so anyway

$\log^s_a(x)\log_a(L_a(a)) = \log_a(L_a(x))$

[jovial rock]

ln(5) - ln(6) = ln(5/6), so what if La(5) / La(6) = La(…..)

[dawn finch]

RoyalBanana

[azure crane]

that would be a different L but ok

I assume La(log_6(5)) bc that's the next step up after division

[jovial rock]

Maybe

Is this possible?: La(logx(a^x)) = La(a)

[azure crane]

?

no

[jovial rock]

I just used this idea to convert division into log

[azure crane]

that would mean contradictions

[azure crane]

Wdym convert lol

[jovial rock]

Like ln(5) -ln(6) is ln(5/6)

[azure crane]

$f(x)/f(y)=f(log_y(x))$?

[jovial rock]

Subtraction into division

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

[azure crane]

if you set y=x, you can see f(1)=1

Ah

if you set x to y^x and rearrange the terms

[jovial rock]

Yeah I see

But that wouldn’t work

[azure crane]

yeah

[jovial rock]

Because commutative

[azure crane]

Just set y=1

[jovial rock]

Not*

[azure crane]

but f(1)=1

so f(x)=1

[azure crane]

technically correct lol

[jovial rock]

Yeah

[white cove]

@royal river Look at this

[royal river]

okay?

[white cove]

Read it and send your thoughts

[royal river]

it is weird?

[white cove]

I don't know

I haven't read it

[jovial rock]

Wouldn‘t that be possible with bases?

And that would mean it‘s equal to L(x)

[jovial rock]

So that should maybe work?

[azure crane]

?

[jovial rock]

La(x)/La(a) = La(loga(x))
La(a^x)/La(a) = La(loga(a^x))
La(a^x)/La(a) = La(x)
La(a^x)= La(x) * La(a)

And that is just this

So La(logx(a^x)) = La(a)

Not sure about that

Maybe I did something wrong

What if L2(2^0) ? As we know this is equal to L2(1) so it‘s 1. that means L2(2) * L2(0) = 1, so L2(0) = 1/L2(2), so L2(0) = L2(2)^(-1)

And I don’t See what this should only hold for 2, so La(0) = La(a)^(-1)

And if I would plug in 1 for a: L1(0) = L1(1)^(-1). We know La(1) and L1(a) = 1, so: L1(0) = 1^(-1), and that means L1(0) = 1

Oh

Right

This just proves that L1(a) = 1

[azure crane]

Oh yeah I already found that mb I forgot to say

[jovial rock]

Nice

[azure crane]

so $L_a({}^xa)=L_a(0)^{-x}$

and $L_a(a^x)=\frac{L_a(x)}{L_a(0)}$

[dawn finch]

Coffey 2.0

[azure crane]

interesting

[jovial rock]

Ohh, cool

[azure crane]

hmm plugging in $a=0$ to the second formula

$x>0: L_0(0)=\frac{L_0(x)}{L_0(0)}$

$x=0: L_0(1)=\frac{L_0(0)}{L_0(0)}$

[dawn finch]

Coffey 2.0

[azure crane]

ofc L_a(1)=1 for any a soo L_0(1)=1

[jovial rock]

Yep

[azure crane]

$x>0: L_0(0)^2=L_0(x)$

$x=0: 1=\frac{L_0(0)}{L_0(0)}$

[dawn finch]

Coffey 2.0

[azure crane]

the second one is good, nice

interesting so

[jovial rock]

Yeah, it‘s just x/x = 1

When x is not 0

[azure crane]

Interesting so for $x>0: L_0(x)=L_0(0)^2$

[dawn finch]

Coffey 2.0

[jovial rock]

Wait, where did you get that from?

[azure crane]

important to note you can't plug in x<0 if a=0 because that would lead to 0^(negative something)=1/0^positive something=1/0

[jovial rock]

Ah okayy, i see

[azure crane]

Plugging in a=0 and x>0 into $L_a(a^x)=\frac{L_a(x)}{L_a(0)}$

[dawn finch]

Coffey 2.0

[azure crane]

(0^x is 0 for x>0)

[jovial rock]

Yeah right

[azure crane]

So for x>0, $L_0(0)=\frac{L_0(x)}{L_0(0)}$

[dawn finch]

Coffey 2.0

[azure crane]

So I just multiplied both sides by $L_0(0)$ to get $x>0: L_0(x)=L_0(0)^2$

[dawn finch]

Coffey 2.0

[jovial rock]

L0(0) can be any number?

Because square Root of L0(x) = L0(0)

[azure crane]

No it just means L_0(x) is constant for x>0

[jovial rock]

But the x is only on one side

That would mean L0^-1(L0(0)^2) = x

[azure crane]

yes

[azure crane]

No, it just means the inverse doesn't exist

it's like.. $f_a(b)=ab$

[dawn finch]

Coffey 2.0

[azure crane]

$f_0(b)=0$ it doesn't have an inverse

[dawn finch]

Coffey 2.0

[azure crane]

when a=0

because any input gives that output

[jovial rock]

Ah

Okay

[jovial rock]

Wouldn‘t that just mean it‘s commutative? So fa(b) = fb(a) ?

[azure crane]

yes

lol

[robust basalt]

well im a bit late but i love you experimenting with things like that, sorry i didnt real all the messages just like 60 or smth, but do u have any conclusion @jovial rock @azure crane or any progress made? I would love to document it and maybe analyse this further myself it thats okay for yall, we could also collaborate if your still interested

[jaunty hornet]

$Fiboboost(6) = 12334445555566666666$

[dawn finch]

Silverstonely

[jaunty hornet]

one 1, one 2, two 3, three 4, five 5, eight 6, thirteen 7s
its fibonacci concatenation timing basically

idk how to use this bot so srry if its wrong

[azure crane]

The formula for this is complicated but it works

[jaunty hornet]

Nice fr?

[azure crane]

$\text{Fiboboost}(x)=\frac{x!}{9^x}\prod_{n=1}^{x}(10^{(\frac{ ({\frac{1+\sqrt{5}}{2}})^n-({\frac{1-\sqrt{5}}{2}})^n}{\sqrt{5}})}-1)$

[dawn finch]

Coffey 2.0

[jovial rock]

oh cool, golden ratio

[azure crane]

yeah it comes up in the formula for the Fibonacci numbers

[jovial rock]

niceee, ye I saw that somewhere on yt I think

hmm, it's quite fast.

sadly only for integers

I was wondering if it is possible to create a function which grows like tetration and accepts non integers

[jovial rock]

Not sure but I think we created many formulas for super logarithmus and tetration but I think the biggest problem is extending it to the reals. I'm open to collaborate, I'm still working on it sometimes in my free time and struggeling with it a bit, but I think I'm making very little progress on it

[jovial rock]

