Assume a function is real-valued

asked 2022-02-27 20:45:00 +0200

ripple_carry
91 ●3 ●11

In the following expression:

var('z')

u = function('u')(z)
v = function('v')(z)

f = u + i * v

show(f.diff(z).conjugate().expand().simplify_full())
# conjugate(diff(u(z), z)) - I*conjugate(diff(v(z), z))

I would like Sage to make the simplification conjugate(diff(u(z), z)) = diff(u(z), z) because $u, v$ are intended to be real-valued. But my first attempt:

assume(u, 'real')

didn't work:

TypeError: self (=u(z)) must be a relational expression

What is the right way to write this assumption?

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answered 2022-02-28 17:22:39 +0200

dsejas
2804 ●14 ●49 ●95

Hello, @ripple_carry! The easiest way I know of achieving that effect is to explicitly declare the imaginary part of the function to be zero. Just use the imag_part_func argument for function as follows:

var('z')

u = function('u',imag_part_func=0)(z)
v = function('v',imag_part_func=0)(z)

f = u + i * v

show(f.diff(z).conjugate().expand().simplify_full())

This outputs

diff(u(z), z) - I*diff(v(z), z)

Hope this helps!

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Asked: 2022-02-27 20:45:00 +0200

Seen: 415 times

Last updated: Feb 28 '22

Assume a function is real-valued edit

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