# piecewise defined function via def

I've made the following experiment with Sage:

    def f(x):

if (0<= x<= 1/2):
return 1
else:
return 0
assume(0<= x<= 1/2)
show(f(x))
show(f(1/3))


However I get outputs 0 and 1 respectively. Can someone clarify please? Thanks.

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Sometimes one should constrain sage to evaluate...

def f(x):
if bool(0<= x<= 1/2):
return 1
else:
return 0

assume(0<= x<= 1/2)
show(f(x))
show(f(1/3))


This gives:

\newcommand{\Bold}{\mathbf{#1}}1
\newcommand{\Bold}{\mathbf{#1}}1


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I think the issue is the double inequality in assume: only the first one is taken into account, as one can check with the function assumptions(), which returns the list of assumptions known to Sage:

sage: assume(0 <= x <= 1/2)
sage: assumptions()
[0 <= x]


Actually, one shall use assume with two single inequalities instead of a double inequality:

sage: assume(0 <= x, x <= 1/2)
sage: assumptions()
[0 <= x, x <= (1/2)]


Then your function will return 1 for f(x), as expected.

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This is clearly a bug, thanks for reporting ! See trac ticket 24726

Note that this is actually not the problem with assume but with symbolic expression involving more than one comparison operator:

sage: 0 <= x <= 1/2
0 <= x

sage: e = 0 <= x <= 1/2
sage: e.operator()
<built-in function le>
sage: e.operands()
[0, x]


Note the following weird behaviour:

sage: assume(0 <= x)
sage: 0 <= x <= 1/2
x <= (1/2)


This is because now 0 <= x is evaluated to True, and True and something returns something.

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