Does i^i^i^i^... converge to a real number?

Question

That's √(-1) to the √(-1) to the √(-1), etc., with infinite exponents.

Here's a few notes on why I think it might, although I have no idea how to approach this rigorously:

(1) i^i is a real number, approximately equal to .20788

(2) x^x^x^x^... converges if 1/e ≤ x ≤ x.

(3) Note that 1/e < .20788 < e.

The second item above should read:

x^x^x^x^... converges if 1/e ≤ x ≤ e.

Geez, rough night... I'll get it right this time: x^x^x^x^... converges if (1/e)^e ≤ x ≤ e^(1/e), or about .06 < x < 1.44. Having said that, I think you guys are right -- the link to the Google calculator certainly makes it look like this series does not converge.

Aspx has to get Best Answer, IMO. He fearlessly called the series convergent when every poster before him said otherwise, and he clearly spent some time surfing the Net to confirm he was right. Heck, he even had to endure two thumbs down votes.

However, Scythian also made a nice contribution, first giving the exact value and then describing the spiral as it converges. The spiral explains why the numbers seem to be jumping all over the place when you try this on a calculator -- you actually have to go a good ways before the convergence becomes clear. Scythian, I owe you a best answer. :)

Answer 1

Sorry guys to interrupt your mathematical assumptions, i just have the programming hunger turns on, when i see similar problems.

with a little small cute matlab program:

Code:

function a=geti()
a=i;
for s=1:200
a=i^a
end

end



i got at the last values of 200 (ie. 190 --> 200) (2 hundred times powering i)

a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
a = 0.4383 + 0.3606i
.
.
.


the expression converges to a = 0.4383 + 0.3606i (as approximation)

which is a complex number.


* * * * * * * * * * * * *
you can also try google also:
search for : i^i^i^i^i .. . .. .. put as many i as you want untill you reach a limit

something like:
i^i^i^i^i^i ^i^i^i^i^i^ i^i^i^i^ i^i^i^i^i^i^i^ i^i^i^i^i^ i^i^i^i^i

you will have to try about i powered 80 times to start seeing my number in google.

* * * * * * * * * *
The First case of ksoileau is periodic:
i^i = 0.2079
(i^i)^i = -i
((i^i)^i)^i = 4.8105
(((i^i)^i)^i)^i = i
((((i^i)^i)^i)^i)^i = 0.2079
.
.
.

The second case converges to a complex number as I explained above which is: 0.4383 + 0.3606i
* * * * * * * * * * *
Okay guys you want it Mathematically go to:

http://mathworld.wolfram.com/PowerTower.html

Take a look at equation number (18), this just proves my answer with the special case h(i).

Answer 2

It converges to a specific complex number, which is the root of the equation i^x = x. The number is approximately 0.438282 + 0.360592 i. The exact value for x is (2i/π) ProductLog(-iπ/2).

Addendum: The successive values of i^...^i create a converging spiral that looks like self-simliar nested triangles.

Answer 3

i'll try. just comment on my approach okay...

let z = i^i^i^i^... (z is a complex number, to be safe cause we are not sure that z is a real number yet, and assuming that i^i^i^i^... is a convergent expression.) ©

z = i^z
taking the log of both sides
log z = z log i.
log z = z(πi/2) .. :-0
... from this expression, if ever z is a real number, it cannot be a positive number.

we can probably approach this again by letting z = re^(iθ)
... or by assuming first that z is a negative real number, i guess it would be simpler to check that first....
would anybody want to finish this. :)

Answer 4

Suppose such a real number x existed. Then it would have to satisfy either x=x^i or x=i^x, depending on how you interpret a tower of exponentials.
In the first case, if x=x^i, then x=x^i=(x^i)^i=x^(i*i)=x^(-1) so then x would have to be either 1 or -1. But 1^i=e^(-2*pi) which is not equal to 1, and (-1)^i=e^(-pi) which is not equal to -1.
In the second case, if x=i^x, then x=e^(pi*i*x/2)
=cos(pi*x/2)+i*sin(pi*x/2). Since x is real, sin(pi*x/2)=0 which implies cos(pi*x/2) is 1 or -1, hence x must be 1 or -1. But 1 is not equal to i^1, and -1 is not equal to i^(-1), so there are no solutions in either case.

Answer 5

e^(i.theta) = cos(theta) + i.sin(theta)
cos(theta)=0
sin(theta) = 1
theta = pi/2

e^(i.pi/2) = i
ln(i) = i.pi/2

i^i = y
ln(y) = i.ln(i) = -pi/2
y = e^(-pi/2)

i^i = e^(-pi/2)

i^(i^i) = i^(e^(-pi/2)) ---- a complex number

I doubt it would converge to a real number. Maybe someone else can provide a rigorous proof.
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i^i^i^i = i^(i^(i^i))) ≠ (i^i)^(i^i)
i^i^i^i is not a real number

i^i^i^i is approximately 0.0500922361 + 0.602116527 i
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ksoileau raised an interesting point
1^i = (e^0)^i = e^0 = 1
1^i = (e^2pi.i)^i = e^(-2pi)
1^i = (e^4pi.i)^i = e^(-4pi)
.
.
.
When dealing with imaginary/complex exponents you have multiple solutions so the series would not converge to a single number. You would get multiple answers (this does not imply that any of them would be real).

Answer 6

No.

i = exp(i * pi/2)

i^i = exp(-pi/2)

i^i^i = exp(-i * pi/2)

i^i^i^i = exp(pi/2)

etc.

Not a rigorous proof, but you get the idea. It does not converge.

Answer 7

you can't have the square root of a negative number. it can't be done.