So Many Nested Roots!

Algebra Level 4

$\begin{cases} \sqrt{ \dfrac{2}{2^{2^0}} + \sqrt{ \dfrac{2}{2^{2^1}} + \sqrt{ \dfrac{2}{2^{2^2}} + \sqrt{ \dfrac{2}{2^{2^3}} + \sqrt{ \dfrac{2}{2^{2^4}}\cdots}}}}} = \sqrt{xy} \\ \sqrt{ 11 + \sqrt{ 13 + 2\sqrt{ 15 + 3\sqrt{17 + 4\sqrt{ 19 +5\sqrt{\cdots}}}}}} = x + y + a \end{cases}$

If the system of equations above holds true for positive real numbers $x$ , $y$ and $a$ , with $y$ being a prime number, find the value of

$\begin{aligned} \sqrt[3]{ a \sqrt[3]{ a \sqrt[3]{a \sqrt[3]{a \cdots}}}}. \end{aligned}$

Try another problem on my set! Let's Practice

The answer is 1.

1 solution

Chew-Seong Cheong
Apr 12, 2017

From the reference on nested radical (eqn. 18) , we have:

$\sqrt{\frac 2{2^{2^0}} +\sqrt{\frac 2{2^{2^1}} +\sqrt{\frac 2{2^{2^2}} +\sqrt{\frac 2{2^{2^3}} +\sqrt{\frac 2{2^{2^4}} +\cdots}}}}} = \sqrt 2$

$\implies x=1$ and $y=2$ .

Using Ramanujan's formula (eqn. 26) below:

$x+y+a = \sqrt{yx+(a+y)^2+x\sqrt{y(x+a)+(a+y)^2+(x+a)\sqrt{y(x+2a)+(a+y)^2+(x+2a)\sqrt \cdots}}}$

Putting $x=1$ , $y=2$ and $a=1$ :

$\begin{aligned} 1+2+1 & = \sqrt{2+(1+2)^2+\sqrt{2(1+1)+(1+2)^2+(1+1)\sqrt{2(1+2)+(1+2)^2+(1+2)\sqrt \cdots}}} \\ & = \sqrt{11+\sqrt{13+ 2 \sqrt {15+3\sqrt{17 + 4\sqrt{19 + 5\sqrt \cdots}}}}} \end{aligned}$

$\implies a = 1$ . Therefore, $\sqrt [3] {a\sqrt [3] {a\sqrt [3] {a\sqrt [3] {a\sqrt [3] {\cdots}}}}} = \sqrt [3] {1\sqrt [3] {1\sqrt [3] {1\sqrt [3] {1\sqrt [3] {\cdots}}}}} = \boxed{1}$

I was expecting that you could give a practical solution for the first equation.. I can't understand the explanation from the website that you've given there.

Fidel Simanjuntak - 4 years, 2 months ago

Coz, i'm wondering, is it true that $\sqrt{ \dfrac{n}{n^{n^0}} + \sqrt{ \dfrac{n}{n^{n^1}} + \sqrt{ \dfrac{n}{n^{n^2}} + \sqrt{ \dfrac{n}{n^{n^3}} + \sqrt{ \dfrac{n}{n^{n^4}}\cdots}}}}} = \sqrt{n}$

For $n$ is a positive real integer..?

Fidel Simanjuntak - 4 years, 2 months ago

I don't think so. If it works, I think the Wolfram MathWorld reference would have included it. The reference is quite thorough. It works for $n=1$ and $n=2$ . But I numerically check for $n=3$ it diverges. The $n=2$ is a special case of the equation below putting $q=\frac 12$ , $n=2$ , $x=1$ and $k=-1$ . You can try with different $q$ to see.

Chew-Seong Cheong - 4 years, 1 month ago

@Chew-Seong Cheong – I am sorry for bothering you. But for me, I can't see it clearly, becoz it's a bit blur and I have a problem with my eyes.. Can you help me?

Fidel Simanjuntak - 4 years, 1 month ago

@Fidel Simanjuntak – You mean you can't see the formula I posted? It is very long and complicated. Perhaps you can ask people around to jot it down for you.

Chew-Seong Cheong - 4 years, 1 month ago

@Chew-Seong Cheong – Yeah, it's very long and complicated. Moreover, it's a bit blur..

Fidel Simanjuntak - 4 years, 1 month ago

@Fidel Simanjuntak – Sending you an enlarged image. I hope that it works. https://imgur.com/a/6IDyS

Chew-Seong Cheong - 4 years, 1 month ago

@Chew-Seong Cheong – Thank you, sir

Fidel Simanjuntak - 4 years, 1 month ago

So Many Nested Roots!

Try another problem on my set! Let's Practice

The answer is 1.

1 solution

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