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Comments on How to apply a unit constraint in SymPy

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How to apply a unit constraint in SymPy

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The solution to this is probably easy, but I haven't been able to find it. Using SymPy, I am trying to solve this equation: $$ {x_1}^2 + {x_2}^2 + {x_3}^2 + 3 = a $$ with this constraint: $$ {x_1}^2 + {x_2}^2 + {x_3}^2 = 1 $$ It is obvious that $a=4$. So, I wrote this code:

import sympy as sp

a, x1, x2, x3 = sp.symbols('a x1 x2 x3')
eq1 = x1**2 + x2**2 + x3**2 + 3 - a
eq2 = x1**2 + x2**2 + x3**2 - 1
sol = sp.solve([eq1, eq2], a) 

print(sol)

Which returns:

{a: x1**2 + x2**2 + x3**2 + 3}

The unit-norm constraint on x is not applied. How can I get SymPy to apply the unit-norm constraint, which results in $a=4$?

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Post
+0
−0

Are the coefficients of eq1 always 1? If so, you can simply subtract one equation from the other.

>>> sp.solve(eq1-eq2, a)
[4]

P.S. I don't have much experience with SymPy, so LMK if I missed some subtlety.

History

1 comment thread

That would work with the example I posted. But that is a minimal working example which I could easily... (2 comments)
That would work with the example I posted. But that is a minimal working example which I could easily...
Trevor‭ wrote 2 months ago · edited 2 months ago

That would work with the example I posted. But that is a minimal working example which I could easily solve that by hand. For even a slightly more complicated example example like this: $$ \sqrt{b x_1^2 + b x_2^2 + b x_3^3} + 3 = a $$ I would expect to get $a = 3 + \sqrt{b}$ and your suggested technique wouldn't work.

wjandrea‭ wrote 2 months ago

Trevor‭ Yeah, fair enough. I wasn't sure what class of equations you were solving. Of course you could solve it by hand but I thought you might have been automating solving trivial equations. Anyway...