General Formula for all arithmetic operations

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

As long as f is continuous

it's interesting it works on the interval where f is continuous

like, if f is discontinuous at x=3 and x=-5, it works for -5<x<3

[azure crane]

Oops typo

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

[dawn finch]

RoyalBanana

[azure crane]

There's a more generalized version btw if you want

$f(x)=\sum_{n=0}^{\infty} f^{(n)}(a)\frac{{(x-a)}^n}{n!}$

[dawn finch]

RoyalBanana

[azure crane]

for any constant a

[jovial rock]

Yeah

[azure crane]

Hmm it's working for the basic sum one so far
f(x+1)=f(x)+g(x)

[azure crane]

I was using $\lim_{N \to \infty} f(N+1)-f(N)$ to approximate the derivative

[dawn finch]

RoyalBanana

[azure crane]

Since $\lim_{N \to \infty} f(N+x)-f(N)=0$

[dawn finch]

RoyalBanana

[azure crane]

that means pretty much any +x doesn't matter for big N

so the difference between the derivative and that is very small

it makes sense if you think about the graph also

since it flattens out

[jovial rock]

Yeah, right

Hmm

So if a function grows faster than e^x, that means that the derivative grows faster

So for functions stronger than e^x, we could use the anti derivative, the integral

Couldn’t we use the integral of some fast growing function to stabelize the growth?

[azure crane]

I mean you could just do the trick thingy

That I did

Instead of $f(x+1)=g(f(x))$ you do $f(x+1)=1/g(1/f(x))$

[dawn finch]

RoyalBanana

[jovial rock]

Maybe this could be helpful

I think this Converges to 0

Or infinity

Not sure

[jovial rock]

I tried to take the ln on both sides and defined Sk(x) = ln(d/dx lnk(x))

Idk if this is more stable

[azure crane]

The right side simplifies to $\frac{1}{\prod_{m=0}^{k-1}ln_m(x)}$

[dawn finch]

RoyalBanana

[azure crane]

That simplifies to $S_k(x)=-\sum_{m=1}^{k} ln_k(x)$

[dawn finch]

RoyalBanana

[azure crane]

Nice!

I think imma just redefine S

$S_k(x)=\sum_{m=1}^{k} ln_m(x)$

So $S_k(x)=-ln(ln_k'(x))$

[dawn finch]

RoyalBanana

[azure crane]

Ooooh

Hold on

[dawn finch]

RoyalBanana

[jovial rock]

Uhh

Crazy

[jovial rock]

The -m cancels out?

[azure crane]

$S_k(x)=\lim_{N \to \infty} \sum_{m=1}^{N} ln_m(x)-ln_{m+x}(x)$

[dawn finch]

RoyalBanana

[azure crane]

I just swapped k for m-k

Thats a sum thing you can do that

you have to switch the top and bottom numbers of the sum too

after

so the bigger one is on top

[jovial rock]

Okay

[azure crane]

So then I got $-\sum_{m=0}^{k-1} ln_{k+1}(x)$

Which simplifies to

[dawn finch]

RoyalBanana

[jovial rock]

Ah I understand

[azure crane]

Which simplifies to $-\sum_{m=1}^{k} ln_{k}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Nicee

[azure crane]

I think

using the f(x+1)=f(x)+g(x) thingy

oh wait typo

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

[dawn finch]

RoyalBanana

[azure crane]

Perfect

see now

[jovial rock]

Cool

[azure crane]

We can use this to get $ln_{1/2}(x)$ or anything

[dawn finch]

RoyalBanana

[jovial rock]

But how?

[azure crane]

Just plug in a non integer value for k

We already have another definition for Sk(x)

[jovial rock]

Yeah true

Can we substitute it back to the original Definition of lnk?

[azure crane]

$-ln(ln_k'(x))=\lim_{N \to \infty} \sum_{m=1}^{N} ln_m(x)-ln_{m+k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Hmm

[jovial rock]

We would need to take the antiderivative

[azure crane]

yeah

But what's the antiderivative of lnk(x)?

antiderivative of ln(x) is xlnx-x

[jovial rock]

Hmm

[azure crane]

Aw man

It needs one of those functions

Like W

That can't be expressed normally

I used Wolfram alpha to see

[jovial rock]

I was messing around with chatgpt but I know that chat got does a lot of wrong things:

[azure crane]

antiderivative of ln(ln(x))=xln(ln(x))-li(x)

[azure crane]

Doesn't work for k=1 lol

[jovial rock]

Okay, yeah it probably doesn’t work

[azure crane]

:/

[azure crane]

^

[azure crane]

Ok so maybe we can get something useful out of this some other way

[jovial rock]

We just need to solve for this right?

[jovial rock]

Maybe there is a pattern

[jovial rock]

Yeah probably

[azure crane]

When I tried ln(ln(ln(x))) it broke Wolfram alpha lmao

[jovial rock]

Ohhh, xD

[jovial rock]

Yo mean ln, not li right?

[azure crane]

No

[jovial rock]

Uh

[azure crane]

^

[jovial rock]

What is li?

