General Formula for all arithmetic operations

[jovial rock]

Oh

No you mean ln^3

[azure crane]

ln^3(a)=ln(a)

ye

[jovial rock]

Okay

[azure crane]

this is right right

[jovial rock]

Yeah

[azure crane]

alr

$ln(a) = \frac{1}{e^{-ln(ln(a))}}?$

[dawn finch]

RoyalBanana

[azure crane]

$ln(a) = e^{ln(ln(a))}?$

[dawn finch]

RoyalBanana

[jovial rock]

Cancels

[azure crane]

$ln(a) = ln(a)?$

[dawn finch]

RoyalBanana

[azure crane]

let's gooo it works

[jovial rock]

Yup

[azure crane]

Cool!

Alr so $ln^n(a) = ln^{(n+1)}_{(-ln^n(ln^n(a)))}(1)$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

Very nice

We already have a general ln formula for all operations

[azure crane]

Let's try to find a formula for $ln^n_k(x)$

[dawn finch]

RoyalBanana

[azure crane]

like we just did for $ln^2_k(x)$

[jovial rock]

Hmm

[dawn finch]

RoyalBanana

[jovial rock]

x - e -e

For k = 2

x - e*k

ln^3_k(x) = x /e^k

Hmm

[azure crane]

Ohh I think I know

hold on

$ln_k^1(x)=?$

[dawn finch]

RoyalBanana

[azure crane]

$ln^1(x)=?$

[dawn finch]

RoyalBanana

[azure crane]

$log^n_a(H(n,a,b))=b$

[dawn finch]

RoyalBanana

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

alr i thought so

$ln^1_1(x)=x-e$

$ln^1_2(x)=x-2e$

$ln^1_3(x)=x-3e$

[dawn finch]

RoyalBanana

[azure crane]

$ln^1_k(x)=x-ke$

[dawn finch]

RoyalBanana

[azure crane]

ooh that's what this was yeah

$ln^1_k(x)=e*-k+x$

$ln^2_k(x)=e^{(-k)}*x$

$ln^3_k(x)={}^{(\ln^4(x)-k)}e$

[dawn finch]

RoyalBanana

[azure crane]

*ln^4 not ln

[azure crane]

I have figured it out

$ln^1_k(x)=e*(ln^2(x)-k)$

$ln^2_k(x)=e^{(ln^3(x)-k)}$

$ln^3_k(x)={}^{(\ln^4(x)-k)}e$

[dawn finch]

RoyalBanana

[azure crane]

$ln_k^n(x)=H(n+1,e,ln^{n+1}(x)-k)$

[dawn finch]

RoyalBanana

[azure crane]

$\ln^n(H(n,e,b))=b$

$ln^n(ln^{n-1}_k(x))=ln^n(x)-k$

[dawn finch]

RoyalBanana

[azure crane]

Nice

[azure crane]

$H(n,e,ln^n(b))=b$

[dawn finch]

RoyalBanana

[azure crane]

So $H(n+1,e,ln^{n+1}(x))=x$

[dawn finch]

RoyalBanana

[azure crane]

$H(n+1,e,ln^{n+1}(x)-k)=ln_k^n(x)$

$H(n+1,e,ln^{n+1}(x))=x$

[dawn finch]

RoyalBanana

[azure crane]

makes sense

Okok so $H(n+1,e,ln^{n+1}(x)-k)=ln_k^n(x)$

[dawn finch]

RoyalBanana

[azure crane]

$H(n+1,e,ln^{n+1}(H(n+1,e,0))-k)=ln_k^n(H(n+1,e,0))$

[dawn finch]

RoyalBanana

[azure crane]

$H(n+1,e,-k)=ln_k^n(H(n+1,e,0))$

[dawn finch]

RoyalBanana

[azure crane]

$H(n,e,k)=ln_{-k}^{n-1}(H(n,e,0))$

[dawn finch]

RoyalBanana

[azure crane]

$H(n,a,b)=log_{a,{-b}}^{n-1}(H(n,a,0))$

[dawn finch]

RoyalBanana

[azure crane]

Using this formula, I did some maths and figured out H(0,a,b)=b+1

yup

b+1

noice!

So that time I inputted n=1 imma do n=0 now!

[azure crane]

You're not gonna like this

It gave me H(-1,a,b)=b+1

I really don't think there's a way around this

[azure crane]

Know what I'm manually finding a way around this

You're right this does not make sense

[jovial rock]

Oh

You did a lot of stuff

Maybe we kinda need to reconstruct the formula

I did some research and I found that 1+b should be 1+ 1+1+1+…1 b times, and this would reduce the unary operation into an operation with kinda nothing. It doesn’t really make sense H(1,a,2) should be H(0,a,a) which is a?a. Now I see, it’s 2 by definition. But what would H(1,a,1) be? It doesn’t really make sense to me that this is a, because the complexity breaks down to addition, we can’t break down addition, because it’s one of the most abstact operations we know (if we don’t think outside the box). We could just continue this to get the successor and then just nothing, (maybe the assigment operation. By definition you’re right, but by addition logic it’s a bit diffrent

1 + 1+1 … b copies?

The formula is made on the fact that you cannot break down addition, but we want to find a general formula

[azure crane]

This formula is wrong

$a+3 \ne a+1+1$
it doesn't work for $n \le 1$

[dawn finch]

RoyalBanana

[jovial rock]

Yeah

I guess

[azure crane]

That means it probably wouldn't work for n=0 either

[azure crane]

Probably

[azure crane]

I think I get why you can't go below b+1

see with like b+a or b*a depending on a, the number of times it's being repeated is different

but with b+1 it's not

you cant have something being repeated turn into b+1 bc that would mean there would be a value for it to change from into b+1 after so many repetitions, but b+1 never changes based on a

lemme give example

a*b can be turned into something being repeated a times bc it depends on a
a+a+a+a... you can change the number of a's around
but how do you do that with b+1? you can't change the a value you can't change the number of a's

anyway what the heck does it mean to have H(2.5,a,b) lol

somewhere inbetween multiplication and exponentiation?

[azure crane]

Actually I suppose I could generalize the way I figured it out but for any n

Nice it worked!

