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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    $begingroup$
    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.
    $endgroup$
    – KReiser
    Dec 25 '18 at 7:17










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    Thanks for the information.
    $endgroup$
    – SAK
    Dec 25 '18 at 7:32
















1












$begingroup$


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











$endgroup$








  • 1




    $begingroup$
    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.
    $endgroup$
    – KReiser
    Dec 25 '18 at 7:17










  • $begingroup$
    Thanks for the information.
    $endgroup$
    – SAK
    Dec 25 '18 at 7:32














1












1








1





$begingroup$


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











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 25 '18 at 7:32







SAK

















asked Dec 25 '18 at 7:14









SAKSAK

83




83








  • 1




    $begingroup$
    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.
    $endgroup$
    – KReiser
    Dec 25 '18 at 7:17










  • $begingroup$
    Thanks for the information.
    $endgroup$
    – SAK
    Dec 25 '18 at 7:32














  • 1




    $begingroup$
    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.
    $endgroup$
    – KReiser
    Dec 25 '18 at 7:17










  • $begingroup$
    Thanks for the information.
    $endgroup$
    – SAK
    Dec 25 '18 at 7:32








1




1




$begingroup$
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.
$endgroup$
– KReiser
Dec 25 '18 at 7:17




$begingroup$
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.
$endgroup$
– KReiser
Dec 25 '18 at 7:17












$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32




$begingroup$
Thanks for the information.
$endgroup$
– SAK
Dec 25 '18 at 7:32










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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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    $begingroup$

    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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      0












      $begingroup$

      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









      $endgroup$
















        0












        0








        0





        $begingroup$

        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









        $endgroup$



        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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 25 '18 at 12:05









        mechanodroidmechanodroid

        27.5k62447




        27.5k62447






























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