[azure crane]

idk

[jovial rock]

Huh

[azure crane]

Probably some integral of something that can't be expressed normally

[jovial rock]

I see, I got the same thing with geogebra

[azure crane]

Let's do that

$-ln(ln_k'(x))=\lim_{N \to \infty} \sum_{m=1}^{N} ln_m(x)-ln_{m+k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

$ln(ln_k'(x))=\lim_{N \to \infty} \sum_{m=1}^{N} ln_{m+k}(x)-ln_m(x)$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=e^{\lim_{N \to \infty} \sum_{m=1}^{N} ln_{m+k}(x)-ln_m(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=\lim_{N \to \infty}e^{ \sum_{m=1}^{N} ln_{m+k}(x)-ln_m(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=\lim_{N \to \infty} \prod_{m=1}^{N} e^{ln_{m+k}(x)-ln_m(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=\lim_{N \to \infty} \prod_{m=1}^{N} \frac{e^{ln_{m+k}(x)}}{e^{ln_m(x)}}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=\lim_{N \to \infty} \prod_{m=1}^{N} \frac{ln_{m+k-1}(x)}{ln_{m-1}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k'(x)=\lim_{N \to \infty} \prod_{m=0}^{N-1} \frac{ln_{m+k}(x)}{ln_{m}(x)}$

[jovial rock]

Nicee

[dawn finch]

RoyalBanana

[azure crane]

Ofc here we can just swap N-1 for N

[jovial rock]

Yeah

[azure crane]

$ln_k'(x)=\lim_{N \to \infty} \prod_{m=0}^{N} \frac{ln_{m+k}(x)}{ln_{m}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

Interesting

[jovial rock]

That’s way more compact

[azure crane]

$ln_k'(x)=\prod_{m=0}^{\infty} \frac{ln_{m+k}(x)}{ln_{m}(x)}$

[dawn finch]

RoyalBanana

[jovial rock]

Yess

[azure crane]

Say

I wonder

Using that formula $\frac{ln_{k+1}'(x)}{ln_{k}'(x)}=...$

[dawn finch]

RoyalBanana

[azure crane]

The ln_m(x)s cancel

in the fraction

$\frac{ln_{k+1}'(x)}{ln_{k}'(x)}=\prod_{m=0}^{\infty} \frac{ln_{m+k+1}(x)}{ln_{m+k}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

and now you see

here

You can write out the terms

they all cancel except one

[jovial rock]

Hmm, so ln(ln(e))/ln(e) = ln(e)?

[azure crane]

Don't forget the derivatives

[jovial rock]

Ah yeah, and it works with them?

[azure crane]

Bc remember $\prod_{n=0}^{\infty}\frac{f(n+1)}{f(n)}=\frac{1}{f(0)}$

[dawn finch]

RoyalBanana

[azure crane]

^

[jovial rock]

Yeah true

I see

[azure crane]

Yesssss! $\frac{ln_{k+1}'(x)}{ln_{k}'(x)}=\frac{1}{ln_{k}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$\frac{ln_{k}'(x)}{ln_{k+1}'(x)}=ln_{k}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Nice

[jovial rock]

Now we need to divide these two things?

[azure crane]

I did yeah

[jovial rock]

Okay

[azure crane]

^

[jovial rock]

Ah yess

[azure crane]

The ln_m(x)s cancel out

[azure crane]

Hmm I also wonder using that formula $\frac{ln_{k+\frac{1}{2}}'(x)}{ln_{k}'(x)}=...$

[dawn finch]

RoyalBanana

[azure crane]

You can get this with the recursive formula

[azure crane]

But this..

Interesting

[jovial rock]

Hmm

[azure crane]

$ln_k'(x)=\prod_{m=0}^{\infty} \frac{ln_{m+k}(x)}{ln_{m}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

$ln_{k+\frac{1}{2}}'(x)=\prod_{m=0}^{\infty} \frac{ln_{m+k+\frac{1}{2}}(x)}{ln_{m}(x)}$

[dawn finch]

RoyalBanana

[jovial rock]

Maybe replace the 1/2 with a variable

[azure crane]

$\frac{ln_k'(x)}{ln_{k+n}'(x)}=\prod_{m=0}^{\infty} \frac{ln_{m+k}(x)}{ln_{m+k+n}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

Interesting

If n is whole it cancels out very nicely

[jovial rock]

Wouldn’t we need to be able to solve for lnk to solve this function?

[azure crane]

?

[jovial rock]

For non integer values

[azure crane]

ye

[jovial rock]

lnk0.5 for example

This makes it kinda useless I guess

[azure crane]

No

[jovial rock]

Uhh

I think I’ll rewatch the factorial vid to understand this

[azure crane]

$\frac{ln_k'(x)}{ln_{k+1/2}'(x)}=\prod_{m=0}^{\infty} \frac{ln_{m+k}(x)}{ln_{m+k+1/2}(x)}$

[dawn finch]

RoyalBanana

[azure crane]

This is a completely new fact

[azure crane]

^

But not this

[jovial rock]

Hmm

Couldn’t we do something like lnm+k(x) = lnm(x), because k gets nothing compared to m when m gets very big?

Ah I see

[azure crane]

lnk(x) doesn't flatten out

[jovial rock]

Yeah

[azure crane]

well, along k anyway

along x it does

[jovial rock]

Can we maybe do something here, because it’s summation, which should behave better right?

[azure crane]

eh

you can use ln(a)-ln(b)=ln(a/b)

[jovial rock]

I’ll try to figure out values for a very big x

I don’t understand how this stuff works in desmos

Hmmmmm…. What is 0- infinity?…

Hmm? What is infinity-ln(infinity)

[azure crane]

?

[azure crane]

infinity, well in the limit

[azure crane]

?

[jovial rock]

I tried finding S1(e^e)

[azure crane]

You can just cancel out the terms

[jovial rock]

And weird stuff happens if you increase the k in lnk

[azure crane]

for k whole

[jovial rock]

uhh

[azure crane]

$S_1(x)=\lim_{N \to \infty} \sum_{m=1}^{N} ln_m(x)-ln_{m+1}(x)$

[dawn finch]

RoyalBanana

[azure crane]

$S_1(x)=ln_1(x)$

[dawn finch]

RoyalBanana

[azure crane]

$S_1(x)=ln(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Uhh

Maybe this works okay

[azure crane]

^

[jovial rock]

And what if I use S2 ?

