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2nd order differential with limit - keep getting zero

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the solution of the following differential in the limit z -> i/2:

    5age3n.jpg

    2. Relevant equations

    Quotient rule, product rule.

    3. The attempt at a solution

    I found the first derivative using the quotient rule, with the answer as two fractions. I then took both fractions and differentiated them again, one of which required the quotient rule and the other of which required both the quotient rule and the product rule.

    However, every fraction in the final answer as a (z - i/2) term in the numerator and no similar terms in the denominator, which makes them all zero when you take the limit. My final answer is this:

    2pt71oh.png

    The answer should be -3i/16 after the limit is taken.
     
    Last edited: May 23, 2010
  2. jcsd
  3. May 23, 2010 #2

    Cyosis

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    The denominator [itex]4z^2+1 \rightarrow 0[/itex] when [itex] \ z \rightarrow i/2[/itex]. As a result you end up with an indeterminate expression, 0/0.
     
  4. May 23, 2010 #3

    gabbagabbahey

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    Hint: [itex]4\left(\frac{i}{2})^2+1=0[/itex]:wink:
     
  5. May 23, 2010 #4
    Hmm. So it is possible to get to -3i/16 from where I've gotten or have I made a mistake in my logic? Cancelling out the brackets that equal zero on the top and bottom of each fraction doesn't seem to help (and I don't think you can do that anyway).

    The reason this differential has to be done is because of the following residue rule:

    8zit5d.png

    With the original equation being:

    in925g.png

    (it has a 3rd order singularity at i/2, and one at -i/2 that can be ignored, so you make z0 = i/2)


    I'm sure I've made a stupid mistake somewhere...
     
  6. May 23, 2010 #5

    gabbagabbahey

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    There are a couple of errors in your expression for [itex]f''(z)[/itex]. To make it easier to calculate, just realize that [itex](4z^2+1)=4(z+i/2)(z-i/2)[/itex] and simplify your expression before differentiating.
     
  7. May 23, 2010 #6
    Ah, thanks for that tip. I now get the correct answer. :D
     
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