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I just watched this YouTube video by Michael Penn about the expression

$$\sqrt{i \sqrt{i \sqrt{i \sqrt{i \dots}}}}$$

and how to evaluate it. At the end of the video, he mentions that if we do not interpret square roots as just being principal square roots, then the solutions take the form $$ e^{\frac{i \pi}{2} + i \pi \bigl( n_0 + \frac{n_1}{2} + \frac{n_2}{4} + \frac{n_3}{8} + \cdots \bigr)} $$

where $n_0, n_1, n_2, ... \in \{0, 1\}$. Because all real numbers in $[0, 2)$ can be expressed as binary numbers of the form $b_0.b_1b_2b_3b_4b_5\ldots\,$, this means that solutions to the original nested radical work out to "any complex number of modulus $1$."

(However, if we do restrict ourselves to the principal square root, we'd get back a single answer.)

I'm trying to interpret exactly what this claim means and was wondering if the following line of reasoning works.

  • Let's start with $\sqrt{i}$. This means "a number that, if squared and divided by $i$, is equal to $1$." There are two such numbers.

  • Now, let's try $\sqrt{i\sqrt{i}}$. This means "some number that, if squared and divided by $i$, then squared and divided by $i$ again, gives $1$."

  • Let's then try $\sqrt{i\sqrt{i\sqrt{i}}}$. That would mean "some number that if squared and divided by $i$, then squared again and divided by $i$, then squared again and divided by $i$, you get $1$."

And more generally, any finite expansion of the nested radical can be interpreted as "some number that, if repeated squared and divided by $i$ a total of $n$ times, gives $1$."

For starters - is this a correct way of thinking about the partial terms of the series?

Assuming that this is the case, I'm struggling to make sense of what the infinitely nested radical would mean. The above process-based formulation breaks down if you repeat this process infinitely many times, since at a first reading it would mean "a number where, if you square and divide by $i$ infinitely many times, yields $i$." That doesn't make sense to me, nor does it feel like a legal strategy for extending the finite case to a limit.

Rather, it seems like what's going on here is that if you repeatedly increase the number of iterations of "square and divide by $i$," you start filling up more and more of the complex unit circle. And the infinite limit working out to "all complex numbers of modulus $1$" then would mean something to the effect of the following:

Let $z$ be an arbitrary complex number of modulus $1$. Then by repeatedly squaring $z$ and dividing by $i$, we can make the number get as close to $1$ as we'd like.

Is this a reasonable conclusion to draw? Or is this not an appropriate way of thinking about the infinitely nested radical?

Thanks!

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    $\begingroup$ The original claim is wrong. You cannot get the square roots to converge to anything except for $1$. This is because $1$ is the only number on the unit circle that is its own square root (principal or otherwise). $\endgroup$ May 9 at 22:54
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    $\begingroup$ @JosephCamacho The original claim is right. And actually, if you choose the principal square root, the nested radicals converge to $i$, not to $1$. $\endgroup$
    – jjagmath
    May 10 at 1:31
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    $\begingroup$ This works best if you think of complex numbers in terms of their lengths and arguments. Multiplying by $i$ adds $\frac{\pi}{2}$ to the argument and taking the square root halves it (and also square roots its length). $\endgroup$
    – TomKern
    May 10 at 2:03
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    $\begingroup$ @templatetypedef It is the Riemann surface associated with the logarithm. See the Wikipedia article here. You can think about the points of this space (surface) as pairs $(r,\theta)$ where $r$ is the modulus and $\theta$ is the phase. These correspond to $r e^{i\theta}$. In this space there are infinitely many "$i$"s: $e^{i\theta}$ where $\theta \equiv \frac{\pi}{2} \pmod{2\pi}$. $\endgroup$
    – Gary
    May 12 at 1:42
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    $\begingroup$ @JosephCamacho Your comment is wrong in a very subtle way. The question is about the sequence of square roots nesting inward (right-infinite), but you are thinking of an iterative process of taking $z \mapsto \sqrt{iz}$ with varying branches of $\sqrt{}$. They are not quite the same, if you think about. If you're familiar with continued fractions, the nesting there is analogous. $\endgroup$
    – Erick Wong
    May 13 at 18:03

3 Answers 3

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As you are trying to interpret what the number $$\sqrt{i\sqrt{i\sqrt{i\sqrt{\cdots}}}}$$ you are going sequentially to understand this, which is a good strategy. However, the direction in which you are looking at this number isn't really the correct direction.

