Why Is (-n)^fractional Invalid ?
Date: 05/26/2003 at 18:01:52 From: Richard Subject: Why is (-n)^fractional invalid ? Dear Dr. Math, I recently ran into an odd problem. The problem is with negative numbers being raised to a fractional exponent. Apparently this is an invalid operation: (-2)^1.999 = invalid|error (-2)^2.001 = invalid|error However a negative number raised to an integer is perfectly fine. (-2)^2 = 4 What is the reason behind this? The most confusing part about it is that I would expect the former two equations to simply give results close to 4, as the later equation does. This in tandem with : 2^1.999 ~= 3.99723 2^2 = 4 2^2.001 ~= 4.00278 I'm baffled.
Date: 05/26/2003 at 19:56:05 From: Doctor Ian Subject: Re: Why is (-n)^fractional invalid ? Hi Richard, Let's stop and think about what a fractional exponent means. If the exponent is rational, we have b/c a which is the same as b 1/c (a ) which is the same as ___ c| b \| a If a is negative, and b is even, then a^b is positive, and we don't have a problem. On the other hand, if a is negative, and b is odd, then a^b is negative, and that's a problem is c is even. When I say 'problem', I mean a problem if we're limited to working with real numbers. But it we extend our number system to include complex numbers, then we have ______ 1/2 2| 1 ___ (-1) = \| (-1) = \| -1 = i Of course, most calculators can't cope with complex numbers... Anyway, does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
Date: 05/26/2003 at 20:03:41 From: Doctor Tom Subject: Re: Why is (-n)^fractional invalid ? Hi Richard, There are valid answers, but the problem is that usually they will be complex numbers. For example, every number has two square roots: 4 has 2 and -2, -4 has 2i and -2i. Similarly, every number has three cube roots. The three cube roots of 1, for example, are 1, (-1 + sqrt(3)i)/2 and (-1- sqrt(3)i)/2. You can work out any of these with Euler's formula: e^(it) = cos(t) + i sin(t), where the sine and cosine are taken in radians. Thus e^0 = e^(2 pi i) = e^(4 pi i) = e^(6 pi i) = ... = 1. The number 1 has an infinite number of valid logarithms in the complex plane. All numbers, except zero, also have an infinte number of complex logarithms. Let's use this idea to take the cube root of 1 and get all possible answers: I want to solve: x^3 = 1, so I'll do it with logarithms: 3 log x = log 1, so log x = (log 1)/3, so x = e^((log 1)/3). But we can use any valid version of log 1: 0, 2 pi i, 4 pi i, 6 pi i, ... So the answer could be e^0, e^((2 pi i)/3), e^((4 pi i)/3), e^((6 pi i)/3), et cetera. And we can work these out using Euler's theorem: e^0 = 1 e^((2 pi i)/3) = cos(2 pi /3) + i sin(2 pi/3) = - 1/2 + i sqrt(3)/2 e^((4 pi i)/3) = cos(4 pi / 3) + i sin(4 pi/3) = -1/2 - i sqrt(3)/2 e^((6 pi i)/3) = cos(2 pi) + i sin(2 pi) = 1 and you'll find that at this point it loops again and again through the three cube roots since the angle has effectively run all the way around the circle. For fifth roots, you'll get 5 answers before the looping starts, etc. In fact, if you plot the n-th roots of 1 on the complex plane, they lie in a circle centered at 0 + 0i of radius 1 in a perfect regular n-gon. The cube roots lie on an equilateral triangle, the fourth roots on a square, the fifth roots on a regular pentagon, etc. If you're not taking the n-th root of 1, but rather of some other number, the n-th roots again lie in a perfect circle about the origin, in a perfect n-gon, but the radius will be the real n-th root of the absolute value of the number. The starting position on the circle will vary, depending on the number. For example the 5-th roots of -2 lie on a regular pentagon, one of whose roots is on the negative real axis and whose radius is the 5-th root of 2 - about 1.148698. Even if you're taking n-th roots of complex numbers they again form a perfect n-gon centered at the origin, with radius the absolute value of the complex number, and with an oddball starting point. Euler's formula can also be used to find these. So, getting back to the problem you mentioned, what is, say, 1^(2/3)? Of course you'll get the standard answer 1, but there are two others that can be obtained by squaring the other two cube roots of 1. In general, to find all the possible values of x^(m/n), you need to find all the n-th roots of x as described above, and then raise each of them to the m-th power. If you have a negative value, like 2^(-2/3), work out all three values for 2^(2/3) and invert them (put them in the denominator of a fraction with 1 as the numerator). If you raise a negative number to a fractional power of m/n, then it may even be that all of the n-th roots will be complex: (-4)^(1/2) has 2i and -2i as the roots, for example. So if you take (-2)^(1.999) that's really (-2)^(1999/1000), so there are 1000 answers, all of which will be complex. Now the answers will, in a sense, be continuous. You know that one of the answers for (-2)^2 is 4, so one of the complex answers will be near 4, maybe something like 3.99 + .01i. By the way, things are even uglier if the exponent is not a fraction but rather an irrational real number. In that case, there is an infinite number of logarithms, and hence an infinite number of results. If you try to punch (-2)^(1.999) into your calculator, it's overwhelmingly likely that the company that built the calculator did not code it up to find all 1000 of the possible complex results; they found it easier simply to display the text: (invalid/error), and now perhaps you know why. By the way, some companies do a little better than that. I have an HP 48GX on my desk and I just evaluated (-2)^(1.999) and obtained: 3.99720864645 -0.0125576426316 i so although it doesn't give all 1000 complex answers, at least it gave me the one that is close to the pure real result of (-2)^2 = 4. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/
Date: 05/26/2003 at 21:28:36 From: Richard Subject: Thank you (Why is (-n)^fractional invalid ?) Dr. Ian and Dr. Tom - many thanks! Dr. Ian : D'oh. Well I was on the right track :) Thanks! Dr. Tom : That is an amazingly concise in-depth answer. Indeed, my old casio and all PC applications that have math processing here (but obviously not with complex numbers) simply throw an error. It's interesting that you do get a result close to what I thought it would have been, but I'm relieved to know that the problem lies in there being more than one (thousands. yikes.) answer. I think the train of thought where values lay at the corners of an n-gon on the complex plane helped the most. Thanks again! Best regards, Richard
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum