Custom Query (1030 matches)

In ccl 1.6 (32-bit) we have

1 > (atanh #c(-2d0 -1d-20))
#C(-0.5493061443340549D0 -1.5707963267948966D0)

but

1 > (atanh #c(-2d0 -0d0))
#C(-0.5493061443340549D0 1.5707963267948966D0)

The spec says that for on the negative real axis less than -1, atanh is continuous with Quadrant III. Since (float-sign -0d0) -> -1d0, both answers should have the same signs and the imaginary part should be negative.

As the subject says, (abs #c(1d300 1d300)) signals an overflow. But in fact, the answer is approximately 1.414d300, which still fits in a double-float.

The implementation is probably just doing sqrt(x^2+y^2). A better answer can be achieved using |x|*sqrt(1+(y/x)^2) if |x| >= |y|, with obvious modifications for |x| <= |y|.

(log #c(1d300 1d300)) signals overflow, but it shouldn't since the absolute value of #c(1d300 1d300) fits in a double-float.

This could be related to ticket:830.