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Thread: All about i

  1. #1 All about i 
    Forum Bachelors Degree Demen Tolden's Avatar
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    I have a few questions about i that I should have asked long ago. This post was originally in my Techniques for Integration thread, but since it was a more general question I thought I would just start a new thread to ask it. This is what I posted:
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    Ok, I have a stupid question, but here goes.

    (0 - 1)<sup>4/5</sup>

    Does this equal 1 or a version of i? It must be a version of i since if I had (-1)<sup>1/2</sup>, which I know should be i, I could turn that into (-1)<sup>2/4</sup>, square the -1 to make 1, and then take the fourth root of 1 to get 1. There must be a rule that says when considering exponents, you must divide first.
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    Also, how would you write something like (-1)<sup>1/9</sup> in terms of i?


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  3. #2  
    Forum Professor serpicojr's Avatar
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    Egad, this reply turned into a short novel... Don't worry if you can read it all at once!

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    Here's the high school algebra answer: consider the denominator first. In other words, if you're raising to the n/m-th power, take the mth root first (if this is defined) and then raise to the nth power. You can always take the mth root for positive numbers, and you can take mth roots for negative numbers if m is odd. The mth root always has the same sign as the original number. So, in your example:

    (-1)<sup>4/5</sup> = ((-1)<sup>1/5</sup>)<sup>4</sup> = (-1)<sup>4</sup> = 1

    Using these conventions, you can use the usual rules for exponents. However, this is kind of unsatisfactory--not all exponents are defined, and some of the results, for example the one I just calculated, are sort of unintuitive.

    The abstract algebra or number theory point of view would probably say this: (-1)<sup>4/5</sup> is the 4th power of a nontrivial 5th root of -1. (-1 is technically a 5th root of itself, but we agree that if we want to talk about the number -1 we can just call it -1.) What is a nontrivial 5th root of -1? It's a root of x<sup>5</sup>+1 other than 1. Using polynomial division and dividing out the factor x+1, we see that these are the roots of x<sup>4</sup>-x<sup>3</sup>-x<sup>2</sup>+x-1. A wonderful result called the Fundamental Theorem of Algebra says that any polynomial completely factors over the complex numbers, so this polynomial has roots in the complex numbers--in fact, it has four distinct roots--so we can talk about this all in the context of the complex numbers. So letting a be one of these roots, we can define (-1)<sup>4/5</sup> to be a<sup>4</sup>. It turns out that these numbers are just the nontrivial 5th roots of 1. (If you raise to the fifth power, you get (-1)<sup>4</sup> = 1.) The algebraist or number theorist would probably go by this interpretation. However, there's a sort of lack of precision here--this number is defined algebraically, and you can't really put your hands on it. So...

    There's the analytic point of view: we can define all exponents in terms of the exponential function. For example, for a and b real, a positive, we have:

    a<sup>b</sup> = e<sup>(ln a)b</sup>

    Okay, so if we can define a complex exponential and natural logarithm, we're good. And we can. But we're not really good. Why? Because the complex logarithm is a pretty ugly function at first glance--you can't define it consistently over all of the complex numbers if you want it to be differentiable. This might sounds like a serious flaw, but it turns out to be a wonderful fact which allows us to prove all sorts of neat things and which is a fundamental piece of many branches of mathematics. I don't want to overwhelm you with detail, so let's just look at some specific cases.

    In a few chapters, you'll be learning about power series, and we can use these to define the complex exponential and logarithm functions. However, we can define the exponential at imaginary numbers without all of this fuss. For any real number r:

    e<sup>ir</sup> = cos r + i sin r

    You can use addition formulas to show, thus defined, the exponential function retains its additive properties. But notice something weird. Cosine and sine are periodic of period 2π. So if I plugged in (r+2π)i instead of ri, I would get the same answer. So the complex exponential is periodic with period 2πi. Weird stuff. So if I want to define the logarithm as the inverse function to the exponential function, I run into a wall--there are infinitely many inverses to a given number! So, for example, we have that, for any z = (2n+1)πi, n ranging over all integers:

    e<sup>z</sup> = -1

    So if I want to define (-1)<sup>4/5</sup> using complex analysis, I get:

    (-1)<sup>4/5</sup> = e<sup>4(2n+1)πi/5</sup> = e<sup>(8n+4)πi/5</sup>

    It turns out that, as n ranged over the integers, this will give us exactly 5 values:

    1, cos π/5 + i sin π/5, cos 2π/5 + i sin 2π/5, cos 3π/5 + i sin 3π/5, cos 4π/5 + i sin 4π/5

    You can check that these are the 5th roots of 1, so the algebraic and analytic points of view give the same thing. And they both give you some sort of ambiguity--there is no number which sticks out as the 5th root of 1, although people typically use cos π/5 + i sin π/5 to represent this.

    It's important to remember that the usual rules of exponentiation start to break down when you treat things in a more sophisticated manner. You have to be careful and remember definitions, or else you'll start proving some things worthy of SolomonGrundy.


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  4. #3  
    Forum Bachelors Degree Demen Tolden's Avatar
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    Thank you for that. I have no intention for attempting to break or challenge any rules of math at this time, just learning them. I think the rebellious side of my personality isn't interested in math. :P
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    Forum Professor serpicojr's Avatar
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    There's nothing wrong with trying to break the rules of math--you just have to make sure you don't break the rules of logic!
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