I have an equation: f = (100*E**(-x*1600)-50), it takes a very long time to give a solution for x but in another online solver, it only takes a few ms. How can I do to boost the performance?
raw code:
f = (100*E**(-x*1600)-50)
solve(f,x)
Expected result:
If you just want the real solution (and not the many symbolic solutions) use nsolve. The bisect solvers is appropriated for this function with a steep gradient near the root. Defining lam = var('lam') gives
If you want a symbolic solution (but not the many imaginary ones) then solveset or the lower level _invert can be used:
>>> from sympy import solveset, var, Reals
>>> from sympy.solvers.solvers import _invert
>>> var('x')
x
>>> eq=(100*E**(-x*1600)-50)
>>> solveset(eq,x,Reals)
{log(2)/1600}
>>> from sympy.solvers.solvers import _invert
>>> _invert(eq,x)
(log(2)/1600, x)
Related
I am studying two different systems in python, looking for fixed points and their stability. Managed to solve completely for the first one, but applying the same method raises an error i dont know how to deal with in the second one.
TypeError: loop of ufunc does not support argument 0 of type Zero which has no callable exp method
I don't really know how to handle it, since when i make an exception for this error I simply skip the answers and i am certain there are possible answers and analytically i see no reasons for them not to exist
from sympy import *
from numpy import *
from matplotlib import pyplot as plt
r = symbols('r', real=True)
x = symbols('x', real =True)
#first one
fx =r*x+((x**3)/(1+x**2)) # DEf. both fet and right side in EQ
fps = solve(fx, x)
print(f"The fixed points are: {fps}")
dfx = lambdify(x,fx.diff(x))
for fp in fps:
stable_interval = solve_univariate_inequality(dfx(fp)<0, r, domain=Reals, relational=False)
unstable_interval = solve_univariate_inequality(dfx(fp)>0, r, domain=Reals, relational=False)
#print(type(stable_interval))
print(f"{fp} is stable when {stable_interval}")
#print(type(unstable))
print(f"{fp} is unstable when {unstable_interval}")
fx2 = r*x+( x* E**x)
fps2 = solve(fx2, x)
print(f"The fixed points are: {fps}")
dfx2 = lambdify(x,fx2.diff(x))
for fp in fps2:
stable_interval = solve_univariate_inequality(dfx2(fp)<0, r, domain=Reals, relational=False)
unstable_interval = solve_univariate_inequality(dfx2(fp)>0, r, domain=Reals, relational=False)
#print(type(stable_interval))
print(f"{fp} is stable when {stable_interval}")
#print(type(unstable))
print(f"{fp} is unstable when {unstable_interval}")
I expected the method i created to be applyable to the second system fx2 but i dont understand the logic behind why this doesnt remain true.
Oscar mentioned in the comment to not mix star imports: that's correct! Let's understand what you are doing:
with from sympy import * you are importing everything from sympy, like cos, sin, ...
with from numpy import * you are importing everything from numpy, like cos, sin, ... However, many things share the same names as sympy, so you are effectively overriding the previous import. Result: a complete mess that will surely raise errors down the road, as your namespace now contains names pointing to numpy and others pointing to sympy. Numpy and Sympy doesn't work well together!
Best ways to resolve the situation. Keep things separated, like this:
import sympy as sp
import numpy as np
Or import everything only from one module:
from sympy import *
import numpy as np
Now, to your actual problem. With this command:
dfx2 = lambdify(x,fx2.diff(x))
# where fx2.diff(x) results in:
# r + x*exp(x) + exp(x)
lambdify created a numerical function that will be evaluated by Numpy: note that this function contains an exponential, which is a Numpy exponential. Then, you evaluated this function with dfx2(fp), where fp is a symbolic object (meaning, it is a Sympy object). As mentioned before, Numpy and Sympy do not work well together.
Easiest solution: asks lambdify to create a function that will be evaluated by Sympy:
dfx2 = lambdify(x, fx2.diff(x), "sympy")
Now, everything works as expected.
Alternatively, you don't use lambdify. Instead, you substitute your values into the symbolic expression. For example:
dfx2 = fx2.diff(x)
for fp in fps2:
stable_interval = solve_univariate_inequality(dfx2.subs(x, fp)<0, r, domain=Reals, relational=False)
unstable_interval = solve_univariate_inequality(dfx2.subs(x, fp)>0, r, domain=Reals, relational=False)
#print(type(stable_interval))
print(f"{fp} is stable when {stable_interval}")
#print(type(unstable))
print(f"{fp} is unstable when {unstable_interval}")
I have a simple equation, trying to solve for using symbolic, however the code gets stuck and I do not get an error for me to debug. How can I do this correctly?
from sympy import *
from sympy import init_printing
init_printing(use_latex = True)
import sympy as sp
from numpy import random
import numpy as np
import math
from decimal import *
timeS = symbols("t")
eq1 = Eq( (-4221.4125*exp(-2750.0*timeS)*sin(6514.4071*timeS) + 10000.0*exp(-2750.0*timeS)*cos(6514.4071*timeS)),8000)
T_off = solve(eq1, timeS )[0]
display("T_off = ",T_off)
Your equation looks like one that doesn't have an analytic solution so you should use nsolve instead. There is a root between -0.001 and 0 so the bisect method gives:
In [59]: nsolve(eq1, timeS, [-0.001, 0])
Out[59]: 3.46762014687136e-5
This is an ill-posed equation. Rescale by dividing both sides by 10**4 and replace timeS with Symbol(x)/10**4 and use nsolve
>>> eq2 = eq1.func(eq1.lhs/10**4, eq1.rhs/10**4).subs(timeS, symbols('x')/10**4)
>>> from sympy import nsolve
>>> nsolve(eq2,0)
0.346762014687135
So timeS ~= 0.35e-4. The sqrt function automatically scales sums so sqrt(eq1.rewrite(Add)) will reduce coefficients of the terms to something less than 1 but such simplification is not part of SymPy simplification of general epxressions at present.
this might be a silly question. But I am desperate. I am a math teacher and I try to generate Math tests. I tried Python for this and I get some things done. However, I am not a professional programmer, so I get lost with MathMl, prettyprint() and whatsoever.
