sympy

问题 It is easy to obtain such rewrite in other CAS like Mathematica. TrigReduce[Sin[x]^2] (*1/2 (1 - Cos[2 x])*) However, in Sympy, trigsimp with all methods tested returns sin(x)**2 trigsimp(sin(x)*sin(x),method='fu') 回答1: The full "fu" method tries many different combinations of transformations to find "the best" result. The individual transforms used in the Fu-routines can be used to do targeted transformations. You will have to read the documentation to learn what the different functions do,

问题 It is easy to obtain such rewrite in other CAS like Mathematica. TrigReduce[Sin[x]^2] (*1/2 (1 - Cos[2 x])*) However, in Sympy, trigsimp with all methods tested returns sin(x)**2 trigsimp(sin(x)*sin(x),method='fu') 回答1: The full "fu" method tries many different combinations of transformations to find "the best" result. The individual transforms used in the Fu-routines can be used to do targeted transformations. You will have to read the documentation to learn what the different functions do,

问题 I am generating an expression with two fractions, and want to pretty print as a whole expression with LaTeX, to then put on a worksheet. E.g. in the form: (5/7) * (3/4). However, when I do the following: fract1 = sympy.sympify(Fraction(5,7)) fract2 = sympy.sympify(Fraction(3,4)) expression = sympy.Mul(fract1,fract2,evaluate=False) It returns 5*3/(7*4) Clearly it is combining the fraction but not actually evaluating, but I want to be able to produce it in a format suitable as a question for a

问题 I am trying to solve for non linear equations in python. I have tried using the solver of the Sympy but it doesn't seem to work in a for loop statement. I am tyring to solve for the variable x over a range of inputs [N] . I have attached my code below import numpy as np import matplotlib.pyplot as plt from sympy import * f_curve_coefficients = [-7.14285714e-02, 1.96333333e+01, 6.85130952e+03] S = [0.2122, 0, 0] a2 = f_curve_coefficients[0] a1 = f_curve_coefficients[1] a0 = f_curve

问题 I'm trying to make a python proram to find derivatives and integrals as well as showing how. I have so far found that there is an integral_steps function which returns the steps used, but I have not found an equivalent for differentiation. Does anyone know if there is an equivalent? If there isn't, do you have any ideas on how to find the steps needed to find a derivative? 回答1: Method 1 (manual) Looking at the code, the Derivative class is where the top-level logic lives. That's only the top

问题 I'm trying to make a python proram to find derivatives and integrals as well as showing how. I have so far found that there is an integral_steps function which returns the steps used, but I have not found an equivalent for differentiation. Does anyone know if there is an equivalent? If there isn't, do you have any ideas on how to find the steps needed to find a derivative? 回答1: Method 1 (manual) Looking at the code, the Derivative class is where the top-level logic lives. That's only the top

问题 There seems to be no equivalent for cosd , sind in sympy (ie cosine and sine for arguments in degrees). Is there any simple way to implement those functions ? For numpy, I did : numpy.cosd = lambda x : numpy.cos( numpy.deg2rad(x) ) Something like : sympy.cosd = lambda x : sympy.cos( sympy.pi/180*x ) works for evaluation, but the expression is printed as : cos(pi*x/180) which is not great for readability (I have complicated expression due to 3D coordinates changes). Is there any way to create

问题 Suppose I define the following expression: poly1 = 6/(25*(x + 3)) + 1/(5*(x + 3)**2) which prints out: 6/(25*x + 75) + 1/(5*(x + 3)**2) I have two questions regarding this expression. Firstly, is there a way to keep the expression in the format in which I entered it? In particular, is there something that I could do in order to leave the denominator of the first term as 25*(x + 3) ? Secondly, if I have the expression in its printed form, what is the best way to break apart the expression,

问题 I would like to be able to simplify mathematical expressions from a string in Python. There are several "commutative" ways of doing it. Is there a non-commutative function for that? I know that sympify from sympy can do some non-commutative jobs, here you have an example: from sympy import * x=Symbol('x',commutative=False) y=Symbol('y',commutative=False) print sympify(3*x*y - y*x - 2*x*y) it will print x y -y x, however if we apply sympify to the string, that is, print sympify('3*x*y - y*x -

问题 I have developed Bicubic interpolation for demonstration to some undergraduate students using Python Programming language. The methodology is as explained in wikipedia, The code is working fine except the results I am getting are slightly different than what is obtained when using scipy library. The interpolation code is shown below in the function bicubic_interpolation . import numpy as np import matplotlib.pyplot as plt from mpl_toolkits import mplot3d from scipy import interpolate import