fraction power

Let $m$ be an integer and $n$ a positive factor of $m$.  If $x$ is a positive real number, we may write the identical equation

$${(x^{\frac{m}{n}})}^n=x^{\frac{m}{n}\cdot n}=x^m$$

and therefore the definition of $n^{\mathrm{th}}$ root (http://planetmath.org/NthRoot) gives the

$\sqrt[n]{x^m}=x^{\frac{m}{n}}.$ (1)

Here, the exponent $\frac{m}{n}$ is an integer.  For enabling the validity of (1) for the cases where $n$ does not divide $m$ we must set the following

Definition.  Let $\frac{m}{n}$  be a fractional number, i.e. an integer $m$ not divisible by the integer $n$, which latter we assume to be positive.  For any positive real number $x$ we define the  fraction power  $x^{\frac{m}{n}}$ as the $n^{\mathrm{th}}$

$x^{\frac{m}{n}}:=\sqrt[n]{x^m}.$ (2)

Remarks

1. The existence of the in the right hand side of (2) is proved here (http://planetmath.org/existenceofnthroot).

2. The defining equation (2) is independent on the form of the exponent $\frac{m}{n}$:  If  $\frac{k}{l}=\frac{m}{n}$,  then we have  ${(\sqrt[n]{x^m})}^{ln}={[{(\sqrt[n]{x^m})}^n]}^l=x^{lm}=x^{kn}={[{(\sqrt[l]{x^k})}^l]}^n={(\sqrt[l]{x^k})}^{ln}$,  and because the mapping  $y\mapsto y^{ln}$  is injective in ${ℝ}_{+}$, the positive numbers $\sqrt[l]{x^k}$ and $\sqrt[n]{x^m}$ must be equal.

3. The fraction power function  $x\mapsto x^{\frac{m}{n}}$  is a special case of power function.

4. The presumption that $x$ is positive signifies that one can not identify all $n^{\mathrm{th}}$ roots (http://planetmath.org/NthRoot) $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$.  For example, $\sqrt[3]{-8}$ equals $-2$ and  $\frac{2}{6}=\frac{1}{3}$,  but one must not

   $${(-8)}^{\frac{1}{3}}={(-8)}^{\frac{2}{6}}=\sqrt[6]{{(-8)}^2}=\sqrt[6]{64}=\mathrm{\hspace{0.33em}2}.$$

   The point is that ${(-8)}^{\frac{1}{3}}$ is not defined in $ℝ$.  Here we have  $l=6$  and the mapping  $y\mapsto y^{ln}$  is not injective in  ${ℝ}_{-}\cup {ℝ}_{+}$.  — Nevertheless, some people and books may use also for negative $x$ the equality  $\sqrt[n]{x}=x^{\frac{1}{n}}$  and more generally  $\sqrt[n]{x^m}=x^{\frac{m}{n}}$  where one then insists that  $\gcd (m,n)=1.$

5. According to the preceding item, for the negative values of $x$ the derivative of odd roots (http://planetmath.org/NthRoot), e.g. $\sqrt[3]{x}$, ought to be calculated as follows:

   $$\frac{d\sqrt[3]{x}}{dx}=\frac{d(-\sqrt[3]{-x})}{dx}=-\frac{d{(-x)}^{\frac{1}{3}}}{dx}=-\frac{1}{3}\cdot {(-x)}^{-\frac{2}{3}}(-1)=\frac{1}{3\sqrt[3]{{(-x)}^2}}=\frac{1}{3\sqrt[3]{x^2}}$$

   The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$’s, although the root functions are not special cases of the power function.

Title	fraction power
Canonical name	FractionPower
Date of creation	2014-09-21 12:12:39
Last modified on	2014-09-21 12:12:39
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	21
Author	pahio (2872)
Entry type	Definition
Classification	msc 26A03
Synonym	fractional power
Related topic	PowerFunction
Related topic	GeneralPower
Related topic	IntegrationOfFractionPowerExpressions
Related topic	NthRootFormulas