PlanetMath: fraction power

(more info)

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fraction power

(Definition)

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

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

and therefore the definition of $n^\mathrm{th}$ root gives the formula

$\displaystyle \sqrt[n]{x^m} = x^{\frac{m}{n}}.$

(1)

Here, the exponent $\frac{m}{n}$ is an integer. For enabling the validity of (1) for the cases where

does not divide

we must set the following

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

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

(2)

Remarks

The existence of the root in the left hand side of (2) is proved here.
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.
The fraction power function $x\mapsto x^{\frac{m}{n}}$ is a special case of power function.
The presumption that is positive signifies that one can not identify all $n^\mathrm{th}$ roots $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$ . For example, $\sqrt[3]{-8}$ equals and $\frac{2}{6} = \frac{1}{3}$ , but one must not calculate
$\displaystyle (-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} = \sqrt[6]{64} = 2.$
The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$ . Here we have 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 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 $\gcd(m,\,n) = 1.$
According to the preceding item, for the negative values of the derivative of odd roots, e.g. $\sqrt[3]{x}$ , ought to be calculated as follows:
$\displaystyle \frac{d\sqrt[3]{x}}{dx} = \frac{d(-\sqrt[3]{-x})}{dx} = -\frac{d(... ...-x)^{-\frac{2}{3}}(-1) = \frac{1}{3\sqrt[3]{(-x)^2}} = \frac{1}{3\sqrt[3]{x^2}}$
The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive 's, although the odd root functions are not special cases of the power function.