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Calculus of variations basic question

  1. May 14, 2006 #1
    this is one of those things that looks like it should be really simple but for some reason i just don't get it :confused: i've looked at a few books and they all start explaining calculus of variations in the same way.

    i'll quote a paragraph from feynmann lectures II (concerning finding the path of least action) just to illustrate my problem:

    "When we have a quantity which has a minimum - for instance, in an ordinary function like temperature - one of the properties of the minimum is that if we go away from the minimum in the first order, the deviation of the function from its minimum value is only second order. At any place else on the curve, if we move a small distance the value of the function changes also in the first order. But at a minimum, a tiny motion away makes, in the first approximation, no difference."

    now, what is meant by first and second order in this context? i honestly don't get what is meant here. surely if you move away from the minimum of a function, the function changes according to itself. e.g. if f(x) = x^2, then if you move away from the minimum x=0 in the x-direction, f(x) changes by x^2. so what is meant by nothing changing in the first order?
  2. jcsd
  3. May 14, 2006 #2
    Here is the Taylor series for the function f(x) about a point a.

    f(x) = f(a) + f'(a)(x - a) + (1/2)f"(a)(x-a)^2 + ...

    so the deviation is:
    f(x) - f(a) = f'(a)(x - a) + (1/2)f"(a)(x-a)^2 + ...

    (x-a) is the first order and therefor so is f'(a)(x - a)
    (x-a)^2 is of second order and therefor so is (1/2)f"(a)(x-a)^2
    The rest of the terms (denoted ...) are of third order and higher.

    If f'(a) = 0 ( a property of the minimum), then
    f(x) - f(a) = (1/2)f"(a)(x-a)^2 + ...
    so the deviation is of second order.
    Last edited: May 14, 2006
  4. May 14, 2006 #3
    thanks, that makes a lot more sense!
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