I'm trying to perform the following integration using sympy;
x = Symbol('x')
expr = (x+3)**5
integrate(expr)
The answer that I'm expecting is:
But what's being returned is:
The following code works in MATLAB:
syms x
y = (x+3)^5;
int(y)
I'm unsure what I'm doing wrong in order to perform this using sympy.
This is actually a common problem seen in Calculus where for these kinds of polynomial expressions, you do get two answers. The coefficients for each of the powers of x exist but the constant factor is missing between them.
As such, there are two methods you can use to find the indefinite integral of this expression.
The first method is to perform a substitution where u = x+3, then integrate with respect to u. Then, the indefinite integral would be (1/6)*(x + 3)^6 + C as you expect.
The second method is to fully expand out the polynomial and integrate each term individually.
MATLAB elects to find the integral the first way:
>> syms x;
>> out = int((x+3)^5)
out =
(x + 3)^6/6
Something to note for later is that if we expand out this polynomial expression, we get:
>> expand(out)
ans =
x^6/6 + 3*x^5 + (45*x^4)/2 + 90*x^3 + (405*x^2)/2 + 243*x + 243/2
sympy elects to find the integral the second way:
In [20]: from sympy import *
In [21]: x = sym.Symbol('x')
In [22]: expr = (x+3)**5
In [23]: integrate(expr)
Out[23]: x**6/6 + 3*x**5 + 45*x**4/2 + 90*x**3 + 405*x**2/2 + 243*x
You'll notice that the answer is the same between both environments, but the constant factor is missing. Because the constant factor is missing, there is no neat way to factor this into the neat polynomial that you are expecting from your output seen in MATLAB.
As a final note, if you would like to reproduce what sympy generates, expand out the polynomial, then integrate. We get what sympy generates:
>> syms x;
>> out = expand((x+3)^5)
out =
x^5 + 15*x^4 + 90*x^3 + 270*x^2 + 405*x + 243
>> int(out)
ans =
x^6/6 + 3*x^5 + (45*x^4)/2 + 90*x^3 + (405*x^2)/2 + 243*x
The constant factor though shouldn't worry you. In the end, what you are mostly concerned with is a definite integral, and so the subtraction of these constant factors will happen, which won't affect the final result.
Side Note
Thanks to DSM, if you specify the manual=True flag for integrate, this will attempt to mimic performing integration by hand, which will give you the answer you're expecting:
In [26]: from sympy import *
In [27]: x = sym.Symbol('x')
In [28]: expr = (x+3)**5
In [29]: integrate(expr, manual=True)
Out[29]: (x + 3)**6/6
Related
I'm making a solver of cubic equations in Python that includes division of polynomials.
from sympy import symbols
# Coefficients
a = int(input("1st coef: "))
b = int(input("2nd coef: "))
c = int(input("3rd coef: "))
d = int(input("Const: "))
# Polynomial of x
def P(x):
return a*x**3 + b*x**2 + c*x + d
x = symbols('x')
# Find 1 root by Cardano
R = (3*a*c - b**2) / (9*a**2)
Q = (3*b**3 - 9*a*b*c + 27*a**2*d) / (54*a**3)
Delta = R**3 + Q**2
y = (-Q + sqrt(Delta))**(1/3) + (-Q - sqrt(Delta))**(1/3)
x_1 = y - b/(3*a)
# Division
P(x) / (x - x_1) = p(x)
print(p(x)) # Just a placeholder
The program returns an error: "cannot assign to operator" and highlights the P(x) after the # Division comment (worded poorly, yes, but I'm from Russia so idc).
What I tried doing was to assign a variable to a polynomial and then dividing:
z = P(x)
w = x - x_1
p = z / w
print(p)
But alas: it just returns a plain old quotient (a = 1, b = 4, c = -9, d = -36):
(x**3 + 4*x**2 - 9*x - 36)/(x - 2.94254537742264)
Does anyone out here knows what to do in this situation (not to mention the non-exact value of x_1: the roots of x^3+4x^2-9x-36=0 are 3, -4, and -3, no floaty-irrational-messy-ugly things in sight)?
tl;dr: Polynomial division confusion and non-exact roots
I am not sure what exactly your question is but here is an attempt at an answer
The line
P(x) / (x - x_1) = p(x)
is problematic for multiple reasons. First of all it's important to know that the = operator in python (and a lot of other modern programming languages) is an assignment operator. You seem to come from more of a math background, so consider it to be something like the := operator. The direction of this is always fixed, i.e. with a = b you are always assigning the value of b to the variable a. In your case you are basically assigning an expression the value of p which does not make much sense:
Python can't assign anything to an expression (At least not as far as I know)
p(x) is not yet defined
The second problem is that you are mixing python functions with math functions.
A python function looks something like this:
def some_function(some_parameter)
print("Some important Thing!: ", some_parameter)
some_return_value = 42
return some_return_value
It (can) take some variable(s) as input, do a bunch of things with them, and then (can) return something else. They are generally called with the bracket operator (). I.e. some_function(42) translates to execute some_function and substitute the first parameter with the value 42. An expression in sympy however is as far as python is concerned just an object/variable.