So I saw this approximation off e^^0.5 and I was very fascinated about it, even though it's pretty basic and I just don't know a lot of math. So what happens with tetration, you apply exponentiation function multiple times, for example exp(exp(exp(x))) = e^e^e^x. What we can do is make a notation like expk(1), which is equal to e^^k. and if we do exp(exp(x)), we get exp2(x). So that mean exp0.5(exp0.5(x)) = e^x. So if we find exp0.5(x). What we need to do is to find f(x) if f(f(x)) = e^x. e^x = 1 + x + x^2/2! + x^3/3 + ... because of the taylor series. now you could just say let f(x) be a + bx + cx^2, and f(f(x)) = a + b * (a + bx + cx^2) + c* (a + bx + cx^2)^2 because we are going to take the first 3 parameters of the e^x taylorseries. We through the bigger things like x^3 and x^4 away so we get: 1 + x + x^2/2! = a + ab + ac^2 + x(b^2 + 2abc) + x^2(bc + 2ac^2 + b^2c). Now we take the parameters and equal them to the parameters in the taylor polynomial: a + ab + ac^2 = 1, b^2 + 2abc = 1, bc + 2ac^2 + b^2c = 1/2. If we solve each of them we get a ≈ 0.4979, b ≈ 0.8781, c ≈ 0.2618 and then we input the parameters for f(x) and we get: f(x) = 0.4979 + 0.8781x + 0.2618x^2 which gives us if we input 1 for f(x): 1.6378. this is an approximation of e^^0.5. It's not very good but it works. And it works even better if we let f(x) be an even bigger polynomial so we have more parameters from the taylor series polynomial. I was wondering if this can be extended to an infinite sum or something to get a percise solution to non integer tetration hights

I heard this is something like a functional squareroot

I tried to approximate it with even more polynomials and these are my solutions:

1.6463542337511945810

[jovial rock]

This is the approximation of the tetration calculator and I think it's pretty percise to my first digits

If there would just be a general formula...

I'm wondering if there is something like a taylor series which lets you compute all the parameters, so you can find the nth parameter and then have something like a1 + a2x + a3x + a4x+...

This reminds me of the form from e = 1/0! + 1/1! + 1/2! + 1/3! + ...

whoa, e^^-0.5 and e^^-1.5

ln(−0.69609) = −0.3629+i3.1416

so thats e^^-2.5

that means that when the height of e^^x is -2 or all the other smaller integers the result is undefined but not for non integers, like in the gamma function. And btw all numbers between e^^-3 to e^^-2 have something with + pi * i

and thats the whole domain of e^^x, when x is a negative number

I think thats very cool because e^^x has such big domain in the negative values

And I guess this calculator isn't capaple of taking the log of negative numbers, but still has a complex number input...

So my challenge now is to find this infinite sum which equals to e^^0.5. No Idea how I should do that

[jovial rock]

So I was messing with this stuff a bit and it's pretty hard to find it out. You would need to find out all the infinite parameters to that infinite equation system and then add them all together, only to get tetration to the height of 0.5

there must be a general formula for this, but I can't find it

But I definetely made progress

If I want to, I could find out e^^(1/3), this would just be solve for f(x), when f(f(f(x))) = e^x

Maybe there could be some deeper connection with the super log

not sure

[jovial rock]

I think I'll stop for now. Still couln't find a formula which could find the exact value of f(1). I made some progress and extended the tetration to values below -2. technically I could find tetration to the reals now, the approximation would just be not very percise. A general Formula could help a lot and I wouldn't need to solve these annoying equation systems.

There probably would need to be a new notation for some function stuff to construct such tetration like functions

[jovial rock]

Btw Sum(infinity, n=1)((e^^x)^n/n!) = e^^(x+1)

I thought this is cool how an infinite sum of negative infinities is equal to 0 and an infinite sum of zeros ist equal to 1

[jovial rock]

Btw

[jovial rock]

uhh, I miss calculated something

Its this

[jovial rock]

nvm, It's completly wrong

I thought that if f(f(x)) = e^x, that f(x) = (e^^0.5)^x what is not right

[jovial rock]

Yesss

I finally found the Definition of f(f(x)) = e^x. It‘s f(x) = e^^(lns(x) + 0.5)

It‘s still a Bit hard, but with an 5. parameter polynomial approximation it works very well

Now I just need to find the Taylor Series for that one

[jovial rock]

So if T(x) = e^T(x-1), If we want to take the derivative we get: T'(x) = T(x) * T'(x-1). T'(x) = T(x) * T(x-1) * T'(x-2). T'(x+1) = T(x+1) * T'(x) --> T'(x) = T'(x+1)/T(x+1). T'(x) = T'(x+2)/T(x+2))/T(x+1) --> T'(x) = T'(x+2)/T(x+2))/T(x+1) --> T'(x) = T'(x+2)/(T(x+1)*T(x+2)) ,
T'(x) = T'(x+3)/(T(x+1) * T(x+2) * T(x+3), T'(x) = T(x+n)/Product(n,k=1)(T(x+k))

[jovial rock]

f^-1(x) = e^^(lns(x) - 0.5)

f^-1(f^-1(x)) = ln(x) = e^^(lns(x)-1)

[jovial rock]

I found this and I think this is really cool. There is even the super log Graph: https://www.desmos.com/calculator/wkjkhlzrm3?lang=de

[jovial rock]

Btw, slogb(a^x) = slogb(b^(logb(a) * x)) = slogb(logb(a) * x) + 1

[trail surge]

Oh my god I can't even scroll to the top to see anything

[azure crane]

5164 messages 🔥

[white cove]

Are we still chatting here?

[jovial rock]

Yes

some stuff:
fa(x) = tet(slog(x)+a) -> fa(e^^x) = e^^(a+x)
fa(x) = tet(slog(fb(x))+a-b)
fa(x)+fb(y) = f-1(fa+1(x) * fb+1(y))
fa(x)-fb(y) = f-1(fa+1(x)/fb+1(y))
fa(x) * fb(y) = f1(fa-1(x)+fb-1(y))
fa(x)/fb(y) = f1(fa-1(x)-fb-1(y))
fa(x)^fb(y) = f1(fa-1(x) * fb(y))
log_fb(y)(fa(x)) = fa-1(x)/fb-1(x) = f1(fa-2(x)-fb-2(y)
fb(y) th root of fa(x) = f1(fa-1(x)/fb(y))

a = slog(fa(x)) -slog(x)

slog(x) = slog(fa(x)) - a
slog(x * y) = slog(ln(x)+ln(y)) + 1
slog(x/y) = slog(ln(x)-ln(y))+1
slog(y^x) = slog(ln(y) * x) + 1
slog(log_y(x)) = slog(f-2(x)-f-2(y)) + 1
slog(y th root of x) = slog(ln(x)/y)+1

Btw f1(x) = exp(x) and f-1(x) = ln(x)
fa(fb(x)) = fa+b(x)

[jovial rock]

And btw: d/dx(fa(x)) = Product(a,k=1,fk(x)) this is very similar to faktorial because for negative a it‘s d/dx(f-a(x)) = Product(a,k=0, f-k(x)^-1)
If we could find out How to Connect this to the gammafunction we could take the darivative of for example f-0.5(x) which should be something with the squareroot of pi

[azure crane]

what's tet(x) and fa(x)?

