# Square root inside a square root

Hi guys I just want to ask how to solve this.

The given is: $$f(x)=\sqrt{x}$$

then solve for the $$(f \circ f)(x)$$

then it becomes. $$(f \circ f)(x)=f(f(x))$$ $$f(x)=\sqrt{\sqrt{x}}$$ Can the expression be simplified?

• $\sqrtx$ could be a good idea, I hope. Think about $\sqrt x=x^{1/2}$ Jun 24, 2014 at 14:17
• $\sqrt{\sqrt{x}}=x^{1/4}$. Is this what you want? Jun 24, 2014 at 14:17
• I just want to know if it can be simplified
– Z'K
Jun 24, 2014 at 14:19
• You mean like a root within a root within a root ? :-) Jun 24, 2014 at 15:06

$\sqrt{x} = x^{1/2}$ so $\sqrt{\sqrt{x}} = (x^{1/2})^{1/2} = x^{1/4} = \sqrt{x}$.

• So if that is the answer. Is the domain [0, + ∞)?
– Z'K
Jun 24, 2014 at 14:21
• That is indeed the domain. Jun 24, 2014 at 14:27

As we have the unwritten index $2$ for the sqare root, we multiply it by the index of the root inside the first root. $$\sqrt{\sqrt{x}} = \sqrt[2\times2]{x}= \sqrt{x}$$

Another example is: $$\sqrt{\sqrt{x}} = \sqrt[3\times4]{x}= \sqrt{x}$$

The nth root of a number $$a$$: $$\;\sqrt[\large n] a = a^{1/n}$$.

The square root of the square root of x is therefore $$\sqrt{\sqrt x} = (\sqrt x)^{1/2} = (x^{1/2})^{1/2} = x^{1/4} = \sqrt[\large 4] x$$

Since the domain of $$\sqrt x$$ is $$[0, + \infty)$$, this is also the domain of $$\sqrt{\sqrt x} = x^{1/4}$$.