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01:00 - 19:0020:00 - 23:00

01:05
Woah
hardcore parkour over here!
@55Cancri You are welcome here any time. Mostly do calculus, and some other funky maths. Feel free to read the transcript as well.
I most certainly will Simply. Thank you for inviting me!
No problem, feel free to enter any time... I shall put this room on my profile page... why isn't it already there? >.<
@55Cancri Also please check out my room as well. it needs a populous.

 The Universe of Quack

Where you are free to do proofwriting and generally hang out.
lol
@Typhon Think it needs to attract a more specific audience to get a good running
01:09
meh
:p
Ooooh thank you so much SBA, I am so excited to watch this and all of his other videos
I will become as knowledeable as Typhon!!!
@55Cancri I suppose I specialize in some combination of calculus, large finite numbers, and large non-finite numbers.
So if you have any questions in those, I can probably answer them
@55Cancri One day you might. Machine learning is definitely an ambitious thing to learn. :-)
Well, I still have to learn what much of that is first before I can even ask question about them!
01:12
perhaps one day you will put into being what I once wrote stories about
Mmm... actually, one of my teachers is interested in machine learning if you'd like to talk to him a bit @55Cancri
granted, if that were to happen I'd probably be quite upset. That would be... evil.
Not sure if he knows much
And what did you write about?
Yes Simply, is he someone on here?
And I don't know any machine learning let
I still need to better my program skills
I am still wrapping my head around the logic
I have html and css down
@55Cancri No, he's on Discord though
01:14
I am currently trying to learn javascript, sql, and node.js
I do Ruby
Horrible Ruby x.x
That is for data analysis right?
or is that R
Nah, I use it for writing programs that would theoretically output extremely large finite numbers
@55Cancri oh just a really twisted story way back when. It was basically a bunch of people on a forum making a story making fun of the forum and me inadvertently calling a forum game about spamming the word duck a "cult" instead of 'spam'. The ultimate plot twist was that it was all a computer simulation on someone's desktop computer. Everyone was just a program.
haha that is awesome Simply!
Like how many exponents worth?
01:15
Also for golfing my code down to the point that its just unreadable jibberish, but short code
lmao Typon!
@55Cancri Oh dear... nowhere near anything I can explain using just exponents I'm afraid...
@55Cancri recursive formulae that literally brute force using a googleplex tower of exponents of googleplexes fails to get past f(5,5)
graham's number?
@55Cancri even bigger
01:17
Oh, the tower thing!
that is smaller than f(1,1) of his function
I can exceed Graham's number in 71 bytes
I once read about that on xdxc article
you like keeping stacking these towers
Ah yes, xkcd...
xD
@55Cancri we're basically making numbers that exceed the collective ego of all presidents and kings in the DC multiverse.
01:17
and it go so large that just "writing" the tower would extend far beyond the sun
@55Cancri Its not quite towers... its a bit worse than exponential towers to say the least
our towers would be so large if expanded that they would wrap the entire universe and beyond.
their egos are quite large so that is an incredible feat
01:18
@55Cancri Aren't you supposed to be taking a leave?
by this i mean the collective current size it has expanded to
the universe has infinite spacial volume to expand into
@55Cancri Actually, it wouldn't fit inside the observable universe
lol, you guys absorbed me back into conversation
I was just going to watch attack on titan
01:19
Would you like to learn about my function?
but as long as you guys have interesting things to say, I am all ears!
A function @Typhon could not exceed xD
certainly!
imagine a titan where every molecule is a googleplex tower of googleplexes
lmao!!!!!
that is quite a large number
incomprehensible.
01:20
@SimplyBeautifulArt not even typhon is that big and yet he is the largest of all giants.
granted, he's also the only giant and most of him is just a rebuilt talos...
@55Cancri Nothing is so large
then again, I never finished that so I cannot technically cite it's size. XD
