N dimensional Numeric integral
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I am trying to evaluate a N dimensional integral in MATLAB, is has a special form as following, does the special form helps me to evaluate my integral faster? simpler?
begin{equation}
int_{-infty}^{infty}... int_{-infty}^{infty} g(boldsymbol{x}) F(boldsymbol{x}) dboldsymbol{x} = int_{-infty}^{infty}... int_{-infty}^{infty} g(||x||^2,sum_{i=1}^N x_i) f(x_1)...f(x_n) dx_1 ...dx_N
end{equation}
where $boldsymbol{x}=[x_1,x_2,...x_N]^T$, and $||.||^2$ is norm of the vector.
integration definite-integrals numerical-methods
asked Nov 23 '16 at 22:33
Alireza
1939
add a comment |
up vote
0
down vote
favorite
I am trying to evaluate a N dimensional integral in MATLAB, is has a special form as following, does the special form helps me to evaluate my integral faster? simpler?
begin{equation}
int_{-infty}^{infty}... int_{-infty}^{infty} g(boldsymbol{x}) F(boldsymbol{x}) dboldsymbol{x} = int_{-infty}^{infty}... int_{-infty}^{infty} g(||x||^2,sum_{i=1}^N x_i) f(x_1)...f(x_n) dx_1 ...dx_N
end{equation}
where $boldsymbol{x}=[x_1,x_2,...x_N]^T$, and $||.||^2$ is norm of the vector.
integration definite-integrals numerical-methods
asked Nov 23 '16 at 22:33
Alireza
1939
Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to evaluate a N dimensional integral in MATLAB, is has a special form as following, does the special form helps me to evaluate my integral faster? simpler?
begin{equation}
int_{-infty}^{infty}... int_{-infty}^{infty} g(boldsymbol{x}) F(boldsymbol{x}) dboldsymbol{x} = int_{-infty}^{infty}... int_{-infty}^{infty} g(||x||^2,sum_{i=1}^N x_i) f(x_1)...f(x_n) dx_1 ...dx_N
end{equation}
where $boldsymbol{x}=[x_1,x_2,...x_N]^T$, and $||.||^2$ is norm of the vector.
integration definite-integrals numerical-methods
asked Nov 23 '16 at 22:33
Alireza
1939
I am trying to evaluate a N dimensional integral in MATLAB, is has a special form as following, does the special form helps me to evaluate my integral faster? simpler?
begin{equation}
int_{-infty}^{infty}... int_{-infty}^{infty} g(boldsymbol{x}) F(boldsymbol{x}) dboldsymbol{x} = int_{-infty}^{infty}... int_{-infty}^{infty} g(||x||^2,sum_{i=1}^N x_i) f(x_1)...f(x_n) dx_1 ...dx_N
end{equation}
where $boldsymbol{x}=[x_1,x_2,...x_N]^T$, and $||.||^2$ is norm of the vector.
