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)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



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

$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$










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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)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



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

$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$










share|cite|improve this question









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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)
$$J[f] = int [f(y)]^p phi{(y)} d{y}$$



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

$$frac{delta{J[f]}}{delta{f(x)}} = p[f(x)]^{p-1} phi(x) spacespacespace (1.12)$$










share|cite|improve this question









New contributor




SAK is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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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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 $J$.






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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
























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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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