I also screwed up and thought talos was the colossus of rhodes
def f(a,b,c=a)a.times{a*=c>0?f(a,b,c-1):b<0?a:f(a,b-1)}&&a end
the latter is just a famous statue. XD
the former is a giant robot
That's my function
62 bytes
01:22
And I have no idea what it is telling me.
Three variables?
f(64,0,1) is larger than Graham's number
Yup, and its written in Ruby
It is basically googleplex level recursion
i have yet to beat it
truly beat it
@55Cancri You aren't supposed to read it, its golfed to fit in as few bytes as possible
@Typhon lol, what's that supposed to mean?
i think forcing it to take itself as input will beat it ultimately
oh I see, like when websites compress code to make the site load faster
01:23
@Typhon Nah, nesting itself is already built into itself
@SimplyBeautifulArt a really large amount of recusion
@55Cancri Well I do a lot of weird things...
Simply that is awesome
@SimplyBeautifulArt i was thinking of nesting the final input back into itself
for another round
maybe do it about 1000 times
01:24
@Typhon Yeah... I already got that covered
but...
you can next itself within itself again
double recursion?
so do you mean the output of that is only 62 bytes?
Nope, already built into the function
01:25
@SimplyBeautifulArt dude I can write f(f(f(f(f(x)))))!
and that will be larger!
@55Cancri No, the program size is 62 bytes. As there are no special characters, it is 62 characters including spaces
@Typhon In the amount of characters it took for you to write that, I can go much bigger
Bigger than you trying to nest if a Graham's number amount of times
Anyways, would you like to see how my function approximately behaves? @55Cancri
i can use string_execute
you two are like my role models.
definitely simply
01:26
and just use itself to determine the amount of recursion
Anyways, so it starts off rather simple.
f(a,-1,0)=a^2^a
and then use that to make another level of recursion.
For clarity, exponents are right to left, so a^2^a=a^(2^a)
@Typhon I told you! I already got that covered!
You could do a Graham's number amount of levels of recursion, but you won't beat me @Typhon
So, for example...
f(3,-1,0)=3^2^3=3^8=6561
What if I did a grahams number of a grahams number recursions?
Its kind of big, but no so big...
@Typhon Nope, dude, I got you way past covered
01:29
What do the -1 and 0 do?
what about f(grahams number) number of recursions?
Nothing. They are the initial values @55Cancri
@Typhon Any sorts of recursion is almost completely covered, trust me
And any sorts of recursions over the amount of recursions is already covered
And recursions over the amount of recursions that are over the amount of recursions!
So now it is just a matter of how large you make a
Believe me, I am good at what I do @Typhon
@55Cancri Oh no, that's not the goal. Making the second and third arguments bigger makes things go crazy
$\text{You got the math rendering?}$
what if I let it run for 24 hours continually doing more recusion at the level of f(graham's number)^f(grahams number) number of recursions per nanosecond and then make it close so that it prints out whatever number it was at that point?
01:31
yep
fair enough
I'll stick to the geometry
@55Cancri I actually know how super mario galaxy works in theory
$$f(a,-1,1)\approx \underbrace{f(f(f(\dots f(}_aa,-1,0),\dots) ,-1,0),-1,0),-1,0)$$
For example...
<=== that is an image of a grid one could theoretically walk on as an approximation of its geodesics. the math is all there. Just haven't made the walker.
@Typhon and yes, I've got you well beat
@SimplyBeautifulArt then I will use induction to find the larger number.
I will take your largest possible output and I shall.... add 1
01:34
$$f(2,-1,1)\approx f(f(2,-1,0),-1,0) = f(2^{2^2},-1,0) = f(16,-1,0)=16^{2^{16}}$$
wait!
I just realized
we are going about it all wrong
we need to find really really really small numbers
and plug them into 1/x
$$f(3,-1,1)\approx f(f(f(3,-1,0),-1,0) ,-1,0)$$
@55Cancri So you can see how this goes?
@Typhon The smallest non-zero real numbers are equivalent to 1/large numbers, where said large numbers are found separately, so that's not a useful idea.
damn, I thought I would figure it out on my own, but unfortunately I still do not see what you mean by initial values or the purpose of -1, 0
01:36
