Yeah, a lot of work is needed in sets.
I see that error on master but not on my PR:
https://github.com/sympy/sympy/pull/17593
There it doesn't raise but just returns unevaluated:
In [3]: simplify(a.intersect(b))
Out[3]: {(t, t) | t ∊ [0, 1]} ∩ {(1 - t, t) | t ∊ [0, 1]}
There is no code to
OK, I wasn't sure if that was the intended way, since using tuples quickly
leads to errors such as:
>>> a = ImageSet(Lambda(t, (t, t)), Interval(0, 1))
>>> a
ImageSet(Lambda(t, (t, t)), Interval(0, 1))
>>> (0,0) in a
True
>>> b = ImageSet(Lambda(t, (1 - t, t)), Interval(0, 1))
>>> b
Yes, you can do that with ImageSet:
In [11]: ImageSet(Lambda(t, (t, t)), Interval(0, 1))
Out[11]: {(t, t) | t ∊ [0, 1]}
In [14]: ImageSet(Lambda(t, (t**2, t, 1-t)), Interval(0, 1))
Out[14]:
⎧⎛ 2 ⎞ ⎫
⎨⎝t , t, 1 - t⎠ | t ∊ [0, 1]⎬
⎩ ⎭
On Tue, 1 Oct
For example if I wanted to represent the line segment from (0,0) to (1,1)
(f(t) = (t, t) for t in (0, 1))- can this be done with an image set? How?
Or a curve in 3d space (for example, f(t) = (t^2, t, 1 - t) for t in (0,
1)).
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I meant to share this at the beginning of Google Season of Docs but I
couldn't find it until now. This is a talk that was shared by the GSoD
admins at some point.
https://www.youtube.com/watch?v=t4vKPhjcMZg
There's also a blog post version of the talk if you don't have time to
watch it.
