Mathematica Asked by Billy Kalfus on October 2, 2021
Mathematica provides the following function to compute the arctangent of a number, preserving quadrant information:
ArcTan[x,y]
for real $x$ and $y$, when $y = 0$ and $x ne 0$, $arctan(y/x) = 0$. However, Mathematica doesn’t perform this simplification:
FullSimplify[ArcTan[x,0], Assumptions->{x [Element] Reals}]
(* ArcTan[x, 0] *)
Is this an error on my part or is there an underlying reason why this simplification isn’t performed?
It is because M does not know the actual numerical value of x
(or rather its sign).
Let say x
was 1
ArcTan[1,0]
But what if x
was -1
?
ArcTan[-1,0]
Correct answer by Nasser on October 2, 2021
Assuming[0 < x < π, Simplify@ArcTan[x, 0]]
Answered by David G. Stork on October 2, 2021
ArcTan simplifications are somehow limited:
Simplify[ArcTan[x, y], {x == y, 0 < y, 0 < x}]
Sadly only gives:
ArcTan[y, y]
Answered by Robert Nowak on October 2, 2021
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