[azure crane]

the sum was made specifically bc it cancels out remember

that form of extension

[azure crane]

if k is whole just use this lol^

[jovial rock]

Okayy I see

So S2(e^e) = e+1 ?

And S3(e^e) = e+1+0 = S2(e^e)

Hmm

There must be a way to create infinitys to calculate with ln(0) is infinity, e^ln(0) should be 0, but it can be infinity too

[azure crane]

I used this on a function we already know the extension of
The blue is this method and the black is the actual derivative
It works for all integer x

[jovial rock]

What function is this?

[azure crane]

(the lines aren't actually lines ofc desmos just rounds the numbers at the top and bottom of sums since normally you can't have nonintegers there)

Really think of it more like this

[jovial rock]

Okayy

[azure crane]

I used the harmonic numbers for this

But lemme test it for others

[jovial rock]

Nicee

[jovial rock]

I was thinking a Bit about the lnk function. ln(0) = infinity doesn‘t really make sense, because it approaches 0 but never really reaches 0, it would reach some infinitesimal but not 0. so I thought what if we try to invent a new number class which has numbers that are a solution to ln(0) and try to calculate stuff like ln(-1). And stuff like ln(ln(0))

Ah wait ln(-1) is iπ

[azure crane]

uh $ln(0)=-\infty$

[dawn finch]

RoyalBanana

[azure crane]

Btw

I figured it out

[jovial rock]

But how?

[azure crane]

the formula for f'(x)

if f(x)=f(x-1)+g(x) and smooths out

$f'(x)=-\sum_{n=x+1}^{\infty}g'(n)$

[dawn finch]

RoyalBanana

[azure crane]

Wait

Oh

Duh

Lol

[jovial rock]

I researched stuff ok the internet and it said it‘s undefined, it approches 0 but never reaches it

[azure crane]

That means the same thing

[jovial rock]

Uhh, nice

[jovial rock]

Yeah but it doesn‘t make sense that e^-infinity = 0

[azure crane]

It does

That's true

Just look at the graph of e^x

[jovial rock]

Lemme think…

what is ln(ln(0)) ?

[jovial rock]

I know

[azure crane]

Not really a useful definition but $ln(ln(0))=\pi i+\infty$

[jovial rock]

Hmm

[dawn finch]

RoyalBanana

[azure crane]

wait

yes

[jovial rock]

Soooo, what if ln(ln(ln(0)))

Uhh

[azure crane]

idk

Desmos says it's undefined

even with complex mode on

[jovial rock]

Huh

[azure crane]

Oh I see

You can't have one part be infinite and the other be finite

[jovial rock]

So it‘s infinity + i ?

[azure crane]

No

[jovial rock]

I mean ln(infinity + i)

[azure crane]

I got there with limits I suppose

[jovial rock]

Because pi + infinity = infinity

[azure crane]

Uh yeah but what's that have to do with it

[jovial rock]

Ah

I see

It’s i * pi

[azure crane]

$\lim_{N \to \infty} \ln(N)+\arcsin(\frac{\pi}{N})i$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

[azure crane]

It doesn't work if you just plug in infinity

then you just get infinity+0i

[azure crane]

I got this with eulers formula btw

[jovial rock]

Is ln(pi*i) the same as ln(infinity) + arcsin(pi/infinity) i ?

[jovial rock]

Okay

[azure crane]

idk lol

[jovial rock]

This is complicated…

[azure crane]

Not really useful

[jovial rock]

I just want to know what happens to lnk if k gets large, so we can calculate the limit

[azure crane]

I mean I guess

But there's no reason to know if it doesn't flatten out

and it clearly doesn't

[jovial rock]

Okay

[jovial rock]

How does this function behave?

[azure crane]

Hmm

[azure crane]

btw do you know about this

[jovial rock]

Uh, no haha

I found it

[azure crane]

you can use the real and imag functions

they show you each part of a number or function

oh wow imag(S_k(x)) is really well behaved

like, its almost perfectly a line

Ah I just realized I'm looking at S_x(x) lol

Oops

[jovial rock]

Wdym?

[azure crane]

Which part lol

[jovial rock]

Is it possible to make 3 d graphs?

[azure crane]

Hmm S_k(2) is very much like a line

[azure crane]

No

Well yea

But not very useful here imo

[jovial rock]

Okay

[jovial rock]

Nicee

[azure crane]

also I don't think there's complex mode on the 3d grapher

:/

[jovial rock]

Do you think we can use that for the infinite Limit thingy?

[azure crane]

Like look at this lol

crazy

Oooh

that's imag(S_k(2))

[jovial rock]

Ah yeah

[azure crane]

here's real(S_k(2))

Wait dude we could use the method used in the factorial video

[jovial rock]

Because of the iterated ln for the complex numbers I guess

[azure crane]

he had a function that approaches a line

[jovial rock]

Yeah Right

[azure crane]

Hold on lemme figure out the slope of these lines

can get a really good approximation

of S_k(2)

[jovial rock]

Just approx infinity for N and there we go

[azure crane]

Lol it's so laggy

alr so $real(S_k(2)) \approx real((S_{N+1}(2)-S_N(2))(x-N)+S_N(2))$

[dawn finch]

RoyalBanana

[azure crane]

Using your basic mx+b formula thingy

Wait

[azure crane]

$S_{N+1}(x)-S_N(x)=ln_{N+1}(x)$

[dawn finch]

