This note is for all of the problems in my Nested Radicals set. Some of them are controversial, so I am adding a discussion page for the set. Below are each of the problems:

Please engage in discussions! I created this set to help others learn, but I want to learn from any mistakes that I made, too.

Note by Blan Morrison
2 years, 4 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Regarding this problem :

Let us assume that : $\sqrt{0 + \sqrt{0 + \sqrt{0 + ...}}} = x$

Then, $\sqrt{0 + x} = x$

$\implies 0 + x = x^2 \implies x^2 - x = 0 \implies x(x - 1) = 0$

So, $\boxed{x = 0 \text{ (or) } 1}$

- 2 years, 4 months ago

Are you saying it is undefined?

- 2 years, 4 months ago

I am saying that there are two possibilities for the answer.

- 2 years, 4 months ago

This is interesting. Can square roots of infinite 0's result in 1.

- 2 years, 4 months ago

I think it is in the form of $0\times \infty$, which is indeterminate.

- 2 years, 4 months ago

(Note: You're making the assumption, same as Blan in his solution, that the limit exists. This needs to be done separately.)

Right, so all that you have shown so far is that "If the limit exists, then the limit is either 0 or 1". You have to continue to show that the limit is one of this (we know it cannot be both because limits are unique.)

For an example that is similar in nature, consider the case where you're playing Sudoku, and you have made some steps to conclude that "This square is either 1 or 2". What does this imply? Well, innately you understand "I need more information/clues/logic before I can conclude which value it is".

Staff - 2 years, 4 months ago

My comment is regarding this question, and the question of whether the answer should be 0 or 1. In addition, 3blue1brown did a video on a very similar question recently. I recommend watching the video, but I’ll summarize the relevant parts.

The question in the video centered around the value of the continued fraction $\frac{1}{1+\frac{1}{1+\dots}}$. Using a little algebra, you can get that the value of this fraction is the golden ratio phi, and/or its lesser known twin, $\frac{1-\sqrt 5}{2}$, or about -.618. The video then raised the question of whether this second solution is legitamite, and provided an argument that phi is the only real solution, which I’ll explain in a moment.

First, I’d just like to draw the requisite parallel to our controversy: using algebra, we’ve obtained two possible values for our continued square root, but it isn’t clear which is legitamite. So, let’s do what 3b1b did: treat the continued root as a function composed to itself repeatedly. To clarify, we are going to pick a starting value, k, and keep applying the function to it, then find what the function approaches when applied infinitely many times.

If we pick k as 0,1 , then the function stays constant; that is why they are candidates. But, if you pick any other (positive) number to start, the repeated square roots will quickly approach 1, but never zero. A similar thing happens in the other problem, where running almost any number through 1 + 1/x repeatly quickly tends to phi.*

So this suggests that, if anything, 1 is more likely the value in our question. However, the video actually ends with a short discussion of other perspectives on the matter, and basically the idea that perhaps there is no ‘right answer’. I am interested to hear arguments for 0, or even for both being equally valid, so comment away!

*If you would like to know why this happens, check out the video; the why of it is the main question. I will hint that it has to do with derivatives.

- 2 years, 3 months ago

I have seen the video, but thank you for helping me make that connection! However, I would like to point out you are thinking of the function $f(k)=\sqrt{k\sqrt{k\sqrt{k\dots}}}$, which can be simplified to $k^{\frac{1}{2\infty}}=k^0$. I like you're argument for this function, and as to why $0^0$ "is more likely" to be 1. As you also point out, Grant mentions perspective. This is key to understanding that there are ways to argue that $0^0=0$. I recommend you see this wiki.

As for your/Grant's method, it is essentially the same as Mr. Lin's and my solution to the problem: just evaluate the truncated/partials terms. This can be applied to all nested radical problems; in fact, it is the easiest way to solve the fourth problem. Thank you for your input!

- 2 years, 3 months ago

As Blan mentioned, remember that the value is defined as the limit of sequential terms. Note: With infinite continued fractions, you will need to remember what the definition is. If you use a different definition ( in particular, if you include the "+1" without adding on the denominator), then you might end up with a different result.

What Ram has shown is that "If the limit exists, then it is either 0 or 1". This doesn't mean that both answers are equally valid. Instead, it means that we haven't done enough work to figure out what the answer is (We know it is unique because it is a limit).

(Edited)

Note: The stable/unstable fixed point argument put forth by 3b1b is irrelevant in determining what the limit should be.

In fact, this is a great example of where the unstable solution (IE 0) is the correct limit whereas the stable solution (IE 1) is the incorrect limit.

Staff - 2 years, 3 months ago

My apologies, but I think I’m still misunderstanding something. f(x), as you’ve defined it, is just the fourth root of x, since the zero does nothing. If you repeatedly take the fourth root of a positive number (or square root, it’s the same in the end), it will converge to one, not zero. Could you clarify what you were saying there?

Also, I was not trying to sugest that 0 and 1 are both the answer. I was simply trying to leave the discussion open if someone more knowledgable wanted to correct me, which seems to have happened :)

- 2 years, 3 months ago

Ah yes, I see what you mean here. Let me go back and edit.

Staff - 2 years, 3 months ago

Great set. No controversies at all.

- 2 years, 3 months ago

That is not true; there are multiple things that can be debated or questioned. Please do not put a comment that does not contribute to productive conversation.

- 2 years, 3 months ago

If you say so. I may be wrong, but what I mean is that I personally did not see anything wrong with it.

- 2 years, 3 months ago