Functional Derivative for Specific Question












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Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$






functional-analysis quantum-field-theory functional-calculus







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    It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
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Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$






functional-analysis quantum-field-theory functional-calculus







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  • 1




    It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
    – KReiser
    5 hours ago










  • Thanks for the information.
    – SAK
    5 hours ago














1












1








1







Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$






functional-analysis quantum-field-theory functional-calculus







share|cite|improve this question









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SAK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Can you help me understanding how author got to equation 1.12, and what is phi(X)function. (https://i.stack.imgur.com/16LOQ.jpg)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



$$frac{delta{J[f]}}{delta{f(x)}} = lim_{epsilonto0}frac{[int [f(y) + epsilondelta{(y-x)}]^p phi{(y)}d{y} - int[f(y)]^p phi{(y)}dy ]}{epsilon}
$$
$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$






functional-analysis quantum-field-theory functional-calculus




functional-analysis quantum-field-theory functional-calculus






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edited 5 hours ago





















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  • 1




    It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
    – KReiser
    5 hours ago










  • Thanks for the information.
    – SAK
    5 hours ago














  • 1




    It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
    – KReiser
    5 hours ago










  • Thanks for the information.
    – SAK
    5 hours ago








1




1




It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
– KReiser
5 hours ago




It is customary on Math.Stackexchange to make questions as self-contained as possible. Please write the equations in the body of your question using MathJax.
– KReiser
5 hours ago












Thanks for the information.
– SAK
5 hours ago




Thanks for the information.
– SAK
5 hours ago










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Use the generalized binomial theorem:begin{align}
frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
&= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
&= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
&= p[f(x)]^{p-1}phi(x)
end{align}



$phi$ is just an arbitrary function used to define the functional $J$.






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    Use the generalized binomial theorem:begin{align}
    frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
    &= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
    &= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
    &= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
    &= p[f(x)]^{p-1}phi(x)
    end{align}



    $phi$ is just an arbitrary function used to define the functional $J$.






    share|cite|improve this answer


























      0














      Use the generalized binomial theorem:begin{align}
      frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
      &= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
      &= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
      &= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
      &= p[f(x)]^{p-1}phi(x)
      end{align}



      $phi$ is just an arbitrary function used to define the functional $J$.






      share|cite|improve this answer
























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        Use the generalized binomial theorem:begin{align}
        frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
        &= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
        &= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
        &= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
        &= p[f(x)]^{p-1}phi(x)
        end{align}



        $phi$ is just an arbitrary function used to define the functional $J$.






        share|cite|improve this answer












        Use the generalized binomial theorem:begin{align}
        frac{delta{J[f]}}{delta{f(x)}} &= lim_{epsilonto0}frac1varepsilonleft(int [f(y) + epsilondelta{(y-x)}]^p phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
        &= lim_{epsilonto0}frac1varepsilonleft(int sum_{k=0}^infty{p choose k} [varepsilon delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dy - int[f(y)]^p phi{(y)},dyright)\
        &= lim_{epsilonto0}left[pint delta(y-x)[f(y)]^{p-1}phi(y),dy + sum_{k=2}^infty{p choose k} varepsilon^{k-1} int [delta(y-x)]^k [f(y)]^{p-k} phi{(y)},dyright]\
        &= pint delta(y-x)[f(y)]^{p-1}phi(y),dy\
        &= p[f(x)]^{p-1}phi(x)
        end{align}



        $phi$ is just an arbitrary function used to define the functional $J$.







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        answered 1 hour ago









        mechanodroid

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