RoyalBanana

[azure crane]

Nicee

alr so $real(S_k(2)) \approx real(ln_{N+1}(2)(k-N)+S_N(2))$

Like, a really good approximation lol

at most it's off by around 1

beyond k=8 it's almost completely accurate

[dawn finch]

RoyalBanana

[azure crane]

k not x lol

Same with imag

alr so $imag(S_k(2)) \approx imag(ln_{N+1}(2)(k-N)+S_N(2))$

[dawn finch]

RoyalBanana

[azure crane]

Ofc by definition $real(x)+i*imag(x)=x$

[dawn finch]

RoyalBanana

[azure crane]

so let's combine these two

$real(S_k(2))+iimag(S_k(2)) \approx real(ln_{N+1}(2)(k-N)+S_N(2))+iimag(ln_{N+1}(2)(k-N)+S_N(2))$

[dawn finch]

RoyalBanana

[azure crane]

$S_k(2) \approx ln_{N+1}(2)(k-N)+S_N(2)$

[dawn finch]

RoyalBanana

[azure crane]

Nice!

Nicee so is S_k(3)

And s_k(4)

And s_k(5)

alr

nice

$S_k(x) \approx ln_{N+1}(x)(k-N)+S_N(x)$

[dawn finch]

RoyalBanana

[azure crane]

Oh with this graph of S_k(x) I can see it now

I inputted S_k(-0.001)

[jovial rock]

Oh niceee

[jovial rock]

Looks very Nicee, so for huge N?

[azure crane]

Yep

[jovial rock]

Nice

[azure crane]

Yup it approaches infinity+0i, weird

ln(ln(ln(0))) I mean

Ohh I understand

Eulers formula with that gives us sin(0)*e^infinity

0*infinity

with the limit they cancel out and equal pi or smth

[azure crane]

Ah this only works if S_k(x) flattens out

But it really approaches a line

I gotta add a term at the end

[azure crane]

Ahh typo lol m+k not m+x

[azure crane]

$S_k(x) \approx ln_{N+1}(x)(k-N)+S_N(x)$

So $S_k(x)-S_{k-1}(x) \approx $ ln_{N+1}(x)(k-N)+S_N(x)-(ln_{N+1}(x)(k-1-N)+S_N(x))$

So $ln_{k}(x) \approx $ ln_{N+1}(x)$

[dawn finch]

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

[azure crane]

Interesting

so ln_k(x) does flatten out

for k>0

[jovial rock]

I thought it converges to infinity?

When k>0

Uh

Idk

[azure crane]

Yes

flattens out and converges are different

[jovial rock]

Hmm

I think i understand

[azure crane]

flattens out just means the amount it increases by slowly goes to 0

[jovial rock]

There will be a slope somewhere that flatterns out

[jovial rock]

Yeah

[azure crane]

at infinity

$f'(\infty)=0$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah, and we can use that slope to use that factorial stategy i guess

[azure crane]

remember the definition thingy
$\lim_{N \to \infty} f(N+x)-f(N)=0$

[dawn finch]

RoyalBanana

[jovial rock]

Yep

But this time it‘s on the y axis

I think

[azure crane]

?

[jovial rock]

Or we just use the derivative and do it for the x axis

[azure crane]

For $k>0: \lim_{N \to \infty} ln_k(N+x)-ln_k(N)=0$

[dawn finch]

RoyalBanana

[azure crane]

I suppose this makes sense

[jovial rock]

The harmonic Series flatterns on the x axis, and when a function grows very fast it flatterns out on the y axis

[azure crane]

I suppose

you can also say 1/f(x) flattens out

[jovial rock]

Yeah true

This would be epic

If we find lnk, we found all operations for tetration i guess

[azure crane]

Yup

Ah what was the method he used in the factorial video

[jovial rock]

I forgot

I‘ll rewatch it again

Hmm

Should we use some constant for x for now to make it easier?

1 variable is less complex than 2

[azure crane]

We were kinda doing that already

[azure crane]

When I tried this with x=2 3 4 5 etc

[jovial rock]

Because for every x the Point where it flatterns is diffrent

[azure crane]

wdym "the point where it flattens"?

[jovial rock]

On the x axis

[azure crane]

it's at infinity

?

(infinity, infinity)

[jovial rock]

Wait, lemme try to show it on desmos

[azure crane]

that's why we're using limits

lol

If it did ever flatten out at a certain point then that would mean it would be undefined after that point

[jovial rock]

The graphs are not apearing for some reason but These two values for x have a different point where it converges to infinity

[azure crane]

you put them in the wrong order

[jovial rock]

In What Order do I need to put them?

[azure crane]

no I mean

your formula

and what you put

[jovial rock]

Uh, How to do it correctly?

[azure crane]

you're graphing $ln_x(1)$ and $ln_x(10000000)$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

Two functions

[azure crane]

Not $ln_1(x)$ and $ln_{10000000}(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Wait

[azure crane]

Just switch 10000000 and x

[jovial rock]

Thanks

Here

[azure crane]

ye

[jovial rock]

Two diffrent points where they converge

[azure crane]

That's..

[jovial rock]

To infinity

[azure crane]

not what I meant

[jovial rock]

Am confused

[azure crane]

lol

I'm talking about flattening out

as you go to the right

[jovial rock]

Yeah

[azure crane]

?

