Sorry expert Googologists, but I have a dummy question: How do you express
E20.55#(...(E20.55#(E20.55#63))...), with E20.55#63 of E20.55# 's
in the simplest hyper-e notation form? Is it E20.55#63#(E20.55#63)?
And also this one,
{E2.79#(...(E2.79#(E2.79#3))...), with {...
{E2.79#(...(E2.79#(E2.79#3))...), with
{E2.79#(...(E2.79#(E2.79#3))...), with E2.79#3 of E2.79# 's} of E2.79# 's}...} of E2.79# 's}, with E2.79#3 of { } 's?
I have read the 'hyper-e notation' article a few times, but still don't quite get it as most examples only use 100's towards the end. Appreciate your help.
Msiajoe74 (talk) 12:40, April 2, 2020 (UTC)
- Your answer of the first question looks correct. The answer of the second question is E2.79#3#1#(E2.79#3), if I am correct. (I note that this is the first time when I saw the definition of this notation, and hence I am not confident about the answer.)
- p-adic 05:12, April 2, 2020 (UTC)
- In general with Hyper-E:
- Ea#b = EE....EEa with b E's
- Ea#b#c = Ea#(Ea#(Ea#...(Ea#(Ea#b)))) with c E's
- Ea#b#c#d = Ea#b#(Ea#b#...(Ea#b#(Ea#b#c))..)) with d E's
- Ea#b#c#d#e = Ea#b#c#(Ea#b#c#...(Ea#b#c#(Ea#b#c#d))...)) with e E's
...and so on...
Username5243 (talk) 10:36, April 2, 2020 (UTC)
Thanks a lot for both of your kind help. Yes, I just expanded this E2.79#3#1#(E2.79#3) and I am pretty sure it represents this,
{E2.79#(...(E2.79#(E2.79#3))...), with {...
{E2.79#(...(E2.79#(E2.79#3))...), with
{E2.79#3#(E2.79#3)} of E's}...} of E's}, with E2.79#3 of { } 's. Thanks again !!
Msiajoe74 (talk) 12:40, April 2, 2020 (UTC)
- I advise you not to iterate multiple ellipses like "{…{}…} with {…} with X {'s {'s" because it is confusing. A better way is to separate the expression in the following two steps: Express E2.79#3#1#n for a positive integer n as E2.79#3(…E2.79#3(E2.79#3#1)…) with n E2.79's, and E2.79#3#m for a positive integer m as E2.79#(…E2.79#(E2.79#(E2.79#3))…) with m E2.79's. Then each expression includes at most single tuple of ellipses.
- So the first iteration "{E2.79#(…E2.79#(E2.79#3)…)} with E2.79#3 E's" can be expressed as E2.79#3#(E2.79#3). The second iteration "{E2.79#(…E2.79#(E2.79#3)…)} with {E2.79#(…E2.79#(E2.79#3)…)} with E2.79#3 E's E's" is "{E2.79#(…E2.79#(E2.79#3)…)} with E2.79#3#(E2.79#3) E's" and hence can be expressed as E2.79#3#(E2.79#3#(E2.79#3)). Similarly, you can prove by the induction on n that the (n-1)-st iteration is E2.79#3#(…E2.79#3#(E2.79#3)…) with n E's, and hence is expressed as E2.79#3#1#n. When we deal with such a recursive notation, it is good to use such inductive comparison.
- By the way, we are required to sign when we write a comment on a talk page or a forum page. You can sign by typing four tildas.
- p-adic 12:08, April 2, 2020 (UTC)
Thanks again, I learnt a lot from both of you! Msiajoe74 (talk) 12:40, April 2, 2020 (UTC)
May I ask, how do you express E3.97#【E3.97#【...【E3.97#【E3.97#(E3.97#3)#(E3.97#3)】】...】】, with E3.97#(E3.97#3)#(E3.97#3) of【】's? Is it E3.97#(E3.97#3)#(E3.97#3)#(E3.97#3)? Thanks in advance for your kind assistance again. TON618 13:47, April 17, 2020 (UTC)
No. It is E3.97#(E3.97#(E3.97#3)#(E3.97#3))#(E3.97#(E3.97#3)#(E3.97#3)).
PS. Did you change your signature?
Zongshu Wu 14:10, April 17, 2020 (UTC)
- Thanks, I see, but is there anywhere to simplify it? E3.97#(E3.97#(E3.97#3)#(E3.97#3))#(E3.97#(E3.97#3)#(E3.97#3)) looks so bulky..
- Yes, I changed my signature.
TON618 14:42, April 17, 2020 (UTC)
No. It is impossible to simplify. Zongshu Wu 14:53, April 17, 2020 (UTC)
- Ok, thanks again.
TON618 22:55, April 17, 2020 (UTC)
- By the way, it is good to use the signature with a link to your profile or talk page like TON618 in order to make others understand who you are.
- p-adic 23:53, April 17, 2020 (UTC)
- Oops, sorry about that. Thanks for reminding.
- TON618 04:58, April 18, 2020 (UTC)
May I ask? Is it possible to have E13.1#(E12#1)##34 as in E13.1#(E12#1)#(E12#1)#...#(E12#1), with 34 (E12#1)'s? I have only seen ## immediately after E, not in second row, as in E100##100#2.
- TON618 10:02, April 20, 2020 (UTC)
- Sure. According to the rule written in the article, you can expand in that way.
- p-adic 10:17, April 20, 2020 (UTC)
- Thanks for your kind reply.
- TON618 11:12, April 20, 2020 (UTC)
May I ask if ((Ea#b)#(Ea#b))#((Ea#b)#(Ea#b)) = (Ea#b)#(Ea#b)#(Ea#b)#(Ea#b) is true? If not, then what is its simplified expression? Thanks for your help.
- TON618 18 00:15, April 24, 2020 (UTC)