I got $H(n-1,a,b)=H(n,a,log^n_a(b)+1)$

[dawn finch]

RoyalBanana

[azure crane]

Lemme test

$n=4: a^b={}^{(log^s_a(b)+1)}a$

[dawn finch]

RoyalBanana

[azure crane]

hmm

$2^4={}^{(log^s_2(4)+1)}2$

[dawn finch]

RoyalBanana

[azure crane]

$16={}^{3}2$

[dawn finch]

RoyalBanana

[azure crane]

yup

[azure crane]

this reminds me of that formula we had

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

[dawn finch]

RoyalBanana

[azure crane]

yup checks out!

[jovial rock]

Nicee

I think I there is a Problem in pit previous calculations

This

Wait

It‘s not (e^^2)^^(log(e^^2)s(e) - 1), it‘s (e^^2)^(log(e^^2)s(e) - 1)

And maybe it’s Not even a super log in there, not sure

[jovial rock]

Nicee, so it‘s a?b = a + (log^1(a)(b) + 1) ?

2?4 = 2 + (2 + 1)

2?4 = 5?

[azure crane]

yup

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

[jovial rock]

So 5?100 = 101

This is just S(100)

[azure crane]

yep!

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

[dawn finch]

RoyalBanana

[azure crane]

ofc

No reason we can't generalize this

$H(n,a,b)=log^{n-1}_{a, k}(H(n,a,b+k)$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

Thats kinda sad because it ignores the a but I guess we can‘t do anything about it

Operations with non integer n would be cool

[jovial rock]

Nicee

How to make a half step, maybe 0.5 of an Operation to another one?

[jovial rock]

This one does a full step

There Must be some value for k that does a Full step and then use it to make a half step, like we saw with lnk and lns

[azure crane]

wdym by step?

Say.. $H(n,a,b)=log^{n-1}_{a, -b}(H(n,a,0))$

[dawn finch]

RoyalBanana

[azure crane]

now I wonder if I can get rid of a too

[jovial rock]

That -1, from Operation to and other integer Operation. Can we reach the non integer operations by adjusting k?

[azure crane]

Sure

But with this you have to know either the log or the H operation to figure out the other

it's impossible to figure out both from knowing nothing

with this formula

[jovial rock]

Yeah true

[azure crane]

We need some formula for H(n+k,a,b)

then we can extend it to non integers

and also H(n,a,b+k)

[jovial rock]

remember ln(e)(e^^e) = lns(e^^e) ?

replacing e for k, we get the next operation

[jovial rock]

Hmm

[azure crane]

Ye that's basically this kinda

[azure crane]

do you see the problem

[jovial rock]

Yeah I know

What if we try to analyze the structure of log. We know ln(e)(e^^e) = lns(e^^e), so what if ln^2(x)(e^^e) = lns(e^^e)

[azure crane]

?

what's x here

[jovial rock]

The value we need to find out

The unknown value

[azure crane]

#1330832852819906661 message

[jovial rock]

I guess ln^2(e)(e^e) = ln(e^e), and ln^2(e^e)(e^e^e) = ln(e^e^e)

[azure crane]

$ln^2_x({}^ee)=e^{(-x)}*{}^ee$

[dawn finch]

RoyalBanana

[jovial rock]

Oh

[azure crane]

Wait did you mean ln3 instead of lns on the second one

so it's one more the left one

[jovial rock]

No, I wanted to know how to jump from multiplication to tetration to find the pattern

[azure crane]

Just jump from multiplication to exponentiation to tetration

[jovial rock]

Hmm, so what would x need to be so it equals e?

[jovial rock]

But don’t we need the steps to find the pattern and then find the non integer operations?

[azure crane]

?

btw why don't we use $H_k(n,a,b)$

[dawn finch]

RoyalBanana

[azure crane]

how would that help though

[jovial rock]

So we know what do we need to input for k in lnk(e) so we get ln^3.5(e) or something

[azure crane]

you'd need to invent a new formula first so do that first

[azure crane]

$x=-ln(e/{{}^ee})$

[dawn finch]

RoyalBanana

[azure crane]

$x=-(ln(e)-ln({}^ee))$

[dawn finch]

RoyalBanana

[azure crane]

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

[jovial rock]

Hm, oh okay

[dawn finch]

RoyalBanana

[jovial rock]

So ln^2(e^^(e-1) - 1)(e^^e) = ln(e)(e^^e)

Right?

[azure crane]

hmm yes

👍

[jovial rock]

Okay

Hmm

[azure crane]

Ooh I see where you're going

[jovial rock]

What would your strategy be to find non integer operations?

[jovial rock]

Nicee

Lemme think for lns it’s 1, for ln it’s e, and for ln^2 it’s e^^(e-1) -1

[azure crane]

I was thinking that since we have a definition of H(n+1,a,b) in terms of H(n,a,b) we could find a definition of H(n+k,a,b) in terms of H(n,a,b), and then just plug in non integer k values to that

[jovial rock]

Oh

Yeah true

So that would mean we could repeat H(n,a,b) k times

Like log with k

?

[azure crane]

yeah!

[azure crane]

^^^^^^^

I wonder if H(-0.5,a,b) would be different than H(0.5,a,b)?

Or would it get stuck in the same way that H(-1,a,b)=H(0,a,b)

[jovial rock]

That would be very interesting, maybe it could be like the factorial, where all negative integers equal infinity but the non integer numbers equal a real number

[azure crane]

interesting

[azure crane]

anyway what do you think of this it would be useful^

Though we would need notation to distinguish which part is being repeated

Like the difference between these

$H(n,a,H(n,a,b))$

$H(n,H(n,a,b),b)$

$H(H(n,a,b),a,b)$

[dawn finch]

RoyalBanana

[jovial rock]

Hm

[azure crane]

Maybe we can just choose one to be the definition Im 99% sure the bottom one is useless and decently sure the middle one is

The top one is the one used in the recursive formula anyway

Oh my god this gives the new problem of $H_\frac{1}{2}(n,a,b)$ lmao

[dawn finch]

RoyalBanana

[jovial rock]

Did you have a formula where n in H decreases?

Or increases

Just changes

[azure crane]

^

[azure crane]

Oh yea

yes

definitely

Remember this

well we can rewrite it

[azure crane]

with this

[jovial rock]

How?

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

now let's plug in the formula again!

$H(n,a,b)=H(n-1,a,H(n-1,a,H(n,a,b-2)))$

[dawn finch]

RoyalBanana

[azure crane]

do you see

Now we can do it again

[jovial rock]

So we could use this for Hk ?

