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Taylor expansion

  1. Dec 24, 2004 #1

    quasar987

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    [tex]V = 2\pi \sigma(\sqrt{R^2+a^2}-R)[/tex]

    Show that for large R,

    [tex]V \approx \frac{\pi a^2 \sigma}{R}[/tex]

    I figured if I could develop the MacLaurin serie with respect to an expression in R such that when R is very large, this expression is near zero, then the first 1 or 2 terms should be a fairly good aproximation. But I can't find such an expression.

    Thanks for your help.
     
  2. jcsd
  3. Dec 24, 2004 #2

    learningphysics

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    The given expression approaches zero for large R.

    I get the given answer, if you find the Taylor series with respect to a. The first derivative term is 0, and the second derivative term is the answer. I think for a rigorous answer, you'll have to show that the condition for convergence is satisfied...
     
  4. Dec 24, 2004 #3

    quasar987

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    I did a few similar exercices since my original post and now I clearly see how to do this one:

    [tex]V = 2\pi \sigma(\sqrt{R^2+a^2}-R) = 2\pi \sigma R(\sqrt{1+a^2/R^2}-1)[/tex]

    The first two terms of the binomial epansion of [itex](1+a^2/R^2)^{1/2}[/itex] give the answer.
     
  5. Dec 24, 2004 #4

    quasar987

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    It's funny how your way works too though... it means there's no difference in how V "behaves" when it's R that's very large from when it's a that's very near 0. :eek: Is this pure coincidence? In which kind of function will this sort of symetry happen ?
     
  6. Dec 24, 2004 #5

    learningphysics

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    Not coincidence... I believe it's the same Taylor expansion. If each of us continued our expansion we'd get the same terms.
     
  7. Dec 25, 2004 #6

    arildno

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    Dearly Missed

    The proper expansion parameter is a/R; hence, it doesn't matter whether a is small or R big. In both cases, a/R is small.
    This is readily seen by rewriting V as:

    [tex]V=\frac{2\pi\sigma{a}^{2}}{R(1+\sqrt{1+(\frac{a}{R})^{2}})}=2\pi\sigma{a}\frac{\epsilon}{\sqrt{1+\epsilon^{2}}},\epsilon=\frac{a}{R}[/tex]
     
    Last edited: Dec 25, 2004
  8. Dec 25, 2004 #7

    quasar987

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    ....I see !
     
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