Kaspaueru2

2021-12-25

Square root inside a square root
Hi guys I just want to ask how to solve this.
The given is:
$f\left(x\right)=\sqrt{x}$
then solve for the $\left(f\cdot f\right)\left(x\right)$
then it becomes.
$\left(f\cdot f\right)\left(x\right)=f\left(f\left(x\right)\right)$
$f\left(x\right)=\sqrt{\sqrt{x}}$
Can the expression be simplified?

vicki331g8

Expert

$\sqrt{x}={x}^{\frac{1}{2}}$
$\sqrt{\sqrt{x}}={\left({x}^{\frac{1}{2}}\right)}^{\frac{1}{2}}={x}^{\frac{1}{4}}=\sqrt{4}\left\{x\right\}$

Rita Miller

Expert

As we have the unwritten index 2 for the sqare root, we multiply it by the index of the root inside the first root.
$\sqrt{\sqrt{x}}=\sqrt[2×2]{x}=\sqrt[4]{x}$
Another example is: $\sqrt[3]{\sqrt[4]{x}}=\sqrt[3×4]{x}=\sqrt[12]{x}$

karton

Expert

The nth root of a number a: $\sqrt[n]{a}={a}^{1/n}$
The square root of the square root of x is therefore
$\sqrt{\sqrt{x}}=\left(\sqrt{x}{\right)}^{1/2}=\left({x}^{1/2}{\right)}^{1/2}={x}^{1/4}=\sqrt[4]{x}$
Since the domain of , this is also the domain of $\sqrt{\sqrt{x}}={x}^{1/4}$

Do you have a similar question?