but finding small numbers will be easier...
The first number is the number of times you plug the function into itself right?
@Typhon So you say
@55Cancri Yup
meh
i dont like large numbers anyways
i like geometry
and number theory as in proofs and stuff
@55Cancri Approximately
you like geometry but not trigonometry?
01:37
large numbers is... boring to me
@55Cancri IKR? xD
@55Cancri what do you mean?
or were you the person who failed trigonometry?
oh yeah
i forgot i mentioned that
i was
01:39
Oh, do say, think you can make a larger number than me without being restrained to programs? @Typhon
@55Cancri Its not nice to laugh at people
I think that is awesome considering how intelligent you are
he wasn't
@55Cancri yeah...
I made the highest grade on my trig exam, but during lecture, I didn't understand a single thing! yet this guy who made lower than me followed along perfectly!
Is @55Cancri a 'he'?
If I may ask
Yes I am
01:40
k, just checking
But that just goes to show, grades and test taking don't matter
because I think that guy will walk away with a much better understanding of the subject than I will
@55Cancri i wouldn't know about that, but whatever. Everyone else did fine in my class.
Well, anyways, f(a,-1,1) gets very large, and it gets large much faster than f(a,-1,0) does
and I was totally envious. The grade felt undeserved and like a slap in the face
I just crammed the night before
I probably just didn't pay enough attention
01:41
okay thats it Simply!
so what does the third number represent?
how does it make the function increase faster?
lol
Well, its quite natural actually...
$$f(a,-1,b+1)\approx \underbrace{f(f(f(\dots f(}_aa,-1,b),\dots) ,-1,b),-1,b),-1,b)$$
see that loses me again
So $$f(3,-1,2)\approx f(f(f(3,-1,1) ,-1,1),-1,1)$$
I see two, so I am looking for two of something
but I dont see the 2 corresponding to anything on the right side
or the -1 for that matter
Well, it sort of works like this
01:45
okay change it to 4 and show me what it looks like then
-1,2 corresponds to -1,1
-1,1 corresponds to -1,0
-1,0 corresponds to a^2^a
you're blowing my mind Simply!
And on each step, you do 'a' amounts of whatever it corresponds to
So while f(a,-1,1) grows very fast, f(a,-1,2) grows much faster
And then f(a,-1,0) << f(a,-1,1) << f(a,-1,2) << f(a,-1,3) << ...
So you are subtracting -1 from the third number
sorry for not understanding this
it is probably something I just haven't come across yet in my math classes
we just do f(x) for now
@55Cancri Yup, but don't forget, we are also doing a lot of the functions
01:48
or f(g) for composition
Note, for example, the following:
what do you think?
f(3,-1,1) is very big, right @55Cancri
I know its big because the function will go into itself 3 times
Then f(3,-1,2) ≈ f(f(f(3,-1,1),-1,1),-1,1)
01:50
I also know the 1 will be subtracted from the third 1
The inside f(3,-1,1) is really big.
Let's call x = f(3,-1,1)
So we have... f(3,-1,2) ≈ f(f(x,-1,1),-1,1)
typhon, how did you make that?
what did you use?
We know x is gigantic, and that's how many times we have to make f(...,-1,0)
So while x is big...
f(x,-1,1) is insane
I see, this is a massive recursion
as its equal to f(f(f(f(........f(x,-1,0),....),-1,0),-1,0),-1,0),-1,0)
emphasis on how much stuff is inside the .... portion
And then we have to do f(f(x,-1,1),-1,1)
Which is crazy larger
And so on and so forth to higher numbers
01:53
my powers of abstraction fail me
And then f(a,0,0) corresponds to f(a,-1,a)
I feel as though you have this very precise and incredible understanding of what is happening here,
but I am unfortunately just not there
xD Well, it happens to be my night now
So I think I'm gonna head to bed now
lol I am so sorry Simply!
You are just too smart for me
No problem xD
01:54
I was really trying to understand
but I just couldn't get my head around those last two numbers
you subtract the second from the third,
but what did that mean?
As I like to say, it helps to try and make a larger number than me to start comprehending what is going on
@55Cancri Try making a large number yourself and see where it gets you :)
g'night
lol alright Simply
enjoy your night
and I hope you top your record in the future
with as few bytes as possible :)
@55Cancri I made it by making function objects and then building a 3D model off of the outputs of the functions.
 
