My #MegaFavNumbers ! :)

This video is part of the MegaFavNumbers project. A bunch of Maths YouTubers have come together to make videos about their favourite numbers bigger than one million, which we are calling #MegaFavNumbers.

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I know, I know, I was supposed to only pick one. But picking two let me tell a cool story. I hope you get some inspiration from this!

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It’s been a while since I’ve done a video, but I’m back at it and full of cool ideas. I’m still setting up my home studio, and … yeah, I know you can see some of the greenscreen over my shoulder because I haven’t gotten that completely ironed out yet (pun TOTALLY intended). I’m also sorting my lighting. But, hey, look at that – g l o w y b a c k g r o u n d. : )

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The Taylor Series aims help people love math. Whether you're a student in a math class or someone who has ever asked the question, "What was the big deal about Calculus?" there's something here for you.

I am aiming to have one new video up every month. I hope you enjoy them!

Also, I named this the Taylor Series after the mathematical construct created by James Gregory and formally introduced by Brook Taylor ... and I picked this one in particular because my name is Derrick Taylor. :)

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@TheTaylorSeries

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Music Credits (from AudioBlocks):
Intro music: Thinking by Patrick Smith
Main background music: Island Fun by Neil Cross
Outro music: Midnight City Lights by Neil Cross

Intro: (0:00)

Context: (0:39)
Googleplex: (1:11)
Graham's Number: (3:16)
Challenge: (9:31)
My entry: (9:58)
Outro: (11:24)

[randomletters]

inconceivably large numbers can be hard to comprehend, but what about absurdly small non-zeros?
Graham's number to the power of negative one perhaps

[syedrahman]

Good topic. When I heard about Graham’s number I also thought of doing recursions to create bigger numbers. Then I heard about Tree (3) but even that wasnt quite humbling. The aha moment for me was when I saw a video on numberphile that discussed what wiuld be larger - G(tree3) or Tree(G64). To explain that they showed a generalization of how fast a series grows. I then realized that the growth rate is more interesting than the number itself. And the outcome was amazing - that G64 shows up pretty early in that sequence of series. While Tree is such a fast growing series that you cant even catch that rate even with ordinal infinities. A must watch

[erik-ictp]

The Busy Beaver Function's cool too. :)

[executeOrder]

This video existing without crashing the universe is final proof that we're not living in the matrix! ^^

[Lopoi]

Im now wondering, could you go the other way arround? like starting from subtraction to division and so on.... and then make a miniscule number near 0 and just invert it to make it gigantic?

So like.. something like number of I wonder if that would be bigger than g64 itself.

Edit: ok, now that I think about it that would just be something like 1/[(3^1/(2^g64))-1]. Ok its too late, gotta sleep. Great video, cya

[madhavmangla]

The time stops everytime I watch one of your videos
U are really great 👍
I really appreciate your work.

[SledgerFromTDS.]

Graham's Number = G(64) = 3 [^G(63)] 3, Which is a 3 followed with G(63) arrows ending with a 3, The G Function is a Fast Growing Hierarchy

[christopherguerra]

Im going enter my submittinace, but i have a question. U said that 3↑↑↑↑3 is G0. Ron, or excuse me, Mr. Graham says that 3↑↑↑↑3 is G1. Isn't or does that make a difference in the computation of G(64) because it creates a different starting point for the iteration of the Knuths ↑s?

[StaffanSAN]

When you have described your largest number, i respond by putting your number between parenthesis and add an exclamation mark at the end.

[JivrajSandhu]

Nice.
I watched the live stream also.
Cool stuff.

[bmwck]

I don't think this creates bigger numbers than up-arrow notation, but it is a nice way to generate BIG numbers. It involves polygons and numbers in them to create repeated exponentiation. Here is how it goes:
The smallest polygon is a triangle. and so, to begin the exponentiantion, a number in a triangle is a number raised to itself. Example: 3 in 1 triangle = 3^3=27

Next comes the square: a number in a square is a number in that number of triangles. Example: 3 in a square --> 3 in 3 triangles --> 3^3 in 2 triangles --> (3^3)^(3^3) in 1 triangle =[(3^3)^(3^3)]^[(3^3)^(3^3)] (be careful with parenthesis). Every unfold of a triangle raises the number to itself. 3 in a square = [(3^3)^(3^3)]^[(3^3)^(3^3)] = [27^27]^[27^27]= 1.71... x 10^40

Next up is the pentagon: a number in a pentagon is a number in that number of squares. Example: 3 in a pentagon --> 3 in 3 squares --> un folding the first square is solving 3 in a square so we are left with 1.71... x 10^40 in 2 squares... aaaand the unfolding blew up. If we unfold the next square, we have now 1.71... x 10^40 in 1.71... x 10^40 triangles aaaand still in 1square!!

And 3 in a hexagon, 3 in a heptagon...


I just love this!

I learnt about this method from a video about Wilhelm Ackermann

NIce to see you around Derrick :)

[SledgerFromTDS.]