Take an example, lets consider the series; $$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots.$$

Now, in order to evaluate this series, we look at it in the following manner;

$$\text{Step 1: }\text{ }\text{ }1+\frac{1}{2}=\frac{3}{2}=1.5$$ $$\text{Step 2: }\text{ }\text{ }1+\frac{1}{2}+\frac{1}{4}=\frac{7}{4}=1.75$$ $$\text{Step 3: }\text{ }\text{ }1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}=\frac{15}{8}=1.875$$ $$\vdots$$

As we continue doing this, we learn that as we keep adding without stopping, the series eventually get closer and closer to $2,$ and;

It can get as close to $2$ as possible, if we add a sufficient amount of times.

So now, we can simply say that as we continue this process infinitely many times, the answer becomes equal to $2.$ Makes sense right? The fact is, we are viewing from the place we started from, and not from the place where we end.

Now, when we consider the 'series'; $$\sqrt{i\sqrt{i\sqrt{i\sqrt{\cdots}}}},$$ we are basically looking at it as starting with $i$ and repeating a particular operation a number of times; Start with $1,$ multiply by $i$ and take the square root; $$\sqrt{i}=e^{\frac{\pi}{4}i}$$

Now, start with $e^{\frac{\pi}{4}i},$ multiply by $i$ and take the square root; \begin{align*} \sqrt{i\sqrt{i}} &=\sqrt{i\cdot e^{\frac{\pi}{4}i}}\\ &=\sqrt{e^{\frac{3\pi}{4}i}}\\ &=e^{\frac{3\pi}{8}i}. \end{align*}

Now, start with $e^{\frac{3\pi}{8}i},$ multiply by $i$ and take the square root; \begin{align*} \sqrt{i\sqrt{i\sqrt{i}}} &=\sqrt{i\cdot e^{\frac{3\pi}{8}i}}\\ &=\sqrt{e^{\frac{7\pi}{8}i}}\\ &=e^{\frac{7\pi}{16}i}. \end{align*}

Now, start with $e^{\frac{7\pi}{16}i},$ multiply by $i$ and take the square root; \begin{align*} \sqrt{i\sqrt{i\sqrt{i\sqrt{i}}}} &=\sqrt{i\cdot e^{\frac{7\pi}{16}i}}\\ &=\sqrt{e^{\frac{15\pi}{16}i}}\\ &=e^{\frac{15\pi}{32}i}. \end{align*}

As we keep repeating this operation over and over again, we can see that the sequence of numbers we are getting is; $$e^{\frac{3\pi}{8}i},e^{\frac{7\pi}{16}i},e^{\frac{15\pi}{32}i},\dots$$

Notice that the exponent is getting closer and closer to $\frac{\pi}{2}$? So this means that as we continue to make the operation of ; "multiply by $i$ and take the square root," we will finally get closer and closer to $e^{\frac{\pi}{2}i}=i.$ As thus, we can 'say' that; $$\sqrt{i\sqrt{i\sqrt{i\sqrt{\cdots}}}}=i.$$ And, this way is how you must look at this series. Not the other way, as you said; And more generally, any finite expansion of the nested radical can be interpreted as "some number that, if repeated squared and divided by i a total of n times, gives 1."

So, answering your question: is this a correct way of thinking about the partial terms of the series?

No it's not. The way you must see it is as what I just described to you. Thanks for reading, hope this helps :)

Comment if you have any further questions.

Edit 1

I saw your comment, and saw what your main question is so I'm going to add it down here. Now, when we were considering the nested series of radicals; $\sqrt{i\sqrt{i\sqrt{i\sqrt{\cdots}}}}=i,$ we are mainly only considering the 'principal' square root of the number we get, as we keep repeating the operation over and over again.

However, you want to understand the possible outcomes when we do not restrict ourselves to a 'principle' square root, which is indeed a great question to ask. So in order to understand it that way, lets first note that we are not going to look at the confusing way you stated in your question; "some number that, if repeated squared and divided by i a total of n times, gives 1." We are again going to look at it from the point of view that we've so far been working on.

So, I am going to try and understand all the possible values of $\sqrt{i\sqrt{i\sqrt{i\cdots}}}$ when we consider any square root (not necessarily the principle square root). Note that all of the possible values we can get, are obtained by choosing either principle square root, or non principle square root in each part of the process.

