Ask Your Question
0

How to find arbitrary complex constants in symbolic expressions?

asked 2020-06-27 09:40:36 -0500

Tony-64 gravatar image

How do I go about finding terms in a symbolic expression containing arbitrary complex constants? I had hoped that the code fragment below would yield something like [2Ix], [Ix], [I], and [2I]. However, instead it yields [], [I*x], [], and []. Is there another way to isolate the terms containing an arbitrary complex constant?

w0 = SR.wild(0)

f = x^2 + 2*I*x + 1
g = x^2 + I*x + 1
h = x^2 + 2*x + I
k = x^2 + x + 2*I

print(f.find(I*w0))
print(g.find(I*w0))
print(h.find(I*w0))
print(k.find(I*w0))
edit retag flag offensive close merge delete

1 answer

Sort by ยป oldest newest most voted
0

answered 2020-06-27 11:21:04 -0500

Juanjo gravatar image

updated 2020-06-27 19:22:10 -0500

Edited. This was the original answer

It seems that, in your expressions, $x$ denotes a real number. If this were the case, you can get the coefficients of I by taking the imaginary part of the expression:

sage: f = x^2 + 2*I*x + 1
sage: g = x^2 + I*x + 1
sage: h = x^2 + 2*x + I
sage: k = x^2 + x + 2*I
sage: assume(x,'real')
sage: for expression in [f, g, h, k]:
....:     print(expression.imag()) 
2*x
x
1
2

New answer. I have improved your approach, based on wildcards. It seems that I is not detected because it is not a symbolic variable. So, we can temporarily transform I into such a variable, find the coefficients of I in the expressions and then restore I:

w0 = SR.wild(0)
var("I")
f = x^2 + 2*I*x + 1
g = x^2 + I*x + 1
h = x^2 + 2*x + I
k = x^2 + x + 2*I
m = sqrt(1-x^2) + I*(1-x^3)^(1/3)
n = sqrt(1-x^2) + 2*I*(1-x^3)^(1/3)
p = x^2 + 2*x - I
q = x^2 + 2*x

expressions = [f, g, h, k, m, n, p, q]
for expression in expressions:
    print("\nExpression: ",expression)
    if expression.has(I):
        if expression.has(w0*I):
            aux = expression.find(w0*I)
            coef = aux[0].match(w0*I)[w0]
        else:
            coef = 1
        print(f"The coefficient of I is {coef}")
    else:
        print("The expression does not contain I")
restore("I")

This is the output:

Expression:  2*I*x + x^2 + 1
The coefficient of I is 2*x

Expression:  I*x + x^2 + 1
The coefficient of I is x

Expression:  x^2 + I + 2*x
The coefficient of I is 1

Expression:  x^2 + 2*I + x
The coefficient of I is 2

Expression:  (-x^3 + 1)^(1/3)*I + sqrt(-x^2 + 1)
The coefficient of I is (-x^3 + 1)^(1/3)

Expression:  2*(-x^3 + 1)^(1/3)*I + sqrt(-x^2 + 1)
The coefficient of I is 2*(-x^3 + 1)^(1/3)

Expression:  x^2 - I + 2*x
The coefficient of I is -1

Expression:  x^2 + 2*x
The expression does not contain I
edit flag offensive delete link more

Comments

Thanks for the answer. You are correct that I meant x to be a real number. I thought of this approach too, but unfortunately, it breaks down when the expression becomes a more complex. Here is an example.

w0 = SR.wild(0)

f = sqrt(1-x^2) + I*(1-x^3)^(1/3)
g = sqrt(1-x^2) + 2*I*(1-x^3)^(1/3)

assume(x, 'real')

print(f.find(I*w0))
print(g.find(I*w0))

print(f.imag())
print(g.imag())

In this example, .find(I*w0) yields the term containing the cube root from f, but not from g. However, .imag() yields complicated expressions for both f and g because it takes into account that abs(x) may be greater than 1.

Tony-64 gravatar imageTony-64 ( 2020-06-27 15:53:35 -0500 )edit

Thanks again. The updated approach works when we know the expressions up front. However, one step more difficult is when they result from a computation. Here is another example. Interestingly enough, when solving the equation for x, the .find(I*w0) approach works. However, when solving for y, it doesn't.

w0 = SR.wild(0)

var('y')
assume(x, 'real')
assume(y, 'real')



eqn = (x^3 + y^3 / 27 == 1)

sols = solve(eqn, x)

f, g, h = [sol.rhs() for sol in sols]

expressions = [f, g, h]

for expr in expressions:
    print(expr.find(I*w0))



sols = solve(eqn, y)

f, g, h = [sol.rhs() for sol in sols]

expressions = [f, g, h]

for expr in expressions:
    print(expr.find(I*w0)
Tony-64 gravatar imageTony-64 ( 2020-06-28 05:59:16 -0500 )edit

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 2020-06-27 09:40:36 -0500

Seen: 67 times

Last updated: Jun 27