Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
History of the function concept wikipedia , lookup
Big O notation wikipedia , lookup
Large numbers wikipedia , lookup
Factorization wikipedia , lookup
Non-standard calculus wikipedia , lookup
Collatz conjecture wikipedia , lookup
Fundamental theorem of algebra wikipedia , lookup
Vincent's theorem wikipedia , lookup
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 created: h2013-03-21i by: hpahioi version: h37340i Privacy setting: h1i hDefinitioni h26A03i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 1 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. 2