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We can use the following to add the number $2$ to itself $5$ times.

$$f(n,k) = \sum_{x=1}^k n = n\cdot k$$

$$2 + 2 + 2 + 2 + 2 = f(2,5) = \sum_{x=1}^5 2 = 2\cdot 5 = 10$$

We can use a similar strategy for multiplying the number $2$ by itself $5$ times.

$$g(n,k) = \prod_{x=1}^k n = n^k$$

$$2\cdot2\cdot2\cdot2\cdot2 = g(2,5) = \prod_{x=1}^5 2 = 2^5 = 32$$

How could I define $h(n,k)$ such that $h(2,5) = 2^{2^{2^{2^2}}}$?

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    $\begingroup$ You just did define it. $\endgroup$ Commented Feb 27, 2017 at 7:54
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    $\begingroup$ By saying "$h(n,k)$ such that it is $n$ raised to itself $n$ iterations". That's all you need. You could do recursion. "$h(n,0) = 1$ and $h(n,k+1) = n^{h(n,k)};k\ge 0$". Important to not, in case you haven't. This not associative and $3^{(3^3)} \ne (3^3)^3$. This assumes you meant the former not the latter. $\endgroup$
    – fleablood
    Commented Feb 27, 2017 at 7:58

2 Answers 2

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Thanks to the tag added in Ivan Neretin's edit, I have found that this operation is formally known as "tetration".

In mathematics, tetration (or hyper-4) is the next hyperoperation after exponentiation, and is defined as iterated exponentiation.

$h(n,k)$ would be defined as follows, using standard tetration format.

$$h(n,k) = ^kn$$

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    $\begingroup$ You could "accept" your self-answer, to "close-the-case" for the MSE-readers $\endgroup$ Commented Mar 1, 2017 at 8:11
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Knuth Arrow notation, best explained here.

It also has accommodations for repeated knuth arrow notations, repeated REPEATED knuth arrows, and so on into the sun.

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