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Introductory perturbation theory

  1. Dec 10, 2007 #1
    I've been reading a paper at the following link:

    www.cims.nyu.edu/~eve2/reg_pert.pdf

    I have several questions:

    In the first example they use the method to approximate the roots for

    x^2 - 1 = "epsilon" x

    I was under the impression - wrongly perhaps - that f(x) had to have continuous derivatives to use the Taylor series. When you go to f'''(a), the coefficient c3 vanishes. And, strictly speaking isn't this a Maclaurin series?

    In the problem for x^2 -1 = "epsilon" e^x
    The coefficient X2 for x = -1 is 3/8e^2.
    I'm sure of this since I went through it several times.

    I don't understand what the second figure - Figure 2 - is referring to at all (why two)??
    And what "3" solutions? Doesn't a quadratic have 2 solutions? What is the solid line in the figure a plot of anyway?

    I apologize that I can't post it verbatim but I'm unable to cut and paste from an Adobe file to this format and I'm really having problems using the LaTex editor. I'd appreciate any commentary. Thank you.
    John
     
  2. jcsd
  3. Dec 10, 2007 #2

    Ben Niehoff

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    There are more than two solutions because

    [tex]x^2 - 1 = \varepsilon e^x[/tex]

    is not a quadratic equation!

    If you graph them, you will see that the functions [itex]f(x) = x^2-1[/itex] and [itex]g(x) = \varepsilon e^x[/itex] cross each other in three places (two near the origin, and one further away).
     
  4. Dec 11, 2007 #3
    3 points?

    Please see attached Excel file.
     

    Attached Files:

  5. Dec 11, 2007 #4

    Ben Niehoff

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    That's not an Excel file...and I never open .doc files I see on the Internet. If you made a graph in Excel, you should be able to export the image and just upload that.
     
  6. Dec 11, 2007 #5

    Ben Niehoff

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    "Dear"? I don't think that was necessary.

    Look, x^2-1 is a parabola, with vertex at (0,-1), opening upwards. [itex]\varepsilon e^x[/itex] is an exponential. The roots are places where these two curves cross:

    For positive epsilon, one of two things can happen:

    1. If epsilon is large enough, then the curves will cross at only one point, at some x less than zero. After that, the exponential is always greater than the parabola, so they will never cross again.

    2. For sufficiently small epsilon, there will still be a root at some x less than zero, but the exponential curve will be shallow enough that it will cross the parabola again. That is, for some x > 0, you will have exponential < parabola.

    But no matter how small epsilon is, the exponential is still of greater order than the parabola. That is, there must be some k such that exponential > parabola for all x > k. Therefore, the two curves must cross yet again, as the exponential will eventually be bigger than the parabola.

    Therefore, you can have up to three roots for positive epsilon.

    For negative epsilon, it should be obvious that the two curves can cross in at most 2 places.
     
  7. Dec 12, 2007 #6
    The parabola vs. the exponential

    My sincerest apologies for my previous rudeness. Things aren't going that well on my end, but that's not your problem. Many times the précis I get in reply to my queries don't help or just aren't addressing the questions I'm asking - or trying to ask. Also, in the vein of "Couldn't see the forest for the trees...", if I had bothered to reason along the line you eloquently presented I would have extended the two functions out further than the region around the origin and seen the third root. Again, I apologize.
    But I'm still in a fog as to what that note on Fig. 2 is talking about. How does he get that solid line curve? And the third coefficient for the second root, X2, is wrong. Unless I don't even know how to do algebra and arithmetic any more - a distinct possibility given my predilection for arrogant stupidity - I'm a New Yawkah doncha' know?
    I get 3/8e^2 for X2. He refers to a "numerically obtained solution" I'm assuming a guy wrote this. See above. How does he get a numerically obtained solution?
    I apologize, sir, for my manners, but I've been running into papers in this and other areas - Quantum's a scream – with so many reader specific typos that it's getting maddening. Thank you for your response. And any help in this matter is and would be greatly appreciated. Thanks.
     
  8. Dec 12, 2007 #7

    Ben Niehoff

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    The two roots being plotted are the positive ones. Notice that for sufficiently high epsilon, there is no positive root. "Numerically" just means he obtained the roots through Newton-Raphson or some other method.
     
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