ExponentiationDate: 08/16/97 at 21:28:11 From: Andrew Pomerance Subject: Exponentiation How do I calculate x^y using only exp, ln, log, and the trigonometric functions? I know exp(y*ln(x)) works, but only for x > 0. Date: 08/21/97 at 16:29:28 From: Doctor Rob Subject: Re: Exponentiation The key is DeMoivre's Law: e^(i*t) = cos(t) + i*sin(t), where i is the square root of -1, that is, i^2 = -1. Even if x and y are real, ln(x) may not be, because x may be negative. What is ln(x)? It is a number z such that x = exp(z). In fact, there may be several values of z that work. From DeMoivre's Law, we see that -1 = cos(Pi) + i*sin(Pi) = e^(i*Pi), so ln(-1) = i*Pi. This is not the only answer, however! It is true that -1 = cos(Pi + 2*Pi*k) + i*sin(Pi + 2*Pi*k) = e^(i*Pi*[2*k+1]), so ln(-1) = i*Pi*(2*k+1), for any integer k. This should remind you of inverse trigonometric functions, which also have multiple values. Using this fact, and the definition you started with, x^y = e^(y* ln(x)), and the properties of logarithms, you should be able to define x^y for all real x and y now. Let's try (-2)^(1/2) = e^((1/2)*ln(-2)) = e^([ln(-1) + ln(2)]/2) = e^([i*Pi + 2*Pi*i*k + ln(2)]/2) = e^(i*Pi/2 + Pi*i*k + ln(2)/2) = [cos(Pi/2 + Pi*k) + i*sin(Pi/2 + Pi*k)]*2^(1/2) = [0 + i*(-1)^k]*2^(1/2) = (-1)^k*i*Sqrt[2] = + or - i*Sqrt[2] This seems to make sense. You even get both signs! Furthermore, this same method can be used to define x^y when both are complex numbers, although it may be a multi-valued "function." i^i = e^(i*ln(i)), = e^[i*(Pi*i/2 + 2*Pi*i*k)], for any integer k (because e^(Pi*i/2) = cos(Pi/2) + i*sin(Pi/2) = i), = e^[-Pi/2 - 2*Pi*k], for any integer k. It seems paradoxical that i^i should be real, but that is correct. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/