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Kirill
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Define a sequence $(\mathbf{y})_{i=0}^N$ in $\mathbb{R}^n$ such that: $$\mathbf{y}_{k+1} = \mathbf{y}_{k} + \lambda \nabla_\mathbf{y} E(\mathbf{y}_k,\mathbf{w}), \quad k=0,1,\ldots,N-1,$$ where $\lambda$ is a constant, $\mathbf{w}\in\mathbb{R}^m$, and $E:\mathbb{R}^{n+m}\to \mathbb{R}$ is some differentiable function.

Let $Q:\mathbb{R}^{n}\to \mathbb{R}$ be a differentiable function and $L=Q(\mathbf{y} _N)$.

Applying the chain rule we have: $$\frac{dL}{d\mathbb{w}} = \sum_{k=1}^N\frac{\partial \mathbf{y}_k}{\partial \mathbf{w}} \frac{dQ}{d\mathbf{y}_k}\qquad (1)$$$$\frac{dL}{d\mathbf{w}} = \sum_{k=1}^N\frac{\partial \mathbf{y}_k}{\partial \mathbf{w}} \frac{dQ}{d\mathbf{y}_k}\qquad (1)$$ and $$\frac{dQ}{d\mathbf{y}_k} = \frac{\partial \mathbf{y}_{k+1}}{\partial \mathbf{y}_{k}} \frac{dQ}{d\mathbf{y}_{k+1}}.\qquad (2)$$

(Source: this paper, equation (12) and the one between (12) and (13).)

My questions: how to obtain $(1)$ and $(2)$?

I can show that if $(2)$ holds then $(1)$ holds. But I cannot see why $(2)$ holds. Consider $k=N-1$ for example:

Applying the chain rule we have: \begin{align}\frac{dQ}{d\mathbf{y}_{N-1}} &= \frac{dQ(\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1}))}{d\mathbf{y}_{N-1}} \\ &= \frac{d\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1})}{d\mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \begin{bmatrix}\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}} & \mathbf{I}\end{bmatrix}\begin{bmatrix}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}} \\ \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\end{bmatrix}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}} + \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}}, \end{align} which has an extra term $\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}}$ compared to $(2)$.

Thank you in advance for your help.

Define a sequence $(\mathbf{y})_{i=0}^N$ in $\mathbb{R}^n$ such that: $$\mathbf{y}_{k+1} = \mathbf{y}_{k} + \lambda \nabla_\mathbf{y} E(\mathbf{y}_k,\mathbf{w}), \quad k=0,1,\ldots,N-1,$$ where $\lambda$ is a constant, $\mathbf{w}\in\mathbb{R}^m$, and $E:\mathbb{R}^{n+m}\to \mathbb{R}$ is some differentiable function.

Let $Q:\mathbb{R}^{n}\to \mathbb{R}$ be a differentiable function and $L=Q(\mathbf{y} _N)$.

Applying the chain rule we have: $$\frac{dL}{d\mathbb{w}} = \sum_{k=1}^N\frac{\partial \mathbf{y}_k}{\partial \mathbf{w}} \frac{dQ}{d\mathbf{y}_k}\qquad (1)$$ and $$\frac{dQ}{d\mathbf{y}_k} = \frac{\partial \mathbf{y}_{k+1}}{\partial \mathbf{y}_{k}} \frac{dQ}{d\mathbf{y}_{k+1}}.\qquad (2)$$

(Source: this paper, equation (12) and the one between (12) and (13).)

My questions: how to obtain $(1)$ and $(2)$?

I can show that if $(2)$ holds then $(1)$ holds. But I cannot see why $(2)$ holds. Consider $k=N-1$ for example:

Applying the chain rule we have: \begin{align}\frac{dQ}{d\mathbf{y}_{N-1}} &= \frac{dQ(\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1}))}{d\mathbf{y}_{N-1}} \\ &= \frac{d\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1})}{d\mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \begin{bmatrix}\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}} & \mathbf{I}\end{bmatrix}\begin{bmatrix}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}} \\ \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\end{bmatrix}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}} + \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}}, \end{align} which has an extra term $\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}}$ compared to $(2)$.

Thank you in advance for your help.