[azure crane]

$H(n,a,b)=H(n-1,a,H(n-1,a,H(n-1,a,H(n,a,b-3))))$

oops typo

[dawn finch]

RoyalBanana

[jovial rock]

Nicee, but it would change n only by -1

[azure crane]

We're getting there

Then we can repeat that formula

but first we gotta put it in H_k notation otherwise it'd be a headache

I think the notation should be like this lemme give an example
$H_3(n,a,b)=H(n,a,H(n,a,H(n,a,b)))$

[dawn finch]

RoyalBanana

[azure crane]

Wait

[azure crane]

How do you write this with h_k the last one is n not n-1..

Wait

Ohh

duh

[jovial rock]

Uhhhh

[azure crane]

Just set that last H(n,a,b-3) as the b in this H_k formula

[jovial rock]

Ahh

[azure crane]

$H(n,a,b)=H_3(n-1,a,H(n,a,b-3))$

[jovial rock]

I think I see

[dawn finch]

RoyalBanana

[azure crane]

See, b in H_k basically gets ignored until the very end

[jovial rock]

Yeah

[azure crane]

bc it's H(n,a,H(n,a...

interesting

[azure crane]

Ofc

3 isnt special here!

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

[dawn finch]

RoyalBanana

[jovial rock]

Interesting

[azure crane]

Oh yeah

Lemme borrow the lnk property

$H_k(n,a,H_m(n,a,b))=H_{k+m}(n,a,b)$

[dawn finch]

RoyalBanana

[azure crane]

neat

[azure crane]

alrr

[jovial rock]

What would be this H(n,a,H(n-1,a,b)) ?

[azure crane]

uhh

well

I don't think there's a nice formula for that tbh

idk

[azure crane]

Soo.. $H(n-1,a,b)=H_k(n-2,a,H(n-1,a,b-k))$

[dawn finch]

RoyalBanana

[azure crane]

Imma just say k=b bc that's simple and makes sense

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

[dawn finch]

RoyalBanana

[azure crane]

H(n,a,0) at least for n whole, is 1 if n isn't 1, if n=1 then it's a

for n whole and n not equal to 1: $H(n,a,b)=H_b(n-1,a,1)$

[dawn finch]

RoyalBanana

[azure crane]

Oh wait

if n=2 then it's 0

for n=1: $H(1,a,b)=H_b(0,a,a)$

for n=2: $H(2,a,b)=H_b(1,a,0)$

for n integer and not equal to 1 or 2: $H(n,a,b)=H_b(n-1,a,1)$

interesting

[dawn finch]

RoyalBanana

[jovial rock]

I see

Hmm

[azure crane]

yup checks out

interesting

[jovial rock]

What is the b on the Hb good for?

[azure crane]

?

[jovial rock]

Here

Ah

Now I See

[azure crane]

H(n,a,0) is very simple

[jovial rock]

It repeats and then gets H(n,a,b)

[azure crane]

yup

pretty cool

[azure crane]

but the point is that chooseing k=b leads to H(n,a,0) which allows this, basically getting rid of the H(n... part

so it's only in terms of H(n-1...

[azure crane]

Very nice

So if we have this formula thing

for n=1: $H(1,a,b)=H_b(0,a,a)$

for n=2: $H(2,a,b)=H_b(1,a,0)$

for n integer and not equal to 1 or 2: $H(n,a,b)=H_b(n-1,a,1)$

[dawn finch]

RoyalBanana

[azure crane]

Let's plug in H(n-1,a,1)

for n-1=1: $H(1,a,b)=H_1(0,a,a)$

for n-1=2: $H(2,a,b)=H_1(1,a,0)$

for n-1 integer and not equal to 1 or 2: $H(n-1,a,1)=H_1(n-2,a,1)$

[dawn finch]

RoyalBanana

[azure crane]

Aka
for n=2: $H(1,a,1)=H(0,a,a)$

for n=3: $H(2,a,1)=H(1,a,0)$

for n integer and not equal to 2 or 3: $H(n-1,a,1)=H(n-2,a,1)$

Interesting

[dawn finch]

RoyalBanana

[azure crane]

checks out

[jovial rock]

What about Hb(n,a,Hb(n,a,0))

Oh am confused

[azure crane]

Ah remember this
#1330832852819906661 message

works the same way the lnk rule works

[jovial rock]

Okay

Hm

[azure crane]

does that make sense

hmm

[jovial rock]

I‘m trying to figure out how to do this for more than only -1

H(n,a,b) = H(n-2,…

[azure crane]

I was gonna use the formula again on H_b(n-1,a,1) which is why I started with using the formula on H(n-1,a,1)

too see if there's a pattern

[azure crane]

We need a formula for $H_k(n,a,b)$ like this..

[dawn finch]

RoyalBanana

[azure crane]

yeah

[jovial rock]

Hmm

I‘ll try to figure it out

I don‘t get it

[azure crane]

What part of it?

[jovial rock]

There would need to be some Hb that equals H(n-1,a,0) i think

[azure crane]

Btw I just thought of something

$H_1(n,a,H_{-1}(n,a,b))=H_0(n,a,b)=b$

so $H(n,a,H_{-1}(n,a,b))=b$

but also $H(n,a,log^n_a(b))=b$

so $log^n_a(b)=H_{-1}(n,a,b)$

[dawn finch]

RoyalBanana

[azure crane]

interesting

[jovial rock]

Yeah maybe

I See

[azure crane]

Wait

ofc $H_{-1}(n,a,H_{-1}(n,a,b))=H_{-2}(n,a,b))$

but also now that we know that $log^n_a(b)=H_{-1}(n,a,b)$

$H_{-1}(n,a,H_{-1}(n,a,b))=log^n_a(log^n_a(b))=log^n_{2a}(b)$

Therefore $H_{-2}(n,a,b))=log^n_{2a}(b)$

[dawn finch]

RoyalBanana

[azure crane]

does that make Sense

This is cool

Waitttttt! Ofc also $H_{-1}(n,a,H_{-2}(n,a,b))=H_{-3}(n,a,b))$

but also now that we know that
$log^n_{a}(b)=H_{-1}(n,a,b)$ and

$log^n_{2a}(b)=H_{-2}(n,a,b)$

$H_{-1}(n,a,H_{-2}(n,a,b))=log^n_a(log^n_{2a}(b))=log^n_{3a}(b)$

Therefore $H_{-3}(n,a,b))=log^n_{3a}(b)$

[dawn finch]

RoyalBanana

[jovial rock]

Isn‘t the b changing with the H-1, H-2,..

