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Dividing by a parameter that is zero

  1. Jul 3, 2013 #1

    I'm studying calculus of variations, this is a bit of a recap of regular calculus.

    You got a function, but you plug a vector that is parameterized so your function becomes parameterized, then you take the derivative in 3.5, and then they say: put the parameter to zero in 3.6, but then you divide by zero, I don't get it.


    Last edited by a moderator: Jul 10, 2013
  2. jcsd
  3. Jul 3, 2013 #2


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    Where are you dividing by [itex]\varepsilon[/itex] then?

    You are just using the condition that
    $$f'(\vec r, 0) = \left. \frac{df}{d\epsilon} \right|_{\epsilon = 0}$$
    i.e. ##\varepsilon = 0## is a stationary point to set the left hand side to zero.
  4. Jul 3, 2013 #3


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    It doesn't set dε to 0, it sets ##\frac{df}{dε}## to 0. The rate of change of f is 0 at ε = 0 because f is stationary at that point.
  5. Jul 3, 2013 #4


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    Are you under the impression that [itex]\frac{df}{de}[/itex] involves dividing by e? It does not.
    [tex]\frac{df}{de}= \lim_{h\to 0}\frac{f(e+h)- f(e)}{h}[/tex]
  6. Jul 12, 2013 #5
    My god.What a mistake.

    It's been a while since I've encountered calculus.
    I better review the basics thoroughly than tackling some more advanced stuff.

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