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Let's try out some variables for a using the formula a ^ (a ^ (b - 1)). We'll also include the b = 0 included expressions.
| 2 ✱ −0 = 2 ^ (2 ^ (−0 - 1)) = 2 ^ (2 ^ −1) = 2 ^ (1/2) ≈ 1.4142135 or ²√2 |
| 2 ✱ −1 = 2 ^ (2 ^ (−1 - 1)) = 2 ^ (2 ^ −2) = 2 ^ (1/4) ≈ 1.1892071 or ⁴√2 |
| 2 ✱ −2 = 2 ^ (2 ^ (−2 - 1)) = 2 ^ (2 ^ −3) = 2 ^ (1/8) ≈ 1.0905077 or ⁸√2 |
| 2 ✱ −3 = 2 ^ (2 ^ (−3 - 1)) = 2 ^ (2 ^ −4) = 2 ^ (1/16) ≈ 1.0442737 or ¹⁶√2 |
| 3 ✱ −0 = 3 ^ (3 ^ (−0 - 1)) = 3 ^ (3 ^ −1) = 3 ^ (1/3) ≈ 1.4422495 or ³√3 |
| 3 ✱ −1 = 3 ^ (3 ^ (−1 - 1)) = 3 ^ (3 ^ −2) = 3 ^ (1/9) ≈ 1.1298309 or ⁹√3 |
| 3 ✱ −2 = 3 ^ (3 ^ (−2 - 1)) = 3 ^ (3 ^ −3) = 3 ^ (1/27) ≈ 1.0415284 or ²⁷√3 |
| 4 ✱ −0 = 4 ^ (4 ^ (−0 - 1)) = 4 ^ (4 ^ −1) = 4 ^ (1/4) ≈ 1.4142135 or ⁴√4 |
| 4 ✱ −1 = 4 ^ (4 ^ (−1 - 1)) = 4 ^ (4 ^ −2) = 4 ^ (1/16) ≈ 1.0905077 or ¹⁶√4 |
| 4 ✱ −2 = 4 ^ (4 ^ (−2 - 1)) = 4 ^ (4 ^ −3) = 4 ^ (1/64) ≈ 1.0218971 or ⁶⁴√4 |
| 4 ✱ −3 = 4 ^ (4 ^ (−3 - 1)) = 4 ^ (4 ^ −4) = 4 ^ (1/256) ≈ 1.0054299 or ²⁵⁶√4 |
As you could tell, as the value of b gets smaller into the negative zone, the answer of the Superpowerition operation gets closer and closer to "1" but it would never be exactly 1. The only way that would happen is if the value of b is a negative infinity number.
You may have also seen a pattern. Let's dig a bit into the formula.
| a ✱ b = a ^ (a ^ ( b - 1)) = a ^ (1/(a ^ ((b × −1) + 1))) = |
| 2 ✱ 0 = 2 ^ (2 ^ (−0 - 1)) = 2 ^ (1/(2 ^ (0 + 1))) = 2 ^ (1/2) ≈ 1.4142135 or ²√2 |
| 2 ✱ −1 = 2 ^ (2 ^ (−1 - 1)) = 2 ^ (1/(2 ^ (1 + 1))) = 2 ^ (1/4) ≈ 1.1892071 or ⁴√2 |
| 2 ✱ −2 = 2 ^ (2 ^ (−2 - 1)) = 2 ^ (1/(2 ^ (2 + 1))) = 2 ^ (1/8) ≈ 1.0905077 or ⁸√2 |
| 2 ✱ −3 = 2 ^ (2 ^ (−3 - 1)) = 2 ^ (1/(2 ^ (3 + 1))) = 2 ^ (1/16) ≈ 1.0442737 or ¹⁶√2 |
| 3 ✱ −0 = 3 ^ (3 ^ (−0 - 1)) = 3 ^ (1/(3 ^ (0 + 1))) = 3 ^ (1/3) ≈ 1.4422495 or ³√3 |
| 3 ✱ −1 = 3 ^ (3 ^ (−1 - 1)) = 3 ^ (1/(3 ^ (1 + 1))) = 3 ^ (1/9) ≈ 1.1298309 or ⁹√3 |
| 3 ✱ −2 = 3 ^ (3 ^ (−2 - 1)) = 3 ^ (1/(3 ^ (2 + 1))) = 3 ^ (1/27) ≈ 1.0415284 or ²⁷√3 |
| 4 ✱ −0 = 4 ^ (4 ^ (−0 - 1)) = 4 ^ (1/(4 ^ (0 + 1))) = 4 ^ (1/4) ≈ 1.4142135 or ⁴√4 |
| 4 ✱ −1 = 4 ^ (4 ^ (−1 - 1)) = 4 ^ (1/(4 ^ (1 + 1))) = 4 ^ (1/16) ≈ 1.0905077 or ¹⁶√4 |
| 4 ✱ −2 = 4 ^ (4 ^ (−2 - 1)) = 4 ^ (1/(4 ^ (2 + 1))) = 4 ^ (1/64) ≈ 1.0218971 or ⁶⁴√4 |
| 4 ✱ −3 = 4 ^ (4 ^ (−3 - 1)) = 4 ^ (1/(4 ^ (3 + 1))) = 4 ^ (1/256) ≈ 1.0054299 or ²⁵⁶√4 |
Looks like we found another translation for the Superpowerition representation!
| Formula | Coding | Same as | Same as |
|---|---|---|---|
| 1 | a ^ (1 / (a ^ ((b × −1) + 1))) | a ↑ (1 ÷ (a ↑ ((b × −1) + 1))) | a(1 ÷ (a((b × −1) + 1))) |
| 2 | a ^ (a ^ (b - 1)) | a ↑ (a ↑ (b - 1)) | a(a(b - 1)) |
| 3 | N.A. | (a ^ (a ^ b)) ROOT a or (a ↑ (a ↑ b)) ↓ a |
a√(a ↑ (a ↑ b)) ora√(a(ab)) |
| Also note that a√(b) or b ↓ a is the same as b(1/a) or b(1 ÷ a), so the above formula could be converted to this below: | 4 | (a ^ (a ^ b)) ^ (1 / a) | (a ↑ (a ↑ b)) ↑ (1 ÷ a) | (a(ab))(1 ÷ a) |
| Because (ab)c = a(b × c), the above formula can be converted to this below: | 5 | a ^ ((a ^ b) * (1 / a)) | a ↑ ((a ↑ b) × (1 ÷ a)) | a((ab) × (1 ÷ a)) |
Let's do some recapping of math operation displays seen here.
Multiplication: use an asterisk (*) for coding. For website displaying, we use a multiplication sign (a × b) since the asterisk looks like the Superpowerition sign that we use as the notation symbol.
Division: use a forward slash (/) for coding. For website displaying, you'll see a forward slash (a / b) or a division sign (a ÷ b).
Exponentiation: use a caret (^) to bring the number to the left of it to the power of the number to the right of it, as in a ^ b. For website displaying, you'll see the caret (a ^ b), an up arrow (a ↑ b) or displayed like ab.
Rooting: for coding, you'll have to use a modified verison of exponentiation where a ^ (1 / b) since the PowerBasic coding I use doesn't have anything except for a SQR function that takes square roots only. On the website, you'll see the bth root of a displayed as a down arrow for b√a, a ROOT word for a ROOT b, or a down arrow for a ↓ b.
Logarithim: It comes in between exponentiation and similar to rooting. Say you need to solve a problem where a and c are known values but b isn't like this: ab = c. What value of b would power the base of a up to become c? You use a logarithim function like this: loga(c) = b. You fill in a and c with 10 and 100 like this: log10(100) = b, and b is the answer of 2 since 102=100 and the 2√100 is 10.
Superpowerition would be using an asterisk, but since that's taken by most computer coding languages that use it as a multiplication sign, we'll use this notation symbol ✱ for a ✱ b.
Inverse operations for superpowerition? I haven't figured them out yet. If a ✱ b = c, then I would have to invent formulas that would find the value of a if b and c are known, and the value of b if a and c are known. I would also have to give names to these operations as well.
More on exponents and root conversions.
ab = 1 ÷ a-b = (1 ÷ b) √ a
a-b = 1 ÷ ab = (1 ÷ -b) √ a
a1 ÷ b = b √ a
a1 ÷ -b = -b √ a
Note: the nunber before the radical sign in a root operation and may not function for the root function if the value of (1 ÷ b) in the equation "(1 ÷ b) √ a" is negative or a non-integer or both depending on the numbers. The root function won't work if the value (1 ÷ b) is zero or a fraction with zero as a divisor.
With this formula:
ab = 1 ÷ a-b = (1 ÷ b) √ a
23 (8)
1 ÷ 2-3 (1 ÷ 0.125 = 8)
(1 ÷ 3) √ 2 (0.33333 √ 2 = 8)
8 = 1 ÷ 0.125 = 0.33333 √ 2
That works if b is a positive number. Let's see what happens when b is a negative number:
2-3 (0.125)
1 ÷ 23 (1 ÷ 8 = 0.125)
(1 ÷ -3) √ 2 = 0.125
All the answers are the same thing.
Do not use the calculator for Windows 10. It doesn't do negative fractional roots of numbers. Use a hand-held calculator instead.
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