Is there a "binomial" theorem for super exp or tetration?

  • Thread starter japplepie
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  • #1
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In the binomial theorem (x+x')^n = a summation of terms.

Is there any way I could express (x+x')^^n in a summation?
 

Answers and Replies

  • #2
Mentallic
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Well let's consider the simplest non-trivial example which would be

[tex]^2(x+y)=(x+y)^{x+y}=(x+y)^x(x+y)^y[/tex]

Now we can see that if x and y are non-negative integers then we can expand that last expression with the application of the binomial theorem twice. If they're not however, then we'd get an infinite expansion and thus attempting to compute one more level in the tetration:

[tex]^3(x+y)[/tex]

Wouldn't be possible because the binomial theorem only works for finite n.

Ok, so we're forced to have x,y be non-negative integers, but also if either of them were 0 then the problem becomes trivial, so we'll restrict ourselves to the natural numbers.

I don't expect there is a way to express it in the form that you're hoping for, but because tetration is simply repeated exponentiation, which means that recursively,

[tex]^n(x+y)=(x+y)^{^{n-1}(x+y)}[/tex]

and so you can iteratively use the binomial theorem as you work your way down the power tower.

[tex]f_n=\sum_{i=0}^{f_{n-1}}\binom{f_{n-1}}{i}x^iy^{f_{n-1}-i}[/tex]
where
[tex]f_k=^k(x+y)[/tex]
 

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