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.

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## Comments

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TopNewestYour 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

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

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

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

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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.

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f(x) = 2sqrt(x)

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

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Could it be written in this fashion?

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

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

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

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