Help needed with partial differentiation equation of matrices

[mikeraj]

With reference to Eq. 16 in the attachment, I need help in understanding how the working is done to obtain the answers in Eq. 17.

Eqs. 10, 11, and 15 are the inputs needed to solve Eg. 16.

I am puzzled how a matrix could be differentiated with respect to another matrix. I would highly appreciate it if an example calculation can be shown. Thanks in advance!

Equations.pdf

[zztop]

With reference to Eq. 16 in the attachment, I need help in understanding how the working is done to obtain the answers in Eq. 17.

 

Eqs. 10, 11, and 15 are the inputs needed to solve Eg. 16.

 

I am puzzled how a matrix could be differentiated with respect to another matrix. I would highly appreciate it if an example calculation can be shown. Thanks in advance!

It is not "differentiated", formula (16) is just a summation.

[mikeraj]

My problem is not with the summation but on the partial differentiation in Eq 16

[zztop]

My problem is not with the summation but on the partial differentiation in Eq 16

What you need is the dfinition of the derivative of a vector wrt a vector. The result is a n x n matrix, see here.

Edited February 28, 2017 by zztop

[studiot]

What you need is the dfinition of the derivative of a vector wrt a vector. The result is a n x n matrix, see here.

 

 

Looks like a jacobian matrix to me.

 

Look here, particularly the examples.

 

https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

[zztop]

 

 

Looks like a jacobian matrix to me.

 

Look here, particularly the examples.

 

https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant

Yes, it does, except the elements are obtained thru trivial divisions.

[mikeraj]

Could someone show me an example calculation of the Jacobian matrix in Eq. 16?

 

I just need to understand the calculation steps. I highly appreciate it. Thanks!

[zztop]

Could someone show me an example calculation of the Jacobian matrix in Eq. 16?

 

I just need to understand the calculation steps. I highly appreciate it. Thanks!

First line of the matrix is: [latex]\frac{v_1(1)}{v_{1st}(1)}, \frac{v_1(1)}{v_{1st}(2)},\frac{v_1(1)}{v_{1st}(3)}[/latex]

 

The second line is: [latex]\frac{v_1(2)}{v_{1st}(1)}, \frac{v_1(2)}{v_{1st}(2)},\frac{v_1(2)}{v_{1st}(3)}[/latex]

 

The third line is: [latex]\frac{v_1(3)}{v_{1st}(1)}, \frac{v_1(3)}{v_{1st}(2)},\frac{v_1(3)}{v_{1st}(3)}[/latex]

Edited March 1, 2017 by zztop

[Bignose]

First line of the matrix is: [latex]\frac{v_1(1)}{v_{1st}(1)}, \frac{v_1(1)}{v_{1st}(2)},\frac{v_1(1)}{v_{1st}(3)}[/latex]

 

The second line is: [latex]\frac{v_1(2)}{v_{1st}(1}, \frac{v_1(2)}{v_{1st}(2)},\frac{v_1(2)}{v_{1st}(3)}[/latex]

 

The third line is: [latex]\frac{v_1(3)}{v_{1st}(1)}, \frac{v_1(3)}{v_{1st}(2)},\frac{v_1(3)}{v_{1st}(3)}[/latex]

Why do you think you can just drop the differentiation operators?

[zztop]

Why do you think you can just drop the differentiation operators?

Because the vectors in his example have CONSTANT components. If you plug in the matrices calculated by the method I suggested, you get the results in the handout. Did you even bother to read the exercise statement? Had you read it, you too could have figured it out. Or maybe not.

Edited March 1, 2017 by zztop

[mikeraj]

Hi zztop, The example in the link you gave did not show how the division involving two 1x3 matrices can be done. If I were to make the denominator into an inverse matrix and multiply it with the numerator, the problem would be the inverse for a 1x3 matrix cannot be determined (determinant cannot be calculated). Thanks

[zztop]

Hi zztop, The example in the link you gave did not show how the division involving two 1x3 matrices can be done.

This is why you need to go back to your class and see what you were told in terms of calculating [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] when [latex]\vec{v_i}, \vec{v_j}[/latex] are BOTH CONSTANTS.(the link I cited deals with the case when the two vectors are VARIABLE). Do you think you can do that? I am willing to bet that the class notes contain the same exact formulas I came up with.

