1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Limit of COMPLEX-valued functions

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is to evaluate:
    +∞
    ∫ exp(-3x- i ωx) dx
    0

    But I don't understand how to compute
    lim exp(-3x- i ωx)
    x->+∞

    The following is the solution presented, but I don't understand it...
    lim exp(-3x- i ωx) =
    x->+∞
    lim exp(-3x) exp(- i ωx) =
    x->+∞
    lim exp(-3x) [cos(ωx) - i sin(ωx)]
    x->+∞
    Now since exp(-3x)->0 as x->+∞ and [cos(ωx) - i sin(ωx)] is bounded, we have that
    lim exp(-3x) [cos(ωx) - i sin(ωx)] = 0
    x->+∞


    Can someone please explain the part in red?
    1) What does it mean for a COMPLEX-valued function [cos(ωx) - i sin(ωx)] to be BOUNDED?
    2) What is the meaning of a LIMIT of a COMPLEX-valued function?
    3) Why exp(-3x)->0 as x->+∞ and [cos(ωx) - i sin(ωx)] bounded => lim exp(-3x) [cos(ωx) - i sin(ωx)] = 0 ?


    I am so confused...everything I've learnt in calculus is in the field of REAL numbers, and I have no background in complex analysis at all, so please explain in the simplest way...

    2. Relevant equations
    N/A

    3. The attempt at a solution
    N/A

    Any help is greatly appreciated! :)
     
  2. jcsd
  3. Nov 14, 2009 #2
    exp(-3x) goes to zero because exp(3x) is in the denominator. As for the cos(wx)-isin(wx), it will just oscillate back and forth as x goes to infinity. So you are multiplying zero and an oscillating function.
     
  4. Nov 14, 2009 #3
    But there is "i" in it. What does it mean to say that the COMPLEX-valued function [cos(ωx) - i sin(ωx)] is BOUNDED?
    cos(ωx) - i sin(ωx) is not even a real number...
     
  5. Nov 14, 2009 #4
    Sine and cosine are bounded because they always return values between -1 and 1. If you have an imaginary component, that doesn't change the value sine will return. If you graph the function on a complex plane it will oscillate as x goes to infinity. It will not blow up or converge.
     
  6. Nov 14, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since the complex numbers is not an ordered field, saying that a complex valued function, f(x), is bounded means that |f(x)| is bounded. In any case, e(a+bi)x= eaxebix= eax(cos(bx)+ i sin(bx)). If a is negative, eax goes to 0 as x goes to infinity while cos(bx)+ i sin(bx) has absolute value 1 for all x. As Legion81 said, that product goes to 0 as x goes to infinity.
     
  7. Nov 14, 2009 #6
    Math doesn't really change when you deal with complex numbers. You can ignore the "i" if you just want to evaluate a limit. Hope that helps.
     
  8. Nov 14, 2009 #7
    1) But how can we visualize complex-valued functions like f(x)=(cos(bx)+ i sin(bx))? What does it mean geometrically that a complex-valued function is bounded?

    2a) But I still don't understand the idea of a limit of a complex-valued function. For example, if we take lim exp(-3x- i wx), the function is complex-valued, so should we get a complex number after taking the limit? Is it possible to get a real number?

    2b) Also, can we treat "i" as a scalar/constant, and pull it out of the limit?

    3a) |cos(wx) - i sin(wx)| <= M for all x
    So here we can take, for example, M to be 1, right? (because |cos(wx) - i sin(wx)| =1 ?)
    i.e. is it correct to say that |exp(-3x) [cos(ωx) - i sin(ωx)]| ≤ |exp(-3x)| = exp(-3x) ?

    3b) When you say that a bounded function times a function that goes to 0 implies that the limit of the whole thing goes to 0, are you using squeeze/sandwich theorem? But does it apply for COMPLEX-valued functions? What is the statement of that theorem for complex-valued functions?

    Thank you for answering!:smile:
     
  9. Nov 14, 2009 #8
    1) If you are trying to visualize what it looks like in a real plane, then the function would only exist when (wx) is n*pi (n=0,1,2...). You would have a discontinuous function. If you consider the complex plane (where the vertical axis is imaginary and the horizontal axis is real), you could trace out a circle of radius 1 with this function. So cosine and sine (considered separately) have a magnitude less than 1. That is what it means to be bounded in a complex plane.

    2a) A complex number has two parts, real and imaginary. If it helps, you can think of every limit you have ever calculated as being a complex limit for a+bi, where b=0. The same concept of a limit for real numbers is used for complex numbers. Yes, you can get a real number, example: take the limit as x goes to infinity of 1+(i/x). This is complex, but the second term goes to zero and all you have is a real number, 1. This 1 is also complex, with the i component being 0.

    2b) No, just as you cannot pull out a 2, 3, 15 or any other numbers. You treat i as any other number.

    3a) Yes and no. You are correct that |cos(wx)-isin(wx)| <= M for all x. That magnitude is not less than 1 though. Consider the time when wx = 7*pi/4 (using unit circle in complex plane).

    3b) I honestly forget what theorem you use (haven't looked at them in a while), but the same logic applies to complex functions.
     
  10. Nov 15, 2009 #9
    2a) When we say lim f(x) = a real number, it makes sense to talk about the function f(x) being CLOSE to that real number in the limit since the real numbers are ordered. But when we say lim f(x) = a complex number, what does it mean to be CLOSE to a complex number?

    2b) I mean...can we do the following?
    lim (ix) = i lim(x)
    [idea similar to lim(ax) = a lim(x) for real number a]
    I think this is OK, but I'm not sure as I really have no background at all in complex analysis. Things like this are never presented properly to me...but somehow I have to know how to evaluate these in another course.

    3a) hmm...I don't understand your point.
    |cos(wx)-isin(wx)| =cos2(wx) + sin2(wx) = 1 for ALL x. It's always equal to 1, so why is it incorrect to say that |cos(wx)-isin(wx)| ≤ 1 for all x? (i.e. M=1)

    3c) exp(-3x)->0 and |[cos(wx) - i sin(wx)]| is bounded IMPLIES
    lim exp(-3x) |cos(wx) - i sin(wx)| = 0
    x->∞
    But I think we are supposed to evaluate
    lim exp(-3x) [cos(wx) - i sin(wx)] , right?? Is this 0 as well?
    x->∞
    i.e. If f(x) is complex valued, then are limf(x)=0 and lim|f(x)|=0 EQUIVALENT?


    Your help is greatly appreciated! :smile:
     
  11. Nov 17, 2009 #10
    hmm...I'm still confused...could someone please help?
     
  12. Nov 17, 2009 #11
    2a) Being close to a complex number is the same as being close to a real number. Your function will approach that value. I'm not exactly sure where your confusion is so it is kind of hard to answer this.

    2b) If your function allows i to be factored out then you can do that. It can be treated as a normal number. In your original problem, i cannot be taken out of the limit.

    3a) I'm not sure what I was thinking on that one! This would have a magnitude of 1. If it was the form r(cos(wx)-isin(wx)) with r>1, then it would have a magnitude larger than 1. Sorry if that caused any confusion.

    3c) Because one part of your function is going to zero, it will make the limit go to zero. Knowing that lets you evaluate the limit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Limit of COMPLEX-valued functions
Loading...