One for the mathematicians

[AlanMiller]

This is a function which has had me bemused for some time. OK, we all know we can't take the square root of a negative number, if we stick within the set of real numbers. But we can take the cube root, fifth root etc. So for what values is the function:

f(x) = a^x

defined when a is negative? Obviously, all the integers and all vulgar fractions with odd numbers in their denominators (and the polite fractions too, I guess). All the repeating decimals, since these are fractions with odd multiples of 9 in the denominator. So some fractions are valid in the domain, others not. What about irrational numbers? Are some included and some not? The function domain is obviously discrete, so... how would one describe the domain for such a function, over the set of real numbers?

Alan

[HansV]

There is some info here: Exponentiation, in particular the sections Powers of minus one, Negative nth roots and Failure of power and logarithm identities.

[AlanMiller]

The section "Negative nth roots" is more or less what I had figured as my partial "discrete" solution so far. I'm still wondering if certain real (irrational) exponents will give real solutions, despite the fact that such solutions are inaccessible using "power and logarithm identities". I'm wondering if Complex logarithms might provide solutions, which may turn out to be (or not) real numbers. Perhaps there is some way to create an Argand diagram of such a function and be able to distinguish the real solutions? I'm very rusty on this stuff and my brain hurts

Alan

[HansV]

For positive a, a^x is a continuous function of x, so we can define a^x for irrational x by approximating x with rational numbers.

If a is negative, the situation is entirely different. The set of rational x for which a^x is defined is dense within the set of all rational numbers, i.e. between each pair of rational numbers we can find a number x for which a^x is defined. The set of rational x for which a^x is not defined is also dense within the set of all rational numbers, i.e. between each pair of rational numbers we can find a number x for which a^x is not defined. In other words, a^x is discontinuous everywhere. So we can't use approximation to define a^x for irrational x. This all holds when we restrict ourselves to the domain of real numbers.

[AlanMiller]

I'll have to give that some thought. I can see where the "dense" solution set comes from, but I can't see how that rules out all irrational reals as solutions, even though it's not possible to "test" them using the methods applicable to rational numbers. My natural "feeling" would be that some irrational exponents would give real results, others not - but this is hardly the mathematical treatment I'm seeking. Thanks the references and info - all food for headaches thought.

Alan

[HansV]

Let's take a concrete example. The value of 2^pi does not have an "intrinsic" meaning the way 2^3 or 2^(1/2) has. We "compute" 2^pi as follows:

Compute 2^3 and 2^4
Compute 2^3.1 [i.e. 2^(31/10) = the 10th power root of 2 to the power 31] and 2^3.2
Compute 2^3.14 and 2^3.15
Compute 2^3.141 and 2^3.142
Compute 2^3.1415 and 2^1416
Compute 2^3.14159 and 2^14160
Compute 2^3.141592 and 2^141593
...

This way, we successively narrow down the value. The values converge to a number that we define as 2^pi.

But for negative a, a^x cannot be computed by approximation, since between each pair of x-es for which we can compute a^x, there will be x-es for which a^x does not exist.

[AlanMiller]

Yep, I see your point here. It seems you can always make the next approximation real or not, as you wish. Even so, if 2^p can be expressed in terms of an infinite series, using logarithms or exponentials, and shown to be real, then I can't see why (in principle, if the method was available) that (-2)^p could not also be expressed this way, and the result adjudged as real or not. Perhaps it's because such a series in not continuous?

Alan

[HansV]

That's the point I've been trying to make.

[AlanMiller]

So is it the lack of continuity that makes it an unanswerable question? That is, two dense, discrete distributions superimposed on top of each other, one with real and one with imaginary numbers?

Alan

Edit - I think I see it clearer now. I'm visualizing it as two convergent sequence of better approximations, both heading for the same magnitude, but one real (using odd denominators) and one imaginary (using even denominators).

[AlanMiller]

OK, I'm still harping on this because I still can't see that it is invalid to say that an irrational power of a negative number is a real number. Looking again at (±2)^p ...

It is possible to get an approximation to 2^p to any precision, using a value of p to any precision... as many significant figures as you wish. I'm going to use a value of p = 3.14159 for the purpose of example. This can be expressed as a rational fraction as shown in the first table row below, and an approximation of 2^p calculated.

Now, this rational fraction can't be used for -2 because the denominator of 100,000 is even. So the next step is to approximate to it, using the pair of fractions in the next two rows below. These give results shown in the 2^d column. Not close enough at the selected precision of the approximation yet, so try the next pair of fractions. Keep going until the result for 2^d is the same as 2^p at the same precision of the approximation.

From all this I conclude that whatever the value of 2^p you wish to approximate to, there is also an approximation of (-2)^p to match it, which is also a real number. More generally it seems that if p is irrational and b is a positive real, then it is valid to say that (-b)^p = -(b^p) to any desired precision for b and/or p.

Alan

[HansV]

If this would hold for ALL irrational numbers, it would hold for all rational numbers too, wouldn't it? After all, your proof uses approximation by rational numbers. Let's apply it to b = 1 and p = 1/2 (0.5). We get (-1)^0.5 = -(1^0.5) = -1. So we have proved that the square root of -1 is -1.

[AlanMiller]

But this is where the discrete distributions come in. There are obviously exact values for which this will not hold as well as exact values for which no approximations are relevant e.g. -8^1/3 and it will hold. But what I'm saying is that by virtue of the dense nature of the discrete distribution for allowable values, any irrational power of a negative number can be expressed as the negative of that same power of its positive counterpart, to any precision desired. The discrete set of non-allowables is the set of rational fractions I have defined. But can irrational values like p and Ö3 be sensibly included in non-allowable values? I'm tending to think not. I can't think of any kind of formal proof either, as a way to address this.

Alan

[HansV]

OK, let's agree to disagree, I don't think we'll get any further with this.

(This reminds me of a discussion between mathematicians and theoretical physicists some 30 years ago. To solve a complicated problem, the physicists had to compute an integral. This ntegral diverged to infinity, so the physicists simply subtracted another integral that diverged to infinity as a "correction factor" to obtain a finite result. The mathematicians stated that this made the whole calculation meaningless since you could produce ANY result that way, but they couldn't convince the physicists...)

[AlanMiller]

I guess it depended on whether it helped solve a practical problem. Getting round various singularities enabled further progress in quantum physics, but may have represented an "illegal move" to the pure mathematician. What I'm thinking about with this problem is to possibly get round the brickwall of encountering such an expression. Without needing (or wanting) to evaluate it, I thought that if one could say that the solution would be or could be "made" real, then the stumbling block is circumvented and equation bashing can continue.

I find your example a bit harder to conceive of, since I've also recently become nicely confused reading a paper discussing how all infinities are not equal, and some are more infinite than others!

Alan