There is a similar algorithm to the eucldian division. Look here:
Division by increasing power order | Mathematical Garden
Regards.
Fernando Revilla
Hi. How do I find the power series for tanh(z) (in the complex plane, if that matters, which it doesn't).
I know the power series for sinh(z) and cosh(z) (same as sin(t), cos(t), but without the sign changes):
Now what? You can't divide series (or, at a minimum, I don't know how to divide series). So what's next?
Thanks,
-J
There is a similar algorithm to the eucldian division. Look here:
Division by increasing power order | Mathematical Garden
Regards.
Fernando Revilla
You can use the same derivation as any other Taylor series:
.
You have to evaluate as well as the first, second, third, etc. derivatives at the point .
You should find that when you substitute the values into your Taylor series formula that you can simplify using Bessel numbers.
An 'a little unconventional' approach: the function is the solution of the following ODE...
(1)
If You suppose that is analytic in [and that is true...] is...
(2)
All right!... now from (1) and (2) You can derive the as follows...
... and so one...
Kind regards
You can find the power series for here. The formula for the coefficient of involves Bernoulli numbers.