Define a sequence $(\mathbf{y})_{i=0}^N$ in $\mathbb{R}^n$ such that: $$\mathbf{y}_{k+1} = \mathbf{y}_{k} + \lambda \nabla_\mathbf{y} E(\mathbf{y}_k,\mathbf{w}), \quad k=0,1,\ldots,N-1,$$ where $\lambda$ is a constant, $\mathbf{w}\in\mathbb{R}^m$, and $E:\mathbb{R}^{n+m}\to \mathbb{R}$ is some differentiable function.

Let $Q:\mathbb{R}^{n}\to \mathbb{R}$ be a differentiable function and $L=Q(\mathbf{y} _N)$.

Applying the chain rule we have: $$\frac{dL}{d\mathbf{w}} = \sum_{k=1}^N\frac{\partial \mathbf{y}_k}{\partial \mathbf{w}} \frac{dQ}{d\mathbf{y}_k}\qquad (1)$$ and $$\frac{dQ}{d\mathbf{y}_k} = \frac{\partial \mathbf{y}_{k+1}}{\partial \mathbf{y}_{k}} \frac{dQ}{d\mathbf{y}_{k+1}}.\qquad (2)$$

(Source: this paper, equation (12) and the one between (12) and (13).)

My questions: how to obtain $(1)$ and $(2)$?

I can show that if $(2)$ holds then $(1)$ holds. But I cannot see why $(2)$ holds. Consider $k=N-1$ for example:

Applying the chain rule we have: \begin{align}\frac{dQ}{d\mathbf{y}_{N-1}} &= \frac{dQ(\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1}))}{d\mathbf{y}_{N-1}} \\ &= \frac{d\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1})}{d\mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \begin{bmatrix}\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}} & \mathbf{I}\end{bmatrix}\begin{bmatrix}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}} \\ \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\end{bmatrix}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}} + \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}}, \end{align} which has an extra term $\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}}$ compared to $(2)$.

Thank you in advance for your help.

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f10w
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Derivative chain rule

Define a sequence $(\mathbf{y})_{i=0}^N$ in $\mathbb{R}^n$ such that: $$\mathbf{y}_{k+1} = \mathbf{y}_{k} + \lambda \nabla_\mathbf{y} E(\mathbf{y}_k,\mathbf{w}), \quad k=0,1,\ldots,N-1,$$ where $\lambda$ is a constant, $\mathbf{w}\in\mathbb{R}^m$, and $E:\mathbb{R}^{n+m}\to \mathbb{R}$ is some differentiable function.

Let $Q:\mathbb{R}^{n}\to \mathbb{R}$ be a differentiable function and $L=Q(\mathbf{y} _N)$.

Applying the chain rule we have: $$\frac{dL}{d\mathbb{w}} = \sum_{k=1}^N\frac{\partial \mathbf{y}_k}{\partial \mathbf{w}} \frac{dQ}{d\mathbf{y}_k}\qquad (1)$$ and $$\frac{dQ}{d\mathbf{y}_k} = \frac{\partial \mathbf{y}_{k+1}}{\partial \mathbf{y}_{k}} \frac{dQ}{d\mathbf{y}_{k+1}}.\qquad (2)$$

(Source: this paper, equation (12) and the one between (12) and (13).)

My questions: how to obtain $(1)$ and $(2)$?

I can show that if $(2)$ holds then $(1)$ holds. But I cannot see why $(2)$ holds. Consider $k=N-1$ for example:

Applying the chain rule we have: \begin{align}\frac{dQ}{d\mathbf{y}_{N-1}} &= \frac{dQ(\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1}))}{d\mathbf{y}_{N-1}} \\ &= \frac{d\mathbf{y}_{N}(\mathbf{w},\mathbf{y}_{N-1})}{d\mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \begin{bmatrix}\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}} & \mathbf{I}\end{bmatrix}\begin{bmatrix}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}} \\ \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\end{bmatrix}\frac{dQ}{d\mathbf{y}_{N}} \\ &= \frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}} + \frac{\partial \mathbf{y}_{N}}{\partial \mathbf{y}_{N-1}}\frac{dQ}{d\mathbf{y}_{N}}, \end{align} which has an extra term $\frac{d\mathbf{w}}{d\mathbf{y}_{N-1}}\frac{\partial \mathbf{y}_{N}}{\partial \mathbf{w}}\frac{dQ}{d\mathbf{y}_{N}}$ compared to $(2)$.

Thank you in advance for your help.