Derivative over double summation?

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SAMUEL AYINDE 2019-1-12

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评论： Walter Roberson 2019-1-14

Derivative over double summation.xlsx

Please, can someone help me find the derivatives in this attachment? You can solve it in the attached excel document or solve it in word document and post the snap shot of the solution. Thank you so much.

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David Goodmanson 2019-1-13

Hi Samuel,

so by dP/dm do you mean a matrix whose ij component is dP/dm_ij?

Also the summand in each of the double sums defining P is identical, so the two double sums are almost identical, differing only in slight chages in the upper limits. Is this what is intended?

SAMUEL AYINDE 2019-1-13

Yes, dP/dm is a matrix whose ij component is dP/dm_ij. Yes, the double sums are almost identical. They differ only slightly in the upper limits. That exactly is what is intended.

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Answer 1

Walter Roberson 2019-1-13

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The derivative of absolute value is not well defined.

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Walter Roberson 2019-1-13

编辑：Walter Roberson 2019-1-13

在 MATLAB Online 中打开

Derivative implies that the m values are varying continuously. However, I recognize the structure of what is being computed, and the m values are not varying continuously, they are discretized because the inputs are intended to be pixel intensities.

If you want to treat the m entries as-if they are continuous, then you need to recognize that different m entries need to be considered independent variables, so dP/dm would need to be understood as a Jacobian rather than a single value.

m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j)

take the derivative with respect to the variable m(i,j) and you get

diff(m(i,j) * rho(i,j), m(i,j)) - diff(m(i+1,j) * rho(i+1,j), m(i,j))

which would be

rho(i,j) - diff(m(i+1,j), m(i,j)) * rho(i+1,j)

which would be

rho(i,j) - 0 * rho(i+1,j)

because m(i+1,j) is an independent variable compared to m(i,j)

and differentiating with respect to m(i+1,j) would give -rho(i+1,j) and differentiating with respect to any other m(p,q) would give 0. As a first pass then the double sum would involve sum() of a subset of rho minus sum of a slightly different range of rho, which would would revolve down to the difference of the two boundary conditions. (Hmmm, this might need to be revisited.)

But you are using abs() of the difference rather than the plain difference. The derivative of abs(a*b,a) is b*sign(a) provided that a is real and non-zero. In the case of abs(m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j)) then with m very likely representing pixel values in context, then unless rho is nonlinear and min(diff(unique(rho(:)) is greater than max(diff(unique(m(:)))) so that the rho "projects" the m values into separate spaces, then it is likely that m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j) will come out as 0 at some points, and then the derivative of the absolute value is undefined.

Under the assumption that is probably unjustifiable that m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j) can never be exactly 0, then the derivative of abs(m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j)) with respect to m(i,j) would be rho(i,j) * sign(m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j)) and the derivative with respect to m(i+1,j) would be -rho(i,j)*sign(m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j))

Hmmm, for any given K,L pair, m(K,L) appears once when i=K, j=L in the first term, giving rho(K,L)*sign(m(K,L) * rho(K,L) - m(K+1,L)*rho(K+1,L)), and appears once when i=K-1, j=L in the second term giving -rho(K,L)*sign(m(K-1,L) * rho(K-1,L) - m(K,L) * rho(K,L)) . The sum of those two would be the derivative with respect to m(K,L) except for the boundary condition. So that would be rho(K,L) * (sign(m(K,L) * rho(K,L) - m(K+1,L)*rho(K+1,L)) - sign(m(K-1,L) * rho(K-1,L) - m(K,L) *rho(K,L))) . This appears to be something that could be calculated in vectorized form except perhaps for the boundary conditions.

However, I would be surprised if m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j) can never equal 0, so I do not think the derivative of abs(m(i,j) * rho(i,j) - m(i+1,j) * rho(i+1,j)) would be well defined.

SAMUEL AYINDE 2019-1-14

编辑：SAMUEL AYINDE 2019-1-14

Hi Walter,

I can go with your approximation of absolute value. Now, I want to find dP/dm_ij in the equation below. Can you please help me? Thank you so much.

Walter Roberson 2019-1-14

The solution is as I gave before with the difference of sign() , but remember you need to be careful on the boundary condition .

The derivative is 0 except for at most two terms because the m variable are independent except the two cases of identical subscripts. So you can break it down to three cases, one for each of the two boundary conditions and one for the centre. You can then vectorize for the centre and vectorize for each of the two boundary conditions . or perhaps it works out to be four boundary conditions . Small number of them anyhow .

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Derivative over double summation?

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