Mixing
Integration over two functions
$begingroup$
I need to integrate a function of the form:
$H(t) = int_0^t f(u)g(t-u) du$
The functions are fixed, but aren't simple and so I'm using numerical methods to calculate the value of $H$. However, I need to evaluate the function at a lot of values of $t$. Is there a way to find a function, $P$ such that:
$H(t+1) = P(H(t))$
This would make my computation much easier.
In other words, can I derive the value of $H(t+1)$ from $H(t)$ without having to recalculate the entire integral from $0$ to $t$. An integral identity would be perfect.
Note: $g$ is a decreasing function
integration functions definite-integrals
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
$endgroup$
add a comment |
$begingroup$
I need to integrate a function of the form:
$H(t) = int_0^t f(u)g(t-u) du$
The functions are fixed, but aren't simple and so I'm using numerical methods to calculate the value of $H$. However, I need to evaluate the function at a lot of values of $t$. Is there a way to find a function, $P$ such that:
$H(t+1) = P(H(t))$
This would make my computation much easier.
In other words, can I derive the value of $H(t+1)$ from $H(t)$ without having to recalculate the entire integral from $0$ to $t$. An integral identity would be perfect.
Note: $g$ is a decreasing function
integration functions definite-integrals
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
$endgroup$
$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54
add a comment |
$begingroup$
I need to integrate a function of the form:
$H(t) = int_0^t f(u)g(t-u) du$
The functions are fixed, but aren't simple and so I'm using numerical methods to calculate the value of $H$. However, I need to evaluate the function at a lot of values of $t$. Is there a way to find a function, $P$ such that:
$H(t+1) = P(H(t))$
This would make my computation much easier.
In other words, can I derive the value of $H(t+1)$ from $H(t)$ without having to recalculate the entire integral from $0$ to $t$. An integral identity would be perfect.
Note: $g$ is a decreasing function
integration functions definite-integrals
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
$endgroup$
I need to integrate a function of the form:
$H(t) = int_0^t f(u)g(t-u) du$
The functions are fixed, but aren't simple and so I'm using numerical methods to calculate the value of $H$. However, I need to evaluate the function at a lot of values of $t$. Is there a way to find a function, $P$ such that:
$H(t+1) = P(H(t))$
This would make my computation much easier.
In other words, can I derive the value of $H(t+1)$ from $H(t)$ without having to recalculate the entire integral from $0$ to $t$. An integral identity would be perfect.
Note: $g$ is a decreasing function
integration functions definite-integrals
integration functions definite-integrals
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
asked Jan 17 at 11:52
Michael BarrowmanMichael Barrowman
11
11
$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54
add a comment |
$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54
$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54
add a comment |
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$begingroup$
You may be able to use Convolution theorem for Fourier transform or Laplace transform
$endgroup$
– Somos
Jan 17 at 12:22
$begingroup$
This is exactly what I was looking for. I knew the formula looked familiar, but couldn't put my finger on the fact that it's a convolution. It's not quite the exact solution, but it's definitely pointed me in the right direction. Thank you!
$endgroup$
– Michael Barrowman
Jan 17 at 19:54