Just been having a look at your problem, of shin(z)/z=n, lets re-arrange
this to:
shin(z)=nz.

I'm assuming n is an integer, in which case I'm pretty sure therre will
be 2 roots. One will be pretty small <1, for n being reasonably high. 
In which case we can use the Taylor series expansion.  The other root we
won't be able to use the series, as it is non-convergent.  the problem
is, unless n is really small, we'll have to carry a lot of the tS to get
a reasonable answer, if n > 100 it should be OK (and we'd just need the
first 3 terms), but otherwise we'll have to solve the TS numerically,
I'd recommend bisection as I'm not sure Newton will work (but I always
use bisection - don't like the idea of a series which might not converge
- I'd prefer to guarantee it works and use a little more processor).  If
you have no idea about what I'm talking about, just email and I'll
explain further.

The Taylor Series is by the way :
  shin(z)/z= 1 + z^2/3! + z^4/5! ..... z^n/(n+1)!


So if we take the first four terms and then sub into the quadratic
formulae, we should get the answer for Z^2, root that and we would get
an answer, which for n>100 would be about 10^-5 accurate.  This is all
very roughly estimated, BTW.

Post a reply to this message