User blog:IAmNotATRex/FGH Array Extension Examples

I have decided to write an entire blog post of examples using my FGH extension to (hopefully) give people a greater understanding of it. To understand this blog post, you need to know the definition of my extension, which can be found here: https://googology.wikia.com/wiki/User_blog:IAmNotATRex/Array-Based_Extension_to_the_FGH. I advise that you keep two tabs open, one with this page and one with the definition of my extension, in case you ever get confused.

\(f[0, 0](2)\)
I'll start off with a simple example: \(f[0, 0](2)\).

If you also understand my old extension, know that this is equal to \(F1_{0}(2)\) using my old extension. However, that isn't very important, considering the fact than my new extension is significantly more powerful than my old extension.

If you expand \(f[0, 0](2)\), you'll find that it's equal to \(f[f[0](2)](2)\), which is simply just \(f_{f_{0}(2)}(2)\) in the FGH. As you can see, this number isn't all that large and can be easily calculated.

\(f_{f_{0}(2)}(2)=f_{3}(2)=2048\)

The answer comes out to be \(2048\), which is (relatively) tiny.

Now, if you think that \(f[0, 0](n)\) is just \(f_{\omega+1}(n)\) in the FGH, the next example should prove you wrong.

\(f[0, 0](3)\)
This is equal to \(f[f[f[0](3)](3)](3)\), which is obviously \(f_{f_{f_{0}(3)}(3)}(3)\), or, more simply, \(f_{f_{4}(3)}(3)\).

As you can see, the subscript of \(f_{4}(3)\) is much larger than \(4\), disproving that \(f[0, 0](n)\) is equal \(f_{\omega+1}(n)\).

While this number is certainly larger than \(2048\) (I think \(f_{4}(3)\) alone is around \(\text{G}2\) or \(\text{G}3\)), it still isn't very large compared to other numbers.

\(f[0, 1](2)\)
If you don't understand that increasing the second element by \(1\) will greatly increase the output, you hopefully should by the end of this example.

\(f[0, 1](2)\) is equal to \(f[f[\ldots[f[f[1](2)](2)]\ldots](2)](2)\).

This is similar to \(f[0, 0](2)\), except that there are now \(f^{2}[0, 0](2)\) "layers." What does this mean? Well, let's calculate the value of the number of layers.

\(f^{2}[0, 0](2)\) is equal to \(f[0, 0](f[0, 0](2))\). We already know that \(f[0, 0](2)\) is equal to \(2048\), so using substitution, we get \(f[0, 0](2048)\).

\(f[0, 0](3)\) is already fairly large, and \(f[0, 0](2048)\) is only larger.

So now we know that the number of layers is \(f[0, 0](2048)\), which is equal to \(f[f[\ldots[f[f[1](2048)](2048)]\ldots](2048)](2048)\) with \(2048\) layers. And if you remember, \(f[0, x](n)\) iterates through the subscript of the function used in the FGH, meaning the result should be a pretty large number.

\(f[0, 1](3)\)
This has This has \(f^{3}[0, 1](3)\) layers. The important thing to note is that the function that defines the number of layers is now iterated three times instead of two, resulting in an even larger increase.

\f([0, 2](2)\)
This has \(f^{2}[0, 1](2)\) layers. I don't feel like explaining it because it's a similar concept to \(f[0, 1](2)\).

\(f[1, 0](2)\)
What happens if we increase the first element by \(1\)?

\(f[1, 0](2)\) is equal to \(f[0, f[0, 0](2)](2)\), which simplifies to \(f[0, 2048](2)\).

As we saw earlier, going from \(f[0, 0](2)\) to \(f[0, 1](2)\) was already quite a large increase. As you can imagine, going to \(f[0, 2048](2)\) is an even larger jump.

\(f[1, 0](3)\)
This is equal to \(f[0, f[0, f[0, 0](3)](3)](3)\). Instead of two layers, there are now three. This makes an even bigger difference.

\(f[0, 0, 0](2)\)
Up until now, we've only looked at examples where the array only has two elements. As you can probably guess, adding another element (or really inserting another element at the front) further increases the magnitude of the output.

\(f[0, 0, 0](2)\) is equal to \(f[f[0, 0](2), 0](2)\), which simplifies to \(f[2048, 0](2)\). I may be repeating myself, but if \(f[0, 0](2)\) to \(f[1, 0](2)\) is a large increase, \(f[1, 0](2)\) to \(f[2048, 0](2)\) is an even larger one.

\(F_{0}[0, 0](2)\)
It's time to move on to my more powerful function.

\(F_{0}[0, 0](2)\) is equal to \(f[\{f[0, 0](2)\}](2)\). This simplifies to \(f[\{2048\}](2)\), which is equal to \(f[3, 1, 5, 9](2)\).

While it's nice that we've added (or inserted) another element to the array, it's "only" one more element.

\(F_{0}[0, 0](3)\)
This number will (hopefully) show the true power of converting integers to arrays.

\(F_{0}[0, 0](3)\) is equal to \(f[\{f[\{f[0, 0](3)\}](3)\}](3)\).

\(f[0, 0](3)\) is much larger than \(2048\), meaning that it will generate a much larger array, resulting in an even larger number. And since this process is repeated again, an even larger array will be creating, and in turn, an even larger number.

\(F_{1}[0, 0](2)\)
As is to be expected, this example will make my previous examples look like nothing.

\(F_{1}[0, 0](2)\) is equal to \(F_{0}[\{F_{0}[\ldots[\{F_{0}[\{F_{0}[0, 0](2)\}](2)\}]\ldots](2)\}](2)\), where the number of layers is \(F_{0}^{2}[0, 0](2)\), or \(F_{0}[0, 0](F_{0}[0, 0](2))\).

The resulting number is much larger than the rest of the examples in this blog post.

Conclusion
I hope I didn't sound overly dramatic, as I was just trying to best explain my extension to the FGH. While the examples in this blog post resulted in fairly large numbers, much bigger numbers could be created by using ordinals along with my extension.