So, first note then in any of these operations, we never move out of the unit circle, and we always stay on it (multiplying by $i$, or taking square root), so the number obtained would always be of the form $e^{\frac{k}{2}\pi i}$ for some $k\in[0,4).$ Thus, for simplicity, we will not consider the actual complex number $e^{\frac{k}{2}\pi i},$ but just the $k$ part of the number. We are basically mapping the unit circle the set $[0,4).$

So for example $i$ will be written as $1,$ $-1$ will be written as $2,$ and $1$ will be written as $0,$ and so on. We will only be using the new notation after this.

Now, lets understand the algorithm in terms of the new notation.

Now how do we need to start?

We need to start with $0.$

What is the operation that needs to be repeated?

First of all, we need to take to original number and add $1$ or subtract $3$ according to which one can keep us in $[0,4)$ (this is basically multiplication by $i$). After the addition/subtraction, we need to either divide it by $2,$ or divide it by $2$ and add $2.$ Its our choice (this is equivalent to taking either the principle root or the non-principle root).

So lets, get started with doing these operations;

Note: We will code the choices (principle or non-principle) as 0 or 1. These choices will then form a sort of a 'binary' code for each possibility of the final number obtained. Notice that this is just like the Base 2 method of encoding each number?

$1$: Start with $0.$ Adding $1$ gives $1,$ and taking the non-principle root gives $\frac{1}{2}+2=\frac{5}{2}.$

$0$: Start with $\frac{5}{2}.$ Adding $1$ gives $\frac{7}{2},$ and taking the principle root gives $\frac{7}{4}$

$0$: Start with $\frac{7}{4}.$ Adding $1$ gives $\frac{11}{4},$ and taking the principle root gives $\frac{11}{8}$

$0$: Start with $\frac{11}{8}.$ Adding $1$ gives $\frac{19}{8},$ and taking the principle root gives $\frac{19}{16}$

We can keep doing these operations, and get various different results. Note that they are best tried on a number line (unfortunately I can't draw one here but you can try it yourself)

Now, after doing these operations I noticed that when we keep trying both of the operations simultaneously, we would reach no where, and the answer would keep moving around like the sequence $1,0,1,0,1,0,\dots$ So if we want to see this process converge to some number, we must finally resort to just one of these operations. If we consider the principle square root all the time, then we'll get the original answer $i$ (or $1$ in the new notation), however, if we consider non-principle square root all the time, we'll get it to converge at $-1$ eventually, (you can check)

In some cases, the sequence doesn't converge and rather becomes cyclical. But that's all not very important. The main idea is, how we can represent the numbers using the 'binary' notation, where some of them correspond to a real complex number and some do not.

Hope this helps:) Please comment for any further questions or ideas or errors in my argument.

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  • $\begingroup$ My confusion stems from the following. Take $\sqrt{4}$. If we interpret $\sqrt{4}$ to mean “the principal square root of four,” then this expression evaluates to $2$. On the other hand, suppose we interpret $\sqrt{4}$ not specifically as the principal square root of four, but as “any square root of four.” That would mean there are two different values that work here: $\pm 2$. My question is more specifically about the latter case, since, as you’ve mentioned and is shown in the linked question, if we restrict ourselves to principal square roots there’s a single limit. Does that make sense? $\endgroup$ May 13 at 15:13
  • $\begingroup$ Put another way - I completely follow the math you have here. What I’m looking for is less a derivation of the limit and more about what happens if we don’t restrict ourselves to principal square roots and instead interpret the nested radical as a characterization of a set of complex numbers, whether it’s even possible to do that at all, and what happens if we do. $\endgroup$ May 13 at 15:19
  • $\begingroup$ @templatetypedef Alright, I see what you mean. Let me just edit my post to address that confusion :) $\endgroup$
    – MathMinded
    May 13 at 16:12
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    $\begingroup$ Your exponents keep missing factors of $i$. $\endgroup$
    – J.G.
    May 13 at 18:33
  • $\begingroup$ @J.G. I am unable to see where I made the mistake, can you edit the post to correct it? $\endgroup$
    – MathMinded
    May 14 at 2:00
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The expression as such is ambiguous and needs some interpretation. When taking a complex square-root you have to make a choice of sign, and iterating the procedure you wind up with an expression like the one you mentioned in the beginning, except that I think it should be: $e^{\frac{i\pi}{4}+i\pi(n_0+\frac{n_1}{2}+...)}$

You may, however, also ask what kind of number you may obtain by starting by 1 and applying $n$ operations (multiplying by $i$ and taking an ambiguous square root). As you say it should be a number $z$ on which the reverse operation iterated $n$ times should yield 1.