Uh

I need to think

[azure crane]

Lemme say it a different way

[dawn finch]

RoyalBanana

[azure crane]

On the left side we plug in log(2a)(n)(b) instead of b and on the right side we plug in H(-2)(n,a,b) instead of b
and we can do that since they're equal

hope that makes sense

[jovial rock]

Hmm, okay

So ln(ln(x)) = log(e^e)(x) ?

[azure crane]

?

[jovial rock]

Oh

No

log(2e)

[azure crane]

H_{-1} is a new thing

I suppose

[azure crane]

Wait where's this coming from?

[jovial rock]

Because it‘s like you Said with the H, H-1(n,e,H-1(n,e,b)) = H-2(n,e,b) = log(2e)(b)

[azure crane]

Wait

Ohh I made a typo

$log^n_a(log^n_a(b))=log^n_{a,2}(b)$

[dawn finch]

RoyalBanana

[azure crane]

Mb

not 2a

[jovial rock]

Ah okay

I See

[azure crane]

So anyway $H_{-3}(n,a,b))=log^n_{a,3}(b)$

[dawn finch]

RoyalBanana

[jovial rock]

I have a quick other question. What would x be if log^4(e,x)(e^e) = log^3(e,1)(e^e)

[azure crane]

Interesting

aka lns(x)(e^e) = e

[jovial rock]

it‘s just lns and lnk, but you Switch the k to lns, so lnsk = ln

[jovial rock]

Yes

Hmm

I guess somewhere between 1 and e, but more near to 1 than e

[azure crane]

neat

e/2?

[jovial rock]

lns1(e^e) = 2, lns0(e^e) = e^e, hmm

[jovial rock]

Hmm, i think not, because for 1 it‘s already 2 and that’s to small

[azure crane]

e/2>1

?

ohh its gotta be between 0 and 1

[jovial rock]

Yeah, but 0 is e^e,

[azure crane]

I see

[jovial rock]

Yeah

[azure crane]

1/2?

[jovial rock]

Hmm

I think not because it‘s stronger than exponetial growth, but it’s possible

Is there a formula for lnsk ?

Or something where you can use lns to calculate its k

[azure crane]

I suppose yeah

One sec

$ln_k^n(b)=H(n+1,e,ln^{n+1}(b)-k)$

[dawn finch]

RoyalBanana

[azure crane]

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

[dawn finch]

RoyalBanana

[jovial rock]

Hmm, there must be something that does n-1

[jovial rock]

Maybe this

[azure crane]

Yeah I was thinking about something with that

[azure crane]

So we can generalize this

$H_{-k}(n,a,b))=log^n_{a,k}(b)$

[dawn finch]

RoyalBanana

[azure crane]

$H_{k}(n,a,b)=log^n_{a,-k}(b)$

[dawn finch]

RoyalBanana

[azure crane]

I wonder if we can combine these two formulas then

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

[dawn finch]

RoyalBanana

[azure crane]

$H(n,a,b-k)=log^n_{a,-1}(b-k)$

[dawn finch]

RoyalBanana

[azure crane]

$H_{k}(n-1,a,c)=log^{n-1}_{a,-k}(c)$

[dawn finch]

RoyalBanana

[azure crane]

Ok

Now to plug everything in

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

$log^n_{a,-1}(b)=log^{n-1}{a,-k}(log^n{a,-1}(b-k))$

[dawn finch]

RoyalBanana

[azure crane]

hmm I can replace -k with k

$log^n_{a,-1}(b)=log^{n-1}{a,k}(log^n{a,-1}(b+k))$

[dawn finch]

RoyalBanana

[azure crane]

Interesting

[jovial rock]

Yeah

[azure crane]

Reminds me of this fact $log_a^n(log_{a,k}^{n-1}(x))=log_a^n(x)-k$

[dawn finch]

RoyalBanana

[jovial rock]

Yup, cancels when 0

[azure crane]

I wonder if this works for other numbers not just -1

[jovial rock]

It should

I think it does yeah

[azure crane]

Lemme test this

[jovial rock]

But you can‘t change n-1 i think

[azure crane]

Ok so with base e for simplicity the formula above is

$ln^n_{-1}(b)=ln^{n-1}{k}(ln^n{-1}(b+k))$

[dawn finch]

RoyalBanana

[azure crane]

So: for example $ln^n_{2}(b)=ln^{n-1}{k}(ln^n{2}(b+k))?$

[dawn finch]

RoyalBanana

[azure crane]

Let's say n is 3

$ln_{2}(b)=ln^{2}{k}(ln{2}(b+k))?$

[dawn finch]

RoyalBanana

[azure crane]

^

So $ln_k^2(ln_2(b+k))=e^{(-k)}*ln_2(b+k)$

[dawn finch]

RoyalBanana

[azure crane]

$ln_{2}(b)=e^{(-k)}*ln_2(b+k)?$

[dawn finch]

RoyalBanana

[azure crane]

uhh

Only works if k=0

[azure crane]

Oooh ok well after looking at graphs this seems to work only with -1 and 1

So if it's one it's $log^n_a(b)=log^{n-1}_{a,k}(log^n_a(b+k))$

[dawn finch]

RoyalBanana

[azure crane]

Interesting

So $log_a^n(log_{a,k}^{n-1}(x))=log_a^n(x)-k$

and $log^n_a(b)=log^{n-1}_{a,k}(log^n_a(b+k))$

[dawn finch]

RoyalBanana

[azure crane]

You could probably use this to find log^(n-1)!

