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Aaron Meurer
@asmeurer
I didn't see any dedicated issue for it so I opened sympy/sympy#18907
Megan Ly
@meganly
The minpoly idea is cool. Is there a canonical form for algebraic functions? If so, what is it?
Aaron Meurer
@asmeurer
I'm not sure if there's a straightforward way to define one.
I think you can do something using the fact that algebraic extensions form a vector space over the extended field. So if you have a basis, then every element of the extension is a linear combination of those basis elements. But there's the question of how to chose that basis in a canonical way.
Megan Ly
@meganly
That's something you could do, but I don't think that works for quotients of algebraic expressions. As far as I know there isn't really a canonical form for algebraic functions like there is for rational
Saanidhya
@Saanidhyavats
After making some changes in pull request #18681 I think it's ready to be merged. It would be helpful if someone could take a look init.
Dhruv Mendiratta
@dhruvmendiratta6
@jksuom @Arpan612 I too am running wsl with the sympy repo in windows. I am not able to run profilers(eg: vprof) I installed in ubuntu on the repo. Also local tests using ./bin/test do not work on ubuntu. So is the solution here simply cloning a wsl version of the sympy repo and keeping its functioning separate from windows?
@Arpan612
Yes, I think this solution will be best for now.
If I get any alternative, I will tell you.
@asmeurer Thank you, Sir.
Maarten van der Velde - (Ideogram)
@ideogram

@jksuom I tried it, and it doesn't seem to make any difference. Everythink works fine, until I introduce a variable I. It then says:

TypeError: can't convert expression to float

Kalevi Suominen
@jksuom
Does the same error occur if you change the name to something else?
Maarten van der Velde - (Ideogram)
@ideogram
No, I tried that. It works fine with X, Y, Z, W, H and L
Kalevi Suominen
@jksuom
What is the difference then? Do you begin by defining I in the same way as I = Symbol('I')? That should be done before creating the expression to which evalf is to be applied so that the expression will involve the symbol I and not the imaginary unit I.
danil179
@danil179
I don't know what is the current state of my PR #17745 as it seems 5 months passed already and I don't see any specific problems with the code right now. In my PR I tried to keep the original printing where possible (and hence my PR keeps the original printing with changes only where necessary and doesn't require code changes in the tests). I think that my solution is near optimal in doing that.
@Arpan612
@asmeurer Sir, could you please go through my query in issue #18882?
BasileiosKal
@BasileiosKal
Hi everyone. I made the pr #18875 a few days ago. I was wondering if anyone could review it? Thank you!
Saanidhya
@Saanidhyavats
I have made some changes in pr #18681 and #18559. It would be helpful if someone could review it
Aaron Meurer
@asmeurer
@danil179 sometimes PRs fall by the wayside. Just ping people if no one has said anything for a while.
Saanidhya
@Saanidhyavats
@asmeurer thank you for suggesting the changes, I have made the changes in pr #18681. Can you review it
Daniel Shapero
@danshapero

Hi all -- I'm looking at how to differentiate expressions involving IndexedBase objects. When I run the following code:

import sympy
n = sympy.symbols('n', integer=True)
M = sympy.IndexedBase('M', shape=(n, n))
x = sympy.IndexedBase('x', shape=n)
i = sympy.Idx('i', n)
j = sympy.Idx('j', n)
k = sympy.Idx('k', n)
energy = 0.5 * M[i, j] * x[i] * x[j]
print(sympy.simplify(energy.diff(x[k])))

I get an expression involving lots of Kronecker deltas. I feel it should be possible to simplify these out but I don't know how. Thoughts?

Maq_Owais
@MaqOwais
@MaqOwais
Sir, as I came to know that ' Handling modular equations ' already integrated with _solveset then why it showed
in gsoc project under solvers module?
do we again have to reintegrate under _transolve ?
and remove that part from _solveset ? to have a reply : )
OmarWagih1
@OmarWagih1
Hello All, i just posted my application on the github wiki for some reviews and comments, could you please look at it? Would very much appreciate it. Thanks!
https://github.com/sympy/sympy/wiki/GSoC-2020-Application-Omar-Wagih:-Benchmarks-and-performance
Hello All, i just posted my application on the github wiki for some reviews and comments, could you please look at it? Would very much appreciate it. Thanks!
Gagandeep Singh
@czgdp1807
@Arpan612
Yes sir, I saw it now. I will do it.
@czgdp1807 Could you please tell me who are the potential mentors for symengine project?
Saanidhya
@Saanidhyavats
I have made the changes in pr #18681. It would be helpful if someone could review it
@Arpan612
@Saanidhyavats Could you please go through my proposal and suggest changes?
I am reviewing your changes. I will put up my comments on Github.
Saanidhya
@Saanidhyavats
Sure
BasileiosKal
@BasileiosKal
Hi all! I have made the changes the reviewer's proposed in my pr #18875. Could you please review it?
Mohit Balwani
@Mohitbalwani26
In my PR #18881 travis has already passed on when i checked on travis offcial websote but it is showing still pending in GitHub. What could be the reason?
mohit
@mohitacecode
Hello Arpan,
I have left comment on the issue.
mohit
@mohitacecode
the files mentioned in the issues are already created you have to identify the test in test_matrices.py and test_commonmatrix.py and move them to their related test file. for example all the solvers related test should be transferred to test_solvers.py.
just a opinion the discussion related to pr should be kept there.I will also post the previous comment there.
Saanidhya
@Saanidhyavats
I have made a pr #18932, it would be helpful if someone could guide me further on it
@Arpan612
I am on it.
Julian
@cod3monk
Is there a simple way to find and divide all terms of relationals by the gcd, if one exists?
This should turn 64*M*N - 104*N - 56 <= 11796480 into 8*M*N - 13*N - 7 <= 1474560. simplify does not do the trick.
mohit
@mohitacecode
maybe something like this.
>>> a = 64*M*N - 104*N - 56 - 11796480 <= 0
>>> factor(a)
8*(8*M*N - 13*N - 1474567) <= 0
@cod3monk
Stefan Corneliu Petrea
sss
Aaron Meurer
@asmeurer
@danshapero how should it simplify?
Daniel Shapero
@danshapero
@asmeurer the result I get is
>>> 0.5*(KroneckerDelta(i, k)*x[j] + KroneckerDelta(j, k)*x[i])*M[i, j]
which mathematically I know to be equal to 0.5 * (M[k, j] * x[j] + M[i, k] * x[i]) at the very least
of course I haven't put in any assumptions about M being self-adjoint so I'm not expecting anything there
Aaron Meurer
@asmeurer
oh I see it simplifies when you expand it
I'm not sure if there is code in sympy to simplify deltas like this yet.