# An Earnest Challenge

There appears to be a lot of expertise amongst the members of this site regarding solving nested radicals, so I thought I'd share a challenging one for your radical enjoyment:

$f(x) = \displaystyle\sqrt{x + \sqrt{\frac{x}{2} + \sqrt{\frac{x}{4} + \sqrt{\frac{x}{8} + \sqrt{\frac{x}{16} + \sqrt{\frac{x}{32} + ......}}}}}}$.

The hope is that there is an exact solution, if only for $f(1)$ if not for $f(x)$ in general. I suppose one interesting feature of this function is that $(f(x))^{2} - x = f(\frac{x}{2})$. I'm sure that there are many more interesting features waiting to be discovered.

Note by Brian Charlesworth
5 years, 2 months ago

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My bet is that there isn't any closed form expression for this, not even in the special case of $x=1$.

- 5 years, 2 months ago

You're probably right, but I'm getting used to seeing rabbits being pulled out of hats so I thought I'd post the problem just in case.

- 5 years, 2 months ago

Your problem does not have closed form but

$\sqrt{ x + \sqrt{\frac{x}{2} + \sqrt{\frac{x}{4} + \sqrt{\frac{x}{16} + \sqrt{\frac{x}{256} + \ldots}}}}}$

Can have a closed form

- 5 years, 2 months ago

It does? Cool. I'll have to figure out what that is, then.

- 5 years, 2 months ago

I was solving your radical and did a mistake and I solved the above radical :p, now I will post a problem on this ;)

- 5 years, 2 months ago

Haha. Well, a lot of "mistakes" have led to interesting discoveries. Ill keep an eye out for your problem.

- 5 years, 2 months ago

Could it be written in this fashion?

$f(x) = \sqrt{\sum_{i=0} \frac{f(x)}{2^{i}}}$

- 5 years, 2 months ago

no.. I think the 'x' terms under summation are missing and 'i' should start from 1 instead of 0.. if I'm not wrong.

- 5 years, 2 months ago

It's a recursive function. I thought "i" should start from 0 since the first term is "x", not "x/2"

- 5 years, 2 months ago

For higher degree of radicals f(x)= nth root of 2x

- 5 years, 2 months ago

f(x) = 2sqrt(x)

- 5 years, 1 month ago