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 | 1 | initial version |
The issue here is that you want to make substitution not into f but into each of its coefficients. Direct approach using .map_coefficients() is given by @FrédéricC in the comments.
However, with the i being the imaginary unit, you don't need define it over RR but work in CC where the imaginary unit is already defined as I. So, you can go directly with
L.<a, b> = CC[]
f = a + I*b
f.map_coefficients(real)
Alternative general approach is to define all a, b, i as polynomial variables and work in the quotient ring:
F.<I,A,B> = RR[]
K.<i,a,b> = F.quotient( [I^2 + 1] )
f = a + i*b
f.lift().subs({I:0})
Instead of f.lift().subs({I:0}) giving result in F, you can use f.lift().specialization({I:0}), which gives result in the polynomial ring of A and B only.
| | 2 | No.2 Revision |
The issue here is that you want to make substitution not into f but into each of its coefficients. Direct approach using .map_coefficients() is given by @FrédéricC in the comments.
However, with the i being the imaginary unit, you don't there is no need to define it over RR but , since one can work in CC where the imaginary unit is already defined as I. So, you can go directly with
L.<a, b> = CC[]
f = a + I*b
f.map_coefficients(real)
Alternative general approach is to define all a, b, i as polynomial variables and work in the quotient ring:
F.<I,A,B> = RR[]
K.<i,a,b> = F.quotient( [I^2 + 1] )
f = a + i*b
f.lift().subs({I:0})
Instead of f.lift().subs({I:0}) giving result in F, you can use f.lift().specialization({I:0}), which gives result in the polynomial ring of A and B only.