[karmic lava]

tet = tetration (I assume)

[robust basalt]

https://www.researchgate.net/publication/391941916_Holomorphic_Extension_of_Tetration_to_Complex_Bases_and_Heights_via_Schroder's_Equation?channel=doi&linkId=682ddd0e026fee1034f9e0da&showFulltext=true

you guys may find this interesting

thats my paper with the title "Holomorphic Extension of Tetration to Complex Bases and Heights via Schröder's Equation"

@jovial rock

[jovial rock]

Ye tet ist tetration. It‘s e^^x, and da ist just f the base a. f_a(x) is as I defined: tet(slog(x)+a) here the base a is the a variable. And it‘s f1(x) = e^x, f2(x) = e^e^x, f-1(x) = ln(x), f-2(x) = ln(ln(x))

[azure crane]

ah

so $f_a(x)=\ln_{-a}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Yes

I wanted to use a general thing for iterated ln and iterated e

So it‘s a bit easier

And f0(x) = x

[jovial rock]

Quite cool

Yeah I saw that e^^infinity is equal to the fixpoint of x=e^x and e^^-infinity is the same but with a negative imaginary Part

[jovial rock]

I Wonder if it would be possible to build a efficient tetration calculator using this

[robust basalt]

http://myweb.astate.edu/wpaulsen/

[robust basalt]

http://myweb.astate.edu/wpaulsen/tetcalc/tetcalc.html

this guy has one

but only for some bases, its pretty efficient too, he wrote a paper about approximating it

[jovial rock]

Yeah I know that one, it has some bases but not a lot. And it doesn‘t work for e^^-2.5 or just values below -2

It says it is on a branch cut or something but it‘s just ln(e^^-1.5), so ln(-0.69602474) = -0.36237 + pi*i

I think it is a very good one but there are things that can be improved

[robust basalt]

yeah

[jovial rock]

Question: what if we analyze (-1)^^x ? It‘s kinda weird because (-1)^^-1 = 0, (-1)^^0 = 1, (-1)^^1 = -1, (-1)^^2 = -1, (-1)^^3 = -1. so after 1 the solution is always -1. my question is if there could be some periodic function like sin or cos that could exactly match (-1)^^x

[karmic lava]

How did you get -1^^-1 = 0 ?

[jovial rock]

log_-1(1) = 0

[jovial rock]

(-1)^x = 1, x = 0

[karmic lava]

ah, makes sense

[jovial rock]

Soo, after analyzing tetration with base -1 I saw that it’s a bit more complicated.
If we take f_a(x) = tet(slog(x) + a) where tet and slog are base -1 we see that for all integer a equal or bigger than 1 we get a periodic function. For example (-1)^(-1)^x is just f(x) = f(x+2) and this is for all (-1)^(-1)^…^(-1)^x.
Btw with my definition of f_a(x) = tet(slog(x) + a), a shows how many times you apply the exponential function b^x and x is just the number on top of the tower. So with b=-1, f2(x) = (-1)^(-1)^x.
So we see at f0(x) we just have x, which doesn’t have a periodic thing. And at f-1(x) we get ln(x)/(i*pi). my question is if the periodic holds for just the positive integer or even the positive reals.
Btw I just tried solving (-1)^x using the f(f(x)) = (-1)^x, polynom method but it didn’t work somehow. There was no solution for the coefficients so I guess I can’t really find a way to calculate non integer values for (-1)^^x, which is a bit frustrating

[jovial rock]

Btw (-1)^^x is multi valued i think because let’s say we find (-1)^^0.5. That means we can find (-1)^^1.5, because (-1)^((-1)^^0.5). And using this we can find (-1)^^2.5. But if we solve (-1)^^2 we get -1 and this is the same as (-1)^^1. So (-1)^(-1) = (-1) so that means f0.5((-1)^(-1)) = f0.5((-1)). So (-1)^^1.5 = (-1)^^2.5 ? The only way I could make sense out of this is that for real heights there are infinite solutions because (-1)^^1.5 should have all the solutions of (-1)^^(n+0.5)

Where n is an integer

[delicate rivet]

My latest submission on arxiv finds a substitution that introduces a family of functions analogous to the W lambert, that solves any equation of form x^^n =a, so im glad you are discussing this topic

I fucking love tetration

“From: Ramdejin tetration” will prove it💀

[jovial rock]

nicee, yeah tetration is really cool. There is still a lot to explore

[jovial rock]

btw 3 identities I thoguht I need to note down:

tet(a+b) = fa(tet(b)) = fb(tet(a))
tet(a-b) = f-b(tet(a))

log_y(x) = tet(slog(f-2(x)-f-2(y)) + 1)

[jovial rock]

thats really interesting do you maybe have a link?

[delicate rivet]

I’ll share the arxiv link if it gets approved lol

But explain what tet(a+b) is

@jovial rock If u use tet() u need a delimiter like “,”because it takes 2 arguments, so tet(a,b) = a^^b or a^(a^(a…..^(a)))…))) b times, so tet(a+b) makes no sense tbh

[jovial rock]

ah, yeah most of the time I just use e as the base for b^^x, so tet(a+b) = e^^(a+b)

but any base works

[jovial rock]

ah cool

[delicate rivet]

What is the RHS here? fatet(b)? This is really bad notation and probably incorrect too

[jovial rock]

Sorry. So fa(x) is repeated exponentiation with x on top of the tower. fa(x) = tet(slog(x) + a). This just means e^e^e^...^e^x with the exp function repated a times. tet is tetration, fa(tet(b)) = fa(x) = tet(slog(tet(b)) + a). slog and tet cancel and it gets tet(a+b) or e^^(a+b). If you ask me why I used this f, it's just for the name function, I just used a small name to make more compact notations, maybe not very good ones but it is something.
I can prove log_y(x) = tet(slog(f-2(x)-f-2(y)) + 1) if you want:
take the ln on both sides
ln(log_y(x)) = ln(tet(slog(f-2(x)-f-2(y)) + 1)), ln is the same as f-1(x), because the inverse of this iterated function fa(x) ist just f-a(x).
ln(log_y(x)) = f-1(tet(slog(f-2(x)-f-2(y)) + 1)), this simplifies to:
ln(log_y(x)) = tet(slog(f-2(x)-f-2(y))), here tet and slog cancel:
ln(log_y(x)) = f-2(x)-f-2(y), f-2 is just ln(ln(x)) so this gets:
ln(log_y(x)) = ln(ln(x))-ln(ln(y))
Now this is easy to prove using:
ln(ln(x)/ln(y)) = ln(ln(x))-ln(ln(y))

It is not incorret, I thought it thought a lot, I mean it should not be incorret, but if you have any questions about this stuff feel free to ask. I'm not very good with notation and stuff😅

[jovial rock]

btw what means RHS ?

[cosmic breach]

Right hand side

[jovial rock]

thx

[jovial rock]

No idea if this is useful but maybe try to derive something like slog(x) * y
so slog(x) * 2 = slog(x) + slog(x). We can rewrite this as slog(fslog(x)). I'm using [...] here to show a in fa(x). slog(x) * 3 should be slog(f[slog(x) * 2]). So slog(x) * y should be slog(fslog(x) * (y-1)). Using this we can also get slog(x)^2 = slog(x) * slog(x) = slog(fslog(x) * (slog(x)-1)). Now we can say e^^(slog(x)^2) = fslog(x) * (slog(x)-1)

[delicate rivet]

This notation is bad like i have been saying, use ^^ for tetration thats about it. If u want to do e^^(a+b), write e^^(a+b) dont define your own nonstandard functions and then compound them, making them readably obsolete and redundant

[jovial rock]

yeah I've been using ^^ here, I just don't know what to use for fa(x)

fa(x) = e^^(slog(x) + a)

what should I do here then?

