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Find the derivative of a function

  1. Jun 23, 2014 #1
    1. The problem statement, all variables and given/known data
    If V=exp [ [itex]\int[/itex][itex]^{T}_{0}[/itex]s(t)dt ]

    2. Relevant equations
    What is dV/ds(k), where 0<k<T
    What does this derivative even mean??

    3. The attempt at a solution
    write
    V=exp(Y)
    dV/ds(k) = dV/dY . dY/ds(k)
    =V.[itex]\int[/itex][itex]^{T}_{0}[/itex]ds(t)/ds(k)dt
    =V because ds(t)/ds(k) = 0 for all t except t=k where it is 1.

    I just want to check that this derivative is correct; and more importantly what it means practically. Thank you very much for the help!

    Regards,
     
    Last edited: Jun 23, 2014
  2. jcsd
  3. Jun 24, 2014 #2
    Hello all,
    I've been thinking about the concept of the functional derivative all night/day but I still can not see the practical meaning of such a measure. Scalar derivative - no problem. It is simply the rate of change wrt the scalar variable. But when you have a function argument, and you take the rate of change with regards to just a point on that function argument, then what are you measuring exactly? Any help much appreciated! Thanks!
     
  4. Jun 25, 2014 #3
    A way to look at it:
    s, f, g ##\in## M the function space; ##V:M\to \mathbb R## a functional.

    - ##< f, g> = \int f(x)g(x)\,dx## is a bilinear, symetric and "almost" non-degenerate form. It associate a fuction with a form: $$f\in M \mapsto <f,\cdot> \,or\,<\cdot,f>\in M^*$$

    - The Dirac delta function ##\delta (x) = \delta_0## or ##\delta(x-y) = \delta_y## ##\in## M, y fixed, is the function that correspond to the form $$<\delta_y,\cdot>\in M^*\quad s.t. \quad <\delta_y,f> = f(y)\in \mathbb R$$

    - The functional derivative of a functional at the point(=function) ##g\in M## is the function ##\frac{\delta V}{\delta g}## or ##\frac{\delta V}{\delta g(x)}## (as the notations f or f(x)) such that: $$<\frac{\delta V}{\delta g},\cdot>\in M^* = \mathbf D_g V(\cdot)$$ ##D_g V(\cdot)## the derivative of V at g. It's a linear map sending a vector h at g (h=function again) to the directional derivative of V at g in the direction h:$$D_g V(\cdot): h\mapsto D_g V(h) = \lim_{\epsilon \rightarrow 0}{\frac{V(g+\epsilon h)-V(g)}{\epsilon}}\in \mathbb R$$

    Then ##<\delta_y,\frac{\delta V}{\delta g}> = \frac{\delta V}{\delta g(y)}## the value of the function ##\frac{\delta V}{\delta g(x)}## at x=y.
     
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