/sci/ - Science & Math

>>7098180
2^11
(2^11)!
211!

2^11 shitwad.

>>7098182
No, just redefine 2^11 to be 2 to its own power 11 times, with parenthesis on the right. Or select any higher-order hyperoperation you want for an arbitrarily large value that fits more or less with current convention.

>>7098180
-1/12 obviously

>>7098180
2^(11!)

>>7098180
What operators can we use?

Because if we can use any, then I define f(x) -> y, with y > whatever you can come up with, and create the number f(211).

>>7098215
Except f(x) is a function and not an operator.

>>7098194
Winrar

>>7098180
<span class="math">\delta(x-211)[\math][/spoiler]

>>7098180
<span class="math">\delta(x-211)[/spoiler]

>>7098218
Operators on S are functions f:S\times S\to S.

>>7098218
g(x, y) -> S(S(S(S...S(x)...)))), y times

Does this seem familiar?

Here's a better game: name the largest natural number you can using only a single post on /sci/. No pictures and no links.

>>7098261
the successor function (operator) is still a recursive function.

>>7098261
>not using Church numerals

>>7098266
<span class="math">9 \uparrow^{9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999}9[/spoiler]

This is already too big.

>>7098276
Too big for what?

>>7098279

Let H(x,y,n)=H(x,H(x,y-1,n-1),n) where H(x,y,1)=x+y. Then H(2,2,n)=4 for all natural n and also H(2,3,999999999999999) is probably one of the bigger numbers you've ever seen. Now let G(x,y,m,n)=H(x,y,G(x,y,m,n-1)) where G(x,y,m,1)=H(x,y,m).

Now think about G(2,3,3,9999999999999999999999999999999999999999999999999999).

That is in the running for one of the biggest numbers anyone has ever expressed ever.

>>7098218
>>>highschool

If we can include any symbols we want, than there is no higheset number. 211!!!!..... you can do that forever.

(2!)^(11!)!

>>7098266
Let S be the set of numbers in this thread.

Let x = sup S.

Obviously if you can use as many symbols as you want, then there is no highest number.

>>7098269
So? If addition is just the application of the successor function, then it is itself a function. It's trivial to show it's an endomorphism even.

>>7098336
Meant without referencing other posts, hence a "single post".

∞2^∞11∞

>>7098367
211!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!...
It's a stupid question, you see.

>>7098367
He said natural number.

>>7098301
> H(x,y,n)=H(x,H(x,y-1,n-1),n)
I think you meant H(x,y,n)=H(x,H(x,y,n-1),n) but it doesn't really matter. I can do much better.

Let <span class="math">H_k(x,y,z_1,z_2,\dots, z_{k-1},z_k)= H_{k-1}(x,y,z_1, z_2,\dots,z_{k-2}, H_k(x,y,z_1,z_2, \dots,z_{k-1},z_k-1 )[/spoiler] with <span class="math">H_k(x,y,z_1,z_2, \dots,z_{k-1},1)=H_{k- 1}(x,y, z_1,z_2,\dots, z_{k-2},z_{k-1})[/spoiler] and <span class="math">H_0(x,y) = x+y[/spoiler]. That way, <span class="math">H_1(x,y,n)=H(x,y,n)[/spoiler] from your post, <span class="math">H_2(x,y,n,m)=G(x,y,n,m)[/spoiler], and <span class="math">H_3(3,3,3,3,3)[/spoiler] is already much, much larger than the number you wrote.

Then you could define <span class="math">f(n) = H_{f(n-1)} ( f(n-1), f(n-1), \dots, f(n-1), f(n-1) )[/spoiler] with <span class="math">f(-1)=3[/spoiler] so that <span class="math">f(0)=H(3,3)=6[/spoiler], <span class="math">f(1) = H_{6} (6, 6, 6, 6, 6, 6, 6, 6)[/spoiler], and <span class="math">f(2)[/spoiler] is so ridiculously humongous compared to anything you can write with your function G that you really don't want me to look at <span class="math">f^{f(9)}(f(9))[/spoiler] or other weird combinations that I can write without redefining other functions.

I win.

>>7098380
No he didn't

>>7098384
Read >>7098266

Unless you were not answering that-.

>>7098381
Yeah I did fuck up, and thanks, that was nice. I hoped someone would feel compelled manufacture that sequence of operations.

>>7098431
No problem, that was fun.

>>7098336
>implying S is bounded

>>7098493
Threads can only have a finite amount of posts, so yes it is.

>>7098194
best post in this thread, thanks for the laugh

>>7098510
Educate yourself, fucking uncultured swine.
http://en.wikipedia.org/wiki/Berry_paradox

>>7098281
Legitimately lmao'd

>>7098194
Best today, mode pin it

>>7098180

The answer is infinite, since whatever function you can come up with, can be re-functioned, to wit:

(2^11)!
((2^11)!)!
(((2^11)!)!)!
etc.

>>7098194
Well fucking memed

>>7098301
>>7098381
Hang on.

If H(x,y,n)=H(x,H(x,y,n-1),n) and H(x,y,1)=x+y then:
H(x,y,2) = H(x,H(x,y,1),2)=H(x,x+y,2) which would mean H(x,y,2) is independent of y.

Is that your intention?

>>7098959
No I made a mistake.

It should be H(x,y,n)=H(x,H(x,y-1,n),n-1).

That is, 2 raised to its own power three times with parentheses on the right is 2 to the power of two raised to its own power two times. Or more simply think of how 2^3=2*(2^2).

>>7098959
Ups yeah, I "fixed" his mistake incorrectly. I'm not sure how much I'd need to fix in my follow-up to be correct but it's probably possible to fix it and I'm too drunk to do it, so I'm going to be lazy and just assume there's a way to make it work.

>>7098194
Well played.

>>7099381
There is, I know what you meant too and it can be done. The definition I last gave was correct.

>>7098194
I'm going through a break up with my girlfriend of 2 years and this made me smile for the first time in days

thanks anon

>>7098276
Still not big enough for your momma

>>7098665
Except sup(S) doesn't require us to have complete information of the English language, merely to be able to list all the numbers in this thread at a given point on a relatively slow-moving board. Since so far each poster has only given finitely many numbers, sup(S)=max(S)= >>7098276

The only issue would be if I was to say, "Let X be the list of powers of two"
Now S is unbounded.

Here faggots :
2/(1-1)

>>7098221
yes!

>>7098180
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!211