Anyway soo using that $ln^s(b)=ln_{k}(ln^s(b+k))$

[dawn finch]

RoyalBanana

[azure crane]

Wait is that really true?

that would be cool

So then $ln^s(b+1)=e^{ln^s(b)}?$

I feel like that's too good to be true

[azure crane]

Ah I only checked if it worked for 1 for n=3 ofc

[azure crane]

Ok ignore this this is wrong

[azure crane]

So this only works with -1

[azure crane]

Wish I could pin things lol

[azure crane]

Soo for example
${}^ba=log_{a,k}({}^{(b+k)}a)$

[dawn finch]

RoyalBanana

[azure crane]

makes sense

[jovial rock]

Whoa

[azure crane]

That's incorrect

[jovial rock]

Hmm

I trying to find something for this one formula with lns and ln^n

1
e
e^^(e-1)
e^^e * e^-1 - 1

This is all I got for the 4 n

I can’t See a pattern that can be generalized

And how could we compute such Numbers

[jovial rock]

Hmm

I think I could continue with this

And then find out the optimal value

And solve for 1

Uh I mean y

I just need to replace Capital Pi with some non integer thingy

[azure crane]

If n is unchanging btw it's literally just some constant to the power of x

so if we ignore x for now, we can extend this formula to non integers to figure it out f(n)=f(n-1)ln(f(n-1))

(Once we figure out what f(n) is here, f(x,n)=f(n)^x)

Interesting

and f(1)=2

[jovial rock]

Yeah true

I think it‘s just f(x) = e(x)^W(1). e(0)^W(1) = 1, e(1)^W(1) = e^W(1), e(2)^W(1) = e^W(e^W(1)),…

But we can‘t just do that if we want to take the nth derivative again

And btw e(infinity)^W(1) = e

But I want to have the General Function of that so I Can take the derivative, and after undestanding the pattern the nth derivative and then Talking the nth derivative n times

[azure crane]

Ye we proved that already lol

can you not use that notation though it's easily confused

[azure crane]

Umm $e^{W(1)} \ne 2$

[dawn finch]

RoyalBanana

[azure crane]

Ooh.. f(x-1)=e^W(f(x))

I see what you're going for

Wait it reaches a stable value at an infinity

we can use the extension thing

hold on

https://youtu.be/9p_U_o1pMKo?si=bi5XVCDyaHNhHdz5
The same one used here

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

[dawn finch]

RoyalBanana

[azure crane]

$f(x+1)=f(x)ln(f(x))$

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

[dawn finch]

RoyalBanana

[azure crane]

s=e^W(e^W(e^W(...
e^W(s)=s
W(s)e^W(s)=W(s)s
s=W(s)s
s=0, or W(s)=1
s=0, or s=e
clearly not 0 so s=e

[azure crane]

so we need a super recursive formula

$f(x+2)=f(x+1)\ln(f(x+1))=f(x)\ln(f(x))\ln(f(x)\ln(f(x)))$

[dawn finch]

RoyalBanana

[azure crane]

ah wait

I can reuse the thing from before

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

[dawn finch]

RoyalBanana

[azure crane]

Nice

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

$f(x+N)=f(x)\prod_{m=1}^{N}(\sum_{k=1}^{m}ln_k(f(x)))$

[dawn finch]

RoyalBanana

[azure crane]

Wait this only works if N is positive

hmm

[azure crane]

hmm imma use like $\omega(x,n)$

[dawn finch]

RoyalBanana

[azure crane]

fancy w

$\lim_{N \to \infty} (\omega(f(x),N)-\omega(1,N))=0$

[dawn finch]

RoyalBanana

[azure crane]

Uhh

Imma do this later my brain is fried rn

[jovial rock]

Wo you mean x * ln(x) = 1?

[jovial rock]

Uhhhhh

[jovial rock]

Mine too if I try to understand this haha

[jovial rock]

Oh wait, now I understand, it aproaches e for infinity

Like √2 ^^ infinity approches 2

[azure crane]

So this only works for sums specifically but there's some way to generalize this for other things I suppose

like f(x+1)=g(f(x))

[jovial rock]

Oh okay, I see

And we need to find at which Point of the nth root of y^x (y is just a constant, for example 2) repeated n times the function crosses 1 to find a new constant

I don‘t know

It‘s probably something with lnk, because it’s repeated ln

But how to take the derivative of a recrusive function?

[azure crane]

It's always something to do with f(x+N) and f(N)

[azure crane]

Youd have to find the extension first

[jovial rock]

Maybe just take the natrual log of it and multiply it to it‘s normal form

[azure crane]

I can say though with confidence that $f'(-\infty)=0$

[dawn finch]

RoyalBanana

[jovial rock]

Like with this

[jovial rock]

Yeah probably

[azure crane]

The extension we're doing relies on the fact that it smooths out which the derivative being 0 means the same thing

[jovial rock]

What would be the inverse of e^W(x) ??

[azure crane]

xlnx

[jovial rock]

Yeah

[azure crane]

^

[jovial rock]

Isn‘t ln(x) just the inverse of e^x ?

[azure crane]

yeah

x*lnx

[jovial rock]

Ah

I See

[azure crane]

you can see bc $e^{W(x)}*ln(e^{W(x)})=W(x)e^{W(x)}=x$

[dawn finch]

RoyalBanana

[jovial rock]

True True

[azure crane]

I suppose this means btw that W(xlnx)=lnx

[jovial rock]

Hm

Isn‘t it something like x * ln ^x ?

[azure crane]

Since xlnx is the inverse of e^W(x) that means e^W(xlnx)=x
so W(xlnx)=lnx

[azure crane]

I'm workin on this

this is interesting

It seems to have to do with the inverse of g

[jovial rock]

Sorry for my handwriting, but I think I found something

[azure crane]

Wait how'd you get ln(4)^^2?

[jovial rock]

ln(4) * ln(ln(4)) = ln(ln(4)^ln(4))

[azure crane]

ooooh

interesting

[jovial rock]

Yess

I just don‘t know how to generalize the recrusive tetration

[azure crane]

This is just really useful in other recursive applications too so imma continue on this

I wish I had paper rn but I'm not at my house

I have to write with the latex bot and desmos lol

[jovial rock]

Oh Nicee. Yeah looks very cool.

[jovial rock]

Nice lol

I think I’ll watch the Video, I haven‘t watched it yet

[jovial rock]

Hmm yeahh

I think the other things get complicated very quickly

[azure crane]

I'll figure it out lol

[azure crane]

Oh it's pretty useful

[jovial rock]

Very interesting video, very nice

[jovial rock]

Yeah lol

[azure crane]

Also their video on factorials is a continuation on that

anyway that's where I got this from

[jovial rock]

Nicee. Factorials are maybe found with the capital Pi thingy

[jovial rock]

Nice

[jovial rock]

I tried it with that one Method but it didn‘t work very well

Hmm, maybe something Like this

[azure crane]

I understand it now

[jovial rock]

How?

[azure crane]

So the reason that extension works

Let's do the example f(x+1)=f(x)+g(x+1)

[jovial rock]

Hmm

[azure crane]

$f(x)=f(0)+\sum_{n=1}^{\infty}(g(x+n)-g(n))$

[dawn finch]

RoyalBanana

[azure crane]

Can you see why this works

If x is whole a bunch of terms cancel out

[jovial rock]

This is for which function?

