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The chain rule?

  1. Oct 15, 2007 #1
    I'm trying to understand the proof for this theorem, and I can't see what they did to get from one step to the next.

    THEOREM: Suppose F(z) is an analytic function and that f(z) = F'(z) is continuous on a domain D. Then for a contour C lying in D with endpoints z1 and z2:

    [tex]\int_{C}f(z)dz=F(z_{2}) - F(z_{1})[/tex]

    PROOF:
    Using the definition of the integral, and assuming z'(t) is continuous

    [tex]\int_{C}f(z)dz=\int_{C}F'(z)dz=\int^{b}_{a}F'(z(t))z'(t)dt[/tex]

    This next step is the one I don't get. They say they used the chain rule?

    [tex]=\int^{b}_{a}\frac{d}{dt}\left[F(z(t))\right]dt[/tex]

    But HOW?

    The rest makes sense:

    =F(z(b))-F(z(a))

    =F(z2)-F(z1)

    Hence,

    [tex]\int_{C}f(z)dz=F(z_{2}) - F(z_{1})[/tex]
     
  2. jcsd
  3. Oct 15, 2007 #2

    Hurkyl

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    Equality is symmetric. Have you tried looking at it in the opposite direction?
     
  4. Oct 15, 2007 #3

    quasar987

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    [tex]F'(z(t))z'(t) = \frac{d}{dt}F(z(t))[/tex]

    Do you disagree that this is the statement that the chain rule makes?

    Or look at it in the opposite direction as Hurkyl suggests. I suppose he put his finger on your problem.

    [tex] \frac{d}{dt}F(z(t))=F'(z(t))z'(t)[/tex]
     
    Last edited: Oct 15, 2007
  5. Oct 15, 2007 #4
    Oh shoot I'm a moron!

    Okay.
     
  6. Oct 15, 2007 #5
    Okay they just did it backwards. I don't even want to tell you what I thought was going on. Thanks to both of you!
     
  7. Oct 15, 2007 #6

    quasar987

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    We've all been there. And now you've learned something.

    The next time you wonder why equation (1) equals (2), you will try looking at why (2) equals (1) before giving up.

    Nice going.
     
  8. Oct 15, 2007 #7
    Something about this whole proof feels really trivial. I was surprised that they had a proof, it seems more like a definition.
     
  9. Oct 15, 2007 #8

    quasar987

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    Well using the definition of a complex integral and how it relates to ordinary integrals of real functions, the thm points out that with appropriate conditions, the fondamental thm of calculus "holds" for complex functions integrated over paths.
     
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