I could use expn but yeah

not sure if it's much better

[delicate rivet]

Describe fa(x)

Again

Because e^^(slog(x)+a) makes little sense, its not reducible further

[jovial rock]

soo, we have fa. a is a variable or a base, I don't know how you call it. What it does is that it shows you the amout of times you take the function exp of x. f1(x) = exp(x), f2(x) = exp(exp(x)). f3(x) = exp(exp(exp(x))), I think you get it. The propeties of slog are that slog(e^x) = slog(x) + 1. so slog(e^e^x) = slog(x) + 2. Now if we input our funcrion fa(x) into slog(x) we get slog(fa(x)) and this can be reduced to slog(e^fa-1(x)) = slog(fa-1(x)) + 1, you can continue shifting this: slog(fa-2(x)) + 2, slog(fa-3(x)) + 3 until you get slog(a-a(x)) + a. fa-a(x) = f0(x) and f0(x) is just like not taking the exp function which means it's just the identity so f0(x) = x, so slog(fa(x)) = slog(x) + a. Now take tetration on both sides: e^^slog(fa(x)) = e^^(slog(x) + a). e^^x is the inverse function of slog so this simplifies to fa(x) = e^^(slog(x) + a)

And I have something to my previous thing with applying a function g(x) to slog(x). I found the identities slog(x) * y = slog(fslog(x) * (y-1)) and slog(x)^2 = slog(fslog(x) * (slog(x)-1)), we could rewrite both identities as slog(x) * y = slog(fslog(x) * y - slog(x)) and slog(x)^2 = slog(fslog(x)^2 - slog(x)). Now we see how when slog(x) * y has a * y, that * y appears also in fslog(x) * y - slog(x) and where slog(x)^2 the ^2 appears in fslog(x)^2 - slog(x). The - slog(x) is just a added term. So With this we can say e^^g(slog(x)) = fg(slog(x)) - slog(x) I can not rigorously prove this but it should be clear that this is true after trying some other functions

[delicate rivet]

If slog(e^x) = slog(x) +1, this implies e^^(slog(x) +1) =e^x, that makes sense because e^^(slog(x) +1) = e^(e^^(slog(x)+1-1)) which becomes e^(e^^slog(x)), but by definition e^^slog(x) = x, hence we get e^(x) in the end, so this is just rewriting the definition of tetration

And your “a” is just adding a level of tetration to cancel

So take a=2

This just means slog(e^(e^x)) = slog(x) +2

This is true because of the same reason, e^^(slog(x)+2) = e^(e^x) but e^^(slog(x)+2) = e^(e^^(slog(x)+2-1) = e^(e^x), and if u reduce this by removing e^ on both sides u get e^^(slog(x)+1) = e^x, which is true cuz we just established slog(e^x) = slog(x)+1

So again, no need for the redundant/obsolete fa(x) and g(x), it adds nothing to the analysis and overcomplicates the basic recursive properties of tetration

[jovial rock]

you're right, I mean yeah it is tetration in some way with the a decribing the layers. I just found this useful because f0.5(x) would be the solution of f(f(x)) = e^x. Tetration just calculates the hight and I wanted a function which can take any terms at the top. For example if you find f0.1(x) you could apply it to everything, not only the stuff on the tetration function. for example we only now e^^0. with f0.1(x) we can get f0.1(e^^0) ) = e^^0.1 and f0.1(e^^0.1) = e^^0.2. I get your point with saying it's something with just rewriting tetration but it has it's individual uses for me

[jovial rock]

yeah, I guess I write this down somewhere else so I don't bother you with that stuff. btw g(x) is any function...

[delicate rivet]

This is mainly due to the fact that slog is repeated log or ln

Take e^e^e for instance

Slog(e^e^e) = 3

[jovial rock]

yeah true

[delicate rivet]

But this comes from the fact that u apply the natural logarithms 3 times, to get it from e^e^e to 1

ln(e^(e^e)) = e^e we used 1 ln, then ln(e^e) = e, we used 2 lns, then finally ln(e)=1, we used 3 lns

This is true for all bases, which is why the functional square root of e^x, or any a^x, can be attributed to a^^0.5

What i mean is

Oh wait, nvm

Not a^x but log base a of x

I see what you are saying, but yeah it doesn’t simplify things much because u will still need to solve f(f(x)) = logbase2(x) to reduce the tower’s height by 0.5

If our base is 2**

For tetration i would recommend working with base 2, its much easier to visualize

Log base a of a^^x = a^^(x-1)

[jovial rock]

I mean we don't need to. f(f(x)) = 2^x, then f(1) = e^^(slog(1)+0.5) slog(1) = 0 so e^^0.5. now we can shift this: e^e^^0.5 = e^^1.5 or ln(e^^0.5) = e^^-0.5

[delicate rivet]

Again, this doesn’t help if e^^0.5 is multivalued

[jovial rock]

it isn't

[delicate rivet]

According to who?

[jovial rock]

Kesner, not sure if this is how you write his name

[delicate rivet]

U do realize there are different extensions to real valuesd heights for tetration right….

Kneser*

Kneser’s method is one way to extend tetration

Its not the official way, there is no “official way”

[jovial rock]

So are there methods where it's multivalued?

[delicate rivet]

It’s fundamentally mutlivalued because there exist multiple ways to extend tetration to real heights

That is why its multivalued for non integer heights

[jovial rock]

yeah true, but it would be the same result, the same number right?

[delicate rivet]

No, again, depends on how you extend it

[jovial rock]

I only know that (-1)^^x for non integers is really multivalued

[delicate rivet]

Not sure about that chief

[jovial rock]

the rest yeah, I just assume and I cannot prove

[delicate rivet]

If u think about it

[jovial rock]

me neither but according to my research I think that

[delicate rivet]

(-1)^^0 can technically be either 1 or -1

And (-1)^^(-1) can be either 0 or -1 as well

It doesn’t break the recursion

[jovial rock]

oh

log base -1 of -1 is 1 right?

everything is multivalued there I think

even integers

or idk

there are probably just infinite branches and (-1)^(-1) has a lot more solutions then just -1

[delicate rivet]

Because think about it, (-1)^^(1) = (-1)^(-1^^0) right? Well since we know (-1)^^1 = -1, then if (-1)^^0 was negative 1 instead of positive 1, it still holds that (-1)^(-1) =-1

[delicate rivet]

Not necessarily

[delicate rivet]

Log base -1 of -1 can be -1 as well!

[jovial rock]

yeahh

[delicate rivet]

Because (-1)^(-1) =-1

[jovial rock]

it's mutlivalued

[delicate rivet]

Right

[jovial rock]

yess

(-1)^^x is messed up

[delicate rivet]

In my research i found that 0^^(x) can be extended by (1+cos(x*pi))/2

If we allow 0^0 =1

[jovial rock]

yeah me too

I also made a meme for that one

[delicate rivet]

But yeah its wrong 🤣

Because the recursion breaks

[jovial rock]

🙂

[delicate rivet]

F(x+1) = 0^(F(x)) will no longer be true

[jovial rock]

yeah true

[delicate rivet]

Yeah but neat correlation for intrgrr heights

I was gonna make a clickbaity youtube video saying 0^^i is approximately 2pi

Because using our equation if u plug x=i we get cosh^2(pi/2) or something

Basically the value is really close to tau

[jovial rock]

yeah, bases like 0 or below bases are just a bit confusing and there are just to many things you need to look for without breaking math rules

[jovial rock]

xD, yeah great Idea, there are not really alot of videos about that kind of tetration

[delicate rivet]

Idw plug myself but i make videos on tetration on YouTube (only 2 vids on tetration so far💀)

If u search derivative of tetration u should see a video by darkmaths (or how to differentiate tetration)

Plan to make atleast 2 more in the coming few weeks

[jovial rock]

Ah cool, nice videos. Some things were a bit confusing for me but all the stuff you explained was very interesting