[azure crane]

and all that's left is f(0)+g(1)+g(2)+...+g(x-1)

[azure crane]

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

[jovial rock]

Uhh, Lemme think

[azure crane]

Try writing out the terms

of the sum

[jovial rock]

What is f(0) ?

[azure crane]

f(0)

[jovial rock]

Okay

[azure crane]

anything it works for any function with this property

[azure crane]

I might've made a typo here, but it's something like this

[azure crane]

$f(1)=f(0)+g(2)-g(1)+g(3)-g(2)+g(4)-g(3)...$

[dawn finch]

RoyalBanana

[azure crane]

ah I did make a typo

[jovial rock]

x = 1, f(0) + g(2) -g(1) + g(3) - g(2)…, ah, f(0) -g(1) + g(3)

[azure crane]

$f(x)=f(0)+\sum_{n=1}^{\infty}(g(n)-g(n+x))$

[dawn finch]

RoyalBanana

[jovial rock]

Nicee

[azure crane]

You see they cancel

[azure crane]

I fixed the typo

[jovial rock]

Yeah

[azure crane]

anyway this only works if it flattens out in the limit bc there is technically like two extra terms left at the end

[jovial rock]

Hmm

[azure crane]

if $\lim_{N \to \infty} f(N+n)-f(N)=0$ it works

[jovial rock]

Intresting

[dawn finch]

RoyalBanana

[azure crane]

for some integer n

[jovial rock]

Yeah

[azure crane]

this is the same as saying it flattens out

[azure crane]

This canceling thing though, can surely be used for more than just sums

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

[dawn finch]

RoyalBanana

[jovial rock]

Uh, but it would need to be something like a * b = 1

But that’s cool

Exponetiation wouldn‘t work i guess, because of the commutative rule

[azure crane]

Slow down lol we'll figure it out for the more general case f(x+1)=g(f(x)) lol

I'm just seeing multiple examples to see if I can notice a pattern

hmm let's do f(x+1)=f(x)g(x+1)

So $f(x)=f(0)\prod_{n=1}^{x}g(n)$

[dawn finch]

RoyalBanana

[azure crane]

So.. the reason this works is bc it cancels out

$\prod_{n=1}^{x}g(n)=\prod_{n=1}^{\infty}\frac{g(n)}{g(n+x)}?$

[dawn finch]

RoyalBanana

[jovial rock]

Okay

Looks complicated haha

But it could work

[jovial rock]

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

Looks very simple

Thats for the nth derivative thingy

This could work

[jovial rock]

Or Just f(x) = ln(f(x-1)^^2)

[azure crane]

The g(x) still depends on f though so we don't know what g(x) is at non integer values of x

[azure crane]

^

[azure crane]

Ohh I see it has the same problem of the extra values at high n

[jovial rock]

I did some stuff

I used that ln trick to make it into addition

[azure crane]

uh

wrong trick

it's the reverse

the product of ln doesn't turn into anything

ln of a product does

unfortunately

[azure crane]

Lemme write out the terms

$g(1)=\frac{g(1)}{g(2)}\frac{g(2)}{g(3)}\frac{g(3)}{g(4)}...?$

[dawn finch]

RoyalBanana

[azure crane]

ahh it has an extra /g(N)

So $\prod_{n=1}^{x}g(n)=\lim_{N \to \infty} g(N)\prod_{n=1}^{N}\frac{g(n)}{g(n+x)}$

[dawn finch]

RoyalBanana

[azure crane]

Interesting

Imma test this

noo why doesn't it work

[jovial rock]

Uhh, i don‘t get it

[azure crane]

$\ln(\prod)=\sum \ln$

[dawn finch]

RoyalBanana

[azure crane]

the reason that works is bc ln(ab)=ln(a)+ln(b)

so ln(abc)=ln(a)+ln(b)+ln(c)

etc

[jovial rock]

Yeah but I didn’t take the ln of the product, I just simplified the equation and then just used the sum thingy to make it more compact

Am confused

[azure crane]

?

[azure crane]

What ln trick?

Ohhh I see you took the ln of both sides ofc

yeah sure that works

[azure crane]

Ah lemme check it for x=2

$g(1)g(2)=g(6)\frac{g(1)}{g(3)}\frac{g(2)}{g(4)}\frac{g(3)}{g(5)}\frac{g(4)}{g(6)}?$

[dawn finch]

RoyalBanana

[azure crane]

ohh I seeee

the number of extras depends on x

[jovial rock]

Yeah

[azure crane]

so it's $\prod_{n=0}^{x-1}g(N+n)$ not $g(N)$

[dawn finch]

RoyalBanana

[jovial rock]

Math is complicated. My Brain is burining rn. I never used the sum and capital pi in my life before

[azure crane]

ahhh but the requirement for the formula was $\lim_{N \to \infty} g(N+n)=g(N)$

[dawn finch]

RoyalBanana

[azure crane]

So we can just say g(N)^x !

[azure crane]

Fair lol

remember to take breaks lol

[azure crane]

Ahh so this is how we get rid of the x being a non integer problem, using the fact that the limit is equal!

So finally $\prod_{n=1}^{x}g(n)=\lim_{N \to \infty} g(N)^x\prod_{n=1}^{N}\frac{g(n)}{g(n+x)}$

[dawn finch]

RoyalBanana

[azure crane]

Yesss it works

progress!

[jovial rock]

Nicee

[azure crane]

Wait we could use this for extending tetration possibly

and lnk

[jovial rock]

Yeah, maybe

[azure crane]

tetration can be written easily as recursion

${}^ba=a^{{}^{b-1}a}$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

[azure crane]

Ok but we gotta solve the general case first ofc

so far we got it for sums and products

[jovial rock]

I think I have an idea. Couldn‘t we use this to Break down tetration into smauler operations and then extend it to the reals?

[azure crane]

These are clues that definitely help

[jovial rock]

Yeah I just learned how to use them effectively

lol

ln2(a^^b) = ln(a^^b) + ln2(a) I guess

[azure crane]

$f(x+1)=g(f(x))$ I mean

[dawn finch]

RoyalBanana

[jovial rock]

Ah

Yeah

[azure crane]

that would instantly solve our f(x+1)=f(x)ln(f(x)) thingy

[jovial rock]

I’m just to stupid to understand it right now, I can’t help you a lot rn

[jovial rock]

This would be crazy

[azure crane]

hey you're not stupid it's complicated even for me lol

[jovial rock]

Ah okayy haha lol

[azure crane]

Wait

no:/

[jovial rock]

I just need a little bit more time to wrap my mind around it

[azure crane]

Yeah it's crazy lol

[jovial rock]

Ah, I mean ln2(a^^b) = ln(a^^b-1) + ln2(a)

[azure crane]

the examples help me wrap my mind around it though lol

[azure crane]

yep!

