A computer algebra system written in pure Python http://sympy.org/ . To get started to with contributing https://github.com/sympy/sympy/wiki/Introduction-to-contributing
Hi developers !
I am Suryam Arnav Kalra , second year undergraduate student in Computer Science and Engineering department at IIT Kharagpur , India.
I am a beginner in open source and i am interested in contributing to Sympy. It would be really great if you could help me get a starting point to begin my journey.
I am Suryam Arnav Kalra , second year undergraduate student in Computer Science and Engineering department at IIT Kharagpur , India.
I am a beginner in open source and i am interested in contributing to Sympy. It would be really great if you could help me get a starting point to begin my journey.
Hi @kunalsingh2002 @suryam35, you guys can start with https://github.com/sympy/sympy/wiki/Introduction-to-contributing for starters, and then you can refer to the docs https://docs.sympy.org/latest/index.html. For the later part, refer to https://github.com/sympy/sympy/wiki/Development-workflow.
The manual integrate module was suggested but the identity I have is for a definite integral
And I don't think manual integrate works with definite integral rules?
@ctr26 could you open an issue for the same? There is a function called heurisch(), which deals with Bessel functions, but it is only for indefinite integrals and is based off of a slow algorithm. I did not find any option which is as good as manual integrate (which, you are right, does not work with rules for definite integrals).
What does it mean when heurisch never returns an output?
Lambda_0*((-r + x/M)*exp((-(-r + x/M)**2 - y**2/M**2)/(2*\sigma_g**2)) - (r + x/M)*exp((-(r + x/M)**2 - y**2/M**2)/(2*\sigma_g**2)))**2/(2*pi*M**2*\sigma_g**6*(exp((-(-r*sin(2*pi/m) + y/M)**2 - (-r*cos(2*pi/m) + x/M)**2)/(2*\sigma_g**2)) + exp((-(-r*sin(4*pi/m) + y/M)**2 - (-r*cos(4*pi/m) + x/M)**2)/(2*\sigma_g**2))))and
16*Lambda_0*(-M*x_0_tau + x)**2*(M*lambda*besselj(1, 2*pi*n_a*sqrt(M**2*x_0_tau**2 + M**2*y_0_tau**2 - 2*M*x*x_0_tau - 2*M*y*y_0_tau + x**2 + y**2)/(M*lambda)) - pi*n_a*sqrt(M**2*x_0_tau**2 + M**2*y_0_tau**2 - 2*M*x*x_0_tau - 2*M*y*y_0_tau + x**2 + y**2)*besselj(0, 2*pi*n_a*sqrt(M**2*x_0_tau**2 + M**2*y_0_tau**2 - 2*M*x*x_0_tau - 2*M*y*y_0_tau + x**2 + y**2)/(M*lambda)))**2/(pi*lambda**2*(M**2*x_0_tau**2 + M**2*y_0_tau**2 - 2*M*x*x_0_tau - 2*M*y*y_0_tau + x**2 + y**2)**3)Trying to integrate either of these across infinity in x
sympy.Integral(integrand,(x,-sympy.oo,sympy.oo)).doit()StackTraces respective:
https://pastebin.com/FTafTjF9
https://pastebin.com/u9NTQPib
Neither converges " @evatiwari
These stack traces after running it for a bit longer
https://pastebin.com/SCHykTyR https://pastebin.com/WRgXuTHu
https://pastebin.com/SCHykTyR https://pastebin.com/WRgXuTHu
Hi, I'm trying to write a parser that converts expressions from my custom expression language to sympy friendly expressions. Does anyone have any resources that would help me? my end goal is to convert expressions written in my language and translate it to something I can evaluate in sympy/numpy.
Hi. I am having some trouble implementing inverse_phi (euler's totient) with prime factorization for very large numbers. Does sympy have any inverse_totient functionality, or could someone possibly point me in the right direction for implementing this using SymPy?
Given (p-1)(q-1) I want to find n, for phi(n) = (p-1)(q-1). I.e: phi(n) = 24 --> n = pq = 35 , except I need it to handle numbers as large as
4529255040439033800342855653030016000000000++
I want to solve a bvp numerically with steps. This is my relation. 1<i, j<5 . the number of equations are huge. How can I make them and then solve with the help of sympy?
Hello all, I was adding a test case for an issue where I encountered assertion error, I do not have much idea about it. Can someone guide me how it can be resolved?
