Radical Radicals

Algebra Level 3

How many positive $x$ satisfy

$\Large \sqrt{x} + \sqrt{\sqrt{x}} + \sqrt{\sqrt{\sqrt{x}}} + \ldots + \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}}=1$

for a given positive integer $n?$

Bonus: What sorts of bounds can you put on these solution(s) in terms of $n?$

0 1

2^n

n

2 solutions

Eli Ross Staff
Oct 12, 2015

Imagine a graph of $y=\sqrt{x} + \sqrt{\sqrt{x}} + \sqrt{\sqrt{\sqrt{x}}} + \ldots + \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}}.$ It is clearly increasing in $x$ and it is continuous. When $x=0,$ $y=0,$ while for $x=1,$ $y=n.$ Thus, for exactly one $x$ between 0 and 1, the equation will be satisfied (imagine the graph intersecting the horizontal line $y=1$ ).

Remark: While this idea is fairly intuitive, it is formalized in Calculus as the Intermediate Value Theorem .

So, can we start to bound what this solution for $x$ might be?

For $x<1,$ the terms in the sum are increasing; e.g. $\underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}}> \cdots > \sqrt{\sqrt{\sqrt{x}}} > \sqrt{\sqrt{x}} > \sqrt{x}.$ Since there are $n$ terms, we must have $n\cdot \sqrt{x} < 1,$ so $x < \frac{1}{n^2}.$ For example, if $\sqrt{x} + \sqrt{\sqrt{x}} + \sqrt{\sqrt{\sqrt{x}}}=1,$ then $0<x<\frac{1}{9}.$

Challenge: Can you find tighter bounds?

One possible method will involve the Applying the Arithmetic Mean Geometric Mean Inequality .

We can use the Arithmetic Mean Geometric Mean (AM-GM) Inequality on the numbers $\sqrt x, \sqrt {\sqrt x} ,\ldots , \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}}.$ The arithmetic mean is $\frac{1}{n}$ since there are $n$ terms that sum to 1. The geometric mean is $\left(\sqrt x \cdot \sqrt {\sqrt x}\cdots \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}}\right)^{\frac{1}{n}} = \left(\prod_{k=1}^{n} x^{\frac{1}{2^k}}\right)^{\frac{1}{n}} = \left(x^{\sum_{k=1}^{n} \frac{1}{2^k}}\right)^{\frac{1}{n}} = \left(x^{1-\frac{1}{2^n}}\right)^{\frac{1}{n}}.$

Since the $n$ numbers are unequal, by AM-GM inequality, the equality can't hold.

$\large \displaystyle \frac{1}{n^n} > x^{1 - \frac1{2^n}} \qquad \Rightarrow \qquad x < n^{-n\cdot \left(1 + \frac{1}{2^n-1}\right)}$

which is tighter than $n^{-n}$ and therefore much tighter than $n^{-2}.$ For $n= 3$ , we get $x < 3^{-3\cdot \left(1+\frac{1}{7}\right)} \approx 0.0231,$ which is much better than $\frac{1}{9} \approx .11.$ As $n$ gets large, this bound becomes quite similar to $n^{-n}.$

Pi Han Goh - 5 years, 8 months ago

Wow! that's nice <3

Nihar Mahajan - 5 years, 8 months ago

Yeah -- it's a much better upper bound than $n^{-2}.$

Can anyone find anything tighter? How about a lower bound tighter than 0?

Eli Ross Staff - 5 years, 8 months ago

@Eli Ross – For the lower bound:

Let $y = \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}} \Rightarrow y^2=\underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n-1 \text{ square roots}}$ .

Notice that $\underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n-1 \text{ square roots}} > \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n-2 \text{ square roots}} > \ldots > \sqrt x$ . So

$1 = \sqrt{x} + \sqrt{\sqrt{x}} + \sqrt{\sqrt{\sqrt{x}}} + \ldots + \underbrace{\sqrt{\sqrt{\sqrt{\cdots\sqrt{x}}}}}_{n \text{ square roots}} < y+ (n-1)y^2$

This is a quadratic inequality, by the Quadratic Formula and using the fact that $y>0,$ we have $x^{1/2^n} = y > \dfrac{-1 + \sqrt{4n-3}}{2n-2} \quad \Rightarrow\quad x > \left(\dfrac{-1 + \sqrt{4n-3}}{2n-2} \right)^{2^n}.$

Combining these inequalities gives $\large \displaystyle \left(\dfrac{-1 + \sqrt{4n-3}}{2n-2} \right)^{2^n} < x < n^{-n\cdot \left(1 + \frac{1}{2^n-1}\right)}$

Pi Han Goh - 5 years, 8 months ago

@Pi Han Goh – Awesome! A few follow-up ideas:

1) The upper bound is similar to $n^{-n}$ as $n$ grows large. How does that lower bound look as $n$ grows large?

2) Which of Pi Han's bounds do people think is "closer" to the actual values of $x?$

3) Anyone have ideas for how we could tighten these further?

Eli Ross Staff - 5 years, 8 months ago

Pi Han Goh
Oct 13, 2015

Let $y = x^{1/2^n}$ , then we have an equation of $y + y^2 + y^4 + \ldots + y^{2^{n-1}} - 1 =0$ . By Descartes' Rule of Signs , there's one positive real root of $y$ , thus there's only $\boxed1$ positive real root of $x$ .

Radical Radicals

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