I suppose you could have some function like RPSqrt, for realpositivesqrt
which maps from non-negative reals to non-negative reals.
It would be an error to type RPSqrt(x) unless x were guaranteed to be
oretty much
explicitly in [0,oo].
Sqrt(x^2) under some conditions might be considered RPSqrt(
On Thursday, August 6, 2020 at 4:07:11 AM UTC-4 Markus Wageringel wrote:
> Even if there are two possible choices, the result of the definite
> integral should be ±8, not 0. It is rather strange to pick the positive
> square root for half the integral and then (discontinuously) the negative
>
Even if there are two possible choices, the result of the definite integral
should be ±8, not 0. It is rather strange to pick the positive square root
for half the integral and then (discontinuously) the negative one for the
other half.
There is a ticket for exactly this integral, by the way:
2020-08-05 18:59:01 UTC, rjf:
>
> There are two square roots. In this (classic) integration
> example/bug, a choice has to be made. You know that 4 has
> two square roots, -2 and 2. The integrand, which also can
> be rewritten as sqrt ( 4-4*cos(x/2)^2) , has 2 square
> roots. Therefore there a
On 2020-08-05 15:49, NicoJG wrote:
> @rjf Isn't the square root defined to be positive?
> Sure: x^2=y <=> x=+/-sqrt(y)
> But I think you would never consider f(x):=sqrt(x) to have the codomain
> of all negative numbers.
> At least I would expect a CAS to interpret a square root to be positive.
>
On Wednesday, August 5, 2020 at 12:49:37 PM UTC-7, NicoJG wrote:
>
> @rjf Isn't the square root defined to be positive?
> Sure: x^2=y <=> x=+/-sqrt(y)
>
But I think you would never consider f(x):=sqrt(x) to have the codomain of
> all negative numbers.
>
With complex numbers, there's no concept
I agree with the general analysis, but I think the statement "Any answer
that supplies only one answer is wrong." goes too far. It may be the case
that sage works inherently in the complex domain, and is unable to
understand that elementary calculus and certain other fields want to remain
in t
@rjf Isn't the square root defined to be positive?
Sure: x^2=y <=> x=+/-sqrt(y)
But I think you would never consider f(x):=sqrt(x) to have the codomain of
all negative numbers.
At least I would expect a CAS to interpret a square root to be positive.
rjf schrieb am Mittwoch, 5. August 2020 um 20:
There are two square roots. In this (classic) integration example/bug, a
choice has
to be made. You know that 4 has two square roots, -2 and 2.
The integrand, which also can be rewritten as sqrt ( 4-4*cos(x/2)^2) ,
has 2 square roots.
Therefore there are two potential different values for t
BTW :
sage: integrate(sqrt(2-2*cos(x)),x, algorithm="fricas")
-2*(cos(x) + 1)*sqrt(-2*cos(x) + 2)/sin(x)
sage: integrate(sqrt(2-2*cos(x)),x, algorithm="mathematica_free")
-2*sqrt(-2*cos(x) + 2)*cot(1/2*x)
Both are visually (on plot) and numerically correct ; both differentiate to
expressions v
Hi,
I discovered a bug, where a definite integral is calculated wrong!
WolframAlpha result for comparison.
Code:
integrate(sqrt(2-2*cos(x)),x,0,2*pi)
Also if I type show() instead of print() SageMathCell just doesn't show
anything.
Also the form in which the indefinite integral is given is not
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