Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solve this question ?

  1. Jun 27, 2008 #1
    (a)to the power x = -infinity , what will the value of x ?
  2. jcsd
  3. Jun 27, 2008 #2
    What is (a)?

    And what the heck is -infinity? Infinity is like zero (opposites), you don't see a -zero do you?
  4. Jun 27, 2008 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    First I know how to "answer" a question or "solve" a problem. I don't know how to "solve" a question.

    Secondly, you will need to define your terms. "a to the power x" makes this look like an equation involving real numbers, but then "-infinity" is not a real number. What number system are you talking about?

    Contrary to what epkid08 said, there is an "extended real number system" which has both -infinity and infinity (but still only one 0!). However, since our regular number operations can't be applied to either infinity or -infinity, you equation still does not makes sense there.
  5. Jun 27, 2008 #4
    I don't have a clue what your question is. Do you?
  6. Jun 27, 2008 #5
    lol, I guess he/she means:

    Find x if:
    ax = -infinity
  7. Jun 27, 2008 #6


    User Avatar
    Science Advisor
    Homework Helper

    And along those lines, there's the IEEE system that has both [itex]\pm\infty[/itex] and [itex]\pm0.[/itex]
  8. Jun 27, 2008 #7
    cshum00 has put the question in a readable way
  9. Jun 27, 2008 #8
    Well i guess it is like epkid08 said. It depends on what a is.

    If a is constant, then the answer is never.
    But if a = log(x) then it is -infinity
  10. Jun 27, 2008 #9


    User Avatar
    Science Advisor
    Homework Helper

    No, not really. We're not working over the real numbers -- thus the negative infinity in the problem. Assuming we're in the extended reals, a = -infinity has x = 1 as a solution.

    And if a = log x, [tex](\log x)^x[/tex] isn't defined in the extended reals for x < 0.
  11. Jun 27, 2008 #10
    Oh my god i can't believe what i am saying. My math is getting so rusty.

    Let me see, step by step:
    ax = -inf
    x = log(-inf)/log(a)
    but log is not defined for any number < 0.

    In my case, a = log(x)
    log(x)x = -inf
    x = -inf/log(x)
    still does not solve the problem at all

    What i was trying to do is to make "a" into something that will make the right hand side equals to x and then therefore x = -inf.
    Last edited: Jun 27, 2008
  12. Jun 28, 2008 #11
    OK, now I get it.

    We're working over the "doubly extended complex numbers (C++)"

    (I just made that name up, so, I get to define it - here goes ---.)

    Let the doubly extended real numbers (R++) = R union {infinity, -infinity} with the expected tolopogy.

    (That is, the topology where "lim f(x) as x -> infinity" and "lim f(x) as x -> -infinity" have their usual meaning.)

    Let C++ = {a+ib|a is in R++ and b is in R++}.

    OK, so much for the basic definitions. Ignore them if you want. I just put 'em in to increase the comfort level of some of the forum-ites. Other definitions are possible. Hopefully the following arguments work for other reasonable definitions.

    First you need to know the formula for the complex logarithm. I will derive it.

    Since for any complex number "c", c = exp(i*arg(c))|c| (that's just writing "c" in polar coordinates), log(c) = log(|c|) + i*arg(c).

    OK, here we go.

    Lemma: log(-infinity) = pi*i


    -1 = exp(pi*i).

    That is "Euler's formula."

    Multiply both sides of Euler's formula by infinity. You get

    -infinity = infinity*exp(pi*i).

    Now take the log of both sides and use log(infinity) = infinity.

    log(-infinity) = infinity + pi*i.

    [It is possible that one needs something like Robinson's "non-standard analysis" to make this proof rigorous.]

    This completes the proof of the lemma. Now let's use it to solve your equation.

    a^x = -infinity.

    Take the log of both sides.

    x*log(a) = log(-infinity).


    Opps! That is as far as forum rules allow me to go. Please complete the argument and post it (the completion of the argument).

    Last edited: Jun 28, 2008
  13. Jun 28, 2008 #12
    I am confused now.

    DeaconJohn, are you saying that in complex numbers
    log(-inf) = - inf ?
  14. Jun 28, 2008 #13
    Cshum00, My bad. I forgot to take the log of the right hand side. I'll edit the post. I noticed a few other things that could use improvement too. DJ
    Last edited: Jun 28, 2008
  15. Jun 28, 2008 #14


    User Avatar
    Staff Emeritus
    Science Advisor

    Deacon John, with that definition of "infinity" you get complex numbers with an "infinite number of infinities"- a different infinity of each end of each straight line through the origin. Topologically, that is the "Stone-Chech compactification" of the complex numbers and it is topologically equivalent to a circle and its interior.

    You can also do the "one point compactification" of the complex numbers by defining a topology so that all "large" complex numbers (large in the sense that the absolute value is large) so that there is only one "infinity". Then you get a set topologically equivalent to the surface of a sphere.

    However, those are purely topological constructions- you haven't said anything about how to do arithmetic with these new "infinities" much less take the logarithm.
  16. Jun 28, 2008 #15

    You're basically right. I'm doing more like "applied mathematics" than pure mathmatics, here. But then, the answer that I gave seemed to me to be in the spirit of the question that was asked.

    Two minor points,

    1) I did say that something like Robinson's non-standard analysis might be required to make the above arguments rigorous, although it seems to me that the intuitive content is clear. Robinson's N-S-A allows one to do normal arithmetic with infinities and infintesimals.

    Actually, another way to do it is just using limits and elementary calculus and throwing in just those infinities that one needs. As you noticed, if one works out the complete solution that I started, one probably needs an infinity at each end of each line through the origin in the plane. Or, maybe each line in the plane. One would have to work out the details to be sure. Of course, that is just exactly what Robinson did. So, if one did that, he would be taking a first step in the direction of what Robinson did.

    The elementary concept of limits can be linked to the topological concept using, for example, the techniques developed for the study of metric spaces in Kelly's General Topology.

    2) a circle at infinity attached to each end of each line in the plane is only a small part of the Stone Check compactification.

    In fact the S.C. c. plane is so big that I've never been able to get an intuitive handle on it.

    The most my intuition has been able to stretch around in the S-C compactification of N. There, for example, you can associate a point in the S-C compactification of N that is in the closure of the even numbers but not the odd numbers. (just take a bounded multiplicative linear functional on l_infinity that is zero on the odd njumber but is non-zero on the even numbers and is not of the form x->f(x) for some x in N (f is in l_infinity) and use the Hahn Banach theorem to extend it to all of N. - Wow -it's been over 30 years since I've actually looked at the definition of the S.C. c. of anything. I hope I got that right.)

    There are similar examples (I'm sure) in the S.C.c. of the plane, showing that there are more points than just the "infinities" attached to the end of each line.

    Neverthe less, the subset of the S-C-c of the plane that you called out is probably the natural topological setting for my arguments.

    And a question - I have the impression that the Stone Chech compactification (of a measure space) is the maximal ideal space of the set of bounded measurable functions when that set (L_infinity or l_infinity) is viewed as a commutative W* algebra acting on the Hilbert space of all square integrable functions of that set. Please let me know if I've got that wrong.


    P.S. One of the many things that's so great about this forum. It's helping me get back in shape. In the last couple of days since I've been on your forum, I've looked up all kinds of things (especially in number theory) that I'd sluffed over as I tried to learn the subject on my own. I've been trying to learn modern number theory off and on - basically on my own - for over ten years now. I still don't have Wiles proof "under my belt," but I get closer all the time.

    I think in the future I'll concentrate on the number theory board.
    Last edited: Jun 28, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Solve this question ?