[azure crane]

oh wow a lotta new stuff has been said lol

[delicate rivet]

Yeah, the main point of it all was to stop using bad mathematical notation when a simple one suffices

[delicate rivet]

This was cancerous to read, but it seems op has the fundamental grasp on tetration down

[delicate rivet]

Also tbis is true, log base -1 of -1 can be: 3 or -3 as well, but usually we look at the principle branch for the multivalued stuff

[queen ledge]

There were already attempts at extending tetration to real and complex values

But really they were just mockeries because there is a fundamental problem with this question

Look them up

[delicate rivet]

Kneser’s method is actually quite robust for fractional heights, the issue is also that i can extend tetration to fractjonal heights with ease, and it won’t violate any rules (it will just be less unique and trash)

Here is one method:

We know a^0 = a^^0

And a^1 =a^^1 = a

So what’s stoping me from saying a^x = a^^x for all x ∈ [0,1]

And now using the recursion rule that a^(a^^x) = a^^(x+1), we can extend tetration to any fractional height (i could be completely wrong abt this🥲🤣)

[queen ledge]

You could, surely, but the problem is that you can extend it in many ways, not just this way

What makes this way better than, say, a^^x = (a-1)x + 1 for all x in [0, 1]?

Sure, it's uglier, but so what?

[delicate rivet]

Actually after some robust research, my method is wrong

[delicate rivet]

f(0.5+0.5) needs to be f(f(0.5))

With my method, this would mean f(1) = 2^^(1) = f(f(0.5)) but 2^(2^0.5) = 2^(sqrt(2)) =2.665 , but we know 2^^(1) =2 and 2!=2.665

[delicate rivet]

And your method would be wrong for a similar reason

[queen ledge]

Where did you get that from

[delicate rivet]

From the recursive definition of tetration, f(2) = 2^^(2) , f(1+1) = f(f(1)) = and we know f(1) = 2, hence f(1+1) = f(f(1)), similarly, f(1+2) = f(2+1) = f(f(f(1))) # (f(x) = 2^^x btw)

supposed to be a comment like in python💀🙏

Bruh

this makes text big wtf the # symbol

[delicate rivet]

Oh wait i might have confused f(0.5+0.5) with f(1+0.5)

No actually I’m right because in tetration each fractional operator adds exactly that fraction of a tower. We call the half tower operator f_0.5 and the full tower operator f_1, To get a full tower you must apply f_0.5 twice to get a full tower

[queen ledge]

What is f(x)?

[cosmic breach]

-# smol

[delicate rivet]

f(x) there is f_0.5(x) the half iterate, a function such that applying it twice gives 2^^x or f(f(x)) = 2^^x

[queen ledge]

Ah

I still don't see why f(1+1) should be f(f(1))

[delicate rivet]

Oh yeah u’re right

Hmmm i was confusing semigroups with iterated functions there

Oh

Wait

[delicate rivet]

If f(x) = 2^^(x) = 2^x between 0 and 1, this means f_0.5(x) = g(x) such that g(g(x)) = f(x), this neccessitates g(x) = 2^^(slog2(2^^(slog2(x)+0.5)) +0.5)), testing at boundaries g(0), this give g(0) equals one, (slog2(0) =-1 , -1+0.5 =-0.5, slog2(2^^(-05)) =-0.5 by definition, this means g(0) = 2^^(-0.5+0.5), = 2^^0 = 2^0, but we need g(g(0)) = 2^^0, not g(0) = 2^^0, this new g(g(0)) = g(1) which is 2, but again, g(g(1)) should be f(1) = 2, not g(1), that’s why my extension is wrong, probably similar working for your extension

Why should g(x) = 2^^(slog2(2^^(slog(x)+0.5))+0.5)? Because this literally simplifies to 2^^(slog(x)+0.5+0.5) = 2^^(slog(x)+1), and this is 2^(2^^(slog(x)+1-1)) = 2^(x), which was the proposed extension between 0 and 1

Also i took long to type this cuz im playing cs2 rn

[karmic lava]

it's like in jupyter markdown

#(with a space behind it) makes the text bigger, ## works too

Header 1

Header 2

Header 3

[delicate rivet]

U really made that math go 1 space above to show how headers work…

Oh well i pinged john via replying he’ll see it

Oh im dumb

G(x) is just 2^^(slog2(x) + 0.5)

Cuz it satisfies G(G(x)) = f(x)

[karmic lava]

It's useful for formatting, sometimes

[delicate rivet]

Damn after all this i actually cannot say why our methods are contradictory, f(1+1) = f(f(1)) is not a requirement and i wrongly equated them. I guess the issue just boils down to the non-uniqueness, but i stand corrected, it is still a valid extension that satisfies rules of tetration

Good convo, brain is toasted cuz i confused g(x) with g(g(x))

https://tenor.com/view/drake-embarrassing-awkward-side-eye-gif-9668135525517859023

[azure crane]

so $f(x)=\log^{4}_{(2,-0.5)}(x)$

[dawn finch]

RoyalBanana

[delicate rivet]

um what is that

unless $\log^{4}_{(2,-0.5)}(\log^{4}_{(2,-0.5)}(x)) = 2\uparrow\uparrow x$ (it doesn't) this is wrong

[dawn finch]

RamDejin

[queen ledge]

You should research tetration first

Thoroughly

To see if there are any useful properties

[delicate rivet]

Trust me, i have done the necessary research

[delicate rivet]

And the answer remains no

Currently got a paper on inverse tetration on hold on arxiv

💀💀😭

[queen ledge]

You've never done enough research, there is always more to do

[delicate rivet]

There’s even more research to do when the subject is actually useful😭

[queen ledge]

Brother

Forget about applications

[delicate rivet]

For almost 1.5 weeks now btw 😭😭 not looking fly money

[queen ledge]

JUST???

JUST 1.5 WEEKS??

[delicate rivet]

Wdym

My tower of hanoi paper got accepted in like

2-3 days

Tops

[queen ledge]

You're living your dream life and you don't even know about it

[delicate rivet]

😭⁉️

I deffo showed it to u i was geeked with joy when my first one got approved

[queen ledge]

We submitted our paper in September, near November or so it was accepted and it is May and it is still not published

IT IS MAY

[delicate rivet]

Arxiv?

[queen ledge]

No, another journal

[delicate rivet]

Oh

[delicate rivet]

I didn’t submit to a journal yet (cap i did but, i decided not to publish it there)

Scirp.org’s AJCM accepted my inverse tetration paper and my tower of hanoi one

But

$599 article processing fee per paper

💀💀💀💀

And they got a reputation for being predatory

So it will probably hurt my credibility

[queen ledge]

Honestly I don't know how much we paid for our paper lol

[delicate rivet]

Would u say its in a similar ball park?

600 benji’s for one paper

[queen ledge]

I mean, we're using different currency

But I think a bit less

[delicate rivet]

Makes sense, thanks for that info

[azure crane]

Wat ? You didn't read what we were working on?

[azure crane]

It does lol

[delicate rivet]

Yeah, no.

But always try again!