4 hours later…
05:35
@SimplyBeautifulArt Looks like you have managed to fill your Calculus chat with large numbers XD is there an Integral and Series chat?
 
7 hours later…
12:19
@BrevanEllefsen xD Idk man!
$$C(\alpha)_0=\{0,1\}\\C(\alpha)_{n+1} =C(\alpha)_n\cup\left\{\gamma+\delta,\gamma \delta,\gamma^\delta,\omega^{\mathrm{CK}}_\gamma, \omega_\gamma,\chi_\gamma(\delta),\psi_\gamma( \eta):(\gamma,\delta,\eta\in C(\alpha)_n) \land(\eta<\alpha)\right\}\\\psi_\beta(\alpha) =\sup\left\{\max\{\gamma_n:( \gamma_n\in C(\alpha)_n)\land (\gamma_n<\omega^{\mathrm{CK}}_{\beta+1})\} :(n\in\mathbb N)\right\}$$
$$D(\alpha)_0=\{0,1\}\\ D(\alpha)_{n+1}=D(\alpha)_n \cup\{\gamma+\delta,\gamma\delta, \gamma^\delta,\omega_\gamma^{CK}, \omega_\gamma, \psi_\gamma(\delta),\chi_\gamma(\eta):(\gamma,\delta,\eta,\pi\in D(\alpha)_n)\land(\eta<\alpha)\} \\\chi_\beta(\alpha)=\sup\{\max\{\gamma_n: (\gamma_n\in D(\alpha)_n)\land (\gamma_n<\omega_{\beta+1})\}:(n\in\mathbb N) \}$$
@BrevanEllefsen Mind you, large infinite ordinals are also of interest
 
1 hour later…
13:26
Good morning @Nilknarf
@SimplyBeautifulArt Hi
Well...
what you wanna do?
Can you help with this?
https://math.stackexchange.com/questions/2326194/proving-the-relationship-between-the-definite-integral-and-area
I posted it as a question for anyone in case you aren't up to it
xD
@Nilknarf the proof involves telescoping sums and the mean value theorem
13:29
Can't find a good duplicate immediately, so...
So, we start with the mean value theorem...
$$f(x)\in C^2\implies \exists c\in[a,b]\implies f'(c)=\frac{f(b)-f(a)}{b-a}$$
This is the mean value theorem
Yes I know
Particularly, I would like you to consider $c\in[a+k\Delta x,a+(k+1)\Delta x]$
Applying the mean value theorem, we end up with...
$$f'(c)=\frac{f(a+(k+1)\Delta x)-f(a+k\Delta x)}{\Delta x}$$
Mhmm...
Note there exists a $\Delta x$ in front of the sum which then cancels with the denominator here
Oh wait, I've done it wrong. Gosh darn it...
$$f(c)=\frac{F(a+( k+1)\Delta x)-F(a+k\Delta x)}{\Delta x}$$
Where $F$ is the anti-derivative of $f$.
13:35
Mmm... I've still done it not quite right....
Anyways... multiply this by $\Delta x$
And then sum it up
and you more or less get a telescoping series
Whereupon you are left with something like $$A=\lim_{n\to\infty}F(a+(n+1)\Delta x)-F(a+\Delta x)$$
And by letting $n\to\infty$, we get $a+(n+1)\Delta x\to b$ and $a+\Delta x\to a$
Hence, we end up with $$A=F(b)-F(a)$$
As claimed
@ÉtienneBézout Hello and welcome to my realm!
Ahh, okay. I hadn't tried using the mean value theorem to reduce it.
:) It reduces quite beautifully I do say...
Yes, it does.
ooh, I was also wondering about some other things
regarding hyperoperations
Haha, imagine calculus without it... we'd all be ripping our hairs out trying to do them integrals
@Nilknarf yes?
Ha, yeah... that would be frustrating expressing every single area as a messy limit of a sum...
Okay
So how is a fractional hyper operation defined? If it even is defined?
13:41
Yeah... we're still working on making tetration extended to fractional heights.
The worst part is that hyper-operations are, in general, non-analytic
...
Well.
Meaning we can't invoke the uniqueness theorem
I was trying to differentiate tetration
I had to define it recursively
13:43
Of course
I started with $f_a(x)=^ax$
32
Q: Is There a Natural Way to Extend Repeated Exponentiation Beyond Integers?

