j0 (and y0) in the range 2 <= x < (p/2)*log(2)
Steve Kargl
sgk at troutmask.apl.washington.edu
Sat Sep 8 04:49:39 UTC 2018
On Fri, Sep 07, 2018 at 10:54:06AM -0700, Steve Kargl wrote:
>
> I'll probably try a different approach tomorrow. j0(x),
> cc, and ss are smoothly varying functions of x. u and v
> appear to be slowly varying (in small intervals), so we can
> have
>
> j0(x) = sqrt(1/pi) * (cc(x) *u-ss(x) *v) / sqrt(x).
> j0(x+e) = sqrt(1/pi) * (cc(x+e)*u-ss(x+e)*v) / sqrt(x+e).
>
> where e is some small perturbution. Two equation and two
> unknowns is easily solved. Simply need to do this across
> the interval to generate u(x) and v(x).
>
I may have have a better approach!
j0(x) = sqrt(1/pi) * (cc(x) * u - ss(x) * v) / sqrt(x)
y0(x) = sqrt(1/pi) * (cc(x) * u + ss(x) * v) / sqrt(x)
j0(x) + y0(x) = 2 * sqrt(1/pi) * cc(x) * u
j0(x) - y0(x) = - 2 * sqrt(1/pi) * ss(x) * v
So, we have
u = (j0 + y0) / (2 * sqrt(1/pi) * cc)
v = (y0 - j0) / (s * sqrt(1/pi) * ss)
Thus, we can find u = r/s or u = 1 + r/s and v = r/s.
--
Steve
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