So basically you could have just written P = a*x**3 + b*x**2 + c*x + d. What your P(x) function is doing is basically taking the expression a*x**3 + b*x**2 + c*x + d, substituting x for whatever you have put in the brackets, and then giving it back in as a sympy expression. (It's important to understand, that the x in your P python function has nothing to do with the x you define later! Because of that, one usually tries to avoid such "false friends" in coding)
Also, a math function in sympy is really just an expression formed from sympy symbols. As far as sympy is concerned, the return value of the P function is a (mathematical) function of the symbols a,b,c,d and the symbol you put into the brackets. This is why, whenever you want to integrate or differentiate, you will need to specify by which symbol to do that.
So the line should have looked something like this.
p = P(x) / (x - x_1)
Or you leave replace the P(x) function with P = a*x**3 + b*x**2 + c*x + d and end up with
p = P / (x - x_1)
Thirdly if you would like to have the expression simplified you should take a look here (https://docs.sympy.org/latest/tutorial/simplification.html). There are multiple ways here of simplifying expressions, depending on what sort of expression you want as a result. To make for faster code sympy will only simplify your expression if you specifically ask for it.
You might however be disappointed with the results, as the line
y = (-Q + sqrt(Delta))**(1/3) + (-Q - sqrt(Delta))**(1/3)
will do an implicit conversion to floating point numbers, and you are going to end up with rounding problems. To blame is the (1/3) part which will evaluate to 0.33333333 before ever seeing sympy. One possible fix for this would be
y = (-Q + sqrt(Delta))**(sympy.Rational(1,3)) + (-Q - sqrt(Delta))**(sympy.Rational(1,3))
(You might need to add import sympy at the top)
Generally, it might be worth learning a bit more about python. It's a language that mostly tries to get out of your way with annoying technical details. This unfortunately however also means that things can get very confusing when using libraries like sympy, that heavily rely on stuff like classes and operator overloading. Learning a bit more python might give you a better idea about what's going on under the hood, and might make the distinction between python stuff and sympy specific stuff easier. Basically, you want to make sure to read and understand this (https://docs.sympy.org/latest/gotchas.html).
Let me know if you have any questions, or need some resources :)
according to this graph: desmos
print(solve('x**2 + x - 1/x'))
# [-1/3 + (-1/2 - sqrt(3)*I/2)*(sqrt(69)/18 + 25/54)**(1/3) + 1/(9*(-1/2 - sqrt(3)*I/2)*(sqrt(69)/18 + 25/54)**(1/3)), -1/3 + 1/(9*(-1/2 + sqrt(3)*I/2)*(sqrt(69)/18 + 25/54)**(1/3)) + (-1/2 + sqrt(3)*I/2)*(sqrt(69)/18 + 25/54)**(1/3), -1/3 + 1/(9*(sqrt(69)/18 + 25/54)**(1/3)) + (sqrt(69)/18 + 25/54)**(1/3)]
I was expecting [0.755, 0.57], but, I got something I cannot use in my future program. I desire to get a list of floats as result, so refer to this post, I did following, but I got some even more weird:
def solver(solved, rit=3):
res = []
for val in solved:
if isinstance(val, core.numbers.Add):
flt = val.as_two_terms()[0]
flt = round(flt, rit)
else:
flt = round(val, rit)
if not isinstance(flt, core.numbers.Add):
res.append(flt)
return res
print(solver(solve('x**2 + x - 1/x')))
# [-0.333, -0.333, -0.333]
Now I am really disappointed with sympy, I wonder if there is an accurate way to get a list of floats as result, or I will code my own gradient descent algorithm to find the roots and intersection.
sym.solve solves an equation for the independent variable. If you provide an expression, it'll assume the equation sym.Eq(expr, 0). But this only gives you the x values. You have to substitute said solutions to find the y value.
Your equation has 3 solutions. A conjugate pair of complex solutions and a real one. The latter is where your two graphs meet.
import sympy as sym
x = sym.Symbol('x')
# better to represent it like the equation it is
eq = sym.Eq(x**2, 1/x - x)
sol = sym.solve(eq)
for s in sol:
if s.is_real:
s = s.evalf()
print(s, eq.lhs.subs({x: s})) # eq.rhs works too
There are a variety of things you can do to get the solution. If you know the approximate root location and you want a numerical answer, nsolve is simplest since it has no requirements on the type of expression:
>>> from sympy import nsolve, symbols
>>> x = symbols('x')
>>> eq = x**2 + x - 1/x
>>> nsolve(eq, 1)
0.754877666246693
You can try a guess near 0.57 but it will go to the same solution. So is there really a second real roots? You can't use real_roots on this expression because it isn't in polynomial form. But if you split it into numerator and denominator you can check for the roots of the numerator:
>>> n, d = eq.as_numer_denom()
>>> from sympy import real_roots
>>> real_roots(n)
[CRootOf(x**3 + x**2 - 1, 0)]
So there is only one real root for that expression, the one that nroots gave you.
