Is there a version of the chain rule that applies to hessian matrices?
$begingroup$
Suppose I have a scalar $J(n)$ and two vectors, $mathbf{w}(n)$ and $mathbf{x}(n)$. Now, suppose that $J(n)$ is a fairly straightforward function of $mathbf{w}(n)$, and $mathbf{w}(n)$ is actually a vector function of $mathbf{x}(n)$ (and a complicated one, for that matter, although $frac{partial mathbf{w}(n)}{partial mathbf{x}(n)}$ is relatively easy to calculate). In order to obtain the gradient of $J(n)$ with respect to $mathbf{x}(n)$, I used
$$ frac{partial J(n)}{partial mathbf{x}(n)} = frac{partial mathbf{w}(n)}{partial mathbf{x}(n)} frac{partial J(n)}{partial mathbf{w}(n)}.$$
Now, I'd like to calculate the Hessian of $J(n)$ with respect to $mathbf{x}(n)$. The thing is, it seems to me it would be a lot easier to calculate $frac{partial^2 J(n)}{partial mathbf{w}^2(n)}$ first and then use the chain rule to obtain $frac{partial^2 J(n)}{partial mathbf{x}^2(n)}$ from the previous result. The problem is that I'm having some difficulty with it. I know that, in single variable calculus, if we have $y = f(u)$ and $u = g(x)$, then
$$frac{d^2y}{dx^2} = frac{d^2y}{du^2}bigg(frac{du}{dx}bigg)^2 + frac{dy}{du} bigg( frac{d^2u}{dx^2} bigg),$$
but how does that apply to matrix calculus? I mean, $frac{partial^2 mathbf{w}(n)}{partial mathbf{x}^2(n)}$ is a tensor, right? How do I deal with this situation? Is it even posible to do what I am thinking?
multivariable-calculus matrix-calculus chain-rule
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
$endgroup$
add a comment |
$begingroup$
Suppose I have a scalar $J(n)$ and two vectors, $mathbf{w}(n)$ and $mathbf{x}(n)$. Now, suppose that $J(n)$ is a fairly straightforward function of $mathbf{w}(n)$, and $mathbf{w}(n)$ is actually a vector function of $mathbf{x}(n)$ (and a complicated one, for that matter, although $frac{partial mathbf{w}(n)}{partial mathbf{x}(n)}$ is relatively easy to calculate). In order to obtain the gradient of $J(n)$ with respect to $mathbf{x}(n)$, I used
$$ frac{partial J(n)}{partial mathbf{x}(n)} = frac{partial mathbf{w}(n)}{partial mathbf{x}(n)} frac{partial J(n)}{partial mathbf{w}(n)}.$$
Now, I'd like to calculate the Hessian of $J(n)$ with respect to $mathbf{x}(n)$. The thing is, it seems to me it would be a lot easier to calculate $frac{partial^2 J(n)}{partial mathbf{w}^2(n)}$ first and then use the chain rule to obtain $frac{partial^2 J(n)}{partial mathbf{x}^2(n)}$ from the previous result. The problem is that I'm having some difficulty with it. I know that, in single variable calculus, if we have $y = f(u)$ and $u = g(x)$, then
$$frac{d^2y}{dx^2} = frac{d^2y}{du^2}bigg(frac{du}{dx}bigg)^2 + frac{dy}{du} bigg( frac{d^2u}{dx^2} bigg),$$
but how does that apply to matrix calculus? I mean, $frac{partial^2 mathbf{w}(n)}{partial mathbf{x}^2(n)}$ is a tensor, right? How do I deal with this situation? Is it even posible to do what I am thinking?
