Extending the hyperoperators to all real or complex values is a known open problem. Some time ago I had a somewhat converse idea on how to approach this - instead of starting with a fast growing function(s) on natural numbers and extending it to more inputs, we can directly construct a sequence of fast growing functions defined for all complex values. I do not claim this is of as much interest as extending hyperoperators, but might nevertheless be interesting for some.

Suppose we have chosen a system of fundamental sequences for countable limit ordinals up to some bound. We now construct a sequence \(f_\alpha:\mathbb C\rightarrow\mathbb C\) which satisfies the following propertes:

  1. \(f_\alpha\) is an entire function,
  2. \(f_\alpha(z)\in(0,\infty)\) for \(z\in\mathbb R\setminus\{0\}\) and \(f_\alpha(0)=0\),
  3. \(f_\alpha\) is increasing on \([0,\infty)\),
  4. \(|f_\alpha(z)|\leq|z|\) for \(|z|<1\).

The definition of the hierarchy is as follows:

  1. \(f_0(z)=z^2\),
  2. \(f_{\alpha+1}(z)=\sum_{n=1}^\infty f_\alpha^n\left(\frac{z}{2^n}\right)\), where \(f_\alpha^n\) means the usual function iteration,
  3. \(f_\alpha(z)=\sum_{n=1}^\infty f_{\alpha[n]}\left(\frac{z}{2^n}\right)\) for \(\alpha\) limit ordinal.

We prove that all these functions have properties 1-4 by transfinite induction on \(\alpha\). Everything is clear for \(\alpha=0\). Suppose now \(\beta=\alpha+1\) is a successor ordinal and \(f_\alpha\) satisfy the above properties. By simple induction, \(f_\alpha^n\) satisfy these properties as well. We show that the series \(\sum_{n=1}^\infty f_\alpha^n\left(\frac{z}{2^n}\right)\) converges absolutely and uniformly on disks - fix any \(R>0\) and arbitrary \(|z|\leq R\). Then, as long as \(2^n>R\), we have \(\left|f\left(\frac{z}{2^n}\right)\right|\leq\left|\frac{z}{2^n}\right|\leq\frac{R}{2^n}\), so by Weierstrass M-test we can conclude the series converges uniformly on disks, hence it converges everywhere to an analytic function, so \(f_{\alpha+1}(z)\) satisfies property 1. Properties 2 and 3 are straightforward and to see the property 4 we note, for \(|z|\leq 1\), \[\left|f_{\alpha+1}(z)\right|\leq\sum_{n=1}^\infty\left|f_\alpha^n\left(\frac{z}{2^n}\right)\right|\leq\sum_{n=1}^\infty\left|\left(\frac{z}{2^n}\right)\right|=|z|.\] Dealing with the limit case looks nearly identical.

Functions in these hierarchy are increasingly fast growing - indeed, for \(z>0\), \(f_{\alpha+1}(z)\geq f_\alpha^n\left(\frac{z}{2^n}\right)\), and we can show, though I won't bother to do it, that \(f_\alpha(z)\geq 2^{n+1}z\) for large \(z\), so we can also say \[f_{\alpha+1}(z)\geq f_\alpha^{n+1}\left(\frac{z}{2^{n+1}}\right)\geq f_\alpha^n(z).\] Similarly, we can see that \(f_\alpha(z)\) grows faster than any \(f_{\alpha[n]}\left(\frac{z}{2^n}\right)\), from which we can deduce \(f_\alpha(z)>f_\beta(z)\) for any \(\beta<\alpha\) and \(z\) large enough - if \(\alpha[n]>\beta\), then \[f_\alpha(z)>f_{\alpha[n+1]}\left(\frac{z}{2^{n+1}}\right)\geq f_{\alpha[n]+1}\left(\frac{z}{2^{n+1}}\right)>f_{\alpha[n]}(z)>f_\beta(z)\] for large \(z\) (note we have used transfinite induction implicitly here).

It's worth noting that, although this construction guarantees these functions to grow very fast on the real line (both in positive and negative direction), it's not the case that these functions go to infinity (in absolute value, say) very fast uniformly in each direction. Indeed, this is never possible - if \(f\) is entire but not a polynomial, then it will take values arbitrarily close to zero for \(z\) with arbitrarily large absolute value.

Arbitrarily fast growing functions

Initially I wanted to make this into another post, but I guess it's too short, so I've decided to put it here. Because of that, it is very loosely related to the construction above.

Here is a quick argument to show that for any function \(f:\mathbb N\rightarrow\mathbb N\) there is an analytic function \(g:\mathbb C\rightarrow\mathbb C\) mapping real line to itself such that \(g(n)>f(n)\) for all \(n\in\mathbb N\). If we find a sequence of functions \(g_n(z)\) which satisfies \(g_n(n)>n\), \(|g_n(z)|<\frac{1}{2^n}\) for \(|z|\leq n-1\) and \(g_n(z)>0\) for \(z\in\mathbb R\), then the sum \(g(z)=\sum_{n=1}^\infty g_n(z)\) will be analytic and will satisfy the properties we ask for. Now it's enough to note that \(g_n(z)=\left(\frac{z}{n-1/2}\right)^{k(n)}\) will work for sufficiently large \(k(n)\in\mathbb N\) (something like \(\max\{n^2,nf(n)\}\) should do).

Two remarks: such a function \(g\), and also the functions \(f_\alpha\) constructed in the earlier section, also have all their derivatives positive and fast growing on \([0,\infty)\), so that they don't exhibit any oscillatory behaviour. I also want to point out that this new construction doesn't really give us any new fast growing functions - we need a function \(f\) to start with before we define \(g\).

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.