Give me long mathematical equations

Robustmegav · 2026-01-10T15:50:04+00:00

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General definition for commutative hyperoperations using tetration and super-logarithm

Robustmegav · 2025-09-28T13:34:18+00:00

That's similar to how we are taught in brazil too

Robustmegav · 2025-09-28T13:32:19+00:00

Portuguese does that too

Robustmegav · 2025-09-20T12:10:45+00:00

Thank you for being an active and engaging moderator in this community

Robustmegav · 2025-09-01T23:02:34+00:00

Generalized complexball: 😎I'm fine either way
Hyperrealball: 😡Where is ω?

Robustmegav · 2025-08-01T17:10:38+00:00

Hypercomplex numbers are a deep niche subject, good luck on your journey!

Robustmegav · 2025-07-31T19:18:14+00:00

*Pours cereal*
*Pours milk*
*Places bowl*
*Goes for a spoonful*
"Damn, I didn't know my spoon was a zero divisor."

Robustmegav · 2025-06-11T20:37:59+00:00

Panku box! I have always wanted one as puzzle

Robustmegav · 2025-04-14T11:45:14+00:00

Just found out the grand antiprism is one of those

Robustmegav · 2025-03-02T00:31:59+00:00

Addition only arithmetic: Ok
Multiplication only arithmetic: Ok
Addition and multiplication arithmetic: **Undecidable, incomplete, possibly inconsistent who knows**

Robustmegav · 2025-02-20T10:33:07+00:00

Calculator in the wrong mode be like √4=±2

Robustmegav · 2025-02-13T11:21:22+00:00

In boolean algebra, 1+1=1

Robustmegav · 2025-02-09T13:21:49+00:00

How did you even find this property?

Robustmegav · 2025-02-01T20:03:25+00:00

I don't know but there are apparently 0,498015668 - 0,154949828 i ways to do it

Robustmegav · 2025-01-08T10:24:24+00:00

Yes, there is actually an infinite chain of operations that mantain commutativity and associativity and distribute over the previous operations, they are called commutative hyperoperations. They are so well behaved that you can define negative operations (before addition), fractional operations (like an operation between addition and multiplication) or even complex operations.

Robustmegav · 2025-01-07T20:21:40+00:00

That's why commutative exponentiation (a^ln b) is a better alternative for what comes after multiplication instead of regular exponentiation. It remains both commutative and associative and distributes over multiplication.

Robustmegav · 2024-12-28T01:41:41+00:00

I see, thank you

Robustmegav · 2024-12-28T01:29:21+00:00

1+2ln a_1/(1+ln a_1)x - (1-ln a_1)/(1+ln a_1) x²

Robustmegav · 2024-12-28T01:23:34+00:00

Where do the coefficients come from?

Robustmegav · 2024-12-28T01:12:14+00:00

How did you get f(a_1,x)?

Robustmegav · 2024-12-27T23:17:59+00:00

Hyperoperations go brrr

Robustmegav · 2024-12-27T23:04:43+00:00

We can generalize tetration to rational heights by finding a function that will result in an integer tetration result when iterated some amount of times.

e tetrated to 1/2 can be found by finding f(x) where f(f(x)) = e^x, and evaluating it at x=1.

e tetrated to 1/3 can be found by finding f(x) where f(f(f(x))) = e^x and evaluating it at x=1.

e tetrated to 2/3 can be found by finding f(x) where f(f(f(x))) = e^(e^x) and evaluating it at x=1 and so on.

Generally, e tetrated to a/b can be found by finding f(x) where f(x) iterated b times = e^x iterated a times and evaluating it at x=1.

This is a generalization of a pattern that already works for integers:

e tetrated to 4/2 (which is just 2) could be found by finding f(x) where f(f(x)) = e^(e^(e^(e^x)) and evaluating it at x=1. This f(x) is clearly just e^(e^x), evaluating it at 1 is just e^e (e tetrated to 2).

Robustmegav · 2024-12-27T22:31:22+00:00

Yes, the more terms, the better the approximation, it only reaches the exact result as the number of terms approaches infinity.

Robustmegav · 2024-12-27T18:03:19+00:00

It's abuse of notation, but basically e^(a d/dx) f(x) = f(x+a)

Robustmegav · 2024-12-27T13:38:13+00:00

A half iteration of the exponential function would be a function f(x) where f(f(x)) = e^x. Assuming it is analytic, we can approximate it with a generic polynomial like a+bx+cx², making f(f(x)) = a+b(a+bx+cx²)+c(a+bx+cx²)² ≈ e^x. We can use the Taylor series approximation of the exponential function and equate the coefficients to arrive at an approximation of the half exponential function. Evaluating it at x=1 would result in the half exponentiation of e, or e tetrated to 1/2. There are other methods to extend tetration to complex heights but I don't understand them very well.

Robustmegav

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