integration definite-integrals numerical-methods
integration definite-integrals numerical-methods
asked Nov 23 '16 at 22:33
Alireza
1939
asked Nov 23 '16 at 22:33
Alireza
1939
asked Nov 23 '16 at 22:33
Alireza
1939
asked Nov 23 '16 at 22:33
Alireza
1939
asked Nov 23 '16 at 22:33
Alireza
1939
1939
Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52
add a comment |
Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52
Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52
add a comment |
1 Answer
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Maybe try writing as a Fourier integral:
$$g(||x||^2,sum x)=int dp dq G(p,q)e^{ip||x||^2+iqsum x}$$
then
$$g(||x||^2,sum x)f(x_1)cdots f(x_n)=int dp dqG(p,q)prod_j e^{ip x_j^2+iq x_j}f(x_j)$$
so that if you can evaluate numerically or otherwise
$F(p,q)=int dx e^{ip x^2+iq x}f(x) $
you can write the total integral as
$$int dp dqG(p,q)F(p,q)^N$$
answered Nov 22 at 13:42
user617446
261
add a comment |
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1 Answer
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1 Answer
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active
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up vote
0
down vote
Maybe try writing as a Fourier integral:
$$g(||x||^2,sum x)=int dp dq G(p,q)e^{ip||x||^2+iqsum x}$$
then
$$g(||x||^2,sum x)f(x_1)cdots f(x_n)=int dp dqG(p,q)prod_j e^{ip x_j^2+iq x_j}f(x_j)$$
so that if you can evaluate numerically or otherwise
$F(p,q)=int dx e^{ip x^2+iq x}f(x) $
you can write the total integral as
$$int dp dqG(p,q)F(p,q)^N$$
answered Nov 22 at 13:42
user617446
261
add a comment |
up vote
0
down vote
Maybe try writing as a Fourier integral:
$$g(||x||^2,sum x)=int dp dq G(p,q)e^{ip||x||^2+iqsum x}$$
then
$$g(||x||^2,sum x)f(x_1)cdots f(x_n)=int dp dqG(p,q)prod_j e^{ip x_j^2+iq x_j}f(x_j)$$
so that if you can evaluate numerically or otherwise
$F(p,q)=int dx e^{ip x^2+iq x}f(x) $
you can write the total integral as
$$int dp dqG(p,q)F(p,q)^N$$
answered Nov 22 at 13:42
user617446
261
add a comment |
up vote
0
down vote
up vote
0
down vote
Maybe try writing as a Fourier integral:
$$g(||x||^2,sum x)=int dp dq G(p,q)e^{ip||x||^2+iqsum x}$$
then
$$g(||x||^2,sum x)f(x_1)cdots f(x_n)=int dp dqG(p,q)prod_j e^{ip x_j^2+iq x_j}f(x_j)$$
so that if you can evaluate numerically or otherwise
$F(p,q)=int dx e^{ip x^2+iq x}f(x) $
you can write the total integral as
$$int dp dqG(p,q)F(p,q)^N$$
answered Nov 22 at 13:42
user617446
261
Maybe try writing as a Fourier integral:
$$g(||x||^2,sum x)=int dp dq G(p,q)e^{ip||x||^2+iqsum x}$$
then
$$g(||x||^2,sum x)f(x_1)cdots f(x_n)=int dp dqG(p,q)prod_j e^{ip x_j^2+iq x_j}f(x_j)$$
so that if you can evaluate numerically or otherwise
$F(p,q)=int dx e^{ip x^2+iq x}f(x) $
you can write the total integral as
$$int dp dqG(p,q)F(p,q)^N$$
answered Nov 22 at 13:42
user617446
261
answered Nov 22 at 13:42
user617446
261
answered Nov 22 at 13:42
user617446
261
answered Nov 22 at 13:42
user617446
261
261
add a comment |
add a comment |
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Have you looked into whether the Jacobian of the function $f(x)=(| x |^2,sum_{i=1}^N x_i,x_3,x_4,dots,x_N)$ is too complicated to be practically useful?
– Ian
Nov 23 '16 at 22:46
@lan very very complicated
– Alireza
Nov 23 '16 at 22:51
Is it really though? It's the determinant of a matrix where the first row is $a_{ij}=2x_j$, the second row is $a_{ij}$ all equal to $1$, and the other rows are just diagonal with a diagonal entry of $1$. Is it that hard to get the determinant of such a matrix? It seems to me that you could just cofactor expand across the last row a bunch of times...
– Ian
Nov 24 '16 at 0:56
@lan unfortunately I don't get your point. You are not considering function $g(.)$ which is a function of $||x||^2$ and $sum x_i$ . And I don't get your definition of $f(.)$ !
– Alireza
Nov 24 '16 at 21:37
My preliminary suggestion was to replace two of your variables, one with the sum of the squares and the other with the sum. Then g would only depend on two of your variables, and the rest of the integration would (hopefully) simplify through not depending on those two variables. But I missed that your domain of integration after this substitution would be complicated. So it was a bad idea.
– Ian
Nov 24 '16 at 21:52