Negative Numbers to Powers
Date: 11/23/2001 at 16:04:43 From: Filoxus Maximus Subject: Powering negative numbers If u is an irrational number and x is a negative number, what is x^u? How do I determine even whether that number is positive or negative? It isn't a problem when x is positive, but when x is negative the situation is a bit complicated. In the end: how much is (-e)^(-e)?
Date: 11/24/2001 at 12:03:57 From: Doctor Rob Subject: Re: Powering negative numbers Thanks for writing to Ask Dr. Math, Filoxus. Your question boils down to evaluating (-1)^u. The way to try to do this is to use Euler's Equation: http://www.mathforum.org/dr.math/faq/faq.euler.equation.html If you do this, you'll find that, for every integer k, -1 = cos(Pi*[2*k+1]) + i*sin(Pi*[2*k+1]), = e^(i*Pi*[2*k+1]), (-1)^u = [e^(i*Pi*[2*k+1])]^u, = e^(i*Pi*u*[2*k+1]), = cos(Pi*u*[2*k+1]) + i*sin(Pi*u*[2*k+1]). Since u is irrational, this means that (-1)^u has infinitely many different values, one for each integer k. If u were rational, there would only be finitely many different values, and one could pick out of this finite set a value for (-1)^u which was the simplest: either a real one or a complex one with the smallest positive argument. In ,the case where u is irrational, there is no convenient way to pick a "simplest" value, since the infinite set of values form a dense subset of the unit circle in the complex plane, none of which is real. If you let k = 0, you get one of the values, (-e)^(-e) = e^(-e)*(-1)^(-e), = e^(-e)*(cos[-Pi*e]+i*sin[-Pi*e]), = e^(-e)*(cos[Pi*e]-i*sin[Pi*e]), which I leave to you to evaluate numerically. If you let k = -1, you get the complex conjugate of the above answer. If you let k = 1, you get a different value, (-e)^(-e) = e^(-e)*(-1)^(-e), = e^(-e)*(cos[3*Pi*e]-i*sin[3*Pi*e]). If you let k have other values, you get other different answers. All have the property that when raised to the power (-1/e), they give the answer -e. Feel free to write again if I can help further. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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