User blog:Nayuta Ito/generalization of arrow notation

First, how many is $$3\uparrow\uparrow1.5$$?

You know these:

$$3\uparrow\uparrow0=1$$

$$3\uparrow\uparrow1=3$$

$$3\uparrow\uparrow2=3^{3\uparrow\uparrow1}$$

And these:

$$3\uparrow\uparrow0=1=3^0$$

$$3\uparrow\uparrow1=3=3^1$$

Look at the previous formulas again. You can see a pattern:

$$3\uparrow\uparrow k=3^k\;(0\leq k< 1)$$

So, tetration is generalized like this:

$$a\uparrow\uparrow b=a\uparrow(a\uparrow\uparrow (b-1))\;(b\geq 1)$$

$$a\uparrow\uparrow b=a^b\;(0\leq b< 1)$$

$$a\uparrow b=a^b$$

Now I can answer the question.

$$3\uparrow\uparrow1.5$$

$$=3\uparrow(3\uparrow\uparrow0.5)$$

$$=3\uparrow(3^{0.5})$$

$$=3\uparrow\sqrt{2}$$

$$=3^{\sqrt{2}}$$

$$=4.7288\cdots$$

This is proper because it is greater than $$3\uparrow\uparrow0$$ and smaller than $$3\uparrow\uparrow1$$.

Next, I will make for n arrows.

It's almost same pattern:

$$a\uparrow^n b=a\uparrow^{n-1}(a\uparrow^n (b-1))\;(b\geq 1)$$

$$a\uparrow^n b=a^b\;(0\leq b< 1)$$

$$a\uparrow b=a^b$$

I think this is not good, but it must be like this by the definition.

Next, I have to think about a half arrow.

$$x\uparrow^0 y=x\times y$$

$$x\uparrow^1 y=x^y$$

Look at these more...

$$x\uparrow^0 y=x^1\times y^1$$

$$x\uparrow^1 y=x^y\times y^0$$

And one more...

$$x\uparrow^0 y=x^{y^0}\times y^1$$

$$x\uparrow^1 y=x^{y^1}\times y^0$$

Now I have done (but too assertive).

I will summarize all definitions:

$$a\uparrow^n b=a\uparrow^{n-1}(a\uparrow^n (b-1))\;(b\geq 1)$$

$$a\uparrow^n b=a^b\;(n>1 \and 0\leq b< 1)$$

$$a\uparrow^n b=a^{b^n}\times b^{1-n} (n\leq 1)$$