| Version 7 |
Version 6 |
| 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 |
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 |
| $$(x^{\frac{m}{n}})^n = x^{\frac{m}{n}\cdot n} = x^m$$ |
$$(x^{\frac{m}{n}})^n = x^{\frac{m}{n}\cdot n} = x^m$$ |
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and therefore the definition of \PMlinkname{$n^\mathrm{th}$ root}{NthRoot} gives the \PMlinkescapetext{formula}
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and therefore the definition of \PMlinkname{$n$th root}{NthRoot} gives the \PMlinkescapetext{formula}
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| \begin{align} |
\begin{align} |
| \sqrt[n]{x^m} = x^{\frac{m}{n}}. |
\sqrt[n]{x^m} = x^{\frac{m}{n}}. |
| \end{align} |
\end{align} |
| For enabling the validity of (1) for the cases where $n$ does not divide $m$ we must set the following |
For enabling the validity of (1) for the cases where $n$ does not divide $m$ we must set the following |
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\textbf{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\, {\em fraction power}\, $x^{\frac{m}{n}}$ as the $n^\mathrm{th}$ \PMlinkescapetext{root}
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\textbf{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\, {\em fraction power}\, $x^{\frac{m}{n}}$ as the $n$th root
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| \begin{align} |
\begin{align} |
| x^{\frac{m}{n}} := \sqrt[n]{x^m}. |
x^{\frac{m}{n}} := \sqrt[n]{x^m}. |
| \end{align} |
\end{align} |
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| \textbf{Remarks} |
\textbf{Remarks} |
| \begin{enumerate} |
\begin{enumerate} |
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| \item 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} = |
\item It's easy to see that (2) is independent on the form of the exponent $\frac{m}{n}$. |
| [(\sqrt[l]{x^k})^l]^n = (\sqrt[l]{x^k})^{ln}$,\, and thus the positive numbers $\sqrt[l]{x^k}$ and $\sqrt[n]{x^m}$ are equal. |
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| \item The fraction power function\, $x\mapsto x^{\frac{m}{n}}$\, is a special case of power function. |
\item The fraction power function\, $x\mapsto x^{\frac{m}{n}}$\, is a special case of power function. |
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\item The presumption that $x$ is positive signifies that on can not identify all $n^\mathrm{th}$ \PMlinkname{roots}{NthRoot} $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$.\, For example, $\sqrt[3]{-8}$ equals $-2$ and\,
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\item The presumption that $x$ is positive signifies that on can not identify all $n$th \PMlinkname{roots}{NthRoot} $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$.\, For example, $\sqrt[3]{-8}$ equals $-2$ and\,
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| $\frac{1}{3} = \frac{2}{6}$, but one \textbf{must not} \PMlinkescapetext{calculate} |
$\frac{1}{3} = \frac{2}{6}$, but one \textbf{must not} \PMlinkescapetext{calculate} |
| $$(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} |
$$(-8)^{\frac{1}{3}} = (-8)^{\frac{2}{6}} = \sqrt[6]{(-8)^2} |
| = \sqrt[6]{64} = 2.$$ |
= \sqrt[6]{64} = 2.$$ |
| The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$.\, Nevertheless, we often see the equation\, $\sqrt[n]{x} = x^{\frac{1}{n}}$. |
The point is that $(-8)^{\frac{1}{3}}$ is not defined in $\mathbb{R}$.\, Nevertheless, we often see the equation\, $\sqrt[n]{x} = x^{\frac{1}{n}}$. |
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| \item According to the preceding item, for the negative values of $x$ the derivative of \PMlinkname{odd roots}{NthRoot}, e.g. $\sqrt[3]{x}$, ought to be calculated as follows: |
\item According to the preceding item, for the negative values of $x$ the derivative of \PMlinkname{odd roots}{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\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{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}}$$ |
\frac{1}{3\sqrt[3]{(-x)^2}} = \frac{1}{3\sqrt[3]{x^2}}$$ |
| The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$'s, although the \PMlinkescapetext{odd} root functions are not special cases of the power function. |
The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$'s, although the \PMlinkescapetext{odd} root functions are not special cases of the power function. |
| \end{enumerate} |
\end{enumerate} |