[azure crane]

Uh, yes

??

wait

weren't you banned

[delicate rivet]

Yeah, and weren’t you upvoting that when it happened

[delicate rivet]

So u can take ur pseudomath and shove it where the sun don’t shine 🙂

[azure crane]

It's not pseudomath😭

[delicate rivet]

Then do the math, instead of giving a terribly formatted math equation and expecting it to make sense😭

[azure crane]

It's not terribly formatted you just haven't learned it yet lol

[delicate rivet]

“you just haven’t learned it yet” thank god i avoided the crank school of crankamatics

[azure crane]

do you know H notation

[azure crane]

? no we invented it lol

[delicate rivet]

Nah, what does that have to do with a dual base argument for the logarithm 😭😭😭

[azure crane]

it probably already exists its just how we notated it

[delicate rivet]

Yeah, okay buddy, and our friend in the other discussion “invented” eidometry, doesn’t make it not crankamatics

[delicate rivet]

If you read ur original equation, its not base a but 2 arguments for the base

And H(n,a,b) is already reserved to show “Hyperoperators” 😭

[trail surge]

How can you say if you literally haven't read all that is there though about what the person in the thread is trying

[azure crane]

Yes that's what it means in that equation

[azure crane]

the second number means how many times the log is done

it's a bit of a mess but it worked

[delicate rivet]

Log^1 base a of (H(1,a,b)) = logbase a of (a+b) and idk who taught you math but

$log_{a}(a+b) \neq b$

Then it’s gibberish notation, log^1 in functional notation either means log applied once or log to the power one, and their both equal to log

[azure crane]

It's different notation

[delicate rivet]

Yeah, trash notation.

Definitely different

[dawn finch]

RamDejin

[queen ledge]

Well, one thing you could try to do is to find an alternative formula for a^^x for some specific non-trivial a

Because how is a^b extended to reals? You get the identity e^(b ln a) and both can be computed for reals

Actually you'd still have the non-uniquity

[delicate rivet]

Yeah, cuz exponentiation is non commutative

[queen ledge]

Does that mean that exponentiation is unique?

[delicate rivet]

The reason exponentiation is easy to extend is because multiplication is commutative

And i can explain how

[queen ledge]

Don't care

[delicate rivet]

Knew it

So wanted to lyk that i could

So consider 2 times 3.5, this can be written in terms of addition as 2(1) +2(1) + 2(1) + 2(0.5)

[queen ledge]

Oh

[delicate rivet]

Exponentiation, say 2^(3.5) can be written as 2^(1) times 2^(1) times 2^(1) times 2^(0.5), in the last two examples, the order of each 2(something) or 2^(something didnt matter)

With tetration however, 2^^(3.5) can be interpreted as (2^^0.5)^(2^^1)^(2^^1)^(2^^1) but depending on where 2^^(0.5) is placed in the tower u get a different value for 2^^(3.5)

[queen ledge]

What?

[delicate rivet]

Wait so there’s a unique extension?

[queen ledge]

Assuming some stuff

[delicate rivet]

Like, im trying to say that when u take 2 times 3.5, u can rewrite as 2(1)+2(1)+2(1)+2(0.5) and same for exponentiation 2^(3.5) = 2^(1) x 2^(1) .. x 2^(0.5) but because addition and multiplication are commutative, the order of where 2(0.5) is placed in the sum or 2^(0.5) is placed in the product is irrelevant because addition and multiplication are commutative

That won’t apply for tetration anymore

U’ll get a unique value based on where u place the 2^^(0.5) in the tower

If u wanted to compute 2^^(3.5)

[queen ledge]

I'd say it'd be reasonable to always place it in the end. It would be well defined to define f(a + b) = exp^a(f(b)), where a is an integer and b is in [0, 1)

[delicate rivet]

Just a thought i had, but ig if we fix 2^(fractional height) to be the last term in the sequence that can be resolved so i am just kinda yapling

Yapping*

[delicate rivet]

Yeah true, my thoughts exactly, and u can just have it be the last term in the sum and product examples

For consistency

Yeah for the longest time it felt like it was exponentiations non commutativity that made it impossible to extend tetration ti all real heights but ay go kneser

[queen ledge]

How did you not even check Wikipedia

[delicate rivet]

I did

I know the website that uses this method

By arkansas state

But i didn’t know its the only unique way

[queen ledge]

There is at least one thing in common between you and the concept of love

You're both blind

[delicate rivet]

Also, in the official university website they say “We have an argument of why this should be the official tetration” 💀💀

Implying there are other methods 😭

http://myweb.astate.edu/wpaulsen/tetcalc/tetcalc.html

Like i said bro, i have done my research

[queen ledge]

What are the other methods?

[delicate rivet]

Any of the previous methods we discussed that follows F(x) = b^F(x-1)……

And f(0),f(-1) = 1,0

There is a technique called regular iteration with abel’s function

[queen ledge]

Oh, right

Abel, son of Adam

Forgot he was a mathematician

[delicate rivet]

U’ve heard of “Abelian” groups right?

[queen ledge]

I am a second year student

[delicate rivet]

Abelian is not a person if you r confused

Undergraduate?

[queen ledge]

Yeah

Wait

[delicate rivet]

Math major?

[queen ledge]

Yeah, math major

[delicate rivet]

Fire

Wait what

[queen ledge]

I will finish bachelor in 2 years

Anyways

[delicate rivet]

I did some googling and the abel iteration method is actually called schroder iteration method💀

So yeah i didn’t get son of adam wrong

[queen ledge]

Very wild assumption

[delicate rivet]

Yeah and then they average that

To cancel out imaginary parts

Average of L and L* is the kneser method for real valued tetration

[queen ledge]

Oooh this paper is actually not that hard to read

But I have other stuff to do

[delicate rivet]

Ciao

[queen ledge]

Also not exactly I think

There's contour integration involved

[delicate rivet]

Right, but i do know the reason we use conjugates is to cancel the imaginary parts

So idk if they average that or sum it a different way

[queen ledge]

Read the damn thing

[delicate rivet]

So yeah, like i was saying, they merge the normal the fixed point L and it’s conjugate to generate real valued solutions……

[azure crane]

not really

[delicate rivet]

Yes really

[delicate rivet]

No wonder it’s non-standard and something “you invented”💀

Remember when u said log^1 is not log “lol”

💀💀💀

[azure crane]

that's true-

you know when creating notation you only have so much choice right? Duplicate notation happens all the time

[delicate rivet]

Or just use existing notation… especially if you’re “new notation” adds nothing to the analysis

[azure crane]

or just don't hassle someone over notation...

[delicate rivet]

And if i humor this for a second, log^1 base a(y) according to your notation is equivalent to y-a

[delicate rivet]

It’s not hassling you if you can’t accept the truth and just get sensitive over it being bad notation 😭

[delicate rivet]

Figured, still bad notation

Just use inverse hyperoperators at that point

[delicate rivet]

Also your notation btw is also just a worse version of $H(-n,H(n,a,b),b) = a$

[dawn finch]

RamDejin

[azure crane]

Is all you do complain?

and pressure?

[delicate rivet]

R u feeling pressured because someone called out terrible notation?