[jovial rock]

Because ln(a^^b) = a^^(b-1) * ln(a)

[azure crane]

b-1

[jovial rock]

Yeah

I forgot

[azure crane]

h

my brains frying

[jovial rock]

Same

[azure crane]

Useful

Imma try $f(x+1)=f(x)^{g(x+1)}$ now

[jovial rock]

Hmm

[dawn finch]

RoyalBanana

[azure crane]

Oh I understand

So it cancels for the product bc division is the inverse of multiplication
It cancels for the sum bc subtraction is the inverse of adding

[azure crane]

soo.. the inverse of exponentiation is log or roots

I think it's roots here bc g is the exponent

Hmm

Need some notation for this

oh yeah $(x^a)^b=x^{ab}$

[dawn finch]

RoyalBanana

[azure crane]

Can use that for now

just to make it easier to write

I'm doing this on desmos and I gotta say I think I'm making progress

good progress!

https://www.desmos.com/calculator/0jgtpv1ifk

[jovial rock]

Uhh, looks Crazy

[azure crane]

True, but you're just not used to the capital pi notation yet

I somewhat understand this

[jovial rock]

Yeah I know

I‘ll get used to it in some time

I‘ll just solve some Problems with it

One of them is my beloved f(x) = f(x-1) * ln(f(x-1))

[azure crane]

yep

[azure crane]

I'm cooking

[jovial rock]

Nicee

[azure crane]

I think I'm onto something here..

[azure crane]

I made this thing F, in this case $F(f,x)=f^{g(x)}$

because $f(x+1)=f(x)^{g(x)}$

[dawn finch]

RoyalBanana

[azure crane]

it's proving very useful

so in every case $f(x+1)=F(f(x),x)$

[dawn finch]

RoyalBanana

[azure crane]

by definition

Let's define $F^{-1}$, in this case $F^{-1}(f,x)=f^{\frac{1}{g(x)}}$

[dawn finch]

RoyalBanana

[azure crane]

$F(F^{-1}(f,x),x)=f$

[dawn finch]

RoyalBanana

[azure crane]

by definition

and also ig

$F^{-1}(F(f,x),x)=f$

[dawn finch]

RoyalBanana

[jovial rock]

Finally something: f(x+k) = f(x) * k! * capitalPi(m)(lnm(x))

[jovial rock]

Probably that’s not even true

Ahhh

[jovial rock]

Uh

[jovial rock]

Ah I See recrusive F

[jovial rock]

Hmm, nicee

Wasn‘t thinking about that

[jovial rock]

Wait I think this is actually True because it‘s ln, ln(f(x)) = ln(f(x-1)) + ln(ln(f(x-1))), so this is addition and this would make Perfect sense for the factorial

The only Problem now is the capital pi which is only of integers

Hmm, could we extend the lnk function to the reals?

[azure crane]

ye

sure why not

I'm almost done with the extension thing btw

soon

soon you'll be able to solve $f(x+1)=g(f(x))$ I think for any g(x)

[dawn finch]

RoyalBanana

[azure crane]

*as long as f(x) flattens out

the limit definition

uhh

As long as $\lim_{N \to \infty} f(N+1)-f(N)=0$

[dawn finch]

RoyalBanana

[azure crane]

Oh interesting

[azure crane]

so this isn't true for tetration

but!

so normal teteration is for example

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

that doesn't meet that limit requirement

but we can do a little trick

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

that is equal to 1/the tetration function

[azure crane]

This is true for 1/tetration function ofc bc the tetration function grows exponentiatially large so 1/it grows less and less

interesting

so we can find the extension of 1/it then do 1/that which cancels out to just give the extension of the tetration function!

Ok more similar notation lol (dw won't be needed for long) $F_n(f,x)=F(F(F...(F(f,x+1),x+2),x+3...,x+n)$

[dawn finch]

RoyalBanana

[azure crane]

Temporary notation

just to make it easier to write

[jovial rock]

Wait this would be pretty useful for lnk

Nicee

[azure crane]

I'm using it for the general case lol

this is kinda interesting how it works lol

[azure crane]

Ahhhh I'm understanding it now

$F^{x}(F^{-1}_N(F_N(f(0),0),x),N)$

[dawn finch]

RoyalBanana

[azure crane]

ok now to rewrite all this notation lol

$f(x+1)=F(f(x),x)$

[dawn finch]

RoyalBanana

[jovial rock]

Hmm

[azure crane]

Oh wait $F_N(f(0),0)=F_{N-1}(f(1),1)=F_{N-2}(f(2),2)=...=f(N)$

[dawn finch]

RoyalBanana

[azure crane]

So $F^{x}(F^{-1}_N(f(N),x),N)$

[dawn finch]

RoyalBanana

[azure crane]

I think that's as far as simplifying goes

[azure crane]

Oh I didn't mention $F^{n}$ is like $F_{n}$ except the terms on the right don't go up they stay the same there's just n terms

[dawn finch]

RoyalBanana

[azure crane]

Let's test this now
$F(f,x)=f^{g(x)}$ because $f(x+1)=f(x)^{g(x)}$

and in this case $F^{-1}(f,x)=f^{\frac{1}{g(x)}}$

[dawn finch]

RoyalBanana

[azure crane]

$F_N^{-1}(f,x)=?$

[dawn finch]

RoyalBanana

[azure crane]

Hmm

$F_2^{-1}(f,x)=f^{\frac{1}{g(x+1)g(x+2)}}$

[dawn finch]

RoyalBanana

[azure crane]

ah

$F_N^{-1}(f,x)=f^{\frac{1}{\prod_{n=1}^{N}g(x+n)}}$

[dawn finch]

RoyalBanana

[azure crane]

I think

So $F^{x}(F^{-1}_N(f(N),x),N)$

[dawn finch]

RoyalBanana

[azure crane]

so then thats equal to..

