I'm trying to make some basic substitutions but SymPy doesn't want to help me out
x, y, z, k = symbols("x y z k", positive=True, real=True)
exp = x**4 + x**3 + x**2 + x
what_im_expecting = simplify(y**(Rational(1/4)) + y**(Rational(3/4)) + sqrt(y) + y)
what_i_actually_get = exp.subs(x**4,y)
exp, what_i_actually_get, what_im_expecting
returns
x + y**(Rational(3, 4)) + sqrt(y) + y
Can anyone help me out?
a more complex example:
The method subs can be trusted to replace the terms that exactly match the given "old" expression, which x**4 here. The replacement of other things related to x**4 is not so certain. (There are many open issues with subs: some say it substitutes too much, some say too little.) There is some substitution logic specific to powers, but x by itself is not formally a power, so it escapes that logic. A workaround: temporarily replace x by x**1, preventing automatic evaluation of that power to x.
x1 = sp.Pow(x, 1, evaluate=False)
subbed = exp.subs(x, x1).subs(x**4, y).subs(x1, x)
Now subbed is y**(3/4) + y**(1/4) + sqrt(y) + y.
But, don't expect human-like ingenuity from subs. With the same workaround, trying to do subs(x**4 - 1, y) results in x**3 + x**2 + x + y + 1: nothing like sqrt(y+1), etc, appears. It's better to substitute in the most direct way possible:
subs(x, (y+1)**Rational(1, 4))
Then you don't need any workarounds.
Related
How do I transform an expression like this:
y = [x*e^(x/2) + e^x + e^(x/2)]^2
in this:
y = {e^(x/2)*[x + e^(x/2) + 1]}^2
using Sympy??
It seems that the factoring doesn't recognize the exponential as well as a power. So convert the exp(x/2) -> y, factor, then back substitute:
>>> eq
(x*exp(x/2) + exp(x/2) + exp(x))**2
>>> factor(eq.subs(exp(x/2),y)).subs(y,exp(x/2))
(x + exp(x/2) + 1)**2*exp(x)
I have a code that calculate some mathematical equations and when I want to see the simplified results, it can not equate 2.0 with 2 inside power, which is logical since one is float and the other is integer. But decision was sympys where to put these two values, not mine.
Here is the expression in my results that sympy is not simplifying
from sympy import *
x = symbols('x')
y = -exp(2.0*x) + exp(2*x)
print(simplify(y)) # output is -exp(2.0*x) + exp(2*x)
y = -exp(2*x) + exp(2*x)
print(simplify(y)) # output is 0
y = -2.0*x + 2*x
print(simplify(y)) # output is 0
y = -x**2.0 + x**2
print(simplify(y)) # output is -x**2.0 + x**2
is there any way working around this problem? I am looking for a way to make sympy assume that everything other than symbols are floats, and preventing it to decide which one is float or integer.
this problem has been asked before by Gerardo Suarez but not with a satisfactory answer.
There is another sympy function you can use called nsimplify. When I run your examples they all return zero:
from sympy import *
x = symbols("x")
y = -exp(2.0 * x) + exp(2 * x)
print(nsimplify(y)) # output is 0
y = -exp(2 * x) + exp(2 * x)
print(nsimplify(y)) # output is 0
y = -2.0 * x + 2 * x
print(nsimplify(y)) # output is 0
y = -(x ** 2.0) + x ** 2
print(nsimplify(y)) # output is 0
Update
As #Shoaib Mirzaei mentioned you can also use the rational argument in the simplify() function like this:
simplify(y,rational=True)
I'm trying to solve an integral with sympy. But it gives me a wrong solution. Why?
import sympy
from sympy import Integral, exp, oo
x, y = sympy.symbols("x y", real=True)
b, u, l, t = sympy.symbols("b u l t ", real=True, positive=True)
Fortet = Integral(exp(-l * t) * (sympy.sqrt(2 * sympy.pi * t)) ** (-1) * exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, oo))
Fortet.doit()
Result (wrong):
Piecewise((-(-b/2 + y)*sqrt(2*l +
u**2)*(-sqrt(pi)*sinh(sqrt(2)*sqrt(b)*sqrt(l +
u**2/2)*sqrt(polar_lift(1 + y**2/(b*polar_lift(b -
2*y))))*sqrt(polar_lift(b - 2*y))) +
sqrt(pi)*cosh(sqrt(2)*sqrt(b)*sqrt(l + u**2/2)*sqrt(polar_lift(1 +
y**2/(b*polar_lift(b - 2*y))))*sqrt(polar_lift(b - 2*y))))*exp(b*u -
u*y)/(sqrt(pi)*(b - 2*y)*(l + u**2/2)), Abs(arg(1 +
y**2/(b*polar_lift(b - 2*y))) + arg(b - 2*y)) <= pi/2),
(Integral(sqrt(2)*exp(-l*t)*exp(-(b - t*u -
y)**2/(2*t))/(2*sqrt(pi)*sqrt(t)), (t, 0, oo)), True))
Expected (correct) solution:
Solution = (exp((-u)*(b - y)) * exp(sympy.sqrt(u**2 + 2*l)*(b-y)))/(sympy.sqrt(2*l + u**2)) #RIGHT solution
Both results are in fact the same. The first one is probably slightly more correct. You tend see these polar_lift functions whenever SymPy tries to do something like square rooting something when it does not know the signs of the things inside (after integrating)
A polar_lift does not appear below, but this basic Gaussian example shows that SymPy tries to be as general as possible:
from sympy import *
x = Symbol("x", real=True)
y = Symbol("y", real=True)
s = Symbol("s", real=True) # , positive=True
gaussian = exp(-((x-y)**2)/(2*(s**2)))
nfactor = simplify(integrate(gaussian, (x,-oo,oo)))
print(nfactor)
You need s to be declared as positive: s = Symbol("s", real=True, positive=True). A similar thing happens with these kinds of polar_lift(b - 2*y) functions in your example. It also happens with the question I reference below.
