Infinite power tower of i?
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 dhokarena56
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Infinite power tower of i?
I was browsing the OEIS (oh, you magnificent timewaster, you!) and came upon Sequence A077589: the digital expansion of i^{i[sup]i...}[/sup], or, if you like, ^{∞}i.
So apparently this is a finite, real, positive number. Anyone care to explain how?
So apparently this is a finite, real, positive number. Anyone care to explain how?
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Re: Infinite power tower of i?
dhokarena56 wrote:I was browsing the OEIS (oh, you magnificent timewaster, you!) and came upon Sequence A077589: the digital expansion of i^{i[sup]i...}[/sup], or, if you like, ^{∞}i.
So apparently this is a finite, real, positive number. Anyone care to explain how?
That’s the decimal expansion of the real part of the complex number ^{∞}i.
By definition of complex exponentiation, a^{b} = e^{b Log(a)}, where “Log” is the principle value of the complex log function. So i^{x} = e^{xiπ/2}. You can do some work with limits to show the power tower converges. After that, finding the value is easy. It must satisfy z = ^{∞}i = i^{z} = e^{ziπ/2}.
But z = e^{ziπ/2} can be solved with the Lambert W function:
1/z = e^{ziπ/2}
1 = ze^{ziπ/2}
iπ/2 = (ziπ/2)e^{ziπ/2}
So by definition of W,
ziπ/2 = W(iπ/2)
z = (2i/π) W(iπ/2)
wee free kings
Re: Infinite power tower of i?
I actually fiddled with the convergence a bit, thinking geometrically, and couldn't get anywhere. Anyone have an easy way to see it?
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.
Re: Infinite power tower of i?
Well, the goal is to show that the map z→e^{z·iπ/2} has an attractive fixed point, with i in the basin of attraction. We’ve already found the fixed point, but I don’t know/recall enough about iterated mappings from ℂ→ℂ off the top of my head to come up with a good way to find its basin of attraction.
The best I’ve come up with in a few (sleepdeprived) minutes are some bounds on the orbit of z→i^{z}. Writing ^{n}i = r_{n}cis(θ_{n}), it’s easy enough to show that e^{π/2}≤r_{n}≤1 and 0≤θ_{n}≤π/2 for all n.
This comes from the expressions:
r_{k+1} = e^((π/2) r_{k} sin(θ_{k}))
θ_{k+1} = (π/2) r_{k} cos(θ_{k})
Combining those, we get a recurrence involving only the polar angle:
θ_{k+1} = (π/2) cos(θ_{k1}) e^(θ_{k} tan(θ_{k1}))
And an expression for modulus in terms of angle:
r_{k} = (2/π) θ_{k+1} / cos(θ_{k})
It’s easy to see from the anglerecurrence that the arg always remains in [0, π/2] (and only hits the endpoints for n = 1 and 2). Going back to the first equation for modulus, we then find the min and max values there. I just hope I didn’t mess anything up, since I’m really tired right now.
The best I’ve come up with in a few (sleepdeprived) minutes are some bounds on the orbit of z→i^{z}. Writing ^{n}i = r_{n}cis(θ_{n}), it’s easy enough to show that e^{π/2}≤r_{n}≤1 and 0≤θ_{n}≤π/2 for all n.
This comes from the expressions:
r_{k+1} = e^((π/2) r_{k} sin(θ_{k}))
θ_{k+1} = (π/2) r_{k} cos(θ_{k})
Combining those, we get a recurrence involving only the polar angle:
θ_{k+1} = (π/2) cos(θ_{k1}) e^(θ_{k} tan(θ_{k1}))
And an expression for modulus in terms of angle:
r_{k} = (2/π) θ_{k+1} / cos(θ_{k})
It’s easy to see from the anglerecurrence that the arg always remains in [0, π/2] (and only hits the endpoints for n = 1 and 2). Going back to the first equation for modulus, we then find the min and max values there. I just hope I didn’t mess anything up, since I’m really tired right now.
wee free kings
Re: Infinite power tower of i?
All right, I’ve made a little progress. The Brouwer fixedpoint theorem guarantees that f(z)=i^{z}=e^{i½πz} has a fixedpoint in the quarterannulus 0≤θ≤½π, e^{π/2}≤r≤1. That follows since the set is homeomorphic to a closed ball, while f is continuous and maps that set to itself.
Now, if we can find a compact set homeomorphic to the closed unit ball on which f′(z)<1, then the contraction mapping theorem implies the function has a unique and attractive fixedpoint in that set. (Note to self: use the phrase “unique and attractive” more often).
