On Tue, Aug 11, 2020 at 6:57 AM My Name <[email protected]> wrote:
>
> Thanks very much for the explanation.
>
> The assumptions issues in Sympy have deterred me from using Sympy almost
> completely. It's not clear to me why the old assumption system had to be
> replaced. It's not clear to me what's unimplemented in the new system.
>
> You say above, "The point about relations not being implemented is that there
> is not currently a way to use the assumption that e.g. x > y but that's not
> needed here so this should work". But your code shows Sympy "using" the
> assumptions y > -∞ ∧ y < ∞, which are relations. I you are using the word
> "use" in a way I don't understanding, and I'm left with this vague warning
> about relations being unusable in Sympy, and I go back to Maple.
The old assumptions only implement a subset of possible intervals,
with adjectives like "positive", "negative", "finite". You can
represent something like x > 0 because that's the same thing as
"positive", but you can't represent x > y because that isn't
representable by the set of things that the old assumptions can
represent. The only way to represent more advanced relations like x >
y is to use a different syntax from what the old assumptions use,
which is the idea behind the new assumptions.
Also, the old assumptions aren't going away. We originally were going
to do that, but the current plan is to keep it around as it works just
fine for those things that it can represent.
Aaron Meurer
>
> On Monday, August 10, 2020 at 3:53:01 AM UTC-7 Oscar wrote:
>>
>> You can use the old assumptions:
>>
>> In [10]: x = Symbol('x')
>>
>>
>> In [11]: y = Symbol('y')
>>
>>
>> In [12]: a = Symbol('a')
>>
>>
>> In [13]: integrate(cos(x*y), (x, 0, a))
>>
>> Out[13]:
>>
>> ⎧sin(a⋅y)
>>
>> ⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
>>
>> ⎨ y
>>
>> ⎪
>>
>> ⎩ a otherwise
>>
>>
>> In [14]: y = Symbol('y', nonzero=True)
>>
>>
>> In [15]: integrate(cos(x*y), (x, 0, a))
>>
>> Out[15]:
>>
>> sin(a⋅y)
>>
>> ────────
>>
>> y
>>
>>
>>
>> The new assumptions and refine can handle nonzero. The point about relations
>> not being implemented is that there is not currently a way to use the
>> assumption that e.g. x > y but that's not needed here so this should work:
>>
>> In [16]: y = Symbol('y')
>>
>>
>> In [17]: integrate(cos(x*y), (x, 0, a))
>>
>> Out[17]:
>>
>> ⎧sin(a⋅y)
>>
>> ⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
>>
>> ⎨ y
>>
>> ⎪
>>
>> ⎩ a otherwise
>>
>>
>> In [18]: refine(_, Q.nonzero(y))
>>
>> Out[18]:
>>
>> ⎧sin(a⋅y)
>>
>> ⎪──────── for y > -∞ ∧ y < ∞ ∧ y ≠ 0
>>
>> ⎨ y
>>
>> ⎪
>>
>> ⎩ a otherwise
>>
>>
>>
>> I guess that this just isn't implemented yet in refine.
>>
>>
>> Oscar
>>
>>
>>
>> On Monday, 10 August 2020 11:06:02 UTC+1, My Name wrote:
>>>
>>> I do this:
>>>
>>> import sympy
>>> sympy.srepr(sympy.integrate(S('cos(x * y)'), S('(x, 0, a)')))
>>>
>>> It returns this:
>>>
>>> "Piecewise(ExprCondPair(Mul(Pow(Symbol('y'), Integer(-1)),
>>> sin(Mul(Symbol('a'), Symbol('y')))), And(StrictGreaterThan(Symbol('y'),
>>> -oo), StrictLessThan(Symbol('y'), oo), Unequality(Symbol('y'),
>>> Integer(0)))), ExprCondPair(Symbol('a'), true))"
>>>
>>> How do I find the value of the Piecewise expression when y is nonzero? I
>>> have tried using sympy.refine with Q.nonzero, but that has not worked.
>>> Indeed the docs for refine warn, "Relations in assumptions are not
>>> implemented (yet)". Does that mean there's no way to find the value of the
>>> integral assuming y is nonzero?
>
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