 

 

 

If I were to make the denominator into an inverse matrix and multiply it with the numerator, the problem would be the inverse for a 1x3 matrix cannot be determined (determinant cannot be calculated). Thanks

 

This has nothing to do with the problem you are asked to solve, this is not about "dividing" two matrices.

Edited March 1, 2017 by zztop

[mikeraj]

This is why you need to go back to your class and see what you were told in terms of calculating [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] when [latex]\vec{v_i}, \vec{v_j}[/latex] are BOTH CONSTANTS.(the link I cited deals with the case when the two vectors are VARIABLE). Do you think you can do that? I am willing to bet that the class notes contain the same exact formulas I came up with.

 

Could you show one worked example? It will be easier for me to understand.

Edited March 1, 2017 by mikeraj

[zztop]

 

Could you show one worked example? It will be easier for me to understand.

Look at post 8.

[mikeraj]

I would appreciate it if someone could provide some helpful inputs to my original question. Thanks

[zztop]

I would appreciate it if someone could provide some helpful inputs to my original question. Thanks

Post 8 is a worked example using your exact data. We don't do the homework for you, you need to put in some effort.

[Bignose]

Because the vectors in his example have CONSTANT components.

Well then something is still grossly wrong.

 

Derivatives of constants are 0.

 

And there is no such thing as differentiation with respect to a constant.

 

Either way, you don't just drop differentiation symbols and leave the original terms there. Something isn't right in the pdf or elsewhere.

 

And, no need to be a jerk: "Had you read it, you too could have figured it out. Or maybe not." isn't really necessary. Just trying to help out.

Edited March 4, 2017 by Bignose

[zztop]

Well then something is still grossly wrong.

 

Derivatives of constants are 0.

 

And there is no such thing as differentiation with respect to a constant.

 

Either way, you don't just drop differentiation symbols and leave the original terms there. Something isn't right in the pdf or elsewhere.

 

And, no need to be a jerk: "Had you read it, you too could have figured it out. Or maybe not." isn't really necessary.

I made it quite clear that the OP needs to go back and find where this exercise comes from. Since his vectors are constants, the only possible interpretation is that differentiation is reduced to division, this was an educated guess confirmed by the results. It also happens that if you made the effort to perform the calculations, you would have found that the resultant matrices match the answers in the linked pdf. .

 

 

 

Just trying to help out.

 

You weren't: I had already explained what vector/matrix differentiation is and I also explained the educated guess I was making for the OP particular case.

Edited March 4, 2017 by zztop

[Bignose]

It also happens that if you made the effort to perform the calculations, you would have found that the resultant matrices match the answers in the linked pdf. .

No. Differentiation of a constant with respect to anything is a 0. The OP's answers are not 0s.

 

Differentiation with respect to a constant doesn't mean anything. OP's answers are something.

 

There is something more wrong than just 'if you had made the effort to perform the calculations, you would have found...'

 

Show me how you differentiate a constant and get something other than 0. Show me how you differentiate something with respect to a constant.

 

There is no calculation to perform. The term as given by OP's pdf doesn't exist.

the only possible interpretation is that differentiation is reduced to division,

I have a lot of years doing math, and I have never seen differentiation reduced to division. This isn't even wrong. This is a gross misunderstanding of what differentiation is. Whether yours or OPs, I don't really care, but it needs to be corrected.

[Bignose]

RE the downvotes to this:

I have a lot of years doing math, and I have never seen differentiation reduced to division. This isn't even wrong. This is a gross misunderstanding of what differentiation is. Whether yours or OPs, I don't really care, but it needs to be corrected.

Lol, you can downvote any and all of my posts all you want. I really don't care about some tally of internet points.

 

But it won't change the fact that differentiation and division are two very different operations, and one cannot be 'reduced' to another. If you disagree, why don't you show us why. I have shown why I disagree: the rules of differentiation with respect to a constant are clear and there is no definition of what it means to differentiate with respect to a constant. This is a discussion forum, let's discuss.

Edited March 5, 2017 by Bignose

[zztop]

RE the downvotes to this:

 

Lol, you can downvote any and all of my posts all you want. I really don't care about some tally of internet points.