The derivative f′(z) = i½πe^{i½π(a+bi)}, so f′(z)=½πe^{½πb}. That is less than 1 when b>(2/π)ln(½π). So it will suffice to find a set with all imaginary parts greater than that threshold, homeomorphic to the closed unit ball, that f maps into itself. I’m still working on that part. Once we prove such a set exists, then:
Spoiler:
Now, if we can find a compact set homeomorphic to the closed unit ball on which f′(z)<1, then the contraction mapping theorem implies the function has a unique and attractive fixedpoint in that set. (Note to self: use the phrase “unique and attractive” more often).
The derivative f′(z) = i½πe^{i½π(a+bi)}, so f′(z)=½πe^{½πb}. That is less than 1 when b>(2/π)ln(½π). So it will suffice to find a set with all imaginary parts greater than that threshold, homeomorphic to the closed unit ball, that f maps into itself. I’m still working on that part. Once we prove such a set exists, then:
Spoiler:
wee free kings
Re: Infinite power tower of i?
Okay, I’ve got a proof of convergence now.
Brouwer guarantees a fixedpoint in the quarterannulus mentioned above. Symbol manipulation and numerical estimation shows there is only one fixedpoint in that region, and gives its location as z_{0} from the previous post.
The norm of the derivative is less than one, f′(u) < 1, on any closed disk D centered at z_{0} with radius less than Im(z_{0})  (2/π)·ln(½π), for reasons shown above. Numerically that critical radius is R≈0.07310599 as per Wolfram Alpha.
So for any z≠z_{0} in such a disk, f(z)  z_{0} = f(z)  f(z_{0}) = ∫[imath]_{z_0}^z[/imath] f′(u)·du ≤ ∫[imath]_{z_0}^z[/imath] f′(u)·du < z  z_{0} where we use first the triangle inequality and second the fact that f′(u)<1 on a straight line from z_{0} to z. Thus f(z) is closer to z_{0} than z was.
Additionally, for any two distinct points u, v ∈ D∖{z_{0}}, we have f(u)  f(v) = ∫[imath]_{v}^u[/imath] f′(q)·dq ≤ ∫[imath]_{v}^u[/imath] f′(q)·dq < u  v. Therefore f is a contraction mapping on D. Now the contraction mapping theorem shows the orbit under f of each point in that disk converges to the fixed point z_{0}.
In particular, that means if at any time the orbit of i under the map f gets within distance R of z_{0}, it will converge to z_{0}. As it turns out, the 20th iteration is within such a disk, as per Wolfram Alpha. Hence the orbit of i under f(z)=i^{z} converges to z_{0}, and therefore ^{∞}i=(2i/π)W(iπ/2).
Brouwer guarantees a fixedpoint in the quarterannulus mentioned above. Symbol manipulation and numerical estimation shows there is only one fixedpoint in that region, and gives its location as z_{0} from the previous post.
The norm of the derivative is less than one, f′(u) < 1, on any closed disk D centered at z_{0} with radius less than Im(z_{0})  (2/π)·ln(½π), for reasons shown above. Numerically that critical radius is R≈0.07310599 as per Wolfram Alpha.
So for any z≠z_{0} in such a disk, f(z)  z_{0} = f(z)  f(z_{0}) = ∫[imath]_{z_0}^z[/imath] f′(u)·du ≤ ∫[imath]_{z_0}^z[/imath] f′(u)·du < z  z_{0} where we use first the triangle inequality and second the fact that f′(u)<1 on a straight line from z_{0} to z. Thus f(z) is closer to z_{0} than z was.
Additionally, for any two distinct points u, v ∈ D∖{z_{0}}, we have f(u)  f(v) = ∫[imath]_{v}^u[/imath] f′(q)·dq ≤ ∫[imath]_{v}^u[/imath] f′(q)·dq < u  v. Therefore f is a contraction mapping on D. Now the contraction mapping theorem shows the orbit under f of each point in that disk converges to the fixed point z_{0}.
In particular, that means if at any time the orbit of i under the map f gets within distance R of z_{0}, it will converge to z_{0}. As it turns out, the 20th iteration is within such a disk, as per Wolfram Alpha. Hence the orbit of i under f(z)=i^{z} converges to z_{0}, and therefore ^{∞}i=(2i/π)W(iπ/2).
wee free kings
Re: Infinite power tower of i?
