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Why doesn't f(alpha) act like a function?

asked 2018-12-23 16:58:12 +0100

deburgess7 gravatar image

updated 2023-01-09 23:59:49 +0100

tmonteil gravatar image

var('t,alpha,beta,gamma')

sage: i [ 0 -I] [-I 0]

sage: f(alpha) = cos(alpha)*i

sage: f(alpha) [ 0 -Icos(alpha)] [-Icos(alpha) 0]

sage: f(0) [ 0 -Icos(alpha)] [-Icos(alpha) 0]

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If we define `f` by

    def f(x, y, z):
        return x * y * z

then `f(2, 3, 5)` returns `30` but `f(2*3*5)` gives:

    TypeError: f() takes exactly 3 arguments (1 given)

produces:

If we define f by

def f(x, y, z):
    return x * y * z

then f(2, 3, 5) returns 30 but f(2*3*5) gives:

TypeError: f() takes exactly 3 arguments (1 given)

Please edit your question to do that.

slelievre gravatar imageslelievre ( 2018-12-24 13:43:09 +0100 )edit

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answered 2018-12-23 21:23:38 +0100

rburing gravatar image

updated 2018-12-23 21:30:44 +0100

This is trac ticket #12075 (since 2011).

This way to define a function is syntactic sugar offered by Sage, but it doesn't yet work for matrix functions.

I suggest defining an ordinary Python function instead (using a lambda for brevity):

sage: var('alpha')
sage: i = matrix(SR, [[ 0, -I], [-I, 0]])
sage: f = lambda x: cos(x)*i
sage: f(0)
[ 0 -I]
[-I  0]
sage: f(alpha)
[            0 -I*cos(alpha)]
[-I*cos(alpha)             0]
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Asked: 2018-12-23 16:58:12 +0100

Seen: 276 times

Last updated: Dec 23 '18