User:Unknown95387

I HAVE A DREAM, THAT I'LL CREATE A NUMBER THAT WILL BE LISTED AT LEAST IN EPSILON LEVEL (cuz i don't want to say something like "\(f_{\varepsilon_{0 }} (3)\) CHALLENGE COMPLETED. i want to get there on my own). i'll write here my progress day by day no matter if you care or nah. and i need to learn BAN.

29. November 2017
I've learn some basics about Bird's Array Notation (I don't mean \(\{a,b,c,d\}\) or something like that). I don't want to start thinking about some fast-growing function till the 2. December (Or 1.).

By the way: Exactly 1 year ago I've started my very first relationship (Nearly 3 moths long). \(\text{Boom}\) yeh yeh (And I didn't have another one when this relationship end)

So that's all for this day. I'll write something new tomorrow.

30. November 2017
Today I've decided to stop learning things about BAN, 'cause I didn't have an idea of how can be BAN difficult for a beginner just like me. So that means, I'll start with making simple fast-growing function tomorrow and I'll learn BAN meanwhile doing that function.

1. December 2017
I've been thinking about the type of function. If it would be something growing faster than FGH, or some type of function, which is used to define Clarkkkson. And I've decided to make something kinda different.

So let's say...

\(a \uparrow b = a^b\) - we all know up-arrow notation. It works the same. Nothing more special.

Now

\(a \Uparrow b = a \uparrow a^{a \uparrow a ^{a \uparrow a^{.^{.^{. }}}}} \), where are \(b\) \(a\)'s

\(a \Uparrow \Uparrow b = a \Uparrow a \Uparrow a \Uparrow . . . \Uparrow a \Uparrow a\), where \(a \Uparrow a\) is repeated \(b\) times.

So let's define, that:

\(a \Uparrow^{n} b = a \Uparrow^{n-1} a \Uparrow^{n-1} . . . \Uparrow^{n-1} a \Uparrow^{n-1} a\), repeated \(b\) times.

Now we will define \(\lambda\) function:

\(\lambda(n) = n \Uparrow^{n-1} n\)

In this case, \(\lambda\) function's growth rate is \(f_{\omega}(n)\) (I think).

That's all for this day. If you saw any mistake here, please, write on my talk page.

2. December 2017
This is an extension of \(\lambda\) function. Let's say, that \(\lambda^{x}(n) = \lambda(\lambda(\lambda . . . \lambda(\lambda(n)))...)))\), where are \(x\) iterations of \(\lambda(\).

Now,

\(\lambda^{\uparrow x}(n)\) = \(\lambda^{x^{\lambda^{x^{\lambda^{x^{(n)}(n)}(n)}(n)}(n)}(n)}\), where are \(x\) iterations of \(\lambda^{x}\).

And also,

\(\lambda^{\uparrow \uparrow x}\) = \(\lambda^{\uparrow x^{\lambda^{\uparrow x^{.^{.^{(n)}(n)}(n)}(n)}(n)}(n)}\), where are \(x\) iterations of \(\lambda^{\uparrow\ x}\).

And so on...

Don't forget, that we can define:

\(\lambda^{\Uparrow x^{\lambda^{\Uparrow x^{\lambda^{\Uparrow x^{.^{.^{(n)}(n)}(n)}(n)}(n)}(n)}(n)}(n)}\), there are \(x\) iterations of \(\lambda^{\Uparrow x}\)

\(\lambda^{\Uparrow x}(n) = \lambda^{x \uparrow x}(n)\) (\(x^x\) iterations)

\(\lambda^{\Uparrow\Uparrow x^{\lambda^{\Uparrow\Uparrow x^{\lambda^{\Uparrow\Uparrow x^{.^{.^{(n)}(n)}(n)}(n)(n)}(n)}(n)}(n)}(n)}\), there are \(x\) iterations of \(\lambda^{\Uparrow\Uparrow x}(n)\)

Aaaand it continues the same.

Now let me say, how fast-growing is this.

We have \(\lambda^{\Uparrow 2}(3)\). That's \(\lambda^{2 \uparrow 2^{\lambda^{2 \uparrow 2 }}} (3)\). I'll begin with computing the first \(\lambda^{2 \uparrow 2}\). That is \(\lambda(\lambda(\lambda(\lambda(3))))\). Let's count the first \(\lambda(3)\)

\(\lambda(3) = 3\Uparrow \Uparrow 3 = 3 \Uparrow (3 \Uparrow 3) = 3 \Uparrow (3 \uparrow 3^{3 \uparrow 3^{3 \uparrow 3 }} ) = 3 \Uparrow (27^{27^{27 }} ) = 3 \Uparrow \{27,3,2\}\). That's \(3\uparrow3^{3\uparrow3^{3\uparrow3^{.^{.^{. }}}}} \). And it's repeated \(\{27,3,2\}\) times! Well, \(\{27,3,2\}\) is approximately \(10^{6.3470460733 \times 10^{38 }} \). Keep on mind, that we didn't even compute the first \(\lambda(3)\).

