What values does this converge for?

Question

kainui · Answer

This is recursively defined:

For $c_0 = c$ and
$$c_n = c_{n-1}^{c_{n-1}}$$

For what values of c does $c_\infty$ converge?

kainui · Answer

It looks kinda weird but for a quick example:

$c = 10$ means $c_0=10$ so if we want to calculate $c_1$ it's:

$$c_1 = c_0^{c_0} = 10^{10}  = 10000000000 $$
Then to calculate $c_2$ it will be:

$$c_2 = c_1^{c_1} = 10000000000^{10000000000} $$ which as we can see as we approach $c_\infty$ will diverge for c=10. So no good!

ganeshie8 · Answer

$e^{-e}\le c\le e^{1/e}$

kainui · Answer

Oh how did you figure this out? I honestly don't know the answer haha

ganeshie8 · Answer

Ahh scratch that, that doesn't work..

kainui · Answer

One way I just thought of is "at infinity" we'll have

$$c_\infty = c_\infty^{c_\infty}$$

Since it's "converged" already. So then I can solve for it:

$$\ln (c_\infty) = c_\infty \ln (c_\infty)$$$$1 = c_\infty$$

Which is kinda like one way of getting an answer c=1 will definitely work, but this doesn't really satisfy my.

The simplest bounds I can come to are  $0 \le c \le 1$ BUT negative numbers or numbers larger than 1 could still work, these are just the values I find kinda 'obvious'.

ganeshie8 · Answer

yeah it converges for c=-1,
in all my innocence initially i thought it is a trick question of tetration/lambertw... but it isn't.. it looks like the mother of tetration..

kainui · Answer

Yeah it's quite weird!

kainui · Answer

I was thinking when you gave that answer at first that you had solved it with lambert w and I was like, "Ahhh he beat me at my own game! How did I not find this?" But now I don't have an answer anymore so I don't know anymore if I should be happy or sad about this haha.

dan815 · Answer

square roots

kainui · Answer

It appears to be that c=1/2 converges to 1.

ganeshie8 · Answer

a quick bruteforce gives it converges for all negative integers!
 $c\in \mathbb{Z^{-}} $ or   $c\in [0,1]$

kainui · Answer

```
public class RecursiveEigenvalue {

	public static double cToC(double c) {
		return Math.exp(c * Math.log(c));
	}
	
	public static double recC(int n, double c){
		for(int i = 0; i< n ; i++){
			c = cToC(c);
		}
		return c;
	}

	public static void main(String... args) {
		System.out.println(recC(1000,0.5));
	}
}
```

That gave me this output: `0.9990053500905812`

So looks good haha I was assuming evrything below 1 would actually converge towards 0.

kainui · Answer

Also, is my code for doing $c^c$ bad @ganeshie8 because I couldn't find a better way to do it off hand haha.

dan815 · Answer

|dw:1439868702577:dw|