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

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Kainui
 one year ago
Best ResponseYou've already chosen the best response.2This is recursively defined: For \(c_0 = c\) and \[c_n = c_{n1}^{c_{n1}}\] For what values of c does \(c_\infty\) converge?

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2It 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
 one year ago
Best ResponseYou've already chosen the best response.1\(e^{e}\le c\le e^{1/e}\)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Oh how did you figure this out? I honestly don't know the answer haha

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Ahh scratch that, that doesn't work..

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2One 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
 one year ago
Best ResponseYou've already chosen the best response.1yeah 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2It appears to be that c=1/2 converges to 1.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1a quick bruteforce gives it converges for all negative integers! \(c\in \mathbb{Z^{}} \) or \(c\in [0,1]\)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2``` 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
 one year ago
Best ResponseYou've already chosen the best response.2Also, is my code for doing \(c^c\) bad @ganeshie8 because I couldn't find a better way to do it off hand haha.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i think 1000th term is good enough... ur code doesn't take much time even if u increase it to a million..

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I did 10 million and got: 0.9999998999996575

dan815
 one year ago
Best ResponseYou've already chosen the best response.0try some other positive fractions i think they all go to 1

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2After 10 million each, just random numbers: .75 > 0.9999998999996742 .9 > 0.9999998999997273 .1 > 0.9999998999996702

dan815
 one year ago
Best ResponseYou've already chosen the best response.0we should allow complex route convergences and turn it into complex valued problem and try to find radius of convergence

dan815
 one year ago
Best ResponseYou've already chosen the best response.0maybe its magnitude of 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1it is weird, it has a natural lower bound : \(c\ge 100\) kai try \(c = 101\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh negative numbers will also produce fractions true

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I'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
 one year ago
Best ResponseYou've already chosen the best response.0how about we think about complex convergences!

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay but i wonder what this means

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i feel it should converge for all negative integers, maybe NaN could refer to very small numbers too..

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2Also I like this idea of looking at complex numbers too

dan815
 one year ago
Best ResponseYou've already chosen the best response.0can u geometrically make sense of a complex root

dan815
 one year ago
Best ResponseYou've already chosen the best response.0is it like an elliptical rotation

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Ok 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.1im thinking of expressing the problem using knuth's uparrow notation, but it looks very tricky..not sure if below is right \[\lim\limits_{n\to\infty} ((c\uparrow^{n1} 2)\uparrow (c\uparrow^n 2))\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2Yeah 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
 one year ago
Best ResponseYou've already chosen the best response.0consider \(b(n)=\log c_n\) 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 $$b(n)=\frac1{n1/b_0}$$and so we have taht $$c_n=\exp(b_n)=\exp\left(\frac1{n1/b_0}\right)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now i'm not sure if that's even mathematically valid for some range of \(b_0\) but it looked cool

Kainui
 one year ago
Best ResponseYou've already chosen the best response.2I like when things look cool @oldrin.bataku :P
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