I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:
let ^{∞}i=z.
z^{i}=^{∞+1}i=^{∞}i=z
z^{i1}=1
(i1)ln(z)=0
i1=/=0
so ln(z)=0
so z=1
All the other values posted here are complex unless I'm mistaken but OP stated that the solution was real. I'm by no means sure that my argument holds, but it seems to be the first answer satisfying all the properties we were given for the solution.
let ^{∞}i=z.
z^{i}=^{∞+1}i=^{∞}i=z
z^{i1}=1
(i1)ln(z)=0
i1=/=0
so ln(z)=0
so z=1
All the other values posted here are complex unless I'm mistaken but OP stated that the solution was real. I'm by no means sure that my argument holds, but it seems to be the first answer satisfying all the properties we were given for the solution.
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Re: Infinite power tower of i?
eSOANEM wrote:I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:
let ^{∞}i=z.
z^{i}=^{∞+1}i=^{∞}i=z
This is wrong. (^{a}b)^{b} ≠ ^{a+1}b.
Re: Infinite power tower of i?
Benoni wrote:eSOANEM wrote:I'm not sure as to the rigour of this proof, but I think that, assuming a value exists, I can show what that value is:
let ^{∞}i=z.
z^{i}=^{∞+1}i=^{∞}i=z
This is wrong. (^{a}b)^{b} ≠ ^{a+1}b.
Really? Doesn't that follow from the definition of tetration? Or is it because a^b=/=b^a?
Edit: I think I see it, it's because exponentiation isn't commutative and because of the direction of its nesting.
Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.
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Re: Infinite power tower of i?
I think there is an easier way to do this:
first assume that i^(i^(i^... converges to some L
L=i^(i^(i^...
We can see that this is equivalent to L=i^L
this is the same as i=L^(1/L)
using the product log function where W(z) is the inverse function of f(W)=We^W and doing some algebra you get L=W(ipi/2)*(2i/pi)
Check out the paper nested radicals and other infinitely recursive expressions by Michael McGuffin. That'll give you a nice ground on working with all sorts of weird things like infinite exponent towers, continued fractions and nested radicals in a very easy to floow and understand way
first assume that i^(i^(i^... converges to some L
L=i^(i^(i^...
We can see that this is equivalent to L=i^L
this is the same as i=L^(1/L)
using the product log function where W(z) is the inverse function of f(W)=We^W and doing some algebra you get L=W(ipi/2)*(2i/pi)
Check out the paper nested radicals and other infinitely recursive expressions by Michael McGuffin. That'll give you a nice ground on working with all sorts of weird things like infinite exponent towers, continued fractions and nested radicals in a very easy to floow and understand way
Re: Infinite power tower of i?
eSOANEM wrote:Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.
The OP is wrong, the OEIS is right, I already addressed this in the very first reply to this thread.
Qaanol wrote:That’s the decimal expansion of the real part of the complex number ^{∞}i.
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 gmalivuk
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Re: Infinite power tower of i?
You mean like how it was done in the very first reply?Mynameisjosh wrote:I think there is an easier way to do this:
Re: Infinite power tower of i?
Qaanol wrote:eSOANEM wrote:Still, OP said the solution was real but the product log solution is complex, I take it that OP and OEIS is wrong in this regard then.
The OP is wrong, the OEIS is right, I already addressed this in the very first reply to this thread.Qaanol wrote:That’s the decimal expansion of the real part of the complex number ^{∞}i.
Sorry, I misread that.
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Re: Infinite power tower of i?
gmalivuk wrote:You mean like how it was done in the very first reply?
Reading comprehension FTW. Well fortunately my first post is indicative of my normal attention span
Re: Infinite power tower of i?
I started looking at the analogous situation for an arbitrary base, so f_{u}(z) = u^{z} = e^{z·Log(u)}. In particular, I wanted to know for which values of u a fixed point of f_{u} is an attractor. I did not try to find the basin of attraction, just whether it has one.
What I found is that f_{u} has a fixed point at z_{0} = W(Log(u)) / Log(u). That is an attractor when f′(z_{0})<1, meaning Log(u)·e^{W(Log(u))} < 1.
So I fired up Matlab and plotted which values of u satisfy that. Turns out to look like this, where the black region has attractive fixed points:
The pixels are sampled every .02 along both real and imaginary axes. The topmost part has imaginary component 2.10, and the bottommost 2.10. The leftmost point has real component 0.38, and the rightmost 2.04. Along the real axis (which is the symmetry axis here), the leftmost point is 0.08 and the rightmost is 1.52.
I don’t recognize that shape, but the formula involved lots of “Log(u)” so I plotted the same thing for Log(u), giving this:
That looks very much like a cardioid, but I’m not sure if it really is. In any case, the leftmost point is 2.70, the rightmost point is at 0.86, and the rightmost point on the real axis is 0.36. The max imaginary value is 2.00. I rather suspect the leftmost point is actually e, and the rightmost point on the real axis is 1/e, but I don’t know for sure.
