## Kainui one year ago What values does this converge for?

1. 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?

2. 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!

3. ganeshie8

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

4. Kainui

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

5. ganeshie8

Ahh scratch that, that doesn't work..

6. 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'.

7. 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..

8. Kainui

Yeah it's quite weird!

9. 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.

10. dan815

square roots

11. Kainui

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

12. ganeshie8

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

13. 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.

14. 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.

15. dan815

|dw:1439868702577:dw|

16. dan815

does that also go to 1

17. dan815

any fraction?

18. ganeshie8

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

19. Kainui

I did 10 million and got: 0.9999998999996575

20. dan815

what a noob

21. Kainui

XD

22. dan815

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

23. dan815

1/k , k>1

24. Kainui

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

25. dan815

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

26. dan815

maybe its magnitude of 1

27. ganeshie8

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

28. dan815

oh negative numbers will also produce fractions true

29. 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)); } 

30. dan815

31. dan815

okay but i wonder what this means

32. dan815

|dw:1439869339358:dw|

33. ganeshie8

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

34. dan815

|dw:1439869406586:dw|

35. 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.

36. Kainui

Also I like this idea of looking at complex numbers too

37. dan815

can u geometrically make sense of a complex root

38. dan815

is it like an elliptical rotation

39. dan815

|dw:1439870380285:dw|

40. 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

41. 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... :\

42. ganeshie8

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

43. 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.

44. 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 $$c_n=\exp(b_n)=\exp\left(-\frac1{n-1/b_0}\right)$$

45. anonymous

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

46. Kainui

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