(Engineered) diehards

Pavgran · Post by **Pavgran** » April 13th, 2022, 11:01 am

ihatecorderships wrote: ↑
April 13th, 2022, 9:27 am

Pavgran wrote: ↑
April 13th, 2022, 7:36 am
And the equations are

m=255977307969537
u_0=7
u_1=u_0+m*121^(u_0)-516
u_2=u_1+m*121^(u_1)-512
u_3=u_2+m*121^(u_2)-508
u_4=u_3+m*121^(u_3)-504
u_5=u_4+m*121^(u_4)-500
u_6=u_5+m*121^(u_5)-496
u_7=u_6+m*121^(u_6)-492
u_8=u_7+m*121^(u_7)-488
v=u_8+m*121^(u_8)-484
T=4*m*121^v+2054

And how large is that in scientific notation? It's hard to get a good idea of how large something is when its expressed in recursive equations like that.

Well, the structure is very regular. Each subsequent term is used as a power of 121, and that gives the main growth speed. Multiplicating that by some number just translates the thing one level above by small additive number, and that gives negligible effect.
So, we can say that
m≈121^6.917756
u_1≈121^13.917756
u_2≈121^(121^13.917756)
u_3≈121^(121^(121^13.917756))
u_4≈121^(121^(121^(121^13.917756)))
u_5≈121^(121^(121^(121^(121^13.917756))))
u_6≈121^(121^(121^(121^(121^(121^13.917756)))))
u_7≈121^(121^(121^(121^(121^(121^(121^13.917756))))))
u_8≈121^(121^(121^(121^(121^(121^(121^(121^13.917756)))))))
v≈121^(121^(121^(121^(121^(121^(121^(121^(121^13.917756))))))))
T≈121^(121^(121^(121^(121^(121^(121^(121^(121^(121^13.917756)))))))))

That is more than 121 ↑↑ 10 but less than 121 ↑↑ 11.
Switching the base to 10, we can say that
T≈10^(10^(10^(10^(10^(10^(10^(10^(10^(10^29.306343)))))))))
≈10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^1.466961))))))))))
≈11^(11^(11^(11^(11^(11^(11^(11^(11^(11^(11^1.391488))))))))))
That is more than 11 ↑↑ 11 but less than 11 ↑↑ 12.
It is also more than 11 ↑↑↑ 2

Macbi · Post by **Macbi** » April 13th, 2022, 11:07 am

Nathaniel wrote: ↑
April 13th, 2022, 10:41 am
ihatecorderships wrote: ↑
April 13th, 2022, 9:27 am
And how large is that in scientific notation? It's hard to get a good idea of how large something is when its expressed in recursive equations like that.
The reason for using recursive equations or tetration notation is that you cannot effectively write the lifespan using scientific notation. It's bigger than 10^999999999999....9, where there are as many "9"s in that exponent as there are atoms in the observable universe.

In other words: if you want to use scientific notation to get an idea of how large this number is, then it's much, much larger than you're thinking it is.

You could say it's about 10^10^10^10^10^10^10^10^10^(2.0258*10^29).

EDIT: Pavgran pointed out to me that the number at the top of that tower should be 2.0246*10^29. I must have made a mistake somewhere. This is either a tiny error or a huge error, depending on how you look at it.

Post by **dvgrn** » April 13th, 2022, 11:32 am

Nathaniel wrote: ↑
April 13th, 2022, 10:41 am
It's bigger than 10^999999999999....9, where there are as many "9"s in that exponent as there are atoms in the observable universe.

Let's see, 10^82-ish atoms in the observable universe, so Nathaniel's Nines Number is ~10^(10^(10^82))...

u_1 is 9.720734e+28, so
u_2 is 9.720734e+28 + 255977307969537*121^9.720734e+28 - 512
... time to throw exact numbers overboard here ...
u_3 is a similarly big multiple of 121^u_2, so it's easily bigger than 121^(121^(10^28)), same general order of size as Nathaniel's Nines Number,
and
u_4 is bigger than 121 raised to that number, so it's big enough that Nathaniel's Nines Number is tiny by comparison.

We still have to do u_5, u_6, u_7, u_8, v, and T, adding another "121^" at each stage. Yikes.

When I was growing up, Skewes' number was thought to be really big, to the point that Isaac Asimov wrote an article called "Skewered!" about it -- he ultimately couldn't come up with a good mental reference for anything that big. And yet that original Skewes' number is just 10^(10^(10^34)), which is somewhat smaller than u_3 and Nathaniel's Nines Number.

Kazyan · Post by **Kazyan** » April 13th, 2022, 7:30 pm

An additional toad via more spacedust: