Solve for variable but variable is still in answer

asked 2014-08-28 19:49:41 +0100

wzawzdb
15 ●1 ●3 ●8

updated 2014-08-29 15:48:16 +0100

I try to solve for variable P3. Sage gives me an answer for P3 but it still contains P3 on the right hand side (see last equation after the solve in the beginning of sqrt?

sage: A1, A2, A3, P1, P2, P3, u1, u2, u3, r1, r2, r3, g, C, M1, M2, M3 = var('A1 A2 A3 P1 P2 P3 u1 u2 u3 r1 r2 r3 g C M1 M2 M3')
sage: eq1=A3*(P1-P3)==A3*r3*u3^2-(A2*r2*u2^2+A1*r1*u1^2)
sage: eq2=A3*u3*r3==A2*u2*r2+A1*r1*u1
sage: eq5a=solve([eq2],r3)
sage: eq5b=u3^2+2*g/(g-1)*(P3/(eq5a[0].rhs()))-2*C==0
sage: eq6=solve([eq5b],u3)
sage: eq7a=eq1/A3
sage: eq7b=eq7a.subs(-A3*r3*u3^2 == -(A2*u2*r2+A1*r1*u1)*u3)
sage: eq7b
P1 - P3 == -(A1*r1*u1^2 + A2*r2*u2^2 + (A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))*(A1*r1*u1 + A2*r2*u2)/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2))/A3

sage: eq7c=eq7b.subs(u3==eq6[0].rhs()) 
sage: eq7=solve([eq7c],P3)
sage: eq7
[P3 == (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1 - (A1*r1*u1^2 + A2*r2*u2^2 + A3*P1)*g - sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/A3]

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It looks like it sees P3 and P3 as different variables?

wzawzdb ( 2014-08-28 22:39:48 +0100 )edit

Trying to recreate this gave me an error because eq2 was not defined.

rws ( 2014-08-29 07:47:53 +0100 )edit

I forgot to copy that line or something, sorry about that

wzawzdb ( 2014-08-29 15:48:42 +0100 )edit

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answered 2014-08-29 19:06:16 +0100

tmonteil
27423 ●31 ●204 ●517 http://wiki.sagemath.o...

updated 2014-08-31 23:50:05 +0100

First, you should notice that eq6 is a list of two solutions for u3:

sage: len(eq6)
2
sage: eq6
[u3 == -(A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2), u3 == -(A3*P3*g - sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2)]

You can choose the first or the second as follows:

sage: eq6[0]
u3 == -(A3*P3*g + sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2)
sage: eq6[1]
u3 == -(A3*P3*g - sqrt(A3^2*P3^2*g^2 + 2*((g^2 - 2*g + 1)*A1^2*r1^2*u1^2 + 2*(g^2 - 2*g + 1)*A1*A2*r1*r2*u1*u2 + (g^2 - 2*g + 1)*A2^2*r2^2*u2^2)*C))/(A1*(g - 1)*r1*u1 + A2*(g - 1)*r2*u2)
sage: eq6[0] == eq6[1]
False

Second, you found the limitation of the solve function, which is not very powerful, so it is sometimes not able to find a solution of the form P3 = .... Youe equation involves squares roots and solve does not handle such equations very well. A simpler example is:

sage: eq = x == sqrt(x)
sage: eq.solve(x)
[x == sqrt(x)]

EDIT It seems that sympy is better than Maxima in solving this, so you can do:

sage: import sympy
sage: sympy.solve(eq7c.rhs()-eq7c.lhs(), P3)
[(A3*(A1*r1*u1**2 + A2*r2*u2**2 + A3*P1) - sqrt(A3**2*(-2*A1**2*C*g**2*r1**2*u1**2 + 2*A1**2*C*r1**2*u1**2 + A1**2*g**2*r1**2*u1**4 - 4*A1*A2*C*g**2*r1*r2*u1*u2 + 4*A1*A2*C*r1*r2*u1*u2 + 2*A1*A2*g**2*r1*r2*u1**2*u2**2 + 2*A1*A3*P1*g**2*r1*u1**2 - 2*A2**2*C*g**2*r2**2*u2**2 + 2*A2**2*C*r2**2*u2**2 + A2**2*g**2*r2**2*u2**4 + 2*A2*A3*P1*g**2*r2*u2**2 + A3**2*P1**2*g**2)))/(A3**2*(g + 1)),
 (A3*(A1*r1*u1**2 + A2*r2*u2**2 + A3*P1) + sqrt(A3**2*(-2*A1**2*C*g**2*r1**2 ...

(more)

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And there are no other options to solve this? I know there are two solutions (quadratic rule) but I take only one with the eq6[0].rhs() right?

wzawzdb ( 2014-08-29 19:21:31 +0100 )edit

yes, eq6 is a Python list that contains all solutions. There are two of them, to get the second one, you just have to look at eq6[1].rhs()

tmonteil ( 2014-08-30 00:23:13 +0100 )edit

Why do you subtract eq7c rhs from lhs **=^

wzawzdb ( 2014-09-02 17:18:30 +0100 )edit

@wzawzdb, from the Sympy solvers documentation Use solve() to solve algebraic equations. We suppose all equations are equaled to 0, so solving x**2 == 1 translates into the following code: Run code block in SymPy Live

>>> from sympy.solvers import solve
>>> from sympy import Symbol
>>> x = Symbol('x')
>>> solve(x**2 - 1, x)
[-1, 1]

JesterEE ( 2015-02-25 22:52:22 +0100 )edit

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