# Power series for tangent function

• Oct 18th 2009, 03:01 PM
shawsend
Power series for tangent function
Derive the power series for the tangent function starting with the defining equation for Bernoulli numbers.

$\displaystyle \tan(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}}{(2n)!} z^{2n-1}$
• Oct 24th 2009, 05:41 AM
shawsend
Thought I should post this in case anyone is interested in the derivation:

Spoiler:

We first consider the generating function for Bernoulli numbers ($\displaystyle B_n$):

$\displaystyle \frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{B_n z^n}{n!}$

and therefore:

$\displaystyle B_n=\lim_{z\to 0}\frac{d^n}{dz^n}\left(\frac{z}{e^z-1}\right)_{z=0}$
We can easily calculate the first three Bernoulli numbers and the remaining numbers can be calculated via contour integration and expressed in terms of the zeta function at the negative integers. We find:

$\displaystyle \{B_n\}=\{1,-1/2,1/6,0,B_4,0,B_6,0,B_8,0,\cdots\}$

where all even Bernoulli numbers are non-zero and $\displaystyle B_{2n+1}=0$ for $\displaystyle n=1,2,3,\cdots$. Now, add $\displaystyle \frac{z}{2}$ to both sides of the generating expression:

$\displaystyle \frac{z}{e^z-1}+\frac{z}{2}=\left(\frac{z}{2}\right)+B_0+B_1 z+B_2\frac{z^2}{2}+\cdots)$

but $\displaystyle B_1=-1/2$ and the remaining odd Bernoulli numbers are zero. Thus:

$\displaystyle \frac{z}{e^z-1}+\frac{z}{2}=B_0+B_2\frac{z^2}{2}+\frac{B_4 z^4}{4!}+\frac{B_6 z^6}{6!}+\cdots)$
We can then write the expression as:
$\displaystyle \frac{z}{e^z-1}+\frac{z}{2}=\sum_{n=0}^{\infty}\frac{B_{2n}z^{2 n}}{(2n)!}$

Now note the left size can be written as:

$\displaystyle \frac{z}{e^z-1}+\frac{z}{2}=\frac{z}{2}\left(\frac{1+e^z}{e^z-1}\right)=\frac{z}{2}\left(\frac{e^{z/2}+e^{-z/2}}{e^{z/2}-e^{-z/2}}\right)$
so therefore:

$\displaystyle \frac{z}{2}\coth(\frac{z}{2})=\sum_{n=0}^{\infty}\ frac{B_{2n}z^{2n}}{(2n)!}$

Letting $\displaystyle z=2iw$ then:

$\displaystyle iw\coth(iw)=\sum_{n=0}^{\infty}\frac{B_{2n}(2iw)^{ 2n}}{(2n)!}$

which simplifies to:

$\displaystyle w\cot(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n}}{(2n)!}$

or:

$\displaystyle \cot(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n}w^{2n-1}}{(2n)!}$

we now use the identity:

$\displaystyle \cot(w)-2\cot(2w)=\tan(w)$ and therefore:

$\displaystyle \tan(w)=\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} w^{2n-1}}{(2n)!}-2\sum_{n=0}^{\infty}\frac{(-)^n B_{2n}(2n)^{2n-1}w^{2n-1}}{(2n)!}$

which then simplifies to:

$\displaystyle \tan(w)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}\left(2^{2n}-1\right) B_{2n} w^{2n-1}}{(2n)!},\quad -\pi/2<w<\pi/2$