Traceback (most recent call last):
File "/home/shardul/sympy/sympy/series/tests/test_series.py", line 208, in test_issue_9173
assert Q.series(y, n=3) == b_2y**2 + b_1y + b_0 + O(y**3)
AssertionError
@sayandip18 You can always fetch from someone's (in your case, your own) fork, and checkout that branch.
This link might help. :)
This link might help. :)
I run the code for this expression in the terminal having installed sympy version 1.7.1 but here the coefficients of y**2 and y are not simplified to final answer. Anyone guide me regarding this issue whether it is done with some purpose or this should be fixed?(Sympy live shell is giving final simplified answer)
>>> from sympy import *
>>> from sympy.abc import y
>>> var('p_0 p_1 p_2 p_3 b_0 b_1 b_2')
(p_0, p_1, p_2, p_3, b_0, b_1, b_2)
>>> Q = (p_0 + (p_1 + (p_2 + p_3/y)/y)/y)/(1 + ((p_3/(b_0*y) + (b_0*p_2 - b_1*p_3)/b_0**2)/y + (b_0**2*p_1 - b_0*b_1*p_2 - p_3*(b_0*b_2 - b_1**2))/b_0**3)/y)
>>> Q.series(y, n=3)
y*(b_0*p_2/p_3 + b_0*(-p_2/p_3 + b_1/b_0)) + y**2*(b_0*p_1/p_3 + b_0*p_2*(-p_2/p_3 + b_1/b_0)/p_3 + b_0*(-p_1/p_3 + (p_2/p_3 - b_1/b_0)**2 + b_1*p_2/(b_0*p_3) + b_2/b_0 - b_1**2/b_0**2)) + b_0 + O(y**3)
>>> simplify(y*(b_0*p_2/p_3 + b_0*(-p_2/p_3 + b_1/b_0)) + y**2*(b_0*p_1/p_3 + b_0*p_2*(-p_2/p_3 + b_1/b_0)/p_3 + b_0*(-p_1/p_3 + (p_2/p_3 - b_1/b_0)**2 + b_1*p_2/(b_0*p_3) + b_2/b_0 - b_1**2/b_0**2)) + b_0 + O(y**3))
b_2*y**2 + b_1*y + b_0 + O(y**3)
4 replies
Hey all! I am a cse fresher from Bits Pilani, and new to open-source. I like the features provided by Sympy and would want to know how can I contribute to it. I have a fair amount of experience with python,c++, and framework Django. If any mentor here can help me with how to start, it would be a great help !
Hi @mohitdmak , you can start with https://github.com/sympy/sympy/wiki/Introduction-to-contributing for starters, and then you can refer to the docs https://docs.sympy.org/latest/index.html. For the later part, refer to https://github.com/sympy/sympy/wiki/Development-workflow.
Can anyone give me some hints on how I can go about debugging a
RecursionError (for specifics, sympy/sympy#9449 )? Any hints you can give would be great as I'm seeing this part of the codebase for the first time. Should I focus on the statements before the recursion happens or during (I have tried printing values during both but still wasn't able to figure anything out)?
1 reply
Hey folks. Possibly n00b question. I'm interested in SymPy for its symbolic simplification features, using it to rearrange complicated expressions built by by sometimes naive automation. I'm leaving everything fully symbolic, so none of the Symbols or Functions have assigned values or concrete implementations. I've had great luck using SymPy this way using
Is there an accepted way to declare a
Thanks for any assistance, and Happy New Year!
parse_expr() with a bit of transformation logic to ensure the expressions are well-formatted in ways SymPy expects. I am stumped by one thing, though. Some of my expressions contain functions are logical (boolean) , so I get expressions like f(x) & g(y). But the default transformations turn these into Functions, which aren't considered Booleans, and thus cause TypeErrors when used as arguments to BooleanFunctions like And() or Not().Is there an accepted way to declare a
Function-like symbol that counts as a Boolean type for these purposes? Is there some better way to achieve this?Thanks for any assistance, and Happy New Year!
1 reply
Hi all, what's the easiest way to perform partial fraction expansion with scypy? I'm trying to do it on H(z) from this paper. So far I've used
prem and pquo to make the numerator order lower than the denominator order. However, the result I get makes no sense, when I subtract (quotient + remainder/denominator) from the original fraction, the answer is not zero as it should be. I've tested on a simple case of (2x + 1)/(x + 1) and the answer was fine there.