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Zeration
#91
Quote:Then, why not make a zerated to the a equal a plus 3 halves as a compromise?

Because that doensn't make any sense.
I know that to someone who is not familiar with the material, all the disccussion about zeration sounds meaningless and arbitrary, but it isn't. I encourage you to read the literature about that. Rubtsov and Romerio gave great motivation and heuristics in their papers. There is also Cesco Reale's article, and many rich threads on this forum. Look for them. The keyword here is mother law.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
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#92
(06/30/2022, 11:32 PM)MphLee Wrote:
Quote:Then, why not make a zerated to the a equal a plus 3 halves as a compromise?

Because that doensn't make any sense.
I know that to someone who is not familiar with the material, all the disccussion about zeration sounds meaningless and arbitrary, but it isn't. I encourage you to read the literature about that. Rubtsov and Romerio gave great motivation and heuristics in their papers. There is also Cesco Reale's article, and many rich threads on this forum. Look for them. The keyword here is mother law.
The Mother Law would make a°a=a+1. The Grand-Mother Law would make a°a=a+2. Why not make a°a=a+1.5, As a compromise zeration?
ฅ(ミ⚈ ﻌ ⚈ミ)ฅ
Please remember to stay hydrated.
Sincerely: Catullus
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#93
Again, because that doesn't make any sense. If you do that you are imposing the failure of both the laws, something that, per se, is a folish move if you don't come up with a better law. And the purpose of the laws is to add constraints to the functional equations in order to fully determine solutions to of hyperoperations where the zeroth-rank is a total function. Again, you need to read more literature and put some effort.

MSE MphLee
Mother Law \((\sigma+1)0=\sigma (\sigma+1)\)
S Law \(\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)\)
Reply


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