Soooo. That fast is growing my \(\lambda\) function with extension. That's all. \(\text{Good} \text{night}\) ('Cause it's 23:45 for me).

3. December 2017
So we have our \(\lambda\) function. Now get ready for more powerful extension.

Let's say, that \(\alpha\) is a random type of extension like I wrote yesterday. For example \(\Uparrow 3\) can be \(\alpha\), and so on

So \(\lambda^{\alpha}_{n}\), where are \(n\) iterations of \(\lambda^{\alpha}\). I'll describe it more.

We have for example \(\lambda^{\uparrow\uparrow 2}_{1}(3)\). The number of iterations of \(\lambda^{\uparrow\uparrow 2}\) is repeated only once. But that's not end.

The number of towers of \(\lambda^{\uparrow 2}(3)\) will be equal \(\lambda^{\uparrow\uparrow 2}(3)\) times. If there was \(\lambda^{\uparrow\uparrow 2}_{2}(3)\), then the number of towers of \(\lambda^{\uparrow 2}(3)\) will be equal \(\lambda^{\uparrow\uparrow 2}_{1}(3)\) times. And so on...

I should make a normal notation out of this, because I planned to use this notation only for this number.

I'll try to compute \(\lambda^{\uparrow\uparrow 2}_{2}(2)\) tomorrow (I'll be wrong anyway tho).

If you saw any mistake, write me on talk page.

4. December 2017
So this will be only about computing \(\lambda^{\uparrow\uparrow 2}_{2}(2)\). Let's start.

There is 2 at the bottom, so there will be \(\lambda^{\uparrow\uparrow 2}_1(2)\) towers of \(\lambda^{\uparrow 2}_1(2)\). But what is \(\lambda^{\uparrow\uparrow 2}_1(2)\) equal to ? The number of towers of that number is \(\lambda^{\uparrow\uparrow 2}(2)\), where the tower is made by stacking \(\lambda^{\uparrow 2}(2)\).

We have no idea what is exactly \(\lambda^{\uparrow\uparrow 2}(2)\) equal to. Yes, I'm talking about \(\lambda^{\uparrow\uparrow 2}(2)\), 'cause we can know what is \(\lambda^{\uparrow 2}(2)\) equal to.

\(\lambda^{\uparrow 2}(2) = \lambda^{2}(\lambda^{2}(2))\)

\(\lambda^{2}(\lambda(\lambda(\lambda(\lambda(2)))))\)

We know, that the \(\lambda(n) > f_{\omega}(n)\). There are 8 \(\lambda\)'s, so it's \(f^{8}_{\omega}(2)\). We didn't end yet.

Now I'll show you the procedure of \(\lambda^{\uparrow\uparrow 2}(2)\):

\(\lambda^{\uparrow 2^{\uparrow 2}(2)}(2) = \lambda^{\uparrow 2^{f^{8}_{\omega}(2)}(2)} = \lambda^{f^{8}_{\omega^{f^{8}_{\omega}(2)}(2) }} \)

If there are \(f^{8}_{\omega^{f^{8}_{\omega}(n)}(n)}\) iterations, there can also be \(f^{8}_{\omega^{\omega }} (n)\) iterations, so \(f^{8}_{\omega^{\omega }} (n) = f^{8}_{\omega^{f^{8}_{\omega}(n)}(n)}\), isn't it ? If yes, then there will be \(\lambda(\lambda(\lambda...(2))))...)))\) with \(f^{8}_{\omega^{\omega }} (2)\) \(\lambda\)'s.

I'm 90% sure that I'm wrong (Write on my talk page if yes).

At least I tried.

14:30
Today I probably won't do nothing, because I'm ill and I've ran out of ideas for \(\lambda\) notation. The time "14:30" is written here, because I said "Probably".

19:30
Okay so I decided to don't write here anything. Not because I'm ill, but I've just ran out of ideas. I have no idea how to make \(\lambda\) function even stronger. I'm also avoiding making salad number of something with something, that does exist (I don't count up-arrow notation). So yeah.

20:19
Who's dying in flat somewhere in Prague ?

Matěj Sál

ok what is wrong with me

6. December 2017
I still haven't got any idea.

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