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Negative based powers

  1. Jul 11, 2009 #1


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    I was trying to visualize/understand what y=(-1)x would look like since my graphing calculator won't graph it because it is discontinuous. I had a bit of luck with the integers, but getting into fractional values of x and irrationals is going to be much more time consuming and complicated.

    I also thought of this:



    But [tex]e^{ix}=-1[/tex] hence, [tex]ix=ln(-1)[/tex]

    So lets choose to the value x=1/2

    y=i however the derivative is [tex]i^2\pi[/tex] which simplifies to a real number [tex]-\pi[/tex]. So while the value does not exist, the derivative does? I'm not understanding how that is possible or if there is any significance to this. Actually, I'm feeling more certain that I made some mathematically infeasible step, such as using:

    [tex]\frac{d}{dx}a^{x}=\frac{a^{x}}{ln(a)}[/tex] only for values a>0, a[tex]\neq[/tex]1

    Please explain this to me :smile:

    edit: fixed derivative
    Last edited: Jul 11, 2009
  2. jcsd
  3. Jul 11, 2009 #2
    could u explain how can we differentiate (-1)^x
    i mean to say by which formula
  4. Jul 11, 2009 #3


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    Well what I know is that [tex]\frac{d}{dx}a^x=a^xln(a)[/tex]
    Don't ask me to prove this though :biggrin:

    Now it just depends on what the restriction to 'a' are. Now, for the real numbers a>0 as the logarithm is only defined for a>0, but what about extending this into the complex numbers? I'm aware that many formal definitions in maths break down when complex numbers are involved, so is using the derivative of (-1)x in this way also broken?

    edit: fixed the derivative
    Last edited: Jul 11, 2009
  5. Jul 11, 2009 #4
    its not true
    the formula is

  6. Jul 11, 2009 #5


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    Thanks for correcting me jontyjashan.
  7. Jul 13, 2009 #6


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  8. Jul 14, 2009 #7


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    It's not clear what your question is. For real numbers, of course, ax is only defined for a> 0. For complex numbers, az is defined for all complex numbers by az= ez ln(a) but, of course, ln(a) is multi-valued. In particular, if [itex]a= -1= (1)e^{-\pi i+ 2n\pi}[/itex] for any integer n so [itex]ln(-1)= -\pi+ 2n\pi= (2n-1)\pi[/itex]. [itex](-1)^z= e^{z(2n-1)\pi}[/itex] where n can be any integer.
    Last edited by a moderator: Jul 15, 2009
  9. Jul 14, 2009 #8
    If you don't want "multi-valued" answers, the best one is the continuous one defined like this...
    [tex](-1)^s = \exp(\pi i s) = \cos(\pi s) + i \sin(\pi s)[/tex]
    As [tex]s[/tex] varies, it goes around the circle, of course reaching [tex]1[/tex] and [tex]-1[/tex] when [tex]s[/tex] is an integer.
  10. Jul 14, 2009 #9
    These steps are good.

    This step is bad.

    The rule is [tex]e^{i \theta} = \cos \theta + i \sin \theta[/tex], with [tex]e^{i \pi} = -1[/tex].

    Be careful. You might quickly assume that [tex]\ln -1 = i \pi[/tex]. But look closer, and you'll see that [tex]e^{-i \pi} = -1[/tex], too. Or more generally [tex]e^{i (\pi +2n \pi)} = -1[/tex] for any integer n. It's periodic. And periodic functions aren't 1-to-1. And functions which aren't 1-to-1 can't be inverted so naively. The "inverse" of [tex]e^x[/tex] is not a single valued function. You have to perform a "branch cut", choosing one of the (infinitely many) values it can take. For example, we define [tex]\ln x = y[/tex] as the smallest, non-negative value for y such that [tex]e^{i y} = x[/tex]. So [tex]ln(-1) = i \pi[/tex].

    Keep in mind, though, that ln and e are only PARTIAL inverses of each other. Just as [tex]\sqrt{x^2}[/tex] need not equal x (because of negative numbers), [tex]ln(e^x)[/tex] need not equal x (because of numbers greater than [itex]2\pi[/itex]. In many cases, it will, but not always.

    Take that point into account and see what you can come up with =-)
  11. Jul 14, 2009 #10


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    I've worked with complex numbers, but didn't know this equality. I've only worked with the RHS so thanks for that.

    ok taking the multi-valued function into account, we instead have: (please don't hesitate to brutalize me for my faults)



    [tex]i\pi(1+2n)=ln(-1)[/tex] *

    * I'm unsure about this step, based on what Tac-tics explained about partial inverses:
    Were my steps correct to take this into account?

    For a>0 where a[tex]\neq[/tex]1 it follows the usual exponential functions, but for a<0 the values are only NOT real (by my logic) when ax has values for x that can only be expressed as [tex]\frac{m}{2n}[/tex] for any n integer.
    In all other cases, the value of ax will keep diverging between positive and negative values. Of course the graph won't be continuous and have infinitely many holes but does that really mean that ax isn't defined for a<0?
  12. Jul 15, 2009 #11
    We'll be gentle =-P

    This is THE fundamental formula for complex numbers. It's called Euler's identity. Richard Feynman called it the most beautiful theorem and mathematics.

