What values does this converge for?

- Kainui

What values does this converge for?

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- Kainui

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

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

\(e^{-e}\le c\le e^{1/e}\)

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## More answers

- Kainui

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

- ganeshie8

Ahh scratch that, that doesn't work..

- Kainui

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

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

Yeah it's quite weird!

- Kainui

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

square roots

- Kainui

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

- ganeshie8

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

- Kainui

```
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

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

|dw:1439868702577:dw|

- dan815

does that also go to 1

- dan815

any fraction?

- ganeshie8

i think 1000th term is good enough... ur code doesn't take much time even if u increase it to a million..

- Kainui

I did 10 million and got:
0.9999998999996575

- dan815

what a noob

- Kainui

XD

- dan815

try some other positive fractions i think they all go to 1

- dan815

1/k , k>1

- Kainui

After 10 million each, just random numbers:
.75 --> 0.9999998999996742
.9 --> 0.9999998999997273
.1 --> 0.9999998999996702

- dan815

we should allow complex route convergences and turn it into complex valued problem and try to find radius of convergence

- dan815

maybe its magnitude of 1

- ganeshie8

it is weird, it has a natural lower bound : \(c\ge -100\)
kai try \(c = -101\)

- dan815

oh negative numbers will also produce fractions true

- Kainui

I'm getting NaN but I don't know if that's legitimate or not since I don't know how Math.log() handles negative numbers. I need to fix this:
```
public static double cToC(double c) {
return Math.exp(c * Math.log(c));
}
```

- dan815

how about we think about complex convergences!

- dan815

okay but i wonder what this means

- dan815

|dw:1439869339358:dw|

- ganeshie8

i feel it should converge for all negative integers, maybe NaN could refer to very small numbers too..

- dan815

|dw:1439869406586:dw|

- Kainui

I was thinking about trying to fix it this way:
```
public static double cToC(double c) {
return Math.exp(Math.signum(c)*c * Math.log(c*c) / 2);
}
```
I don't know if that works or not or if it's the fastest way to do this.

- Kainui

Also I like this idea of looking at complex numbers too

- dan815

can u geometrically make sense of a complex root

- dan815

is it like an elliptical rotation

- dan815

|dw:1439870380285:dw|

- Kainui

Ok so thinking about what dan said and that fixed point method I did earlier, I tried to solve for r and \(\theta\) for an arbitrary complex number:
\[re^{i \theta } = re^{i \theta re^{i \theta }} \]
keep in mind that our angle has a period of \(2\pi\) so we could perhaps substitute in \[\theta = \phi + 2 \pi n\] to get infinitely many solutions.
Ok now to solve this thing...
\[re^{i \theta } = re^{i \theta re^{i \theta }} \]\[\ln r + i\theta = re^{i \theta }(\ln r + i \theta) \]\[0 = (-1)(\ln r + i\theta) + re^{i \theta }(\ln r + i \theta) \]\[0 = ( re^{i \theta }-1)(\ln r + i \theta) \]
So we can solve for the roots of this thing. :D

- Kainui

The right hand part \(\ln r + i\theta = 0 \) is solved only when \(r=1\) and \(\theta = 0\) like we had previously. However we also get this equation:
\[re^{i \theta} = 1\]
So this is also only solved when \(r=1\) and \(\theta = 2 \pi n\). Not really that exciting of a fixed point since it's basically the exact same thing... :\

- ganeshie8

im thinking of expressing the problem using knuth's uparrow notation, but it looks very tricky..not sure if below is right \[\lim\limits_{n\to\infty} ((c\uparrow^{n-1} 2)\uparrow (c\uparrow^n 2))\]

- Kainui

Yeah good idea I don't know either, I feel like those arrows are like a giant black box to me still, I gotta go back and check since I think this would be a good place to figure this out.
I was already typing this, so here goes.
We can look at the general trend. Given \(c > 0\) we can see how the next term compares in general:
\[(-c)^{-c} = \frac{(-1)^c}{c^c}\]
So if c>1 we have
\[\left| \frac{(-1)^c}{c^c}\right| < |c|\]
I guess complex numbers get kinda weird or messy in here so I don't know. I'm going to get a snack and come back.

- anonymous

consider \(b(n)=\log c_n\) so we have $$b(n+1)=e^{b(n)} b$$now notice we also know that $$b(n+1)=e^{d/dn}b$$so in other words we have that \(b(n)\) is an eigenvalue of \(d/dn\) for \(b(n)\) itself, suggesting $$\frac{db}{dn}=b^2\\b^{-2}\frac{db}{dn}=1\\-\frac1b=n+C\\b(n)=-\frac1{n+C}$$now given \(b(0)=b_0\) we see $$b(n)=-\frac1{n-1/b_0}$$and so we have taht $$c_n=\exp(b_n)=\exp\left(-\frac1{n-1/b_0}\right)$$

- anonymous

now i'm not sure if that's even mathematically valid for some range of \(b_0\) but it looked cool

- Kainui

I like when things look cool @oldrin.bataku :P

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