hello,
why is the below code plotting a flat function rather than a box one?
renato
def box(x,c):
if abs(x) c:
return 1
else:
return 0
var('x')
plot(box(x,1),(x,-3,3))
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On Fri, Feb 4, 2011 at 11:37 AM, Renato Budinich renn...@gmail.com wrote:
hello,
why is the below code plotting a flat function rather than a box one?
When you do,
plot(box(x,1),(x,-3,3))
it evaluates box(x,1) which returns 0 because the variable x is not
always less than 1. You need to
hello,
why is the below code plotting a flat function rather than a box one?
There are two things going on. First, in the line
plot(box(x,1),(x,-3,3))
box(x,1) is actually being evaluated when the line is executed, and
not thereafter. IOW you're computing box(x, 1), which is 0, so the
above
On Fri, 4 Feb 2011 19:40:34 +0800
D. S. McNeil dsm...@gmail.com wrote:
hello,
why is the below code plotting a flat function rather than a box
one?
There are two things going on. First, in the line
plot(box(x,1),(x,-3,3))
box(x,1) is actually being evaluated when the line is
plot(lambda x: box(x,1), (x, -3, 3))
but why does this way the execution of the function get delayed?
because lambda is a way to define a function.
This works more or less like the following :
def MyFunction(x)
return box(x,1)
plot(MyFunction,(x,-3,3))
See for example
