Derivative over double summation?

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.

[David Goodmanson]

Hi Samuel,

Since m_ij is an object with two indices, what is the meaning of m_k in the expression dP/dm_k?

[SAMUEL AYINDE]

Hi David,

Thank you so much. I have seen my mistake. I think it is just supposed to be dP/dm. I just need the gradient of P with respect to m and the gradient of P with respect to rho, dP/drho. I will appreciate your help. Also, eventually, after I get the derivative, I want to write a matlab code which will be updating P with respect to the derived gradients. Thank you so much.

[David Goodmanson]

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]

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.

Answer 1

[Walter Roberson]

The derivative of absolute value is not well defined.

[Walter Roberson]

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.

[Walter Roberson]

I see your revision that gives an approximation for absolute value, but the approximation you post is not differentiable at 0.

diff(sqrt((1 + 2*e)*x^2) - e, x);

(1 + 2 e) x

-------------------

(1/2)

/ 2\

\(1 + 2 e) x /

limit(%, x = 0);

undefined

[SAMUEL AYINDE]

Hi Walter,

Thank you so much for this input. I have included an approximation for the absolute value, in order to make it differentiable. Can you please help me now. I will appreciate it if you can solve it and attach it as a picture format. Thank you so much.

[SAMUEL AYINDE]

Hi Walter,

I want to compute dP/dm_i and dP/drho_i. What do you think about the approximation for the absolute value? Is the differentiation continuous in this case if I am just computing dP/dm_i and dP/drho_i?

[Walter Roberson]

"Picture or it didn't happen" ?

Your absolute value approximation is not useful.

Perhaps you mean sqrt(x^2 + 2*e) - e ? The limit of the derivative of that as x approaches 0 is 0. The value of the function itself as x approaches 0 is sqrt(e)*sqrt(2).

[Walter Roberson]

dP/dm_i is 0 because m_i does not exist in the equations. You have m_ij but not m_i .

If you are trying to imply derivative with respect to an entire row of m, then that is still with respect to a vector of variables rather than a single variable.

The analysis I gave above reached

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)))

sign(0) is defined as 0, same as the approximation reached by sqrt(x^2 + 2*e) - e in the limit of e approaches 0, so you could substitute sign() provided that if you write this up as a formal analysis that you demonstrate that you considered the differentiability and substituted sqrt(x^2 + 2*e) - e and then took the limit as e approaches 0 get sign().

[SAMUEL AYINDE]

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]

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 .