Kelvin functions

Remarks

The functions $J_\nu(x e^{3\pi i/4})$ and $K_\nu(x e^{\pi i/4})$ solve $x^2 w'' + x w' - (i x^2 + \nu^2) w = 0$, since $\left( x e^{3\pi i/4} \right)^2 = -i x^2$. Example: for $\nu = 0$, $J_0(z) = \sum_m \frac{(-1)^m (z/2)^{2m}}{(m!)^2}$ and $(-1)^m e^{3\pi i m/2} = i^m$, so the real part keeps the terms with $m = 2k$ and gives $\operatorname{ber}_0(x) = \sum_{k \ge 0} \frac{(-1)^k (x/2)^{4k}}{((2k)!)^2}$. The odd terms give $\operatorname{bei}_0(x) = \sum_{k \ge 0} \frac{(-1)^k (x/2)^{4k+2}}{((2k+1)!)^2}$. A non-example: $J_0(x)$ on the real axis is not $\operatorname{ber}_0(x)$; its series has the term $-\frac{x^2}{4}$, which $\operatorname{ber}_0$ lacks, because the argument is not rotated to the ray $\arg z = \frac{3\pi}{4}$. When $\nu = 0$ the index is usually dropped: $\operatorname{ber} = \operatorname{ber}_0$, and so on.