[jovial rock]

Ah

Uh

I wanted to make a function with k as the variable

[azure crane]

Oh the undefined point of $ln_k(x)$ is ${}^{k-2}e$

or smth

[jovial rock]

So you can see at which Point it turns negative infinity

Now I understand what you meant

[azure crane]

It's not really complicated

[jovial rock]

You wanted to make it for x

So you can find a General formula

[dawn finch]

RoyalBanana

[jovial rock]

And than coming it I guess

[azure crane]

Just think about it lol

It's undefined when it's 0

ln(ln(x)) is undefined when ln(x)=0

ln(ln(ln(x))) is undefined when ln(ln(x)) is 0

etc

[jovial rock]

Yeah

[azure crane]

^

[jovial rock]

Hmm

[jovial rock]

For every x?

[azure crane]

?

I'm saying the undefined point of the function $ln_k(x)$ is at $x={}^{k-2}e$

[dawn finch]

RoyalBanana

[jovial rock]

Ah

Whoah, i guess this is True

[azure crane]

ye

The undefined point of $ln_k(x)$ is x that satisfies $ln_{k-1}(x)=0$

[dawn finch]

RoyalBanana

[azure crane]

Aka e^^(k-2)

so you can see that if k<=0 it's undefined bc theres no undefined point for e^x and e^(e^x) etc

[jovial rock]

My goal is to find the slope of e^^(k-2) + 1/N

When N aproaches to infinity

[azure crane]

-infinity

[jovial rock]

And then use that slope to get the whole function

[azure crane]

lol

[jovial rock]

Oh yeah, typo

[azure crane]

I see what you mean tho

you can do the 1/ trick tho

$f(x+1)=\frac{1}{g(\frac{1}{f(x)})}$

[dawn finch]

RoyalBanana

[jovial rock]

Oh Yeah, I See

[azure crane]

this is equal to 1/normal f(x)

[azure crane]

oh yeah also for this you need this new f(0) to be 1/normal f(0) ofc

You can see this easily if you write out the terms, it's cool

[jovial rock]

Uhh, is this Right?

Ignore the green function

[azure crane]

What are you asking?

[jovial rock]

The 1/x, Trick to get the flatter out on the x axis

[azure crane]

nono

It's 1/f(x)

[jovial rock]

Ah

Okay

[jaunty hornet]

it can be written two ways rigght, like for example H₈ (a, b) and H(8, a, b)?

oh alr

[azure crane]

Well

Not really anymore

bc we invented a new notation

that uses that space already

[jovial rock]

Whoaa, what is this oder thingy

This curve

[jovial rock]

Yeah, but what about these non integer values for n, or negative values

[azure crane]

bc f(3,x)=0 at some point

[jovial rock]

You Can read a lot about that stuff in this Channel, we discussed a lot

[azure crane]

you can do like 10/(f(x)+10) or something

to get rid of that

[jovial rock]

Ah okay i see

[jaunty hornet]

Imagine non-integer hyperoperations like Hπ(a, b)

[jovial rock]

This slope on that y axis is the same as the one for 1/f(x) on the x axis?

[jaunty hornet]

nice

[azure crane]

1/f(x) is just a way to see the slope and have it not be infinite

[jovial rock]

Yeah I know, we are trying to find the formula for that

[jovial rock]

Okay

[azure crane]

hmm

I have an idea though

[jaunty hornet]

maybe H1.5 could be the average between h1 and h2 of a, b but this probably dont make sense

[azure crane]

so for f(3,x)
try 1/(3,e-1/x)

[severe comet]

Yall playing with hyperoperations? That's cool, i used to do that, its fun so good luck

[jaunty hornet]

yesss its so fun

[jovial rock]

Not sure, I think it’s not proportional

[jaunty hornet]

yeah

[azure crane]

*1/(3,1/(e-x))

[jovial rock]

Hmm

[azure crane]

basically moves the slope

to infinity

[severe comet]

I tried creating a hyperoperation that can represent everything from basic operations to smth like the Graham number, i stopped working on it but maybe you can do smth with that

[jaunty hornet]

imma look into graham number

[azure crane]

So then its a function that satisfies the limit thing

f(x+N)-f(N)->0

[severe comet]

i was only able to denote g1, so good luck with g64

[jovial rock]

Nicee

[azure crane]

so basically you transform the function so it satisfies the limit thing, figure out the extension of that, then undo the transform

[jovial rock]

I see, am just a bit confused with the e-x

[jaunty hornet]

what if u use g to ur advantage like g(H₈₀₀₀₀₀₀(g64, g64))
i dont really know about grahams number so im probably wron

[azure crane]

well the point that it's undefined at is at x=e, so 1/(e-x) moves it to infinity

Wait

uhh

I think

[jovial rock]

Uh

[severe comet]

I wanted to make my notation simpler, like H(a,b,c,d), where abcd all correspond to different operations neatly written, maybe you can get inspired from this.

like say a is the number, b is the deegre of the initial operation, addition, multiplication, exponention tetration etc, c is how many times you repeat that, d is how many times you repeat the entire operation by computing the entire thing

[jovial rock]

So 1/lnk(x) approaches one specific value on the x axis right?

[jaunty hornet]

oh nice

[azure crane]

Yeah we had $H_k(n,a,b)$

[dawn finch]

RoyalBanana

[severe comet]

You may also want to explore fractional operations, like say:
1: Addition
2: Multiplication
3: exponentation.

So like lets say we go 2 and 3, what would an operation like 1.5 do to the numbers

[azure crane]

ye

that's what we're doing

[severe comet]

Really? well, you may want to explore how fractional operations sometimes give smaller results than any normal operation, like if you draw a function for addition, multip, exponentan, for 3,3 you get 3+3, 3*3 and 3^3

so

1,6
2,9
3,27

for the operation 1.5, the result is 5.625, which means the mix between addition and multiplication is sometimes less than addition, in some cases, its negative.

