# Question about exponents and i .

1. Jun 28, 2012

### cragar

how would I simplify or figure out $i^{i^{i..}}$
this keep going an infinite tower of i's

Last edited: Jun 28, 2012
2. Jun 28, 2012

### math man

(a^b)^c = a^(bc)

i^i^i^... = i^(i*i*i*...)
$$= \lim_{n→∞}i^{(n\,i)} = \lim_{n→∞}(i^{\,i})^{n}$$
$$i^{\,i} = e^{-\pi/2}$$
$$\lim_{n→∞}(i^{\,i})^{n} = \lim_{n→∞}\frac{1}{e^{n\pi/2}} = 0$$

3. Jun 28, 2012

### cragar

why does (a^b)^c equal a^(bc)
x^x^x does not equal x^(x^2)

4. Jun 28, 2012

### chiro

Hey cragar.

In terms of this I don't think you are going to get a closed-form solution but there is a way to go from say i^i using eulers identity.

To do this we use i^i = e^(i * ln(i)) = e^(itheta) where theta = ln(i). We use the principle branch for ln(i), which gives cos(ln(i)) + isin(ln(i)) = e^(-pi/2) = 0.207879576.

Another way is to use i^i = ((-1)^(1/2))^i = e^(1/2*i^2*pi) = e^(-pi/2), since i has principal argument of pi/2 (cos(pi/2) + isin(pi/2) = i).

Applying this again gives us (e^(-pi/2))^i = e^(-ipi/2) = cos(-pi/2) + isin(-pi/2) = -i. But now are back to the imaginary i in negative form!

I'm guessing (but you will have verify) that we will eventually get back to i after doing this process again (kind of like a spinor).

Because of this, the value in the infinite limit will not exist in the sense of something fixed and something that converges to a definitive value.

5. Jun 28, 2012

### Mute

(a^b)^c = a^(bc), but that is not what you have. You have a^(b^c), which is not equal to a^(bc). Exponentiation is not associative.

As for the actual limit, the result is apparently quoted on this wikipedia page: http://en.wikipedia.org/wiki/Super-exponentiation#Extension_to_complex_bases

There is a formula for a number x raised to itself infinitely many times in terms of the Lambert-W function (which is also quoted on that page), but there does not appear to be a proved extension for complex numbers.

Last edited: Jun 28, 2012
6. Jun 28, 2012

### haruspex

If the limit exists, it is a solution of iz = z.
Writing z = re:
r π cos(θ) = 2θ
r π sin(θ) = -2 ln(r)
The solution at http://en.wikipedia.org/wiki/Super-e..._complex_bases [Broken] does indeed satisfy these equations.

Last edited by a moderator: May 6, 2017
7. Jun 29, 2012

### HallsofIvy

This is true but cragar i talking about $a^{b^c}$