# Is this a good reasoning on fractional tetration?

Searching stuff on Wikipedia somehow I got to tetration, and got really interested on how could an interpretation of fractional tetration be given. So I did the following analysis

$$\left(a \cdot \frac{1}{2} \right) + \left(a \cdot \frac{1}{2} \right) = a$$

$$\left(a \wedge{} \frac{1}{2} \right) \cdot \left(a \wedge{} \frac{1}{2} \right) = a$$

Using this as a patern I could assume...

$$\left(a\ t \ \frac{1}{2} \right) \wedge{} \left(a \ t \ \frac{1}{2} \right) = a$$

Where $t$ stands for tetration so...

$$a\ t \ \frac{1}{2} \equiv ^{\frac{1}{2}}a$$

For example

$$\left(5 \cdot \frac{1}{2} \right) + \left(5 \cdot \frac{1}{2} \right) = 5$$

$$\left(5 \wedge{} \frac{1}{2} \right) \cdot \left(5 \wedge{} \frac{1}{2} \right) = 5$$

$$\left(5 \ t \ \frac{1}{2} \right) \wedge{} \left(5 \ t \ \frac{1}{2} \right) = 5$$

Knowing that multiplication is a short hand for addition i.e.

$$5 \cdot 3 = 5 + 5 + 5$$

But still we can't write "fractional addition", I mean we can't write

$$5 \cdot \frac{1}{2}$$

As adding 5's, yet we can ask "Which number added to itself can give me a 5", so

$$a + a = 5 \rightarrow a = 5 \cdot \frac{1}{2}$$

The same goes for fractional exponentiation, yet you can ask for roots.

$$a \cdot a = 5 \rightarrow a = 5 ^\frac{1}{2}$$

So I can interpet half tetration of 5 as asking, which number $a$ satisfies

$$a^a = 5$$ which approximately is $$a \approx 2.1293$$

Now I could ask for the general case $^{a/b}C$, (at least for $a,b,C \in \mathbb{R}^+)$ let's make another pattern analyzis

$$C \cdot \frac{a}{b} = C \cdot \underbrace{(\frac{1}{b} + \frac{1}{b} + ... + \frac{1}{b})}_{add\ a\ times\ 1/b} = \underbrace{C \cdot \frac{1}{b} + C \cdot \frac{1}{b} + ... + C \cdot \frac{1}{b}}_{a \ times\ C\ \cdot\ 1/b}$$

Next for exponentiation

$$C \hat{} \frac{a}{b} = C \hat{} \underbrace{(\frac{1}{b} + \frac{1}{b} + ... + \frac{1}{b})}_{add\ a\ times\ 1/b} = \underbrace{C \hat{} \frac{1}{b} \cdot C \hat{} \frac{1}{b} \cdot \ ...\ \cdot C \hat{} \frac{1}{b}}_{multiplication \ a \ times\ of\ C\ \hat{}\ 1/b}$$

Now for tetration I assume...

$$C \ t \ \frac{a}{b} = C\ t\ \underbrace{(\frac{1}{b} + \frac{1}{b} + ... + \frac{1}{b})}_{add\ a\ times\ 1/b} = \underbrace{C \ t \ \frac{1}{b} \hat{} C \ t \ \frac{1}{b} \hat{} \ ...\ \hat{} C \ t \ \frac{1}{b}}_{exponentiation \ a \ times\ of\ C\ \ t \ 1/b}$$

For example

$$^{4/3}3 = 3\ t\ \frac{1}{3} \hat{} \ 3\ t\ \frac{1}{3} \hat{}\ 3\ t\ \frac{1}{3} \hat{}\ 3\ t\ \frac{1}{3}$$

So we get $3\ t\ \frac{1}{3}$ or $^{1/3}3 \$ (I used the letter t to indicate tetration and the "computer notation" for the exponent to make more "obvious" the patterns) which is the question "Which number $a$ can satisfy

$$a^{a^{a}} = 3$$

Using Wolfram Alpha to solve the equation we get $a \approx 1.63507 \rightarrow ^{1/3}3 \approx 1.63507$, so

$$^{4/3}3 = a^{a^{a^a}} \approx 4.37135$$

What do you think? :)

• Seems to be a nice generalization of the tetration. But since my interest in tetrations is mainly in the context of very large numbers, I would tend to avoid it. That does not mean that I deny that it might be useful. – Peter May 10 '17 at 9:10
• I calculated $x=3 \uparrow \uparrow \frac{1}{3} \approx$ 1.434378 using Kneser's algorithm. Then $y=3 \uparrow \uparrow \frac{4}{3} = 3^x \approx$ 4.8347307. The Op's equation for $y$ in terms of $x$ using a is incorrect. The Op's equations don't follow the basic tetration rule $y=3^x$, using the Op's values for $x$ and $y$. $\forall z,\;\;\; 3 \uparrow \uparrow ( z+1) = 3 ^ { (3 \uparrow \uparrow z)}$ – Sheldon L May 11 '17 at 12:15