[azure crane]

Dude check your dms

[jovial rock]

Right?

[severe comet]

👍

fair enough

[jovial rock]

Maybe go to an other forum

It’s a very interesting subject but my brain is fried rn

[azure crane]

Anyway

[jovial rock]

Because lnk(x) aproaches infinity on the y axis

Or just this e^^k-2 number

[jaunty hornet]

imagine what g(H²⁰⁰ (2, 3)) would be

the fact it can still get bigger

[jovial rock]

Hmm, but we would just extend ln to real numbers and we already now that right?

I wanted to know the slope of lnx(2) or lnx(e)

[jovial rock]

It’s not that big if you compare it to other numbers

[jaunty hornet]

yeah

[jaunty hornet]

can you explain this notation or reply to a message that explains it

[jovial rock]

Do you know lnk(x)

?

[jaunty hornet]

nope

[jovial rock]

k represents how many times you take the natrual log of x

[jaunty hornet]

does x mean any number

[jovial rock]

And we need non integer numbers

[jovial rock]

Yeah

[jaunty hornet]

oh

[jovial rock]

For k

lnk(e) = e^^-k

If we know lnk we instantly know all the tetration

[jaunty hornet]

ye

[jovial rock]

That’s the goal

For now

[jaunty hornet]

thats cool

[azure crane]

@jovial rock okok lets get back working on this now lol

[jovial rock]

Yeah

Can we express it with k?

I think it converges less quickly when x is a big number

Oh

No I mean k

[azure crane]

Wdym?

Wait so what are you trying to do

[azure crane]

I was assuming you were trying to do this

[jovial rock]

Yeah I know, I tried it but I didn’t completely understand it so I’m trying it with the normal function

[azure crane]

Sorry I couldn't focus before can we try again

[jovial rock]

Yeah

Wait

I think I know what you mean

We want to se how 1/f(x) flatterns

So it would be something like (N, f(N)) and the next point (N+1,f(N+1))

So f(N) ≈ f(N+n) for extremely big N

I think 1/f(x) aproaches 1/e^^(k-2)

[azure crane]

?

how so

wait f(x) is lnk(x) right

some constant k

[jovial rock]

Yeah

[azure crane]

So then that would approach 0

as x->infinity

sorry, I'm kinda confused

[azure crane]

yeah ln(ln(x)) approaches infinity

just insanely slowly lol

[jovial rock]

Ah

I forgot to use the 1/f(x)

[azure crane]

Remember, ln(ln(e^e^(2727382)))=2727382

so it does get big

[azure crane]

what's the problem you're trying to solve?

[jovial rock]

Yeah I think it’s approaching 1/e^^k-2

Not sure

[jovial rock]

Trying to find a way to find the slopes for lnk

[azure crane]

You can use the derivative then

[jovial rock]

Wait

I think we had the derivative of lnk

This right?

[azure crane]

Yeah

That's not the simplified version of it tho

[jovial rock]

You simplified it a bit I think

Yeah

[azure crane]

nono

I mean there's a simplified version of that for integer k

[jovial rock]

Wrong word sry

[azure crane]

and there's also a generalized version too

[jovial rock]

Yeah

[azure crane]

so which do u need rn?

[jovial rock]

I guess the one for the integers so I can find a slope which can be used to use the factorial vid strategy

[azure crane]

$ln_k'(x)=\frac{1}{\prod_{m=0}^{k-1}ln_m(x)}$

[dawn finch]

RoyalBanana

[azure crane]

I hate how the bot does that to the capital pi thing but whatever lol

[jovial rock]

Yeah

But niceee

f(N+1) -f(N) = 1 here?

[azure crane]

no

It's 0

[jovial rock]

Oh okay

[azure crane]

remember the limit thing

[jovial rock]

Yeah

[azure crane]

Wait

Ohhh

I see where you're coming from

$f'(k)=\lim_{N \to \infty} \frac{f(N+1)-f(N)}{\prod_{n=k}^{N-1}g'(f(n))}$

[dawn finch]

RoyalBanana

[azure crane]

With this?

[jovial rock]

Yeah

[azure crane]

But no sadly

[jovial rock]

Okay

[azure crane]

In this case g(x)=ln(x) so g'(x)=1/x

[jovial rock]

Okay, I see

[azure crane]

Ah I used k a different way here a little confusing lol

wait no

nvm

lol

Okok

Oh I understand

I see

[jovial rock]

I‘ll try to find the integer values for this

[azure crane]

This is with respect to x

[azure crane]

this is the derivative with respect to k

[jovial rock]

Yeah

[azure crane]

$\frac{d}{dk}ln_k(x)$

[dawn finch]

RoyalBanana

[azure crane]

Lemme test the formula so

$\frac{d}{dk} ln_k(x)=\lim_{N \to \infty} (ln_{N+1}(x)-ln_N(x))\prod_{n=k}^{N-1}ln_n(x)?$

[jovial rock]

Okayy, yeah

[dawn finch]

RoyalBanana

[jovial rock]

So for k=1, and x= e, it‘s 1/e ?

[azure crane]

Yep

[jovial rock]

Nicee

[azure crane]

ln'(x)=1/x

[jovial rock]

Okay

[azure crane]

Say

Thats kinda similar to ln^2(x)

I wonder if lns'(x) is similar to ln(x)

that would be nice lol

[jovial rock]

Yeah, this would be very cool

Wait, for k=2, x=e, 1/e * 1, because 1/ln(e)?

And for k=3, it’s infinity because 1/e * 1 * 1/0 ?

