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

 $\displaystyle\sqrt[n]{x^{m}}\;=\;x^{\frac{m}{n}}.$ (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 $n^{\mathrm{th}}$

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

Remarks

1. 1.

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

2. 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 $\mathbb{R}_{+}$, the positive numbers $\sqrt[l]{x^{k}}$ and $\sqrt[n]{x^{m}}$ must be equal.

3. 3.

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

4. 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

 $(-8)^{\frac{1}{3}}\;=\;(-8)^{\frac{2}{6}}\;=\;\sqrt[6]{(-8)^{2}}\;=\;\sqrt[6]{% 64}\;=\;2.$

The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$.  Here we have  $l=6$  and the mapping  $y\mapsto y^{ln}$  is not injective in  $\mathbb{R}_{-}\cup\mathbb{R}_{+}$.  — 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. 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 FractionPower 2014-09-21 12:12:39 2014-09-21 12:12:39 pahio (2872) pahio (2872) 21 pahio (2872) Definition msc 26A03 fractional power PowerFunction GeneralPower IntegrationOfFractionPowerExpressions NthRootFormulas