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

arrows of integer numbers

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{3}

=3^{\sqrt{3}}

=6.7049\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.

arrows of non-integer numbers

Old version

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>1)

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

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

New version

There is a problem on the old version. It is that 2\uparrow^n 2 is not 4 when n is not an integer. Is can be 3.7683... when n is around 0.53. This is a gap of 8% and it is not ignoble.

I have thought how to fix it for a week, and I have an answer: linear interpolation.

In math language, a\uparrow^n b=a^{1+n(b-1)}\times b^{1-n} (n\leq 1)

If I want to make 2\uparrow^n 2 always four, it must be like this.

comparizon between old one and new one

(not be written yet)

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