# logarithmic iterative limits

ark13
Posers and Puzzles 25 Apr '07 00:29
1. 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 02:442 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 08:383 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. 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. DeepThought