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fraction power∗
pahio†
2013-03-21 19:38:38
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
m
m
(x n )n = x n ·n = xm
and therefore the definition of nth root gives the formula
√
m
n
xm = x n .
(1)
Here, the exponent 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 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
m
number x we define the fraction power x n as the nth root
√
m
x n := n xm .
(2)
Remarks
1. The existence of the root in the right hand side of (2) is proved here.
2. The defining equation (2) is independent
on the
√
√ form of the exponent
n
m
k
m
m )ln = [( n xm )n ]l = xlm = xkn =
:
If
=
,
then
we
have
(
x
n√
l
n
√
l
l
[( xk )l ]n = ( xk )ln , and√because the mapping y 7→ y ln is injective in
√
l
R+ , the positive numbers xk and n xm must be equal.
m
3. The fraction power function x 7→ x n is a special case of power function.
4. The presumption
that x is positive signifies that one√can not identify all
√
1
nth roots n x and the powers x n . For example, 3 −8 equals −2 and
2
1
6 = 3 , but one must not calculate
p
√
1
2
6
(−8) 3 = (−8) 6 = 6 (−8)2 = 64 = 2.
∗ hFractionPoweri
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The point is that (−8) 3 is not defined in R. Here we have l = 6 and
the mapping y 7→ y ln is not injective in R− ∪ R+ . — Nevertheless,
some
√
1
people and books√may use also for negative x the equality n x = x n and
m
more generally n xm = x 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, e.g. 3 x, ought to be calculated as follows:
√
√
1
2
1
1
d3x
d(− 3 −x)
d(−x) 3
1
= √
=
= −
= − ·(−x)− 3 (−1) = p
3
3
2
dx
dx
dx
3
3 (−x)
3 x2
√
3
The result is similar as ddxx for positive x’s, although the odd root functions are not special cases of the power function.
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