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Posers and Puzzles

Posers and Puzzles

  1. Standard member ark13
    Enola Straight
    25 Apr '07 00:29
    This is just something I've been working on and I need some help.

    For any number, n, taking iterations of f(n) = ln (n) will eventually produce a complex numbers with imaginary components, the first of which being some number + pi*i (which seems relatively expectable given e^(pi*i) = -1).

    But with enough iterations, both the real and imaginary components approach a limit. The real component, I've discovered, is 1/pi, but I can't figure out if there is a significance to the imaginary component. It is equal to approximately 1.337235701. Can anyone help in identification?

    Also, because of the way this complex number, let's call it R, arose, R = e^R = ln R. I believe this is the only number with these properties, is that correct?

    Any other base of logarithm will too produce a limit, though it will be different from the one produced by base e. This knowledge may be helpful if a relationship between the real and imaginary components of different log based can be found.

    Just some things to think over. Any help is appreciated. I can't seem to find much about this online.
  2. 25 Apr '07 02:44 / 2 edits
    Typing while I think:

    So you're looking at the sequence a(n) = ln(a(n-1)), right? So it should converge to some (complex) number x such that x = ln(x). Taking e to both sides:

    x = e^x

    Let x = a+bi

    a+bi = e^(a+bi) = e^a * e^bi = (e^a * cos(b)) + i*(e^a * sin(b))

    That is, we want real numbers a and b such that

    a = e^a*cos(b)
    b = e^a*sin(b)

    It looks like the numbers you're getting satisfy these equations.

    If you can solve this equation for a and b you have your answer, but it's not clear to me how to do that.

    If we assume you're right that a = 1/pi, then we have

    cos(b) = (1/pi) / (e^(1/pi))
    and b = arccos((1/pi) / (e^(1/pi)))
    which, though not helpful, is at least a closed form expression of the imaginary component...

    Another tack:

    if a + bi = e^a * (cos(b) + i * sin(b)), then both sides of the equation must have the same magnitude. That gives us

    a^2 + b^2 = (e^a)^2 = e^(2a)

    Unfortunately I don't think this tells us anything new, because with this equation and the second equation above (b = e^a*sin(b)) I can derive the first equation above (a = e^a*cos(b)) implying that with the first two equations we could have derived this last one already.

    In the same spirit we could say that both sides of that same equation must have the same angle from the x axis, so we can write an equation for the tangent of that angle:

    b/a = tan(b)

    Maybe something useful could be derived from this.
  3. 25 Apr '07 06:40
    Originally posted by ark13
    This is just something I've been working on and I need some help.

    For any number, n, taking iterations of f(n) = ln (n) will eventually produce a complex numbers with imaginary components, the first of which being some number + pi*i (which seems relatively expectable given e^(pi*i) = -1).

    But with enough iterations, both the real and imaginary components ...[text shortened]... ings to think over. Any help is appreciated. I can't seem to find much about this online.
    You have to be very careful about what you mean by the logarithm of a complex number. There's more that one. So when you write

    log(i) = i*pi/2

    why not have log(i)= i*pi*5/2 etc. etc. instead?

    In particular you can't just go between z=log(z) and z=e^z as those they're equivalent statements (one implies the other but not conversely).
  4. 25 Apr '07 08:38 / 3 edits
    http://www.research.att.com/~njas/sequences/A059526
    http://www.research.att.com/~njas/sequences/A059527

    Maybe this is a faster way to calculate this number:

    Start from a random x, and then iterate
    f(x) = x(lnx-1)/(x-1)
  5. Standard member ark13
    Enola Straight
    25 Apr '07 18:33
    Originally posted by GregM
    Typing while I think:

    So you're looking at the sequence a(n) = ln(a(n-1)), right? So it should converge to some (complex) number x such that x = ln(x). Taking e to both sides:

    x = e^x

    Let x = a+bi

    a+bi = e^(a+bi) = e^a * e^bi = (e^a * cos(b)) + i*(e^a * sin(b))

    That is, we want real numbers a and b such that

    a = e^a*cos(b)
    b = e^a*sin(b)
    ...[text shortened]... angent of that angle:

    b/a = tan(b)

    Maybe something useful could be derived from this.
    Very clever. Thanks. I suspected something with the trigonometric functions. I'll keep working with it, but you helped a lot.

    What is the relationship between a and b here, and the a and b obtained using log base 10?
  6. Standard member DeepThought
    Losing the Thread
    25 Apr '07 23:03
    Assuming that this procedure works for the whole complex plane apart from z = 0, and the series is convergent, you must be able to reconstruct the whole complex plane (apart from z = 0) by repeatedly exponentiating points in any neighbourhood of the solution to z = ln(z). Which means that there's a map from any neighbourhood of the solution to z = ln(z) to the whole Argand diagram (apart from z = 0).
  7. 26 Apr '07 05:55
    Originally posted by David113
    http://www.research.att.com/~njas/sequences/A059526
    ...which unfortunately differs from 1/pi in the third decimal place, so that theory is shot.

    We may have to accept the fact -- annoying though it is -- that there's no way of expressing this number with elementary functions. In general, if you have an equation of the type f(x) = g(x) where f and g are fundamentally different functions, you can only approximate the solutions. For instance, x^2 and sin(x) are both well-behaved functions, but there's no nice expression for the nonzero root of x^2 = sin(x).