Timeline for Can numerical differentiation be applied to tensor derivatives?
Current License: CC BY-SA 4.0
16 events
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| Oct 9 at 17:52 | comment | added | Malkoun | I expect the derivative of a tensor of dimensions $n$ by $n$ to be a tensor of dimensions $n$ by $n$ by $n$. Then, if you want a derivative of the original tensor in a certain direction $v$, you just contract $v$ with the first index of the derivative. There are issues with how to define this on a curved space, i.e. a manifold, but this is resolved using connections. I may be misunderstanding your question, in which case I apologize. | |
| Oct 9 at 17:47 | comment | added | Malkoun | I understand that one may want to differentiate a tensor numerically in a certain direction. Why is differentiating a tensor with respect to another tensor of the same dimensions, the way you defined it, natural? What is the motivation? | |
| Oct 9 at 17:30 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Math Jaxed and formatted
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| Oct 9 at 13:52 | review | Close votes | |||
| Oct 14 at 3:08 | |||||
| Oct 9 at 2:34 | comment | converted from answer | user1432369 | Interesting question. I am looking answers about accuracy, any existing implementations doing this calculation in matlab/python. Running a for loop for the array of function might look unrealistics for large scale implementation. What will be the choice of step size, e.g. it should be a uniform for all components or should it be an array of step sizes which would affect accuracy? I need this for applications in continuum mechanics like energy derivatives, derivative of second order with second order. | |
| Apr 22 at 14:15 | comment | added | Jesse Feng | Ok thank you. My numerical estimate is way different from the analytical counter-part, and more likely to be wrong. I was expecting the numerical estimate to be closer to the correct answer. The issue I am having is either a linear approximation is not good enough, or the quantities are wrong then. | |
| Apr 22 at 14:09 | comment | added | Carlo Beenakker | certainly, you just have many functions that depend on many parameters, "tensor" for your purpose is just a word to arrange these functions into an array, you might as well arrange them in a single string. | |
| Apr 22 at 14:03 | comment | added | Jesse Feng | It sounds like you agree that the derivative tensor components can indeed be extended from a simple y = f(x) case since it is just an array of functions? | |
| Apr 22 at 13:52 | comment | added | Carlo Beenakker | a tensor is just an array of functions, I really do not understand the question. | |
| Apr 22 at 13:34 | history | edited | Jesse Feng | CC BY-SA 4.0 |
added 179 characters in body
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| Apr 22 at 13:29 | comment | added | Jesse Feng | In any case, my question is mainly whether if it can be extended to a tensor like that | |
| Apr 22 at 13:22 | comment | added | Jesse Feng | @CarloBeenakker you are correct. What I was thinking is if my step size is small, every point in the derivative can be approximated as linear, so I am simply calculating a linear slope with that. | |
| Apr 22 at 6:01 | comment | added | Carlo Beenakker | what does this mean: "dy/dx where dy = y2 - y1 and dx = x2 - x1" ? the discretization of the derivative $dy/dx$ of the function $y(x)$ is $(1/2)[y(x+1)-y(x-1)]$ | |
| Apr 22 at 0:33 | review | Close votes | |||
| May 23 at 19:27 | |||||
| S Apr 21 at 23:46 | review | First questions | |||
| Apr 22 at 0:13 | |||||
| S Apr 21 at 23:46 | history | asked | Jesse Feng | CC BY-SA 4.0 |