$F^{x}(f(N)^{\frac{1}{\prod_{n=1}^{N}g(x+n)}},N)$

[dawn finch]

RoyalBanana

[azure crane]

ok..

now $F^{n}(a,b)=?$

[dawn finch]

RoyalBanana

[azure crane]

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

[dawn finch]

RoyalBanana

[azure crane]

$F^{n-2}(({a^{g(b)})}^{g(b)},b)$

[dawn finch]

RoyalBanana

[azure crane]

Aka

$F^{n-2}(a^{{g(b)}^2},b)$

[dawn finch]

RoyalBanana

[azure crane]

Oh I see

$F^{n}(a,b)=F^{n-k}(a^{{g(b)}^k},b)$

[dawn finch]

RoyalBanana

[azure crane]

Let's do k=n ofc so we can get rid of F

$F^{n}(a,b)=a^{{g(b)}^n}$

[dawn finch]

RoyalBanana

[azure crane]

interesting

Ok

$F^{x}(f(N)^{\frac{1}{\prod_{n=1}^{N}g(x+n)}},N)$

[dawn finch]

RoyalBanana

[azure crane]

so then that's equal to..

${(f(N)^{\frac{1}{\prod_{n=1}^{N}g(x+n)}})}^{{g(N)}^x}$

[dawn finch]

RoyalBanana

[azure crane]

aka

$f(N)^{\frac{{g(N)}^x}{\prod_{n=1}^{N}g(x+n)}}$

[dawn finch]

RoyalBanana

[azure crane]

ok

there

so this should work

in theory

wait what was f(N)?

ah yes

$f(0)^{\prod_{n=1}^{N}g(n)}$

[dawn finch]

RoyalBanana

[azure crane]

So we can plug that in here..

If $f(x+1)=f(x)^{g(x)}$ and f flattens out as x gets bigger, then

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

[dawn finch]

RoyalBanana

[azure crane]

My brains fried

[azure crane]

Ah I made a typo f(0) not f(N) lol

[jovial rock]

I do not understand haha

[azure crane]

that's ok

but the important part is it works lol

fair lol

anyway I just realized that specific way doesn't work anyway

for $f(x+1)=g(f(x))$ you would need to know $g_n(f(0))$ for non whole n which obviously if you knew that then you wouldn't need this method lol

[dawn finch]

RoyalBanana

[azure crane]

I'm gonna try something else idk

[azure crane]

hmm

So $f(x+n)=g_n(f(x))=g_x(f(n))$

[dawn finch]

RoyalBanana

[azure crane]

hmm

I have an idea

so basically g "steps" 1 to the right on the graph

g_n "steps" n to the right

hmm

Two steps of 1/2 is the same as a step of 1

$g_{\frac{1}{2}}(g_{\frac{1}{2}}(x))=g(x)$

[dawn finch]

RoyalBanana

[azure crane]

so we need a function that satisfies f(f(x))=g(x)

[jovial rock]

Wdym?

Like ln1/2(x) ?

[azure crane]

yeah

in that case g(x)=ln(x)

[azure crane]

I wonder if this has just one solution or multiple solutions

[jovial rock]

I was Messing around with my recrusive function f(x) = f(x-1) * ln(f(x-1) and through I rerange some stuff to get f(x) = e^(f(x-1)) + f(x-1), this function is grows very strong for positive values, but is very stable for negative values. Maybe I could use something with x aproaches negative infinity

Idk

[jovial rock]

Hmm, not sure

[jovial rock]

I think it should have multiple

[azure crane]

I was thinking

about the solution in that video

if you get just one segment 1 long then it solves the whole function

So if the function flattens out then we can get that segment

that's the idea

normally though they get it by doing some other method that doesn't work here but I think I have an idea

[azure crane]

Can you show the steps you did to get that?

[jovial rock]

Yeah true

[jovial rock]

Nvm my shuffeling around with my function isn‘t true sadly, I forgot the ln(f(x)), instead of the f(x)

[jovial rock]

I was interested in the stuff of the video and I was thinking about solving f(x) = x^f(x-1), which is the same like the sum and factorial, but with exponetiation. During the process of solving it, I found it’s pretty much iterated logarithms

This would solve so much: lnk(x) = ln(lnk-1(x))

[jovial rock]

Maybe there could be some other capital letter for exponetiation, the only problem is that it’s not commutative

[azure crane]

I had an idea

so we have this requirement $\lim_{N \to \infty} f(N+x)-f(N)=0$

[dawn finch]

RoyalBanana

[azure crane]

For any x

[jovial rock]

Yeah

[azure crane]

so that ofc means $\lim_{N \to \infty} f'(N)=0$

[dawn finch]

RoyalBanana

[jovial rock]

Only if it flatterns out

[azure crane]

I'm pretty sure that also means though $\lim_{N \to \infty} f''(N)=0$

[dawn finch]

RoyalBanana

[azure crane]

If it flattens out

[jovial rock]

Hmm

[azure crane]

I'm pretty sure that also also means though $\lim_{N \to \infty} f'''(N)=0$

[dawn finch]

RoyalBanana

[jovial rock]

Derivative?

[azure crane]

If the derivative of the function is going to 0

that means the derivative of the derivative is also going to 0

etc

so all of the derivatives are 0

So $\lim_{N \to \infty} f^{(n)}(N)=0$ for any n

[dawn finch]

RoyalBanana

[azure crane]

Now

[jovial rock]

Yess

[azure crane]

We can get f'(x-1) if you take the derivative of both sides of the recursive formula and rearrange a bit

so we can get f'(x-n) for some whole integer n

[azure crane]

meaning we can take some huge number N, estimate the derivative there and backtrack, finding all of the derivatives before it with this

So we can take the limit as N goes to infinity and the estimation of the derivative will get better so the derivatives will get better and eventually they should be exactly right

so we can figure out the all the derivatives of the points

right?

[azure crane]

now there's no reason we can't do that again and get a formula for f''(x-n)

so we can do that again

get the 2nd derivative of all the points

etc

we can get every derivative

Taylor series!

lol it would be very complicated though

but I think it'd work

[jovial rock]

Hmm

Sounds interesting, I just don‘t completely understand it yet

Taylor series is something with iterative derivative i guess

[azure crane]

With taylor series you just need all the derivatives of a function at one point

and that's all

[jovial rock]

Ah okay

Sounds cool

And you just take the sum of all These derivatives?

Ah I See where you‘re going with this

Isn‘t there something with the factorial?

[jovial rock]

Hmm, taylor Series with the tetration Sounds interesting to me