[azure crane]

No, I'm feeling pressured because you're pressuring me about it

[delicate rivet]

Because hopefully it’ll just prevent you from making the same mistake🤷🏽‍♂️

[dire scaffold]

<@&775784618955505685>

[delicate rivet]

Equivalently you could do $H(-n,a,H(n,a,b)) = b$

[dawn finch]

RamDejin

[real verge]

whT

[dire scaffold]

Hi mega can you keep an eye on this convo? It's teetering on the edge of needing a cool down

[delicate rivet]

Bruh

[real verge]

i can keep an eye for a while, but its close to lunch time, i'll be called anytime

[delicate rivet]

Don’t worry, im leaving this convo, i made my point and if it hurts anybody im sorry

[real verge]

@dire scaffold give ncladus perks to send messages in this channel, he cant rn

[real verge]

we never said u hurt anybody

¯_(ツ)_/¯

[delicate rivet]

Pressured* but okay

Not u the other guy

[dire scaffold]

Alr I believe it is done

@vocal wolf how does it look

[vocal wolf]

does 👍

thanks

[dire scaffold]

Sweet deal

@delicate rivet I think it's fine to have disagreements about notation, I was moreso concerned about the way in which both you and @azure crane were interacting that's all. I don't think anyone did anything against the rules but it was getting close to someone breaking #1

[queen ledge]

These managers smh

[delicate rivet]

Does the og author need to close the thread

How do i remove this from my list of discussions

[queen ledge]

Right click in channel tab > Unfollow post

[jovial rock]

every notation was invented one day...

what happened here...

so now we can use "bad" notations again: )

yay

[delicate rivet]

Go crazy! 🤪

[jovial rock]

Integral(-2,0,2^^x) = -2 * Integral(0,2,f0.5(x))

[delicate rivet]

I like this a lil more

Nvm i m missing an ln(n) term

[jovial rock]

btw not true, just wrote some nonsense

maybe functional equation

solve for f(x) where f0.5(x) is the functional squareroot

idk

[jovial rock]

Here is a code I made one time for calculating tetration. Nothing special, just the taylor polynomial. Could be good for approximating Integrals

[jovial rock]

for base e btw

[queen ledge]

Where did you get taylor series for that?

[jovial rock]

https://en.citizendium.org/wiki/tetration

[queen ledge]

Oh

Ok

[delicate rivet]

This looks wrong

It shouldn’t be smooth on the negative side

Oops nvm

[jovial rock]

hmm

yeah true

looks wrong

[proud solstice]

Hello

[jovial rock]

hi

[jovial rock]

Nothing Special but I thought this it interesting:
e^x = e^^(slog(x)+1), so 2^x = e^(x+ln(2)) = e^^(slog(x * ln(2))+1)
And this works for every base:
e^^(slog(x * ln(a))+1) = a^x

[delicate rivet]

That first line is wrong

$2^{x} \ne e^{x+ln(2)}$

[dawn finch]

ReverseDash

[delicate rivet]

$2^{x} = e^{xln(2)}$

[dawn finch]

ReverseDash

[karmic lava]

that's a very obvious thing, no?

[delicate rivet]

Yeah especially when OP simplifies the xln(a) part to be just ln(a^x) lol

[jovial rock]

Uh yeah, idk how I got that +, probably missclicked

[jovial rock]

Idk if this is something new but: when g(x) = (x^^(h) -1)/h where h approaches 0 and f(x) = (x^^(1+h) - x)/h where h approaches 0, then g(x) * x*ln(x) = f(x)

g(x) is just a function that gives you the Derivate of tetration at 0, with the base x

I was wondering if g(x) can be represented with elementary functions or/and simple integrals

It looks similar to (x^h -1)/h where h approaches 0, and the function is just ln(x)

This could be useful because x^0 = x^^0

Btw I already calculated g(2) = 0.889364955, g(e) = 1.091767351, g(10) = 1.684289028
Using the tetration calculator from the web

[jovial rock]

And g(1) should be 0, because 1^^x is constant at 1

This looks similar to the logarithm or super logarithm, maybe there are connections there

I though maybe it could be super logarithm with be base x, where x is the solution to g(x) = 1

But not sure, just an idea

[jovial rock]

A quick Definition to make Notation easier: x^^h = T_x(h)
So slog_x(T_x(h)) = h. Taking the derivative on both sides: d/dh(slog_x(T_x(h))) = slog_x‘(T_x(h)) * T_x‘(h), d/dh(h) = 1. so slog_x‘(T_x(h)) * T_x‘(h) = 1. Setting h = 0 we get slog_x‘(1) * T_x‘(0) = 1. T_x‘(0) is just g(x), because it takes the derivative of tetration at base x so we get
slog_x‘(1) * g(x) = 1,
g(x) = 1/slog_x‘(1)

[jovial rock]

Now we Can easily find slog_x‘(0) with slog_x‘(0) = ln(x)/g(x) or ln(x)/T_x‘(0)
explanation:
T_x(h+1) = x^(T_x(h))
T_x‘(h+1) = x^(T_x(h)) * ln(x) * T_x‘(h) (d/dh, not d/dx)
h = -1 —> T_x‘(0) = ln(x) * T_x‘(-1)

Going back to slog_x‘(T_x(h)) * T_x‘(h) = 1
Setting h = -1:
slog_x‘(T_x(-1)) * T_x‘(-1) = 1, slog_x‘(0) * T_x‘(-1) = 1
Remember: T_x‘(0) = ln(x) * T_x‘(-1)
T_x‘(0)/ln(x)= T_x‘(-1)
So: slog_x‘(0) * T_x‘(0)/ln(x) = 1
slog_x‘(0) = ln(x)/T_x‘(0)
using my calculated values for g(x):
slog_2‘(0) = 0.779 373 166
slog_e‘(0) = 0.915 946 057
slog_10‘(0) = 1.367 096 178

[jovial rock]

Btw with this we can get:
slog_x‘(0)/slog_y‘(0) = log_y(x) * T_y‘(0)/T_x‘(0)
so if we say slog_e(x) is lns(x):
slog_x‘(0)/lns‘(0) = ln(x) * T_e‘(0)/T_x‘(0)

[jovial rock]

So T_x‘(-1) = (x^^(h-1))/h where h approaches 0 and T_x‘(0) = (x^^h - 1)/h where h approaches 0
We know that T_x’(-1) * ln(x) = T_x’(0)
So: (T_x’(0) = (x^^(h-1) * ln(x))/h where h approaches 0. We can change that to: (x^^(h-1) * (x^h - 1)/h^2
maybe this could be interesting.

[jovial rock]

slog_x(x^h) = slog_x(h) + 1, taking the derivative on both sides:
slog_x‘(x^h)*x^h * ln(x) = slog_x‘(h)
Now if we set x = e and h = 0 we get:
slog_e’(1) * 1 * 1 = slog_e’(0).
Is this a contradiction?
We Can See that it is a Identity by taking the identities we had before:
T_x’(0) = 1/slog_x’(1)
slog_x’(0) = ln(x)/T_x’(0)

rearranging: slog_x’(1) = 1/T_x’(0)
with slog_x’(0) = ln(x)/T_x’(0) we see the ln(x) term. If we input e for x we would get exactly 1 and we would get 1/T_x’(0) which is the same as slog_x’(0) and shows that slog_e‘(1) = slog_e‘(0). This only holds for the base e, and this is what gives the constant e a special place in tetration for me```

So: slog_x’(0)/slog_x’(1) = ln(x)

[karmic lava]

please format you text using ` three times (at the beginning and end)

some of your underscores disappeared

because ff discord's formatting

underscore

_underscore_

[jovial rock]

Of the whole message?

Like this?

