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Violation of LHoptial

  1. Apr 5, 2012 #1
    Last edited: Apr 5, 2012
  2. jcsd
  3. Apr 5, 2012 #2
    No, the L'Hopital applies only if both g(z) and f(z) go to zero or infinity. In this case downstairs does not.
     
  4. Apr 5, 2012 #3
    so your saying I shouldn't of used L'Hopital. How else should I of solved it?
     
  5. Apr 5, 2012 #4

    Office_Shredder

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    Simple poles in this form have a very easy way to calculate the residue. Consider f(z)/z. Suppose f(z) is holomorphic at z=0 (note f(z) might be, for example/ 1/(1+z), so you CAN write your problem in this form).

    Then the Laurent series for f(z)/z is
    [tex] f(0)/z+ f'(0) + f''(0) z /2 +....[/tex]

    just taking the Taylor series for f(z) and dividing it by z. Is it clear now what the residue is?
     
  6. Apr 5, 2012 #5

    D H

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    L'Hôpital's rule is perfectly valid here. The function f(z) has two poles, one at i/3 and the other at -i/3. Looking at the latter (your example), we need to calculate
    [tex]\lim_{z\rightarrow -i/3} (z+i/3)f(z)
    = \lim_{z\rightarrow -i/3}\frac{(z+i/3)z}{9z^2+1}
    = \lim_{z\rightarrow -i/3}\frac{z^2+i/3\,z}{9z^2+1}[/tex]
    At the limit we have something of the form 0/0, so L'Hopital's rule applies. Taking the derivatives of the numerator and denominator yields
    [tex]\lim_{z\rightarrow -i/3} (z+i/3)f(z)
    = \lim_{z\rightarrow -i/3}\frac{2z+i/3}{18z} = \frac 1 {18}[/tex]
    At the limit, the numerator and denominator are -i/3 and -18i/3, respectively, so the residue is 1/18.

    The same analysis works for the other pole as well.
     
  7. Apr 5, 2012 #6

    Ray Vickson

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    If we have a function of the form [tex] F(z) = \frac{f(z)}{(z-a) g(z)} [/tex] with [itex] g(a) \neq 0, [/itex] the residue of F at a is just [tex] \text{Res}(F)(a) = \frac{f(a)}{g(a)}. [/tex] There is no need to use L'Hospital's rule here, because the work performed by the rule has already been done. Here is what I mean: you could define [itex] \text{Res}(F)(a) = \lim_{z \rightarrow a} (z-a) F(z).[/itex] If you do that, notice that we have
    [tex] (z-a) F(z) = (z-a) \frac{f(z)}{(z-a)g(z)} = \frac{f(z)}{g(z)}, [/tex] which has a perfectly nice limit equal to [itex] f(a)/g(a),[/itex] with no need to use L'Hospital.

    In fact, you are looking at the problem backwards: the whole point of L'Hospital is to "discover" the powers m and n (if any) in the expression [tex] \frac{N(z)}{D(z)} =
    \frac{(z-a)^n u(z)}{(z-a)^m v(z)}, \text{ where } v(a) \neq 0.[/tex] In the current problem you don't need to do that, because it has already been done for you.

    RGV
     
  8. Apr 5, 2012 #7
    Ok, thanks. I understand, haven't looked at the Laurent series in detail but looks it looks ok.
    While we are on the subject of limits I have got

    [tex]Lim_{z \rightarrow ia} \frac{(z-ia)}{(z^2 + a^2)(z^2+b^2)}e^{imz}[/tex]

    Is there a quick way to find it's residue without multiplying out then taking the derivative of upstairs and downstairs until something happens?

    edit: what is with this new itex. Itallic tex?
     
    Last edited: Apr 5, 2012
  9. Apr 5, 2012 #8

    Ray Vickson

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    Your expression looks OK to me, but maybe you don't like [itex] Lim, [/itex] and would rather have [itex] \lim [/itex] instead? If so, write "\lim" to override the math italics (same as saying "\exp" instead of "exp", "\sin" instead of "sin", etc.)

    Anyway, you can write [itex] z^2 + a^2 = (z - ia)(z + ia), [/itex] then cancel out the [itex] (z - ia)[/itex] factor in both the numerator and denominator. You can just put z = ia in what is left. So, basically, you can just write down the answer with hardly any work at all.

    RGV
     
  10. Apr 5, 2012 #9
    Thanks, I didn't spot that :)
     
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