multivariable-calculus matrix-calculus chain-rule
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
$endgroup$
$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55
add a comment |
$begingroup$
Suppose I have a scalar $J(n)$ and two vectors, $mathbf{w}(n)$ and $mathbf{x}(n)$. Now, suppose that $J(n)$ is a fairly straightforward function of $mathbf{w}(n)$, and $mathbf{w}(n)$ is actually a vector function of $mathbf{x}(n)$ (and a complicated one, for that matter, although $frac{partial mathbf{w}(n)}{partial mathbf{x}(n)}$ is relatively easy to calculate). In order to obtain the gradient of $J(n)$ with respect to $mathbf{x}(n)$, I used
$$ frac{partial J(n)}{partial mathbf{x}(n)} = frac{partial mathbf{w}(n)}{partial mathbf{x}(n)} frac{partial J(n)}{partial mathbf{w}(n)}.$$
Now, I'd like to calculate the Hessian of $J(n)$ with respect to $mathbf{x}(n)$. The thing is, it seems to me it would be a lot easier to calculate $frac{partial^2 J(n)}{partial mathbf{w}^2(n)}$ first and then use the chain rule to obtain $frac{partial^2 J(n)}{partial mathbf{x}^2(n)}$ from the previous result. The problem is that I'm having some difficulty with it. I know that, in single variable calculus, if we have $y = f(u)$ and $u = g(x)$, then
$$frac{d^2y}{dx^2} = frac{d^2y}{du^2}bigg(frac{du}{dx}bigg)^2 + frac{dy}{du} bigg( frac{d^2u}{dx^2} bigg),$$
but how does that apply to matrix calculus? I mean, $frac{partial^2 mathbf{w}(n)}{partial mathbf{x}^2(n)}$ is a tensor, right? How do I deal with this situation? Is it even posible to do what I am thinking?
multivariable-calculus matrix-calculus chain-rule
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
$endgroup$
Suppose I have a scalar $J(n)$ and two vectors, $mathbf{w}(n)$ and $mathbf{x}(n)$. Now, suppose that $J(n)$ is a fairly straightforward function of $mathbf{w}(n)$, and $mathbf{w}(n)$ is actually a vector function of $mathbf{x}(n)$ (and a complicated one, for that matter, although $frac{partial mathbf{w}(n)}{partial mathbf{x}(n)}$ is relatively easy to calculate). In order to obtain the gradient of $J(n)$ with respect to $mathbf{x}(n)$, I used
$$ frac{partial J(n)}{partial mathbf{x}(n)} = frac{partial mathbf{w}(n)}{partial mathbf{x}(n)} frac{partial J(n)}{partial mathbf{w}(n)}.$$
Now, I'd like to calculate the Hessian of $J(n)$ with respect to $mathbf{x}(n)$. The thing is, it seems to me it would be a lot easier to calculate $frac{partial^2 J(n)}{partial mathbf{w}^2(n)}$ first and then use the chain rule to obtain $frac{partial^2 J(n)}{partial mathbf{x}^2(n)}$ from the previous result. The problem is that I'm having some difficulty with it. I know that, in single variable calculus, if we have $y = f(u)$ and $u = g(x)$, then
$$frac{d^2y}{dx^2} = frac{d^2y}{du^2}bigg(frac{du}{dx}bigg)^2 + frac{dy}{du} bigg( frac{d^2u}{dx^2} bigg),$$
but how does that apply to matrix calculus? I mean, $frac{partial^2 mathbf{w}(n)}{partial mathbf{x}^2(n)}$ is a tensor, right? How do I deal with this situation? Is it even posible to do what I am thinking?
multivariable-calculus matrix-calculus chain-rule
multivariable-calculus matrix-calculus chain-rule
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
asked Dec 5 '18 at 15:16
D. TigleaD. Tiglea
61
61
$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55
add a comment |
$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55
$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55
add a comment |
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$begingroup$
It's an interesting question, and I suspect your approach using single variable calculus is the best to try to work out a matrix formulation in the multivariable case. Perhaps the case of two variables will show if there is a tractable expression to be found.
$endgroup$
– hardmath
Dec 5 '18 at 15:32
$begingroup$
Given that it is a tensor, you just have to contract over the correct pairs of indeces
$endgroup$
– Federico
Dec 5 '18 at 15:33
$begingroup$
And how exactly do I do that? I'm new to tensors and tensor calculus, so if you could recommend a good reference on the subject, I would appreciate it quite a lot.
$endgroup$
– D. Tiglea
Dec 5 '18 at 16:55