Functional Derivative for Specific Question
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
New contributor
add a comment |
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
New contributor
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
add a comment |
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
New contributor
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
New contributor
New contributor
edited 5 hours ago
New contributor
asked 5 hours ago
SAK
83
83
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New contributor
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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.
– KReiser
5 hours ago
Thanks for the information.
– SAK
5 hours ago
add a comment |
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
add a comment |
1 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 $J$.
add a comment |
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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 $J$.
add a comment |
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 $J$.
add a comment |
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 $J$.
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 $J$.
answered 1 hour ago
mechanodroid
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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