[azure crane]

Yep

[azure crane]

The 1/0 is in the denominator tho

[jovial rock]

Oh

So it equals 0

[azure crane]

yup!

[jovial rock]

This is pretty cool

Uh wait

[azure crane]

Wait so for k>2 all the derivatives are 0

neat

[jovial rock]

I don’t understand

ln2(e) = 0

And then it’s 1/0 ?

[azure crane]

?

it's the derivative tho

[jovial rock]

It would be 1/e * 1/1 * 1/0 ?

[azure crane]

$ln(ln(x))'=\frac{1}{xlnx}$

[dawn finch]

RoyalBanana

[jovial rock]

But here

[azure crane]

$ln(ln(ln(x)))'=\frac{1}{xln(x)ln(ln(x))}$

[dawn finch]

RoyalBanana

[azure crane]

Wait ?

ohh

you already took the 0 out of the denominator

ok

[jovial rock]

Yes

[azure crane]

yeah it's undefined

[jovial rock]

But what if k = 4?? It’s 1/e * 1 * 1/0 * 1/-infinity

[azure crane]

Take the limit

as x->e

lemme look on desmos

yep it's -infinity

[jovial rock]

Yeah

What is 1/0 * 1/-infinity

[azure crane]

You can't do that tho

you have to take the limit instead

[jovial rock]

Okay

[azure crane]

have you learned limits yet?

[jovial rock]

I’m at the quadratic formula rn in school…

[azure crane]

oohh

it's pretty simple

and I mean actually simple lol

[jovial rock]

I watched some vids and I think I understand it a bit

But not completely

[azure crane]

if you can't calculate f(x) at for example x=1 because of some weird problem like this instead you figure out f(1.0001) basically

you choose a number really really close to 1

[jovial rock]

Oh okay

Yeah

[azure crane]

the closer you get the more accurate

[jovial rock]

Oh Nicee, I understand

[azure crane]

cool!

for lim N->infinity you choose a really big number instead

[jovial rock]

Yeah, right

[azure crane]

and the bigger the more accurate

[jovial rock]

Okay,

Hmm

[azure crane]

So here you can't figure out f(e) bc of that, so you can figure out the limit

$\lim_{x \to e} f(x)$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah, so just use some value like 2.7

Or 2.8

[azure crane]

^

[jovial rock]

Yeah

[azure crane]

yea

f(e+.000001) is like -70 thousand

f(e+.00000001) is like -5 million

so you can see it goes to -infinity

[jovial rock]

Oh okay, so extremely strong slope

[azure crane]

yep

[jovial rock]

Hmm

[azure crane]

it's going from left to right upwards so the slope is positive

[jovial rock]

It’s still hard to find a good pattern for this slope

[azure crane]

the slope is clearly infinite bc it does that lol

It's like vertical lol

[jovial rock]

Yeah

[azure crane]

so it's positive infinity

[jovial rock]

And for the negative it’s basically e^^k

[azure crane]

true

[jovial rock]

This is just the e^^k, but we use it’s negative side for the positive side

[azure crane]

$ln_{-k}(x)={}^ke$

[jovial rock]

It’s flipped

[dawn finch]

RoyalBanana

[jovial rock]

Ye

e^^-2 = undefined

Hmm

[azure crane]

yep

[jovial rock]

How can we fix this?

[azure crane]

wdym

[jovial rock]

Like this it’s vey hard to find a flattering pattern

We need something that flatterns so we can use f(N+n) -f(N) = 0

[azure crane]

f(k)=1/ln_{-k}(x)?

tetration grows exponentially so 1/it flattens out

[jovial rock]

Okay

Yeah maybe

1/e^^k

Oh I remeber

[azure crane]

1/x makes bigger numbers smaller so instead of increases more and more and more it increases less and less and less

[jovial rock]

1/x is for the harmonic series, so we just use this for e^^k

[jovial rock]

Yeah, right

[jovial rock]

We could use the f(N-n) -f(N) = 0 for this I guess

Like in the video

[azure crane]

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

[jovial rock]

Hmm

[azure crane]

Orr

If you want N->infinity

f(k)=1/ln(1/f(k-1))

So this is equal to 1/ln_(-k)(x)

So the new f(0) is 1/x

so this function flattens out as k->infinity

yeah

[jovial rock]

Oh yeah right

I need to find out how to graph this on desmos

[azure crane]

f(x)=1/ln(1/f(x-1))

f(0)=1/n

this is equal to 1/ln_(-x)(n)

[azure crane]

Oh I guess that simplifies to f(x)=-1/ln(f(x-1))

Btw if you wanna know how I'm graphing the recursive functions

It offers you a table but there's a way better way

just say some variable idk let's say p, p=[-500...500]

then do (p,f(p))

[azure crane]

Oh ofc desmos doesn't know how to go backwards hold on-

Okok

[jovial rock]

I’ll try…

[azure crane]

That doesn't work, but this does f(x)=e^(-1/f(x-1))

mb

[jovial rock]

Okay

[azure crane]

anyway that's $1/ln_{(-x)}(n)$

[dawn finch]

RoyalBanana

[jovial rock]

Uh, is this right?

[azure crane]

yeah

[azure crane]

^

1/f(-x) is ln_x(n), but remember we rearranged it so it flattens out to the right

[jovial rock]

Okay, yeah

[azure crane]

Wow it becomes very flat very fast lol

f(3) is like 0.00000026 lol

[jovial rock]

Yeahh

[azure crane]

With n=1

[jovial rock]

Uh, what is this value?

[azure crane]

?

idk

Not related tho

[jovial rock]

Okay

[azure crane]

Oh you mean the number it converges to?

