Function bessel_jy_real 

pub fn bessel_jy_real<P, E, V, const NH: usize, const NE: usize, const NO: usize, const FN: LargeInt, const FD: LargeInt>(
    nu: V,
    x: V,
    x_lo: V,
    t: &LogGamma1p<E, NE, NO>,
) -> (V, V)
where
    E: FloatElement,
    V: FloatVector<Element = E> + TranscendentalMathWithPolicy + SpecializedTranscendentalMath<E> + SpecializedSpecialMath<E>,
    P: Policy,

$(J_\nu(x), Y_\nu(x))$ at arbitrary real order, over the whole positive axis.

The region select over the four arms. Everything above this line is a piece. This is the function.

Three regions, chosen to avoid overlap rather than to minimise cost

x	J	Y
<= 2	series_j_nu	temme_y_nu + upward recurrence
2 .. gate	steed_jy_nu	same pass
>= gate	hankel_jy_nu	same pass

gate is hankel_usable_from. The ascending series is usable to about x = 6 and Steed from about 0.5, so the 2 boundary sits inside both their ranges. It is where Steed stops being cheap (its CF2 needs 42 iterations at x = 2 and 5392 at 0.01) rather than where it stops being right. Boost splits at the same place for the same reason.

Every lane pays for every region any lane is in. Each arm therefore receives the mask of lanes that actually want it, so a lane bound elsewhere cannot extend an iteration. See iterate.

Negative order, and why it is a rotation rather than a sign

At non-integer $\nu$, $J_\nu$ and $J_{-\nu}$ are linearly independent (not a sign apart, as they are at whole orders), so the pair rotates:

J_{-\nu} = J_\nu\cos\nu\pi - Y_\nu\sin\nu\pi, \qquad
Y_{-\nu} = J_\nu\sin\nu\pi + Y_\nu\cos\nu\pi

Everything is computed at $\lvert\nu\rvert$ and rotated once at the end. At whole orders $\sin\nu\pi$ vanishes and this collapses to the familiar $(-1)^n$.

Order reduction, and only where it is needed

temme_y_nu requires $\lvert\nu\rvert \le 1/2$. So in the small-x region the order is split as $\nu = m + u$ with $m$ whole and $\lvert u\rvert \le 1/2$, Temme evaluated at $u$, and $Y$ walked up $m$ steps by $Y_{k+1} = (2k/x)Y_k - Y_{k-1}$. That direction is stable because $Y$ is the dominant solution, the mirror of $J$, where the same direction is the unstable one.

The other two regions need no reduction. Steed and the Hankel expansion take $\nu$ directly.