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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
Associated Topics:
College Exponents
College Imaginary/Complex Numbers
High School Exponents
High School Imaginary/Complex Numbers

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