Fun With Math-Superpowerition: Inventing a Super Power Operation -- Integer a > 1 ✱ Decimal 0 < b

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Integer a > 1 ✱ Decimal 0 < b < 1

On this page, let's make the a in "a ✱ b" an integer > 1 and the b a decimal number between 0 and 1 inclusive.

Let's try out some variables for a using the formula a ^ (a ^ (b - 1)). Let's let b go from 1 to 2 in 0.25 steps.

Let's let a = 2.

2 ✱ 0.00 = 2 ^ (2 ^ (0.00 - 1)) = 2 ^ (2 ^ (-1.00)) = 2 ^ 0.5000000 ≈ 1.4142135

2 ✱ 0.25 = 2 ^ (2 ^ (0.25 - 1)) = 2 ^ (2 ^ (-0.75)) ≈ 2 ^ 0.5946035 ≈ 1.5100904

2 ✱ 0.50 = 2 ^ (2 ^ (0.50 - 1)) = 2 ^ (2 ^ (-0.50)) ≈ 2 ^ 0.7071067 ≈ 1.6325268

2 ✱ 0.75 = 2 ^ (2 ^ (0.75 - 1)) = 2 ^ (2 ^ (-0.25)) ≈ 2 ^ 0.8408964 ≈ 1.7911627

2 ✱ 1.00 = 2 ^ (2 ^ (1.00 - 1)) = 2 ^ (2 ^ ( 0.00)) = 2 ^ 1.0000000 = 2

As you can see, 2 ✱ 0.00 results in the square root of 2, or the 2nd root of 2. 2 ✱ 0.00 works out to 2.

Let's let a = 4 and b = go from 0 to 1 in 0.25 steps, using the formula a ^ (a ^ (b - 1)).

4 ✱ 0.00 = 4 ^ (4 ^ (0.00 - 1)) = 4 ^ (4 ^ (-1.00)) = 4 ^ 0.2500000 ≈ 1.4142135

4 ✱ 0.25 = 4 ^ (4 ^ (0.25 - 1)) = 4 ^ (4 ^ (-0.75)) ≈ 4 ^ 0.3535533 ≈ 1.6325267

4 ✱ 0.50 = 4 ^ (4 ^ (0.50 - 1)) = 4 ^ (4 ^ (-0.50)) ≈ 4 ^ 0.5000000 = 2

4 ✱ 0.75 = 4 ^ (4 ^ (0.75 - 1)) = 4 ^ (4 ^ (-0.25)) ≈ 4 ^ 0.7071067 ≈ 2.6651438

4 ✱ 1.00 = 4 ^ (4 ^ (1.00 - 1)) = 4 ^ (4 ^ ( 0.00)) = 4 ^ 1.0000000 = 4

A coincidence is that 2 ✱ 0.00 = 4 ✱ 0.00. 2 ✱ 1.00 = 4 ✱ 0.50. 2 ✱ 0.50 is very close to equalling 4 ✱ 0.25.

Also note that 4 ✱ 0.00 results in the answer of the fourth root of 4, which matches the square root of 2.

Let's let a = 16 and b = go from 0 to 1 in 0.25 steps, using the formula a ^ (a ^ (b - 1)).

16 ✱ 0.00 = 16 ^ (16 ^ (0.00 - 1)) = 16 ^ (16 ^ (-1.00)) = 16 ^ 0.0625000 ≈ 1.1892071

16 ✱ 0.25 = 16 ^ (16 ^ (0.25 - 1)) = 16 ^ (16 ^ (-0.75)) ≈ 16 ^ 0.1250000 ≈ 1.4142135

16 ✱ 0.50 = 16 ^ (16 ^ (0.50 - 1)) = 16 ^ (16 ^ (-0.50)) ≈ 16 ^ 0.2500000 = 2

16 ✱ 0.75 = 16 ^ (16 ^ (0.75 - 1)) = 16 ^ (16 ^ (-0.25)) ≈ 16 ^ 0.5000000 = 4

16 ✱ 1.00 = 16 ^ (16 ^ (1.00 - 1)) = 16 ^ (16 ^ ( 0.00)) = 16 ^ 1.0000000 = 16

We discover that 4 ✱ 1.50 = 2 ✱ 2.00 = 16 ✱ 1.00 with a value of 16.

Also, 16 ✱ 0.75 = 4 ✱ 1.00 = 2 ✱ 2.00 with a value of 4.

16 ✱ 0.50 = 4 ✱ 0.50 = 2 ✱ 1.00 with a value of 2.

16 ✱ 0.25 = 4 ✱ 0.00 = 2 ✱ 0.00 with a value of the square root of 2, the fourth root of 4 and the eighth root of 16. Do you see a pattern there?