Nested Radicals (3)

Algebra Level 2

As we saw in the previous problem , Rad Chad's model isn't entirely accurate. The next couple of problems in this set will explore this model and its accuracy.

Evaluate $\sqrt{0+\sqrt{0+\sqrt{0+\cdots }}}.$ To clarify, we want to evaluate $\displaystyle\lim_{n\to\infty}a_n$ , where $\{a_n\}_{n=1}^{n=\infty}$ satisfies the recursive relation, $a_{n+1} = \sqrt{0 + a_n}$ , with $a_1=0$ .

See the whole set .

Undefined 0 0.5 1

1 solution

Blan Morrison
Sep 16, 2018

Let's label this as a sequence, and find the limit of the sequence. Define $a_n=\sqrt{0+\sqrt{0+\sqrt{0+\dots \sqrt{0}}}}$ such that there are $n$ square root symbols. Let's find the finite values of $a_n$ : $a_1=0$ $a_2=0$ $a_3=0$ $\dots$ $\displaystyle\lim_{n\rightarrow \infty} a_n=\boxed{0}$ $\beta_{\lceil \mid \rceil}$

This is controversial. Because of that, I have created a discussion thread about this problem for any questions/rebuttals.

I think it is better to ask the staff to get the correct answer.

Ram Mohith - 2 years, 8 months ago

You can tag them through @Calvin Lin @Brilliant Mathematics

Blan Morrison - 2 years, 8 months ago

Replying directly here regarding your comment in his note.

You're making the assumption (same as Blan in this solution) that the limit exists.
Note that your implication signs are unidirectional, meaning that all that we can conclude is that if the limit exists, then it is either 0 or 1.

As an example, suppose you wanted to solve the (yes, really simple) equation $x = 1$ .
You decide to take the folllowing steps:

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

Does this imply that $x = -1$ ?

Well, you are familiar with the idea that "When we square an equation, we potentially introduce extraneous solutions, so we have to check them to see if it fits the original equation". In your solution, you did indeed square both sides of $\sqrt{ 0 + x } = x$ , so let's "check these solutions".

Calvin Lin Staff - 2 years, 8 months ago

That's a good point; we are assuming the limit does exists. Does that mean the expression is undefined/indeterminate? After all, it is in the form of $0\times \infty$ .

Your sudoku example also points this out; you can have 2 different sudoku puzzles that only differ by 2 squares. The analogy can be applied to this expression because you can have 2 different answers depending on your method. This is also the case with $0^0$ . Could you please change the answer to "Undefined?"

Thank you for your clarification!

Blan Morrison - 2 years, 8 months ago

@Blan Morrison – This expression is equal 0. It is not undefined nor 1. (So no, I'm not going to change the answer.)

Yes, the solutions are incomplete.

Hint: The solution is really simple. Calculate initial terms of the truncated radical (which is how the infinite radical is defined).

I edited my original comment to be more explicit.

Calvin Lin Staff - 2 years, 8 months ago

@Calvin Lin – I'm confused. My solution is that process. Are you saying it should be $a_{\infty}$ instead of $\displaystyle\lim_{n\rightarrow \infty}a_n$ ?

Blan Morrison - 2 years, 8 months ago

@Blan Morrison – Sorry, I was only referring to Ram's solution (and types like his).

I got confused as to why you were insisting that the answer is undefined, which implies that you thought your solution was incorrect. Your solution is correct.

Your sudoku example also points this out; you can have 2 different sudoku puzzles that only differ by 2 squares

That statement is not true. I believe they need to differ by at least 8 squares (in the solved version). That would likely make a good problem.

Calvin Lin Staff - 2 years, 8 months ago

@Calvin Lin – Sorry for the confusion; thank you!

Blan Morrison - 2 years, 8 months ago

@Calvin Lin – Actually, the answer is just four. If these four squares contain only two distinct numbers, there are two otherwise identical solutions.

I think expanding the question to 3+ boards might prove interesting, as I’m not sure of a simple way of finding the answer as with 2. A follow-up question perhaps?

Jason Carrier - 2 years, 8 months ago

@Jason Carrier – Yes, you would be correct. I'll let you make the problem since you came up with the solution first.

Blan Morrison - 2 years, 8 months ago

@Blan Morrison – Thank you, but @Calvin Lin actually already did here . As of right now, he still has 8 as the answer, but I’ll put in a report.

Jason Carrier - 2 years, 8 months ago

@Jason Carrier – You should leave the solution!

Blan Morrison - 2 years, 8 months ago

@Jason Carrier – Very good point.

Calvin Lin Staff - 2 years, 8 months ago

Nested Radicals (3)

1 solution

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