{"id":81,"date":"2007-10-25T23:54:23","date_gmt":"2007-10-26T04:54:23","guid":{"rendered":"http:\/\/www.moroha.net\/blog\/?p=81"},"modified":"2017-03-01T11:16:24","modified_gmt":"2017-03-01T16:16:24","slug":"i-i-i","status":"publish","type":"post","link":"https:\/\/www.moroha.net\/blog\/archives\/81","title":{"rendered":"i ^ i ^ i ^ … ?"},"content":{"rendered":"

Today on the internet I saw the calculation for i<\/em>^i<\/em>, where i<\/em> is the square root of -1. It can be solved pretty easily using Euler’s formula, where any number z in the complex plane can be expressed as
\n\\(\\textstyle{}\\)
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\\(\\displaystyle{z=r\\,e^{i\\,\\theta}}\\)<\/center>
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\\(\\displaystyle{}\\)<\/center>
\nwhere \\(r\\) is the absolute value of the number, \\(e = 2.718281828\\ldots\\) is the base of the natural logarithm, \\(i\\) is of course \\(\\sqrt{-1}\\), and \\(\\theta\\) is the angle in radians of the number measured counterclockwise from the positive real axis, essentially giving the complex number in polar coordinates. So for example \\(1\\,e^{\\textstyle{i\\:0}}=1\\), \\(1\\,e^{\\textstyle{i\\:\\pi}}=1\\), \\(1\\,e^{\\textstyle{i\\frac{\\pi}{2}}}=i\\), etc. We know that \\(i^i\\) is going to be some<\/em> number in the complex plane, so we can express that number as \\(\\textstyle{r\\,e^{i\\,\\theta}=e^{\\text{ln}(r)}e^{i\\,\\theta}}= e^{\\text{ln}(r)+i\\,\\theta}=e^z\\), where \\(z=\\text{ln}(r)+i\\,\\theta\\) is itself a complex number.
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\nThis gives us an equation
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\\(\\displaystyle{i^i = e^z}\\)<\/center>
\nwhere all that we have to do is solve for z. First we take the log of both sides<\/p>\n

\\(\\displaystyle{\\text{ln}\\left( i^i\\right)=z}\\)<\/center>
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\\(\\displaystyle{i\\,\\text{ln}\\left( i\\right)=z}\\) .<\/center><\/p>\n

We use the formula above for \\(i\\) to substitute it into the logarithm<\/p>\n

\\(\\displaystyle{i\\,\\text{ln}\\left( e^{i\\,\\frac{\\pi}{2}}\\right)=i\\:i\\,\\frac{\\pi}{2}=i^2\\,\\frac{\\pi}{2}=z}\\)<\/center><\/p>\n

and we know of course that \\(i^2=-1\\), so we have \\(z=-\\frac{\\pi}{2}\\).<\/p>\n

So for the answer we have :<\/p>\n

\\(\\displaystyle{i^i=e^{-\\textstyle{\\frac{\\pi}{2}}}\\approx 0.20788\\;.}\\)<\/center>
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\\(\\displaystyle{}\\)<\/center>
\nPretty neat, huh? So then I started thinking to myself, what if you took \\(\\textstyle{i^{i^i}}\\)? And then \\(\\textstyle{i^{i^{i^i}}}\\), and then again and again? What if you did it an infinite number of times? Would it diverge to some complex infinity? Would it converge to \\(0+0\\,i\\), or perhaps to some other value? It turns out the answer is surprising, to me at least.<\/p>\n

I’m not enough of an esoteric math whiz to try and find an analytical solution to this. If it were an infinite series or infinite product I might take a stab at it, but infinite power? I have never heard of such a thing before. On the real axis, any number \\(N > 1\\) diverges to infinity when taken to it’s own power an infinite number of times, and any number between 0 and 1 converges to 1 when taken to its own power an infinite number of times. But negative numbers and then complex numbers? Not so clear. A few trials showed me they can jump all over the complex plane. So I did the easy thing and wrote a program. This graph shows the results as a line connecting each successive iteration.<\/p>\n

\"i^i^i^_ad_nauseum\"<\/p>\n

You’ll see that the iterations actually converge on a single point! The value of the number comes out to z = 0.438283 + 0.360593 i<\/em>. I wasn’t able to get any more significant digits because even Mathematica choked on the calculation after that.<\/p>\n

So then my next question is, what numbers on the complex plane converge to a finite value when taken to their own power an infinite number of times, and what numbers diverge? I bet this has been studied before by someone, because such a question is very similar to the definition of the Mandelbrot Set<\/a>, the main difference being that Julia Sets (of which the Mandelbrot Set is a subset) and such have a \\(z^n + c\\) type of formula for the iteration.<\/p>\n

For some Mandelbrot Set extra goodies, here is a deep dive into the Mandelbrot Set where the zoom ratio from the beginning to end is bigger than the known universe!
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