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Strange invocation of Taylor series

  1. Jul 28, 2015 #1
    Hi all,

    I was working through a chapter on Lagrangians when I cam across this:

    "Using a Taylor expansion, the potential can be approximated as
    ## V(x+ \epsilon) \approx V(x)+\epsilon \frac{dV}{dx} ##"

    Now this looks nothing like any taylor expansion i've seen before. I'm used to

    ## f(x) = \sum\frac{f^{(i)}(a)(x-a)^i}{i!} ##

    What am I missing here?
     
  2. jcsd
  3. Jul 28, 2015 #2

    SteamKing

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    You just need to expand the definition of the Taylor series over the first couple of terms. Remember to account for x + ε as the argument of V(x).
     
  4. Jul 28, 2015 #3
    What are the first two terms with a = x + ξ ?
     
  5. Jul 28, 2015 #4
    Looks like it's ## f(x) \approx f(x + \epsilon) - f^1(x + \epsilon)\epsilon ##

    Oh, ok, I see now. So the ##\frac{dV}{dx}## is really a ##\frac{dV(x + \epsilon)}{dx}## ??
     
  6. Jul 28, 2015 #5

    TeethWhitener

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    You don't even need to use a Taylor series. Just use the definition of a derivative:
    [tex] \frac{dV}{dx}= \lim_{\epsilon \to 0} \frac{V(x+\epsilon)-V(x)}{\epsilon}[/tex]
    Therefore, for smaller and smaller [itex]\epsilon[/itex], the above becomes a better and better approximation. The expression from your first post just comes from taking the above equation, dropping the limit, and isolating [itex]V(x+\epsilon)[/itex].
     
  7. Jul 29, 2015 #6
    Sorry, do what SteamKing suggested but also take a=x as where the approximation is taken from. That should give the same answer.
     
  8. Jul 29, 2015 #7
    Ok I finally figured it out. I was wrong in my previous comment. ## x \mapsto x + \epsilon ## and ## a = x ##
    Thanks everyone.
     
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