Kainui
  • Kainui
What values does this converge for?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Kainui
  • 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
  • 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
  • ganeshie8
\(e^{-e}\le c\le e^{1/e}\)

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Kainui
  • Kainui
Oh how did you figure this out? I honestly don't know the answer haha
ganeshie8
  • ganeshie8
Ahh scratch that, that doesn't work..
Kainui
  • 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
  • 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
  • Kainui
Yeah it's quite weird!
Kainui
  • 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
  • dan815
square roots
Kainui
  • Kainui
It appears to be that c=1/2 converges to 1.
ganeshie8
  • ganeshie8
a quick bruteforce gives it converges for all negative integers! \(c\in \mathbb{Z^{-}} \) or \(c\in [0,1]\)
Kainui
  • 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
  • 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
  • dan815
|dw:1439868702577:dw|
dan815
  • dan815
does that also go to 1
dan815
  • dan815
any fraction?
ganeshie8
  • ganeshie8
i think 1000th term is good enough... ur code doesn't take much time even if u increase it to a million..
Kainui
  • Kainui
I did 10 million and got: 0.9999998999996575
dan815
  • dan815
what a noob
Kainui
  • Kainui
XD
dan815
  • dan815
try some other positive fractions i think they all go to 1
dan815
  • dan815
1/k , k>1
Kainui
  • Kainui
After 10 million each, just random numbers: .75 --> 0.9999998999996742 .9 --> 0.9999998999997273 .1 --> 0.9999998999996702
dan815
  • dan815
we should allow complex route convergences and turn it into complex valued problem and try to find radius of convergence
dan815
  • dan815
maybe its magnitude of 1
ganeshie8
  • ganeshie8
it is weird, it has a natural lower bound : \(c\ge -100\) kai try \(c = -101\)
dan815
  • dan815
oh negative numbers will also produce fractions true
Kainui
  • 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
  • dan815
how about we think about complex convergences!
dan815
  • dan815
okay but i wonder what this means
dan815
  • dan815
|dw:1439869339358:dw|
ganeshie8
  • ganeshie8
i feel it should converge for all negative integers, maybe NaN could refer to very small numbers too..
dan815
  • dan815
|dw:1439869406586:dw|
Kainui
  • 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
  • Kainui
Also I like this idea of looking at complex numbers too
dan815
  • dan815
can u geometrically make sense of a complex root
dan815
  • dan815
is it like an elliptical rotation
dan815
  • dan815
|dw:1439870380285:dw|
Kainui
  • 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
  • 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
  • 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
  • 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
  • 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
  • anonymous
now i'm not sure if that's even mathematically valid for some range of \(b_0\) but it looked cool
Kainui
  • Kainui
I like when things look cool @oldrin.bataku :P

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