 

But it won't change the fact that differentiation and division are two very different operations, and one cannot be 'reduced' to another. If you disagree, why don't you show us why. I have shown why I disagree: the rules of differentiation with respect to a constant are clear and there is no definition of what it means to differentiate with respect to a constant. This is a discussion forum, let's discuss.

I have not replaced differentiation with division, I have a PhD in Physical Mathematics and I teach at a very famous university. Apparently you do not understand the notion of "convention". The same way two vectors are "divided" by taking the partial differences of their components (if the vectors are variable) one can define the vector division for the case of constant vectors by dividing them component by component. You also fail to understand the notion of "educated guess". I explained (several times) that I reversed engineered the convention from the results of the exercise.

[Bignose]

I have not replaced differentiation with division, I have a PhD in Physical Mathematics and I teach at a very famous university. Apparently you do not understand the notion of "convention". The same way two vectors are "divided" by taking the partial differences of their components (if the vectors are variable) one can define the vector division for the case of constant vectors by dividing them component by component. You also fail to understand the notion of "educated guess". I explained (several times) that I reversed engineered the convention from the results of the exercise.

but you do say 'reduced to' without really clearing up what reduced to is supposed to mean. I took it as replacement for (which is obviously trivially wrong):

 

the only possible interpretation is that differentiation is reduced to division,

As a PhD, and as a 'teach[er] at a very famous university', you really ought to know how important it is to precisely define terms and operations, yes?

 

Simply 'reverse engineering' the results don't actually fix grossly wrong uses of nomenclature and symbols by the OP, does it? Making sure that OP really meant differentiation there and not division would have really cleared this up...

 

In the meantime, you had written thing like:

 

This is why you need to go back to your class and see what you were told in terms of calculating [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] when [latex]\vec{v_i}, \vec{v_j}[/latex] are BOTH CONSTANTS.

... which again is really poorly communicating. Because again, [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] really has no meaning when the vectors are constant because differentiation of a constant is 0 and with respect to a constant isn't defined.

 

This is all I am really arguing for, when symbols and terms are used, that we try to use them as precisely as they are defined, with their limitations and restrictions in place. That was my first objection... asking why the partial differentiation symbols were dropped for no apparent reason. If you explained that those seemed to be there in error and I missed it, I apologize. But I see OP asking for help with the partial differentiation, and I see you saying "the elements are obtained thru trivial divisions." without any real explanation why given the definition in the original pdf clearly includes differentiation symbols. Some clearer communication probably would have cleared this all up; that includes, again, the apparent tone of treating people who try to discuss things with you like idiots.

Edited March 6, 2017 by Bignose

[zztop]

Because again, [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] really has no meaning when the vectors are constant because differentiation of a constant is 0 and with respect to a constant isn't defined.

According to this logic, all the matrices in the exercise would have resulted to being 0. Which, according to the answers to the exercise, they clearly aren't. Sometimes , you need to be able to read BETWEEN the lines.

 

 

the apparent tone of treating people who try to discuss things with you like idiots.

 

I didn't treat anyone "like idiots", I put in a lot of effort trying to help the OP.

Edited March 6, 2017 by zztop

[Bignose]

According to this logic, all the matrices in the exercise would have resulted to being 0. Which, according to the answers to the exercise, they clearly aren't.

Not 0, but completely undefined, because differentiation with respect to constants is meaningless. But, yes, this is exactly my point. Being precise with the definitions and pointing out that something as written has no meaning is very important, in my opinion. Not creating some 'reduction' of differentiation to division. The processes here are as important as the final result. As a teacher, do you award full points where someone makes a gross process error but still gets the right answer?

 

[math]\frac{26}{65} = \frac{2 \rlap{/}{6}}{\rlap{/}{6} 5}=\frac{2}{5}[/math] ?? I would hope this would not get full credit, despite the left and right hand sides being equal.

 

I didn't treat anyone "like idiots", I put in a lot of effort trying to help the OP.

Forgive me if I misread tone in your posts to me, then. Might I suggest thinking about your word choice a little more, because I got a lot of negative vibes. If it was unintended, my apologies.

Edited March 6, 2017 by Bignose

[mikeraj]

I would appreciate it if someone could provide some helpful inputs to my original question. The equations in my question can be found in this reference Journal of Applied Physics 79, 7148 (1996). Thanks in advance.