[jovial rock]

Yeah

[azure crane]

well, s=1/e^s using the recursive equation

se^s=1

s=W(1)

[jovial rock]

Supersquareroot of e : o

oh

No

I thought

e^W(1)

[azure crane]

It's ln(supersquareroot(e))

[jovial rock]

Okay

Nice

[jovial rock]

So back to this…

This converges to 0 very quickly

[azure crane]

Well

No

It's not 0

Just so close to 0 desmos just rounds it

[jovial rock]

I know but for 1/e^^N, it converges 0

[azure crane]

Remember it's f(5)=1/e^^5 (for n=1) not 0

[jovial rock]

Yeah

[azure crane]

Yeah

[jovial rock]

How can we use this?

[azure crane]

idk

[jovial rock]

This is hard ngl

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

Hmm $f'(x)=\frac{f'(x-1)}{f(x-1)^2}e^{-\frac{1}{f(x-1)}}$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

[azure crane]

Ah nice so $f'(x)=\frac{f'(x-1)}{f(x-1)^2}f(x)$

[dawn finch]

RoyalBanana

[azure crane]

So $f'(x-1)=\frac{f'(x)f(x-1)^2}{f(x)}$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

I‘m confused

How Can lnk be transformed so it‘s useful?

[azure crane]

Idk lol

I'm trying to get a derivative out of this lol

If you figure out just 1 derivative then you figure them all out

[jovial rock]

And whats with this one?

[azure crane]

?

Wait

no

[azure crane]

That's with respect to x not k

[jovial rock]

Hm

Wait

d^2/dx^2 ln(x) = -1/x^2

I lost every single brain cell…

Ahh

[azure crane]

Yeah

Same

[jovial rock]

I tried calculating the second derivative, and this is nuts

[azure crane]

Lol

[jovial rock]

I‘m not gonna continue calculating that

[azure crane]

Can't be thattttt complicated

[jovial rock]

Ye

Probably

I just don‘t get it haha

[azure crane]

same

[jovial rock]

Hmm

[jovial rock]

What about this

With k instead of -k

We could say for big k, g(k+1) ≈ g(k)

And then take the inverse

To get the function back

[azure crane]

?

Why tho

that's got nothing to do with tetration

I'm experimenting with this function h
it's 0 if x<0 or x>=1, otherwise it's some function

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

$f(x)=\sum_{n=-\infty}^{\infty} g_n(h(x-n))$

[dawn finch]

RoyalBanana

[azure crane]

the only property of h I know really is that when x=0 it's equal to f(0)

oh also

f is continuous so

the left sided derivative of h(x) at 1 has to be equal to the right sided derivative of g(h(x-1)) at 1

[jovial rock]

Oh, ye i messe something up, i mean instead of -k, g(k-1)

Sry

[jovial rock]

Uhh, where is h() coming from?

[azure crane]

It's just like a slice of a function

the part between 0 and 1

and you can ofc use the recursive formula to get the rest of the function with that slice

[jovial rock]

Uh

I don‘t know…

[jovial rock]

I think I’ll try to use this

[azure crane]

It like in the harmonic series video

you need that one segment and you solve the whole function

[jovial rock]

Oh okay, i see

[azure crane]

Basically I'm just expressing that one segment as a function

[jovial rock]

Okay, interesting

[azure crane]

so you can think of it as all of the solutions to that one segment

[jovial rock]

Yeah Right

I think I have an idea

[jovial rock]

Wait, why does it have to be -1?

[azure crane]

That was just simplifying

$\frac{1}{e^{\frac{1}{f(x-1)}}}$

[dawn finch]

RoyalBanana

[azure crane]

you see

I figured out a cool thing

if you define a new function instead of $f(x+1)=g(f(x))$ let's say $F(x)$ and some other random function idk $r(x)$

$F(0)=r(f(0)), F(x+1)=r(g(r^{-1}(f(x))))$

You can see it if you write out the terms, $F(x)=r(f(x))$

[dawn finch]

RoyalBanana

[azure crane]

In this case I did r(x)=1/x

[azure crane]

So basically $f(x+1)=\frac{1}{e^{\frac{1}{f(x)}}}$ is 1/tetration

[dawn finch]

RoyalBanana

[azure crane]

Since $f(x+1)=e^{f(x)}$ is tetration

[dawn finch]

RoyalBanana

[azure crane]

Ah I made a typo, I meant to type r(g(r^-1(F(x)))), F not f

it's basically a way to write r(f(x)) in recursive form

if you know what the recursive form of f(x) is

[jovial rock]

Ah, I understand okay

[jovial rock]

I do not completly understand it but it Looks helpful

[jovial rock]

Okay, so to get 1/tetration you need to take the recipocal of f(x) and the whole thing e^(1/f(x))

[azure crane]

Heres the thought process I had:

$F(0)=r(f(0))$

[dawn finch]

RoyalBanana

RoyalBanana

RoyalBanana

[azure crane]

neat, huh?

[azure crane]

so you can see that F(x)=r(f(x))

[jovial rock]

Okay niceee, I think I get it

[azure crane]

I just had an idea
all of the segments are connected, and we're assuming f(x) is continuous, so that means that the left sided derivative of the point on the right side of a slice and the right sided derivative of the point on the left side of the next slice have to be equal

And the second derivative

And the third

Etc

Hmm how do I write that with math?

Left sided derivative of $g_n(h(x))$ at $x=1$, = right sided derivative of $g_{n+1}(h(x))$ at $x=0$

[dawn finch]

RoyalBanana

[azure crane]

Ah I made a mistake

[azure crane]

If g(0) isn't 0 that doesnt work

h