Is there anybody who can supply me a complete example that I can execute? It may just contain one small silly equation, that does not matter. I just want to see how I can get it into a Word document. After that, I can use that as a basis. I work on a Mac.
I hope anyone can help me out. Thanks in advance!
Best regards, Johan
This works for me:
from sympy import *
from docx import Document
from lxml import etree
# create expression
x, y = symbols('x y')
expr1 = (x+y)**2
# create MathML structure
expr1xml = mathml(expr1, printer = 'presentation')
tree = etree.fromstring('<math xmlns="http://www.w3.org/1998/Math/MathML">'+expr1xml+'</math>')
# convert to MS Office structure
xslt = etree.parse('C:/MML2OMML.XSL')
transform = etree.XSLT(xslt)
new_dom = transform(tree)
# write to docx
document = Document()
p = document.add_paragraph()
p._element.append(new_dom.getroot())
document.save("simpleEq.docx")
How about the following. The capture captures whatever is printed. In this case I use pprint to print the expression that I want written to file. There are lots of options you can use with pprint (including wrapping which you might want to set to False). The quality of output will depend on the fonts you use. I don't do this at all so I don't have a lot of hints for that.
from pprint import pprint
from sympy.utilities.iterables import capture
from sympy.abc import x
from sympy import Integral
with open('out.doc','w',encoding='utf-8') as f:
f.write(capture(lambda:pprint(Integral(x**2, (x, 1, 3)))))
When I double click (in Windows) on the out.doc file, a word equation with the integral appears.
Here is the actual IPython session:
IPython console for SymPy 1.6.dev (Python 3.7.3-32-bit) (ground types: python)
These commands were executed:
>>> from __future__ import division
>>> from sympy import *
>>> x, y, z, t = symbols('x y z t')
>>> k, m, n = symbols('k m n', integer=True)
>>> f, g, h = symbols('f g h', cls=Function)
>>> init_printing()
Documentation can be found at https://docs.sympy.org/dev
In [1]: pprint(Integral(x**2, (x, 1, 3)))
3
(
? 2
? x dx
)
1
In [2]: from pprint import pprint
...: from sympy.utilities.iterables import capture
...: from sympy.abc import x
...: from sympy import Integral
...: with open('out.doc','w',encoding='utf-8') as f:
...: f.write(capture(lambda:pprint(Integral(x**2, (x, 1, 3)))))
...:
{problems pasting the unicode here, but it shows up as an integral symbol in console}
I have an equation, as follows:
R - ((1.0 - np.exp(-tau))/(1.0 - np.exp(-a*tau))) = 0.
I want to solve for tau in this equation using a numerical solver available within numpy. What is the best way to go about this?
The values for R and a in this equation vary for different implementations of this formula, but are fixed at particular values when it is to be solved for tau.
In conventional mathematical notation, your equation is
The SciPy fsolve function searches for a point at which a given expression equals zero (a "zero" or "root" of the expression). You'll need to provide fsolve with an initial guess that's "near" your desired solution. A good way to find such an initial guess is to just plot the expression and look for the zero crossing.
#!/usr/bin/python
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import fsolve
# Define the expression whose roots we want to find
a = 0.5
R = 1.6
func = lambda tau : R - ((1.0 - np.exp(-tau))/(1.0 - np.exp(-a*tau)))
# Plot it
tau = np.linspace(-0.5, 1.5, 201)
plt.plot(tau, func(tau))
plt.xlabel("tau")
plt.ylabel("expression value")
plt.grid()
plt.show()
# Use the numerical solver to find the roots
tau_initial_guess = 0.5
tau_solution = fsolve(func, tau_initial_guess)
print "The solution is tau = %f" % tau_solution
print "at which the value of the expression is %f" % func(tau_solution)
You can rewrite the equation as
For integer a and non-zero R you will get a solutions in the complex space;
There are analytical solutions for a=0,1,...4(see here);
So in general you may have one, multiple or no solution and some or all of them may be complex values. You may easily throw scipy.root at this equation, but no numerical method will guarantee to find all the solutions.
To solve in the complex space:
import numpy as np
from scipy.optimize import root
def poly(xs, R, a):
x = complex(*xs)
err = R * x - x + 1 - R
return [err.real, err.imag]
root(poly, x0=[0, 0], args=(1.2, 6))
sympy does not find the two real roots of this rather straightforward equation:
import sympy
x = Symbol('x')
solve(-2*x**2*exp(1 - x**2) + exp(1 - x**2),x)
Returns "[oo]"
Is this a bug or is my command ill formed ?
Thanks!
Markus
It seems to work fine for me, so perhaps you just need to upgrade?
>>> from sympy import Symbol, solve, exp
>>> x = Symbol('x')
>>> solve(-2*x**2*exp(1 - x**2) + exp(1 - x**2),x)
[-sqrt(2)/2, sqrt(2)/2]
>>> sympy.__version__
'0.7.2-git'