[karmic lava]

Doesn't really matter if it's just the maths or the text too

[jovial rock]

Okay 👍

[jovial rock]

Btw: if you take the 2nd derivative and set x = e and h = 0 and simplify you get:
slog_e’’(0) - slog_e’’(1) = slog_e’(1)
or
slog_e’’(0) - slog_e’’(1) = slog_e’(0)

[jovial rock]

Following slog_x‘(h) = ln(x) * x^h * slog_x‘(x^h)
We get:
slog_x‘(-1) = ln(x)/x * slog_x(1/x)
And with x = e:

slog_e‘(-1) = 1/e * slog_e‘(1/e)

[jovial rock]

We have slog_x’(h) = ln(x)*x^h*slog_x’(x^h). So we need to find h so that x^-h = ln(x), because we this would cancel with the ln(x).
Solving this gives us h = -(ln(ln(x))/ln(x)
so:
slog_x’(-(ln(ln(x))/ln(x)) = slog_x’(x^(-(ln(ln(x))/ln(x)))

this just shows how for every base there is a Part where the gradient is the same as the known part```

[jovial rock]

For example slog_2‘(0.52876637…) = slog_2‘(1.442695…)

Or: slog_e^(W(2)/2)(2) = slog_e^(W(2)/2)(e^W(2))
The base is e^(W(2)/2)

[karmic lava]

By the way, what is slog_x() ? is it some number s * log_x() ?

[jovial rock]

No, it’s actually the superlogarithm, the inverse of tetration. It’s defined as: slog_b(b^x) = slog_b(x) + 1

[karmic lava]

ah, makes sense

[jovial rock]

slog_2(2^2) = slog_2(2) + 1 = 2
slog_e(2^2) = slog_e(e^(2*ln(2))) = slog_e(2*ln(2)) + 1```

[jovial rock]

But I’m mostly using x for b and h for x in my calculations above because I’m more used to h, h for height

[karmic lava]

Doesn't that make it less readable for others?

Nevermind

[jovial rock]

Maybe, but I think I shouldn’t really, because the variables are at the same place and they are just placeholders for numbers

[jovial rock]

Let’s start from the start:

slog_x’(T_x(h)) * T_x’(h) = 1   h = 1
slog_x’(x) * T_x’(1) = 1

slog_x‘(h) = ln(x) * x^h * slog_x‘(x^h)
slog_x‘(x^h) = ln(x)^-1 * x^-h * slog_x‘(h)

ln(x)^-1 * x^-1 * slog_x’(1) * T_x’(1) = 1
ln(x)^-2 * x^-1 * slog_x‘(0) * T_x‘(1) = 1
T_x‘(1) = (ln(x)^2 * x * 1)/slog_x‘(0)

h = 2
ln(x)^-1 * x^-x * slog_x’(x) * T_x’(2) = 1
ln(x)^-2 * x^-x * x^-1 * slog_x’(1) * T_x’(2) = 1
ln(x)^-3 * x^-x * x^-1 * slog_x’(0) * T_x’(2) = 1
T_x‘(2) = (ln(x)^3 * x^x * x)/slog_x‘(0)

If we continue like this we can see how to product 1 * x * x^x * x^x^x is emerging.
We can now isolate this and get:
(T_x’(h) * slog_x’(0))/(ln(x)^(h+1)) = Product(h,n‎ = 0, T_x(n))
For example let’s find the product from T_2(0) to T_2(4) using this formula:
First using the tetration calc from the web we get T_2’(4) = 1722153.09779. slog_2‘(0)I already calculated in one of my calculations above: 0.779373166. Now we just have to divide by ln(2)^5 and we get: T_2‘(4) * slog_2‘(0) * ln(2)^-5 = 8388608.010838
The expected solution: 2 * 4 * 16 * 65536 =8’388’608
Works pretty well, the values of slog_2’(0) and T_2’(4) weren’t precise enough but it works.
At first glance this looks a bit stupid to use complicated numbers instead of just multiplying these normal numbers, but this could be useful when we have a very long multiplication of power towers but a small base too so the result doesn’t instantly explode. We could just find the derivative very far and calculate the whole product instantly.
And slog_x’(0) we can easily get from T_x’(0) using the formula:
slog_x’(0) = ln(x)/T_x’(0)
If we rearrange it so: 1/T_x’(0) = slog_x’(0)/ln(x)
And replace it in
(T_x’(h) * slog_x’(0))/(ln(x)^(h+1)) = Product(h,n‎ = 0, T_x(n)) 
to get:
(T_x’(h))/(ln(x)^h * T_x’(0)) = Product(h,n‎ = 0, T_x(n)) 
Which could make things easier```

[jovial rock]

Useful paper

[jovial rock]

This is the first version of the Tetration Calculator I'm developing. It works okay, but doesn't have the best precision yet. It gets 5 digits right with 10^^(1/2) and gets up to 10 digits right with 2^^(1/2). So the bigger the base, the less digits it gets right. I'm working on extending it to heights smaller than -2, and I'll improve some things later probably. 2^^(-2.5) for example is about: −0.19050915414263895125508662432599 + pi/ln(2) * i There is pi/ln(2) * i, because log_2(negative number) = some real number + pi/ln(2) * i

Just type in the terminal: python TetrationCalculator_v1.py 2 0.5 --size 30 --dps 60
Where the first number is the base and the second the height. Only bases over e^(1/e) work good, bases below get tricky.
You may need to pip install mpmath but I'm not sure

I wouldn't recommend changing the size of the carleman-matrix because sometimes the precision gets worse when increasing N.

btw, most of this is made with the new ChatGPT agent mode, because it's really hard for me to wrap my mind around these carleman-matrix things and the other methods

[jovial rock]

Would be really cool to learn These things by myself, but idk how. I couldn‘t find anything on YouTube to the carleman-matrix and other things

Tf

Someone explain me how this is Crank

[karmic lava]

See it as a badge of honour ig

[jovial rock]

Uh, okayy. I guess this server isn‘t really into tetration😅

[jovial rock]

I guess I‘ll create a server specialized to tetration and continue there researching tetration, isn’t much to do here I guess haha

[cosmic breach]

Mostly because of the title

Also no one knows what you’re doing here atp

[delicate rivet]

Untrue

[delicate rivet]

So what does your tetration calculator output for 2^^(1/2) btw

Because if it outputs ssrt(2) or 1.559610 then it is probably wrong

@jovial rock Which is why kneser took the complex and conjugate branch merge idea to extend tetration to all reals

Well, (-2,inf] technically cuz there is a branch cut at height = -2

[delicate rivet]

ssrt() being the super square root analogous to slog() which is the superlogarithm as you’ve described

ssrt(a) = x—>a= x^x

[jovial rock]

Truee, it hurts inside me whenever someone tells me ssrt(2) ist equal to 2^^(1/2)

[jovial rock]

I think it has like a 7-10 digit accuracy with base 2, I need to check rn

1.458781816

9 digits I think

With the 1 at the start 10 digits

But for higher bases it only has like 4-5 digits accuracy

I‘m trying to figure out how to construct the Cross Track method from paulsen, because I think it gives like 50 digits accuracy

[jovial rock]

Yess

[jovial rock]

Not 100% sure but I think this is a series expasion for Tetration. L ist the fixedpoint e^L = L. A is schröders function for tetration at 1, where schröders equation is f(e^z) = L*f(z) where L ist still the fixedpoint (the series expansion is in picture 2). Still working on this but I thought this could be a inspiring idea to expand tetration into a series, symplify it and find new identities and stuff

this is all base e, cuz e is easy to work with

[queen ledge]

What a mess

[karmic lava]

What's nL^a_b

[delicate rivet]

Agreed, now it’s really getting crank territory imo

[karmic lava]

I think he really likes tetration

I believe it's not only important to take steps forward, but to also take a step back and rework/revisit older concepts and ideas, clean up notation, etc.