A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Kainui

  • one year ago

What values does this converge for?

  • This Question is Closed
  1. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  4. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ahh scratch that, that doesn't work..

  6. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Yeah it's quite weird!

  9. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    square roots

  11. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  12. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  13. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 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.

  14. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439868702577:dw|

  16. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    does that also go to 1

  17. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    any fraction?

  18. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  19. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I did 10 million and got: 0.9999998999996575

  20. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    what a noob

  21. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    XD

  22. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  23. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1/k , k>1

  24. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  25. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  26. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    maybe its magnitude of 1

  27. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  28. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh negative numbers will also produce fractions true

  29. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how about we think about complex convergences!

  31. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    okay but i wonder what this means

  32. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439869339358:dw|

  33. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  34. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439869406586:dw|

  35. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Also I like this idea of looking at complex numbers too

  37. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can u geometrically make sense of a complex root

  38. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it like an elliptical rotation

  39. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439870380285:dw|

  40. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  46. Kainui
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  47. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.