It appears that for maps of the form f_{u} = u^{z}, the values of u that have an attractive fixed point correspond to values of Log(u) within a cardioid. Of course, I only looked at the principle branch of the Lambert W function, so there could be other uvalues for which additional fixedpoints exist and are attractive.
What I found is that f_{u} has a fixed point at z_{0} = W(Log(u)) / Log(u). That is an attractor when f′(z_{0})<1, meaning Log(u)·e^{W(Log(u))} < 1.
So I fired up Matlab and plotted which values of u satisfy that. Turns out to look like this, where the black region has attractive fixed points:
The pixels are sampled every .02 along both real and imaginary axes. The topmost part has imaginary component 2.10, and the bottommost 2.10. The leftmost point has real component 0.38, and the rightmost 2.04. Along the real axis (which is the symmetry axis here), the leftmost point is 0.08 and the rightmost is 1.52.
I don’t recognize that shape, but the formula involved lots of “Log(u)” so I plotted the same thing for Log(u), giving this:
That looks very much like a cardioid, but I’m not sure if it really is. In any case, the leftmost point is 2.70, the rightmost point is at 0.86, and the rightmost point on the real axis is 0.36. The max imaginary value is 2.00. I rather suspect the leftmost point is actually e, and the rightmost point on the real axis is 1/e, but I don’t know for sure.
It appears that for maps of the form f_{u} = u^{z}, the values of u that have an attractive fixed point correspond to values of Log(u) within a cardioid. Of course, I only looked at the principle branch of the Lambert W function, so there could be other uvalues for which additional fixedpoints exist and are attractive.
wee free kings
Re: Infinite power tower of i?
A bit more numerical work, and some pretty pictures. First, the points in white are those uvalues for which the orbit of u under f(z) = u^{z} diverges to infinity (or at least has absolute value great enough to overflow Matlab) at or before the 1000th iteration. The images go from 3.5 to 6.5 on the real axis, and 7.5 to 7.5 on the complex axis. This region shows most of the “interesting” behavior. Click to view fullsize, where pixels are sampled every 0.01 units.
Exponential fractal: white points numerically
calculated to diverge by 1000th iteration,
black points have not diverged
Cycle length after 1000 iterations
Black = Chaotic, or cyclelength greater than 10
Blue = 1cycle (fixedpoint attractor)
Green = 2cycle (2point attractor)
⋮
Yellow = 10cycle
White = Numerically observed to diverge
Here are the same things, but drawing the result for z at the point log(z) (technically, drawing at point z the result for exp(z), so these are 2πiperiodic). That is, I have simply applied the map Log(z) to the complex plane, thereby warping the locations of points, but not changing any of the values in the iterated exponential map.
As you can see, it very strongly appears that the locus of fixedpoint attractors, shown in blue in the last image, has natural log in the shape of a cardioid. The apparent cusp is at 1/e, and the opposite side is at e. The top and bottom seem to be at c±2i for some real number c I haven’t tried to identify.
To find the cyclelengths, I calculated the 1000th through 1010th iterations, and compared them pointwise to see when the orbit returned to its value from the 1000th iteration, within a tolerance of 0.001. To visualize how the cycles go, here are the 1000th through 1010th iterations in order, with arctanpolar coloration:
Edit: it also looks like the red area (period 3) in the logplot has bounds that approach ±π/2 in the limit of large real component.
Exponential fractal: white points numerically
calculated to diverge by 1000th iteration,
black points have not diverged
Cycle length after 1000 iterations
Black = Chaotic, or cyclelength greater than 10
Blue = 1cycle (fixedpoint attractor)
Green = 2cycle (2point attractor)
⋮
Yellow = 10cycle
White = Numerically observed to diverge
Here are the same things, but drawing the result for z at the point log(z) (technically, drawing at point z the result for exp(z), so these are 2πiperiodic). That is, I have simply applied the map Log(z) to the complex plane, thereby warping the locations of points, but not changing any of the values in the iterated exponential map.
As you can see, it very strongly appears that the locus of fixedpoint attractors, shown in blue in the last image, has natural log in the shape of a cardioid. The apparent cusp is at 1/e, and the opposite side is at e. The top and bottom seem to be at c±2i for some real number c I haven’t tried to identify.
To find the cyclelengths, I calculated the 1000th through 1010th iterations, and compared them pointwise to see when the orbit returned to its value from the 1000th iteration, within a tolerance of 0.001. To visualize how the cycles go, here are the 1000th through 1010th iterations in order, with arctanpolar coloration:
Spoiler:
Edit: it also looks like the red area (period 3) in the logplot has bounds that approach ±π/2 in the limit of large real component.
wee free kings
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