Math Is Fun Forum

Jabuto

Member

Registered: 2017-07-29

Posts: 11

Seed operator

I once thought this,

if multipication is just stacked addition,
and if exponentation is just stacked multipication,
you can get the next operation (tetration) by stacking exponentation,
and repeat infinitely to n.

That would be the seed operator. Ultimately the starter of it all.

After week or so I conviniently stepped over the ackerman function. Such a miracle!
But ackerman only accepts psoitive integers! Why cannot we define new function for same integer results but also accept any complex value?

I mean, e^(pi*i) doesen't make sense either, how can you "multiply anything pi times" or even "imaginary times"? It makes as much sense. I can't see any reason not to.

Even if universe really had special place for only addition, multipication and exponentantion,
a resolve for this all would make my day. or week. whole month.

---

You will die knowing being right. My only cost is your fullfilment.
--- The Devil

zetafunc

Moderator

Registered: 2014-05-21

Posts: 2,436

Website

Re: Seed operator

But ackerman only accepts psoitive integers! Why cannot we define new function for same integer results but also accept any complex value?

Good question: why can't you? Well, nothing is stopping you from doing so. What you are describing is usually called the principle of analytic continuation. As you suggest, this is performed so that we can extend the domain of a function in a way which might be useful to us. This isn't always too difficult to do: however, we might often ask why we might want to do such a thing. In some cases, it is extremely useful: famous examples include the analytic continuation of the Riemann zeta function, and perhaps lesser known examples such as the fractional Fourier transform (FrFT).

One of the easiest examples of analytic continuation to understand is simple: the geometric series. We know that, for all values of

such that

, we have:

However, as we can see, this is only true for values which are bounded in modulus by

. So suppose we want to define it for values which are bigger than

. How do we do this? Well, one way is to simply define

. That's an analytic function everywhere, except for two points (called poles), at

. It agrees with our original function at all of the points where it is defined.

Great! We've just found an analytic continuation of the geometric series, so now we can start plugging in any value we like into the infinite series... or can we? One might naively plug in the value

into our analytic continuation and conclude that:

But this obviously cannot be true! This is where the subtlety of the analytic continuation is important: we have found a new function which agrees with the original at all points inside the unit disc, but this does not mean we can equate the infinite series to our analytic continuation,

. Ramanujan famously wrote this:

This is of course not true: but what it was getting at was the analytic continuation of the Riemann zeta function, extended outside of the usual definition of its domain.