To make precise this interpretation, write $i=\exp(i\pi/2)$ and $z=\exp(i\theta)$. Then the operation on $z$ you describe is $f(z)=z^2/i = \exp(i (2 \theta-\pi/2))$. After $n$ iterations you get $$f^n(z)= \exp\left (i (2^n \theta -\frac{\pi}{2}(2^{n-1}+\cdots +1))\right) $$ This should equal one so the angle should be a multiple of $2 \pi$. For $n\geq 2$:: $$f^n(z)=1 \Leftrightarrow 2^n \theta + i\pi/2=2\pi k \Leftrightarrow \theta=\frac{2\pi (k-1/2)}{2^{n}} , \ \ k\in {\Bbb Z}.$$ Of these angles $\theta_k=\theta_m$ iff $k-m$ is a multiple of $2^n$ so there are precisely $2^n$ distinct complex numbers $z=\exp(i \frac{2\pi (k-1/2)}{2^{n}})$, $k=0,...,2^n-1$ on the circle that satify the requirement. They are equidistantly distributed on the circle and as $n$ increase they become densely distributed and any point on the circle may be approximated by considering a suitable sequence of such points (which is the subject of the video, though formulated differently). Incidentally, writing a binary expansion of $k$ yields an expression fairly similar to the first.

For your last 'reasonable conclusion' that does not quite work in the way you describe. The simplest counter-example is the angle $\pi/2$ or $z=i$ for which $f(i)=i$, so it stays fixed. However, given any $z$ on the circle and $\delta>0$ you may find some $z'$ in the $\delta$-neighborhood of $z$ which after some possibly large, but finite number of iterations maps to 1. In fact if $2\pi/2^n<\delta$ you may find such $z'$ which after at most $n$ iterations maps to 1.

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I'd like to add a small discussion of the other answers, and how they relate. I won't comment on which interpretation is more natural, since these kinds of questions are mostly for fun anyway.

The possible values of the expression depend crucially on the way we interpret/define the limiting process. One iterative way is: Let $a_0 = 1$; For $k\ge1$, choose $a_k$ as one of the values of $\sqrt{ia_{k-1}}$. Then we have $$a_k = \exp\left(i\pi\cdot\left(\frac12 - \frac{1}{2^{k+1}} + n_k + \frac{n_{k-1}}{2} + \cdots + \frac{n_1}{2^{k-1}}\right)\right),$$ where $n_j\in\{0,1\}$. This is similar to Michael Penn's expression, but note that the most significant "digit" is $n_k$ here, as opposed to his $n_0$. We're going in the other direction so to speak. This means that, as $k\to\infty$, the sequence converges if and only if $(n_k)_k$ is eventually constant, i.e. $n_k=0$ or $n_k=1$ for all large enough $k$. In both of those cases the limit is $e^{i\pi/2}=i$. Essentially, we're restricted by not being able to change "past choices". This aligns with MathMinded's answer.

Note incidentally that the values of $n_k$ don't correspond directly to choosing the (non)principal branches. That is to say, it's not like $n_k=0$ will always mean taking the principal branch for example.


H. H. Rugh and Micheal Penn (of the video) seem to have the following interpretation instead. Consider the sequence $$ a_0 = 1, a_1 = \sqrt i, a_2 = \sqrt{i\sqrt i}, a_3 = \sqrt{i\sqrt{i\sqrt i}}, \ldots, $$ where $a_k$ is taken to be any of the $2^k$ values of a $k$-levelled $\sqrt{i\sqrt{i\cdots\sqrt i}}$. In this case, $$a_k = \exp\left( i\pi \cdot\left(\frac12 - \frac{1}{2^{k+1}} + \frac{m_k}{2^{k-1}}\right) \right),$$ where $m_k\in\{0, \ldots, 2^k-1\}$. (Actually, for $k\ge2$, we can even write the more compact $a_k=e^{ i\pi \cdot\left(\frac{m_k}{2^{k-1}} - \frac{1}{2^{k+1}} \right)}$, since the $1/2$ is swallowed by the $m_k$). This matches the possible values of $a_k$ according to the first interpretation, but we no longer require $a_k^2 = ia_{k-1}$. With this, we can indeed choose a sequence of $a_k$ (or equivalently $m_k$) so that it converges to any number on the unit circle. We are basically doing a binary expansion, or approximating by dyadic rationals depending on the viewpoint.

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