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Non-rational Exponents?

  1. Jun 1, 2012 #1
    General Question: I don't have a good grasp of what it means to use a non-rational exponent, such as xe or xi.

    For irrational powers, the best I have is to look at xe (for example) as a series xn where n converges to e using rational n's. That's an analytical approach that feels right, but still doesn't make much intuitive sense to me.

    And that doesn't explain complex powers. I don't think there are any ways to construct a sequence of real numbers that converges to an imaginary number. Is this true?

    Are there any intuitive approaches to what is happening here? Or is analysis the only way to crack this nut?

    Thanks in advance...
     
  2. jcsd
  3. Jun 1, 2012 #2
    That's how I see irrational powers as well. It stems from introducing irrational numbers into the real number line. An irrational number is typically thought of as a limit of rational numbers (really a cauchy sequence). There is a good intuitive notion of exponentiation for rational exponents, and to keep the function continuous, you define irrational powers as limits of rational powers. Maybe it would help to think of irrational numbers themselves as a limit of rational numbers.

    Complex powers are tricky. To define them, you actually need to talk about complex logs, and use euler's formula that ##e^{ix} = cos(x) + isin(x)##.

    A complex power is no longer "filling in gaps" as it was for irrational numbers. Hence limits of real numbers aren't useful. You have to found some other construction that preserves as many of the important exponent properties as you can. I still find euler's formula quite mystical, and it is through that link that one defines general complex exponentiation.
     
  4. Jun 1, 2012 #3

    mathman

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    ab = eblna. If b is complex, you need to use Euler's formula. Further complications arise because lna is not single valued, so you may have to use principal vaue or something similar.
     
  5. Jun 1, 2012 #4
    To get an intuitive feel for irrational powers, I like to think about it as follows:

    I think it is helpful to first consider the question 'why is anything raised to the 0th power 1?' The answer is that, if you look at f(x) = y^x defined for all non-zero x, you will note that f(x) becomes arbitrarily close to 1 as x becomes arbitrarily close to 0. So, f(x) = 4^x and g(x) = 12345567^x, both approach 1 as x approaches 0. So, when extending the domain of f(x) to include x=0, the natural choice is to set f(0)=1. We are basically just filling in a hole with the natural choice. Also note that this is the only choice for f(0) that ensures that f(x) remains continuous on its entire domain after extending the domain to include x=0.

    In what we did above, we assumed that f(x) was defined for all non-zero x, which includes irrationals. But now lets turn to your question and start only with the assumption that f(x) is defined on the rationals. For example, consider the function
    [tex]f(n/m) = 2^{n/m}[/tex]
    defined for all rationals n/m. Now imagine a plot of this function over its entire domain (which is only the rational numbers at the moment).

    It is worth noting that this function is continuous over its current domain, which consists only of the rationals (see the definition of continuity on wikipedia and note that it says "For any epsilon > 0...there exists delta such that for all x in the domain of f..."). So our plot is smooth (in the sense that close values of x map to close values of f(x), so there are no 'jumps').

    Now, our plot currently has a ton of holes in it (one hole at each irrational) and we want to fill in all of those holes. Let's fill it in at x=pi, for example. To fill it in, we just do the same thing we did when trying to define f(0); we make the 'natural choice' by requiring that the function remain continuous after its domain is extended. With this requirement, there is only one choice for f(pi); all other choices will result in a discontinuity at x=pi. So we choose f(pi) so as to keep the function continuous. This amounts to filling in the hole at x=pi with the 'natural choice.'

    To define f(x) for any other irrational x, we just do the same thing we did for x=pi.
     
    Last edited: Jun 1, 2012
  6. Jun 1, 2012 #5

    HallsofIvy

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    This is really echoing AcidRainLiTE. If x is an irrational number then there exist a sequence of rational numbers, [itex]\{b_n\}[/itex] converging to x. We define [itex]a^x[/itex] to be the limit of the sequence [itex]a^{b_n}[/itex]. Of course, it is necessary to prove that the sequence converges and two different sequences converging to x gives the same result. On "practical" result of that is that if, say, "1.23433453" is a rational number approximating x, then [itex]a^{1.23433453}[/itex] is an approximation to [itex]a^x[/itex].
     
  7. Jun 1, 2012 #6
    You could also think of it in the following way. Instead of basing exponents on repeated multiplication, you can think of exponential functions as the primitive concept. We shall call f(x) an exponential function if it changes at a rate proportional to its value. Let's also fix the convention that f(0)=1.

    More concisely, f'(x) = k f(x), and f(0)=1.

    From these basic facts, you can deduce the fact that f(x+y)=f(x)f(y). So if you let b=f(1), then you find that f(n)=f(1)*...*f(1)=b*b*....*b (n times). Therefore, the repeated multiplication aspect of exponents follows from the more fundamental property concerning their rate of growth.

    [EDIT] You could conceive of complex exponents in the same way, however, the concrete/geometric interpretation is harder to understand unless you have a good grasp of complex numbers and their geometry.
     
    Last edited: Jun 1, 2012
  8. Jun 1, 2012 #7

    micromass

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    So we can define the exponential on the rational numbers just fine. We wish to extend this to the real numbers. The thing is that there exists only a unique extension of the exponential function to the real numbers if we want that extension to be continuous. So the question you should really ask is why do we want continuity.

    Also, it is easy to see that the exponential on the rational numbers is increasing. There exists only a unique extension to the real numbers that keep the function increasing. This extension is

    [tex]e^x=\sup\{e^r~\vert~r\in\mathbb{Q},~r<x\}[/tex]

    and agrees with your definition.

    So basically, if we want our real exponential to have the same good properties as the rational exponential, then there is only one choice we can make.

    The extension to the complex numbers is the same story, but a bit different. There are many extensions to the complex numbers which preserve continuity, but there is only one extension which makes the exponential differentiable. So if we want the complex exponential to be continuously differentiable, then there is only one possible choice we can make.
     
  9. Jun 1, 2012 #8

    micromass

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    And yes, this is true. If [itex](x_n)_n[/itex] is a sequence of real numbers, then it will converge to a real number. It can never converge to a complex number a+bi such that b is nonzero. The mathematical term for this is that "[itex]\mathbb{R}[/itex] is closed in [itex]\mathbb{C}[/itex]"
     
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