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?
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2$\begingroup$ $\sqrt[4]x$ could be a good idea, I hope. Think about $\sqrt x=x^{1/2}$ $\endgroup$– Claude LeiboviciJun 24, 2014 at 14:17
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$\begingroup$ $\sqrt{\sqrt{x}}=x^{1/4}$. Is this what you want? $\endgroup$– Pranav AroraJun 24, 2014 at 14:17
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$\begingroup$ I just want to know if it can be simplified $\endgroup$– Z'KJun 24, 2014 at 14:19
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1$\begingroup$ You mean like a root within a root within a root ? :-) $\endgroup$– LucianJun 24, 2014 at 15:06
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3 Answers
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$\sqrt{x} = x^{1/2}$ so $\sqrt{\sqrt{x}} = (x^{1/2})^{1/2} = x^{1/4} = \sqrt[4]{x}$.
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$\begingroup$ So if that is the answer. Is the domain [0, + ∞)? $\endgroup$– Z'KJun 24, 2014 at 14:21
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As we have the unwritten index $2$ for the sqare root, we multiply it by the index of the root inside the first root. $$\sqrt{\sqrt{x}} = \sqrt[2\times2]{x}= \sqrt[4]{x}$$
Another example is: $$\sqrt[3]{\sqrt[4]{x}} = \sqrt[3\times4]{x}= \sqrt[12]{x}$$
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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 $\sqrt{\sqrt x} = x^{1/4}$.
