sympy/sympy

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

Daniel Sears

@highpost

Thanks. I'll look at that.

sajen sarvajith

@sajenjeshan1222

Hi I am sajen, I would like to develop documentation as a technical writer by participating in GSoD 2021. I have completely read the project proposal page for sympy . Can I develop the documentation for it ?

sajen sarvajith

@sajenjeshan1222

can mentors say how should I submit proposal and apply for writing docs ?

Aaron Meurer

@asmeurer

See https://groups.google.com/g/sympy/c/PUh3g_NPy-c for instructions on applying for GSoD

Megan Ly

@meganly

Why does cancel no longer make the leading coefficients of the polynomials are positive? In Sympy 1.6 cancel((1-x)/(x**2-3)) is -(x-1)/(x**2-3) whereas in Sympy 1.7 it is (1-x)/(x**2-3)

1 reply

Also why does factor(10+5*I) no longer factor to 5*(2+I)?

sajen sarvajith

@sajenjeshan1222

See https://groups.google.com/g/sympy/c/PUh3g_NPy-c for instructions on applying for GSoD
Ok

oostben

@oostben

I'm trying to make the expand_trig function work for tanh (see #21365). I have a hacky solution that works for tanh(x+y), tanh(2x). However, I modeled my solution off of the tan, cosh, and sinh _eval_expand_trig methods and I'm running into problems, because the tan function seems to use (1+ iz)**coefficient, and then takes the real and imaginary parts of it, if the coefficient on the expression being multiplied is greater than 1. However this doesn't work for tanh. I'm assuming this is why it was never implemented for tanh. Does anyone have a better way of expanding tanh(coefficientx) for arbitrary coefficient>1?

Aaron Meurer

@asmeurer

Can you use the fact that

\tanh ix = i\tan x

Richard Pausch

@PrometheusPi

I am trying to simplify a general Taylor series of function f:

f = sym.Function('f')
x, x_0, h = sym.symbols("x x_0 h", real=True)
# works fine:
ts = sym.series(f(x),x, x_0, 4)

# but substituting x-x_0 by h
ts.subs(x-x_0,h)
# just returns 𝑂(1;𝑥→𝑥0)

Why is the (x-x0) in the Taylor series not substituted correctly?

Parseval Gramalot

@PGramalot_gitlab

I thought people answered questions here. Seems like a ghost town. Good day.

erentar2002

@erentar2002:matrix.org

[m]

yeah, happens

good luck for finding an answer

Kalevi Suominen

@jksuom

@PrometheusPi subs will replace a sum like x - x_0 (= x + (-1)*x_0) in an expression only if that is also a sum that contains the same arguments:

In [19]: (x - x_0 + 1).subs(x - x_0, h)                                         
Out[19]: h + 1

Otherwise, one must substitute x alone:

In [24]: print(ts.subs(x, x_0 + h))                                             
f(x_0) + h*Subs(Derivative(f(_xi_1), _xi_1), _xi_1, x_0) + h**2*Subs(Derivative(f(_xi_1), (_xi_1, 2)), _xi_1, x_0)/2 + h**3*Subs(Derivative(f(_xi_1), (_xi_1, 3)), _xi_1, x_0)/6 + O(h**4)

Richard Pausch

@PrometheusPi

@jksuom Thanks a lot for your fast answer. That helped tremendously. (I was not aware of the fact, that subs requires the substitution to be part of a sum. And I did not think of just replacing a single variable instead.)

Aaron Meurer

@asmeurer

@PGramalot_gitlab your question is a bit lengthy. You may be better off asking it on the mailing list.

Megan Ly

@meganly

How can I modify the latex printer the print the standard basis vectors as \mathbf{i}, \mathbf{j}, and \mathbf{k}? Currently the vectors get printed as follows

>>> C = CoordSys3D('C')
>>> i, j, k = C.base_vectors()
>>> latex(i)
'\\mathbf{\\hat{i}_{C}}'

Kalevi Suominen

@jksuom

@meganly The formatting of base vectors is hard-coded and cannot be given as a parameter. I would implement a subclass redefining the formatting. Something like this (untested):

class MyCoordSys3D(CoordSys3D):
    def __init__(self):
        self.latex_vects = [r'\mathbf{%s}' % n for n in self.vector_names]

Aaron Meurer

@asmeurer

There's a parameter in the CoordSys3D that lets you set these

Also if you use a custom printer that should take precedence over the defaults.

I don't like how the vector module defines the printers on the objects. SymPy library classes should define printing on the printer classes. I also don't like how it precomputes the latex form at construction time.

Vishesh Mangla

@Teut2711

import sympy as sm

s, Kc, tau_i = sm.symbols("s K_c tau_i")

Gc = Kc*(1+1/(tau_i*s))
Gv = 1
Gp =1/(s**2 + 2*s + 2)
Gm = 1

eq = 1 +Gc*Gv*Gp*Gm

How to get the eq's numerator with leading coefficient as 1?

Megan Ly

@meganly

Thanks for the help @jksuom and @asmeurer . I tried

C= CoordSys3D('C', vector_names = ["\mathbf{i}", "\mathbf{j}", "\mathbf{k}"])

and now the latex for the base vector i is '\mathbf{\hat{\mathbf{i}}_{C}}'.
However, implementing a subclass to do custom printing should work.

thorek1

@thorek1

under which conditions does the follwoing simplify to y? simplify((y^(1/(1-alpha)))^(1 - alpha))

Aaron Meurer

@asmeurer

Look at the docstring of powdenest.

You need assumptions on either y or alpha.

thorek1

@thorek1

@vars alpha positive = true
@vars y real = true

still doesnt work

it seems i would need sympy to understand that 0 < alpha < 1

but i dont know how

powdenest((y^(1/(alpha)))^(alpha),force=true) works finde

Aaron Meurer

@asmeurer

I don't think that is sufficient. Consider y = -1 and alpha = 1/2

thorek1

@thorek1

true, in fact y > 0 and 0<alpha<1

docstring of powdenest doesnt help much in this case it seems

Aaron Meurer

@asmeurer

SymPy doesn't currently have a way to tell powdenest that alpha is less than 1. Using force=True is the best way, after manually verifying that the simplification is mathematically correct.

thorek1

@thorek1

ok, what is the best workaround then?

Aaron Meurer

@asmeurer

force=True

thorek1

@thorek1

doesnt get me there

@vars alpha y positive = true
powdenest((y^(1/(1-alpha)))^(1 - alpha))

Aaron Meurer

@asmeurer

only other option is to do a change of variables that lets you use the assumptions that sympy supports, like positive=True

oh hmm

thorek1

@thorek1

hmm, manually catch all int +/- var cases and replace them

Aaron Meurer

@asmeurer

that's a bug I guess

This seems related sympy/sympy#19627

I guess you can do something with replace with a pattern.

>>> a, b, c = Wild('a'), Wild('b'), Wild('c')
>>> expr.replace((a**b)**c, a**(b*c))
y

thorek1

@thorek1

yeah thats probably a good way to go

I get the solution assuming alpha is an integer but that doesnt seem very safe

thorek1

@thorek1

thanks for your quick reply!

laolux

@laolux:privacytools.io

[m]

I have some issue with integrating the Heaviside function. Usually I would expect integrate(Heaviside(x,0), (x,-1,1) =1, but instead sympy returns Integral(Heaviside(x, 0), (x, -1, 1)). Works fine if I omit the second argument of Heaviside, but I think it should work in any case, as the value at x=0 is irrelevant for the integration (set of measure zero).

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