    It's more than just a formula. It allows us a totally different way to represent complex numbers. By the trig identity [tex]cos^2 x + sin^2 x = 1[/tex], we see that no matter what the value of [tex]\theta[/tex] that [tex]abs(e^{\theta i}) = 1[/tex]. That is, EVERY complex with modulus ("length") 1 can be written in this form. Furthermore, ANY complex at all can be written in the form [tex]r e^{i \theta}[/tex] for some real numbers r and theta, where r is the "length" of the complex number and [tex]\theta[/tex] is the "direction". This form makes it very easy to multiply two complex numbers (you simply multiply the r's and add the [/tex]\theta[/tex]s). You can convert back and forth between this form and the a + bi form without much trouble.

    This step is right. Sort of. Really, you need to do a branch cut on ln. Pick an integer for n. The value ln(-1) is then equal to [tex]i\pi(1+2n)[/tex], subbing in the constant value you picked for n. The ln function on a TI-83 or TI-89 calculator or in Mathematica is the result when n=0, so [tex]ln(-1) = \pi i[/tex].

    This sounds about right. It's only at isolated points, spaced a constant distance away that the result of [tex](-1)^x[/tex] is going to be a real number. The rest of the time, it will have an imaginary part.

    What makes this topic a little difficult to discuss is that everyone coming out of first-year calculus has this mistaken assumption that there is this thing called "raising to a power" that has a single, unambiguous meaning. When the exponent is a positive integer, the definition is simple. When you allow for zero and negative integers, you have to make some judgment calls. We say that [tex]x^{-1} = \frac{1}{x}[/tex] and [tex]x^0 = 1[/tex] because it's convenient to define it that way -- it leads to very nice identities such as [tex]x^a x^b = x^{a + b}[/tex] and [tex](x^a)^b = x^{ab}[/tex].

    When we allow for fractional exponents, though, things get ugly. Suddenly, our identities begin to fail us. [tex]((-1)^2)^\frac{1}{2} /neq -1[/tex]. If we allow [tex]0 ^ 0 = 1[/tex], exponentiation is no longer a continuous function. It all goes to hell! As we might expect, things get even worse when we allow for complex numbers, causing logs to become multi-valued.

    The complex numbers and exponentiation are both miraculous things, but they are sometimes clunky. They are like a beautiful work of an engineer -- they get the job done, but they do have their rough spots.
  13. Jul 15, 2009 #12


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    Tac-Tics, I very much liked your explanation :smile:

    So basically, ax is only defined for a>0 because - even though there are infinitely many real values for a=0 and a<0 - the fundamental definitions for exponentiation breaks down when lurking into this territory? That makes way more sense!

    That example you gave actually shocked me a bit.

    [tex]((-1)^2)^\frac{1}{2}\neq -1[/tex] only if the order of operation is taken in each bracket first. Otherwise, whether by using the exponent rule or flipping the exponents around and taking them in opposite order, it will equal -1.
    I suppose this has to do with the fact that the base a<0 and the exponent rules have now broken down right?

    Is this also why my opening question in this thread has an imaginary value coupled with a real derivative?

    Does the multi-valued aspect of [tex]e^{i\theta}[/tex] matter anymore? I mean, it has already been shown that it doesn't work for a<0, so is there any point endeavoring for an answer?
  14. Jul 15, 2009 #13

    You brought up the definition of exponentiation, which is important. Let me put it more directly.

    How do you define exponentiation?

    If you're going to make reasoned arguments about something, you better have a solid idea of what you're talking about! You have built up several informal notions of what exponentiation is in your algebra and calculus classes, but at no point did anyone give you a formal definition for it.

    If it weren't 3am here, I would take the time to lay out something close to a good definition, but it's late.

    But for the time being, keep this in mind. When you see [tex]x^y[/tex], know that what you're really looking at is a function of two variables (x and y). When you're playing around with a potential definition of exponentiation, you're really just tweaking with the definition of a boring old function that obeys useful properties you are interested in.

    Parentheses always determine the order of operations. And, as it turns out, you cannot switch the order of the exponents here. The rule is that [tex](x^a)^b = x^{ab}[/tex] only when x is positive. That disclaimer is easy enough to forget, but forgetting about it can lead to incorrect results. In fact, there is a famous invalid proof that is based off this technique: http://en.wikipedia.org/wiki/Invalid_proof#Version_1

    The rule that [tex]\frac{d}{dx}a^x = e^x ln(a)[/tex] does not work for [tex]a \leq 0[/tex]. When a < 0, the result of the logarithm would be non-real, as you seem to have discovered. Annoyingly, none of the web sites I found on Google bothered to mention this restriction. But it's there. It's similar to the restriction on changing the order of exponents..... easily forgotten, but important nonetheless.

    It's actually ln(x) that is multi-valued. Luckily, the function [tex]e^x[/tex] has a straightforward definition. [tex]e^x = \Sigma_{n=1}^\infty \frac{x^n}{n!}[/tex]
    It works for both real and complex numbers. It works all the time. Memorize it ;-)

    The confusion with exponents comes when the base is zero, negative, or when you have to "undo"an exponent with a logarithm. But the exponential function [tex]e^x[/tex] is straightforward, easy, and incredibly useful in all areas of math.
    Last edited by a moderator: Jul 15, 2009
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