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TREE(Graham's Number) is just comically inefficient Googology. It makes me wince just to look at. Graham's Number isn't even on the same plane of existence as
by Eliezer 7y ago
TREE(Graham's Number) is just comically inefficient Googology. It makes me wince just to look at. Graham's Number isn't even on the same plane of existence as TREE. You'd just use TREE(TREE(3)) if you wanted to move up in the world. Somebody needs to tell them about the fast-growing hierarchy and how Graham's Number lives around \omega+1 while TREE hangs out with the small Veblen ordinal.
- wodenokoto 7y agoI have absolutely no idea what you are trying to say. The video is mostly about asking which series, TREE or Graham, grows quicker and how we can show it. It does this by starting out asking, which is bigger, TREE(G(64)) or G(TREE(64)). I fail to see how that is comically inefficient or how introducing TREE(TREE(3)) is what they should have done instead. >Somebody needs to tell them about the fast-growing hierarchy and how Graham's Number lives around \omega+1 while TREE hangs out with the small Veblen ordinal. Somebody should watch the video before speaking out about what the video should contain.
- chii 7y agothough the video actually didn't say why TREE grows faster than the epsilon function.
- wodenokoto 7y agoThat’s actually a fair point, they don’t explain why TREE grows faster than anything in the f-series. The argument for where in the f-series Graham’s series lies is also a bit hand-wavy.
- chii 7y ago> where in the f-series Graham’s series lies that is actually explained - fω+1(n) is grahams, because of the way grahams defines as iterated versions of up-arrows.
- pas 7y agoIs it even known why?
- Ravengenocide 7y agoMaybe I'm missing something obvious here, but that's what the video is about?
- deleted 7y ago[deleted]
- speakeron 7y agoThe video is a basic introduction to the fast-growing hierarchy. They do state where Graham's Number sits in the hierarchy, but the explanation doesn't go as far as to say where TREE sits.
- empath75 7y agoIt does actually. It says it grows faster than anything in the hierarchy.
- SAI_Peregrinus 7y agoFaster than anything in the Wainer hierarchy (which ends at $f_{\epsilon_{0}}(n)$. But it's reasonably easy to define a fast-growing hierarchy that uses the Veblen hierarchy of ordinals for the subscripts. Then TREE is $f_{svo}(n)$ where $svo$ is the limit of the finitary veblen sequence as the number of elements goes to $\omega$ (aka the small veblen ordinal).