Ben Blum-SmithThis question has been in my mind since high school. We can get multiplication of natural numbers by repeated addition; equivalently, if we define $f$ recursively by $f(1)=m$ and $f(n+1)=f(n)+m$, then $f(n) = m \times n$. Likewise, we get exponentiation by repeated multiplication. If $g(1)=m$ a...

25
Q: What is the derivative of ${}^xx$

Ali CaglayanHow would one find: $$\frac{\mathrm d}{\mathrm dx}{}^xx?$$ where ${}^ba$ is defined by $${}^ba\stackrel{\mathrm{def}}{=}\underbrace{ a^{a^{\cdot^{\cdot^{\cdot^a}}}}}_{\text{$b$ times}}$$ Work so far The interval that I am working in is $(0, \infty)$. It doesn't make much sense to consider neg...

And then
$$\ln f_a(x)=^{a-1}x\ln x$$
and I basically ended up with
$$f'_a(x)=^ax\bigg[f'_{a-1}(x)\ln(x)+\frac{^{a-1}x}{x}\bigg]$$
Oh thanks
np
25
Q: $n^{th}$ derivative of a tetration function

SindbadI stumbled upon this very peculiar function last summer, namely: $f(x)=x^{x^{x^{...^{x}}}}$, where there is a number $n$ of $x$'s in the exponent, I tried to find the derivative for the function and I was successful, it turned out not to be the most elegant formula but it worked. (Firstly, I inve...

Wow
Interesting reading :)
I'm curious about something else too (sorry :P)
How do people use polar coords to evaluate integrals? What is special about polar coords that makes it easier to evaluate?
Sometimes you have a region
and you want to find that area
Sometimes you are given $y=f(x)$, and you can integrate with respect to $x$
13:53
Sometimes you are given $x=f(y)$ and $y$ is most clearly not writable as a function of $x$, so you integrate with respect to $y$
Sometimes it is most clearly not a function with respect to $x$ or $y$, but it could be seen as a function with respect to $\theta$
Hmm. Okay.
And converting it makes it somehow easier to antidifferentiate?
Well, you need a function to even have a change at integrating (not antidifferentiating)
Ah, right
14:17
Okay, another thing
I was playing around with inverses
And I was wondering
If I have
$$h(x)=f(g'(x))$$
then can I conclude that
$$h^{-1}(x)=g^{-1}\bigg(\int f^{-1}(x)dx+C\bigg)$$
For some $C$?
It seems not
Take derivatives of both sides
And it should become clear that this is not true
Yes...
However, did you know that if you can find the antiderivative of $f(x)$ and $f^{-1}(x)$ exists, then you can find the antiderivative of $f^{-1}(x)$?
14:24
Yeah, there's a formula
So... wanna do the Gamma function reflection formula, complex analysis, something else, or nothing at all?
Perhaps
In the middle of a weird inequality question at the moment though
Yeah
@SimplyBeautifulArt Can't figure out how to rigorously prove that the first one is greater until $x=e$, and then it is smaller
14:45
I've done it on $0\le x\le1,x=e$
Bad question, but a fun problem :P
Can you give me a probability problem that uses an integral?
I don't do probabilistic integrals.
:(
Okay then, let's do some analysis
@SimplyBeautifulArt Or perhaps we should do the Gamma function thingy
Mmm, its a hard inequality
And I have to go sadly
Oh okay. Bye!
15:11
@SimplyBeautifulArt Where is TheGreatDuck? I was going to read some of his stuff about step-functions but I can't find his profile (even by searching).
 
3 hours later…
01:00 - 19:0020:00 - 23:00

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