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I have the following equation:

$$ f = \mathbf{m^{T}Bm} $$

where B is:

$$ \mathbf{B} = \frac{1}{3(m_{x}^{2} + m_{y}^{2})}\begin{pmatrix} 1 & 0.5 & 1\\ 0.5 & 1 & 0.5\\ 1 & 1 & 0.5 \end{pmatrix} $$

and m is:

$$ \mathbf{m} = \begin{pmatrix} m_{x}\\ m_{y}\\ m_{z} \end{pmatrix} $$

also there is another equation:

$$ \mathbf{m} = a\mathbf{x} + b\mathbf{y} + c\mathbf{z} $$

UPDATE:

I extract the equation from a more complicated one but here

$m_x = a$, $m_y = b$ and $m_z = c$

My question is How can I calculate the derivative of f with respect to a ? Thanks very much.

Johnnylin
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  • Please clarify the equation $\mathbf{m} = ax + by + cz$ given that $\mathbf{m}$ is a vector. – mlc Mar 21 '17 at 18:47
  • @mlc I have updated the post. $\mathbf{m}$ is a vector. – Johnnylin Mar 21 '17 at 18:55
  • The last equation is still confusing to me even after the edit. Do you mean is x supposed to represent the unit vector in the x direction, and likewise for y and z? That would make the most sense to me. But in this case $m=(a,b,c)$, and so the middle equation with $m_x$, etc is irrelevant. – Nick Alger Mar 21 '17 at 19:26
  • I am still confused: $\mathbf{m} = ax+by+cz$ has a vector on the LHS; what is on the RHS? – mlc Mar 21 '17 at 19:35
  • @NickAlger I have updated the post. you are correct. – Johnnylin Mar 22 '17 at 01:17
  • @mlc I have updated it. – Johnnylin Mar 22 '17 at 01:18

1 Answers1

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Let $$\eqalign{ P &= \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 1 \\ 2 & 2 & 1 \end{bmatrix},\,\, &Q = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 0 \end{bmatrix} \cr\cr m &= [\,a\,b\,c\,]^T \cr M &= mm^T,\,\,\, &f = \frac{P:M}{Q:M} \cr }$$ where $\,:\,$ represents the inner/Frobenius product, i.e. $A:B=\operatorname{tr}(A^TB)$

Find the differential of the function, then the gradient $$\eqalign{ df &= \frac{(Q:M)\,P:dM-(P:M)\,Q:dM}{(Q:M)^2} \cr &= \Bigg(\frac{(Q:M)\,P-(P:M)\,Q}{(Q:M)^2}\Bigg):dM \cr &= R:dM \cr &= R:(dm\,m^T+m\,dm^T) \cr &= R:dm\,m^T+R:m\,dm^T \cr &= R:dm\,m^T+R^T:dm\,m^T \cr &= (R+R^T)\,m:dm \cr \cr g=\frac{\partial f}{\partial m} &= (R+R^T)\,m \cr \cr }$$ The derivative wrt $a$ is the first component of the gradient $$\eqalign{\frac{\partial f}{\partial a} &= g_x \cr\cr\cr}$$ Update

The gradient of the gradient can also be found. To make that calculation easier, let's use symmetric matrices $$\eqalign{ P &= \begin{bmatrix} 4 & 2 & 4 \\ 2 & 4 & 3 \\ 4 & 3 & 2 \end{bmatrix},\,\, &Q = \begin{bmatrix} 12 & 0 & 0 \\ 0 & 12 & 0 \\ 0 & 0 & 0 \end{bmatrix} }$$ This won't change the function value because the skew parts are annihilated by the inner product with the symmetric matrix $M$. This change makes $R$ symmetric, too.

Let's also define some scalar coefficients to simplify the expression for $R$ $$\eqalign{ R &= \alpha P-\beta Q,\,\,\,&dR&= P\,d\alpha-Q\,d\beta \cr \alpha&=(Q:M)^{-1},\,\,\,&d\alpha&=-\alpha^2(Q:dM)=-2\alpha^2\,m^TQ\,dm \cr \beta &= \alpha^2(P:M) = \alpha f,\,\,\,&d\beta&=\alpha\,df + f\,d\alpha \cr &\,&\,&= \alpha g^T\,dm - 2f\alpha^2\,m^TQ\,dm \cr }$$ Now find the differential and gradient of $g$ $$\eqalign{ g &= 2Rm \cr\cr dg &= 2\,(R\,dm + dR\,m) \cr &= 2\,(R\,dm + Pm\,d\alpha - Qm\,d\beta) \cr &= 2\,(R\,dm + Pm\,d\alpha - Qmf\,d\alpha - Qm\alpha\,df) \cr &= 2\,(R\,dm + (Pm-fQm)\,d\alpha - \alpha Qmg^T\,dm) \cr &= 2\,(R\,dm - Rm\,\frac{d\alpha}{\alpha}- \alpha Qmg^T\,dm) \cr &= 2\,(R\,dm - Rm(2\alpha\,m^TQ\,dm) - 2\alpha Qmm^TR\,dm) \cr &= 2\,\big(R -2\alpha Rmm^TQ-2\alpha Qmm^TR\big)\,dm \cr &= 2\,\big(R -2\alpha RMQ -2\alpha QMR\big)\,dm \cr\cr \frac{\partial g}{\partial m} &= 2R - 4\alpha(RMQ+QMR) \cr\cr }$$ So that's the second derivative, but please double-check the algebra before trusting it.

As a quick check, $(Q,M,R)$ are all symmetric matrices, therefore the second derivative is also a symmetric matrix -- as it should be.

greg
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