# Thread: Power series for tanh?

1. ## Power series for tanh?

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):
$\sinh(z)=\sum_{j=0}^\infty \frac{z^{2j+1}}{2j+1!}$
$\cosh(z)=\sum_{j=0}^\infty \frac{z^{2j}}{2j!}$

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

2. Originally Posted by MSUMathStdnt
Now what? You can't divide series (or, at a minimum, I don't know how to divide series). So what's next?
There is a similar algorithm to the eucldian division. Look here:

Division by increasing power order | Mathematical Garden

Regards.

Fernando Revilla

3. You can use the same derivation as any other Taylor series:

$\displaystyle f(x) = \sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n$.

You have to evaluate $\displaystyle \tanh{x}$ as well as the first, second, third, etc. derivatives at the point $x = a = 0$.

You should find that when you substitute the values into your Taylor series formula that you can simplify using Bessel numbers.

4. An 'a little unconventional' approach: the function $y(x)= \tanh x$ is the solution of the following ODE...

$\displaystyle y^{'} = 1 - y^{2}, y(0)=1$ (1)

If You suppose that $\tanh x$ is analytic in $x=0$ [and that is true...] is...

$\displaystyle \tanh x = \sum_{n=0}^{\infty} a_{n}\ x^{n} = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (2)

All right!... now from (1) and (2) You can derive the $a_{n}$ as follows...

$\displaystyle y(0)=0 \implies a_{0}=0$

$\displaystyle y^{'}= 1-y^{2} \implies y^{'}(0)= 1 \implies a_{1}=1$

$\displaystyle y^{''} = -2 y y^{'} \implies y^{''} (0)=0 \implies a_{2}=0$

$\displaystyle y^{(3)} = -2 y y^{''} -2 y^{'\ 2} \implies y^{(3)} (0)= -2 \implies a_{3}= -\frac{1}{3}$

$\displaystyle y^{(4)} = -2 y y^{(3)} - 6 y^{'} y^{''} \implies y^{(4)}=0 \implies a_{4}=0$

$\displaystyle y^{(5)} = -8 y^{'} y^{(3)} -6 y^{''\ 2} -2 y y^{(4)} \implies y^{(5)}(0)= 16 \implies a_{5} = \frac{2}{15}$

... and so one...

Kind regards

$\chi$ $\sigma$

5. You can find the power series for $\tanh z$ here. The formula for the coefficient of $z^{2j+1}$ involves Bernoulli numbers.