Note: the answer that solve gives is an exact solution to the cubic equation and it can't figure out definitively which ones are a solution to the equation so it returns all three. If you evaluate them you will find that only one of them is real. But since you don't need the symbolic solution, just stick to nroots.
I want to solve this equation witht the following parameters:
gamma = 0.1
F = 0.5
w = 0
A = symbols('A')
a = 1 + w**4 -w**2 + 4*(gamma**2)*w**2
b = 1 - w**2
sol = solve(a*A**2 + (9/16)*A**6 + (3/2)*b*A**4 -F**2)
list_A = []
for i in range(len(sol)):
if(type( solutions[i] )==float ):
print(sol[i])
list_A = sol[i]
However, as supposed, I am getting some real and complex values, and I want to remove the complex ones and only keep the floats. But this condition I implemented is not valid due to the type of sol[i] is either sympy.core.add.Add for complex or sympy.core.numbers.Float for floats.
My question is, how can I modify my condition so that it works for getting only the float values?
In addition, is there a way to speed it up? it is very slow if I put it in a loop for many values of omega.
this is my first time working with sympy
When it is able to validate solutions relative to assumptions on symbols, it will; so if you tell SymPy that A is real then -- if it can verify the solutions -- it will only show the real ones:
>>> A = symbols('A',real=True)
>>> sol = solve(a*A**2 + (9/16)*A**6 + (3/2)*b*A**4 -F**2)
>>> sol
[-0.437286658108243, 0.437286658108243]
Basically I'm trying to run the following code.
import sympy as sp
alpha = sp.Symbol(r'\alpha')
x = sp.Symbol('x')
sp.Q.is_true(alpha != -1)
sp.integrate(x**alpha, x)
This results in the following Piecewise function.
Since I specify global assumptions that alpha != -1, I expected it will simply give me the first expression. So two questions I have:
how do you properly define the assumptions so that the sp.integrate does not ignore them;
is there a way to access (extract) the first (or second) expression from the Piecewise function?
Thanks in advance!
PS. Defining conds='separate' in the sp.integrate only returns the first expression for some reason. So if I needed the second part of the piecewise function, I wouldn't be able to get it.
PPS. In case this matters, I have python 3.8.0 and sympy 1.4.
There is no way to give a specific value as an assumption for a symbol so it can be used in the integration. The best you can do is specify positive, negative, etc... But as for extracting the desired expression from the Piecewise, you can either get it as the particular argument or else feed in a dummy value for x that would extract it. Like the following:
>> from sympy.abc import x
>> from sympy import Piecewise, Dummy
>> eq = Piecewise((x + 1, x < 0), (1/x, True))
>> eq.args[0]
(x + 1, x < 0)
>> _.args[0]
x + 1
>> d = Dummy(negative=True)
>> eq.subs(x, d)
d + 1
>> _.subs(d, x)
x + 1
For a certain project, I'm using sympy to calculate expressions modulo another function. These functions all have binary coefficients (so x^2 + 2x = x^2$). Their application is in Galois Fields.
My issue is that when using the the sympy rem function with inverses (for example x**-1), is simply returns the inverse of the number (so in this case the answer is 1/x) rather than returning the modular inverse.
Due to the below comment, here is some further clarification. An oversimplified version of what I'm doing is:
from sympy import *
x = symbols('x')
f = raw_input() #here f = '(x^3 + x)*(x + 1)^2 + (x^2 + x)/(x^3) + (x)^-1'
expand(f)
>>> x**5 + 2*x**4 + 2*x**3 + 2*x**2 + x + 2/x + x**(-2)
#this is what I'm currently doing
rem(expand('(x^3 + x)*(x + 1)^2 + (x^2 + x)/(x^3) + (x)^-1'), 'x^2')
>>> x + 2/x + x**(-2)
#not the answer I am looking for, as I want all the degrees to be positive
This remainder function doesn't act as a mod function (ie doesn't keep things as positive powers of x), and I'm trying to find a replacement for it. I want to avoid parsing through the expression searching for inverse mods and just have the function deal with that on it's own. I might be missing a parameter, or just looking at a completely different function.
PS: I know that the ability to compute an expression mod another expression, while simultaneously treating inverses as modular inverses exists in sympy since I did it while testing that sympy will be enough for our purposes but didn't save the code back then.
First off, it's better to immediately convert your string into a SymPy expression with sympify. Passing strings to SymPy functions is bad practice.
When you use a polynomial like x + 1/x, SymPy treats this as a polynomial in x and 1/x.
In [73]: Poly(x + 1/x)
Out[73]: Poly(x + (1/x), x, 1/x, domain='ZZ')
I believe ratsimpmodprime does what you want. You should also be able to pass domain=GF(2), but it seems there are some bugs that prevent that from working.