I have no idea why, but N(polar_lift(x)) for any float or int x gives x again; yet, SymPy does not simplify nicely with symbolic x. Turns out if you keep on simplifying, you get nicer and nicer looking answers. I couldn't find anything about polar_lift related to pure math so I don't know what it actually does.
Remember for the simple example above how it gave a piece-wise? Same thing here. So we just take the first piece since the second piece is an un-evaluated integral.
In the code below, I use this question to remove the piece-wise function and then I simplify twice. And finally, I manually remove the polar_lift.
import sympy as sp
x, y = sp.symbols("x y", real=True)
b, u, l, t = sp.symbols("b u l t ", real=True, positive=True)
Fortet = sp.integrate(sp.exp(-l * t) * (sp.sqrt(2 * sp.pi * t)) ** (-1) *
sp.exp(-((b - u * t - y) ** 2) / (2 * t)),
(t, 0, sp.oo), conds='none')
incorrect = Fortet.simplify().simplify()
correct = eval(str(incorrect).replace("polar_lift", ""))
correct = correct.factor()
print(correct)
The result is:
exp(b*u)*exp(-u*y)*exp(-sqrt(2*l + u**2)*sqrt(b**2 - 2*b*y + y**2))/sqrt(2*l + u**2)
That is close enough to your expression. I couldn't make SymPy simplify the sqrt(b**2 - 2*b*y + y**2) to Abs(b-y) no matter how hard I tried.
Note that either SymPy is still wrong or you are wrong since the powers in the numerator are opposite. So I checked on the Desmos for a numeric answer (top one is yours):
Suppose I have the code below. I want to get either the right side of the equation (C1 +x...). How do I do that?
My problem is I have some boundary conditions for derivatives of f(x) at specific points, so I want to calculate those and find out the constants. I have also different values for w(x), so the final code will begin defining a variable called wx instead of having the function w(x).
from __future__ import division
from sympy import *
x, y = symbols('x y')
w, f = symbols('w f', cls=Function)
init_printing(use_unicode=True)
diffeq = f(x).diff(x,x,x,x)-w(x)
expr = dsolve(diffeq, f(x))
print diffeq
print expr
results:
-w(x) + Derivative(f(x), x, x, x, x)
f(x) == C1 + x**3*(C4 + Integral(w(x)/6, x)) + x**2*(C3 - Integral(x*w(x)/2, x)) + x*(C2 + Integral(x**2*w(x)/2, x)) - Integral(x**3*w(x)/6, x)
expr.lhs and expr.rhs will give you the left- and right-hand sides of the equation.
I am writing a code which takes a "code" (Coding Theory) as an input and I have calculated the weight enumerator of it. I want to find the weight enumerator of dual code using MacWilliams Identity.
I have W(z), the weight enumerator of the code, and I want to convert it to W((1-z)/(1+z)) using python.
If you are going to be working with polynomials in an extended fashion, you should be working with a proper computer algebra system (CAS). sympy is great for manipulations like these. For example:
from sympy import poly
from sympy.abc import x,z
W1 = poly(x**2 + 3*x)
# Evaluate the polynomial at x=(1-z)/(1+z)
W2 = W1((1-z)/(1+z))
print W1
print W2
>>> Poly(x**2 + 3*x, x, domain='ZZ')
>>> (-2*z**2 - 2*z + 4)/(z**2 + 2*z + 1)
You can simplify any answer with sympy.simplify and .expand for example:
import sympy
expr = (4*x**2 - 4*x + 4)*(x**3 + 3*x**2 + 3*x + 1)/(4*(x**2 + 2*x + 1))
print sympy.simplify(sympy.expand(expr))
>>> x**3 + 1