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.

$\tan(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1} 2^{2n}(2^{2n}-1) B_{2n}}{(2n)!} z^{2n-1}$
• 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 ( $B_n$):

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

and therefore:

$B_n=\lim_{z\to 0}\frac{d^n}{dz^n}\left(\frac{z}{e^z-1}\right)_{z=0}$
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:

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

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

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

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

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

Now note the left size can be written as:

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

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

Letting $z=2iw$ then:

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

which simplifies to:

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

or:

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

we now use the identity:

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

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

which then simplifies to:

$\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