So this Number is gonna be based off
The Knuths Up Arrow Notation as it is:
- Level 0: Counting (+1)
- Level 1: Addition (+)
- Level 2: Multiplication (×)
- Level 3: Exponential (^)
- Level 4: Tetration (^^)
- Level 5: Pentation (^^^)
- Level 6: Hexation (^^^^)
So let's start off with 2^5 = 2 × 2 × 2 × 2 × 2 = 32, 2^^5 = 2^(2^2^2^2) = 2^(65536), 2^^^5 = 2^^(2^^(2^^2^^2)) = 2^^(2^^(65536))
2^^^^5 = 2^^^(2^^^(2^^^2^^^2)) = 2^^^(2^^^(2^^(65536)))
HBUN(1) = 2^^^^5
HBUN(2) = 2(HBUN(1)^)5
HBUN(3) = 2(HBUN(2)^)5
Tip: Keep going on and on until you reach
HBUN(100) = 2(HBUN(99)^)5
Note: HBUN means Hector's Binary Upper Number
Function: HBUN(C) = 2(HBUN(C - 1)^)5
Size: This number is way bigger than G(64)

[MartOosterhoff]

my favourit number is a googol bc me and my friend always went around at school back when we were like 7 ttelling kids about howmany zeros the number had. the small kids were always amazed so easily lol

[MrxstGrssmnstMttckstPhlNelThot]

Graham's number was generated for solving a specific problem. What was that problem again?

[lumipakkanen]

My constest entry is the the number at the end of this video, but with the 0th operation "counting" replaced by repeated "g subscripting".

[rolandrickphotography]

I love what you’re doing with your channel. Is it also possible without (far too loud) background music? After 3 minutes watching, it had absorbed all attention and I saw your lips moving, but didn’t get any longer one word.

[GripFreak]

What about RAYO'S number? Isn’t this the largest known number? Btw, infinity is not a number. It's an idea.

[therealEmpyre]

Numberphile has a video about a number called Tree[3] that is vastly larger than Graham's number. Then they made another video comparing g_Tree[3] to Tree[g_64]. Apparently, the Tree function grows even faster than the g function, far far faster, so Tree[g_64] dwarfs g_Tree[3].
So, my ridiculously large number is Tree[Tree[ ... g_64 ... [Tree[g64]]]]...

[omlett]

I took you up on that challenge and it was really fun coming up with something, so first of all, thanks for that. I wanted to approach things a little differently, so i thought to myself, what if i just went backwards on the Chain of operations? So the Negative 1st operation would be subtraction, Negative 2nd operation is division, negative 3rd operation "rooting". "What would be the negative 4th operation?", i asked myself. Well, i called it "Root-Rooting" because i couldn't come up with a better name. And it's notated with a down-arrow, because i thought that would be fitting. So i started notating ∛3 as 3 ↓ 3, similar to how you wrote 3^3 as 3 ↑ 3. Then i went to "Root-Rooting" and thought "What would 3 ↓ ↓ 3 look like?" Basically, 3 ↓ ∛3, or ∛∛3, or at least that's how i defined it would be, hope it's okay with you. Then i played around with roots and such and found out that, 1) [ n ↓ n > n+1 ↓ n+1 ] and 2) [ n ↓ n > n ↓ n ↓ n ] and 3) [ n ↓ ↓ n > n+1 ↓ ↓ n+1]. So then i thought, what would [ n ↓ ↓ n ] approach if n = some really large number like *g64*. Did some testing again and found a pattern that hinted at such an operation approaching 1 from the positive side. That was my starting point from where i finally knew where i wanted to take this. If i did [ n ↓ ↓ n ] - 1, i'd get a number very close to 0, but still larger than it. You might see now where i'm going with this. Now if i use said "0-approaching" number and use it to divide a huge number like g64, for convenience sake, the resulting number will be massively bigger than before. So the number i was working with now was represented by this: [ g64/(g64 ↓ ↓ g64) - 1], which is what i called Reverse G0 (or "rG0" for short). Then i thought "Why not use double down-arrow on this number and see what happens?" What that looks like is this:
[ g64/(g64 ↓ ↓ g64) - 1) ↓ ↓ g64/(g64 ↓ ↓ g64) - 1 ] or just [ rG0 ↓ ↓ rG0 ] . The output of this will, just like g64 ↓ ↓ g64, approach 1 from the positive side, but will be much closer to 1 than g64 ↓ ↓ g64 is. With that, i put rG0 into the operation i used to get to rG0 in the first place, i.e.
[ (rG0/(rG0 ↓ ↓ rG0) - 1], which i called "rG1". Then do [ (rG1/(rG1 ↓ ↓ rG1) - 1] to get to rG2, and repeat until you arrive at "rG64" because ending it there just feels right, put rG64 into the formula one final time and you get this:
[ rG64/(rG64 ↓ ↓ rG64) -1 ]. That is my number.
Basically i focused on making a really really small number that's still bigger than 0, then use fractions to blow it out of proportions. I have no clue if all the math i did was right, if i made any mistakes please correct me. Also i have absolutely no idea how big this number actually is, and i'm fine with that. I hope i explained my method at least decently well, and i hope my approach to this challenge was entertaining to read. Thanks again for proposing this challenge :)

[irri]

I've always thought that like the mandelbrot set Infinity comes in packets. Kind of like saying" good enough ". Seeing the universe engineered with a modicum of slop.. bumpy🤔🤗😎