On Wed, Apr 20, 2022 at 6:37 AM Jonathan Gutow wrote:
>
>
> On Apr 19, 2022, at 10:48 PM, Andre Bolle wrote:
>
> Here's what I did. You will notice that the second derivative couldn't see
> the 'x', which had been replaced by psi.
>
> Yes, that is a “feature” of sympy, which assumes that all sym
> assumes that all symbols that do not have an explicit dependence on the
variable of differentiation are constants
`idiff` will allow you to do the differentiation of symbols without
functions, e.g. `dydx for idiff(2*x - y**2, y, x) -> 1/y`
/c
On Wednesday, April 20, 2022 at 7:37:33 AM UTC-5
On Apr 19, 2022, at 10:48 PM, Andre Bolle
mailto:andrebo...@gmail.com>> wrote:
Here's what I did. You will notice that the second derivative couldn't see the
'x', which had been replaced by psi.
Yes, that is a “feature” of sympy, which assumes that all symbols that do not
have an explicit depe
This might be relevant -
https://physics.stackexchange.com/questions/32296/introduction-to-differential-forms-in-thermodynamics
If you find anything of interest in the link you should check it out
with someone in the math department. The limit to my knowledge of
differential forms is that dx
Since I have no rant control I proselytize the following graphics
software (free) for generating publication quality graphics whenever I
have a likely target -
https://galgebra.readthedocs.io/en/latest/
look at the "3D Graphs" and "WebGL" galleries. I especially like this
one (you can zoom,
On Apr 19, 2022, at 12:20 PM, Alan Bromborsky
mailto:abrombo...@gmail.com>> wrote:
I don't think sympy can return f for the integral of (df/dx)dx without first
differentiating and then integrating.
Alan,
Yep, that is essentially the sticking point. It relates to the fact that there
is no co
Look at -
https://galgebra.readthedocs.io/en/latest/
gradient (geometric derivative) and associated operators implemented
with the same interface. Nothing to do with integration implemented.
Attached is the code for Dop.py and the test code in Jupyter notebook.
The problem with the differe
Alan,
I have thought about this a little too. I have not had time to work on it
recently. The issue I ran into is that to make this work well in SymPy you
really need the concept of an infinitesimal dx, dy, dz, etc. Things got
circular when I tried to implement that using the sympy definition o
When I differentiate the function
ψ = exp(I*(k*x))
I get
i*k*exp(I*(k*x)).[which is i*k*ψ]
Is there a way to get exp(I*(k*x)) substituted with ψ, in order to get the
shorter expression i*k*ψ ?
Thanks,
André
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