Deriving the power series for $\cot(x)$ and $\coth(x)$

 In the previous post, we have successfully derived the power series for $\dfrac{x}{e^x-1}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_nx^{n}}{n!}$. We know that the coefficients for this power series are the Bernoulli numbers. This result is important because it allows us to arrive at the power series of $\cot(x)$ and $\coth(x)$. In this post, we will see how the Bernoulli numbers are related to the expansion of $\cot(x)$ and $\coth(x)$.

(added in Feb, 11, 2021):

The derivation of both functions are particularly important to me, for it helps me to derive the summing formula of zeta function at even integers. At first, I knew that there was a connection between the power series of $\dfrac{x}{e^x-1}$, then I tried to rewrite $\cot(x)$ and $\coth(x)$ in exponential forms. After tinkering for a while, I was unable to make a connection between them and $\dfrac{x}{e^x-1}$, so I looked up on wikiproof. Their approach is exactly what I was doing, so I continued, only looking at wikiproof when I was unable to go on. Eventually the formula arrived. I was able to derive them. I believe that my symbolic manipulation ability is still very weak and deplorable, but doing these kinds of derivations certainly improve it. I fumbled at $(1)$ (see below) and re-indexing the summation formula. After seeing wikiproof, I totally understand it.

Deriving power series for $\cot(x)$:

We have proved here that $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$. By definition:

$$\cot(x)=\dfrac{\cos(x)}{\sin(x)}=\dfrac{e^{ix}+e^{-ix}}{2}\cdot\dfrac{2i}{e^{ix}-e^{-ix}}=i\left(\dfrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}\right)\tag{1}$$

Multiply the numerator and denominator for $e^{ix}$, we have:

$$i\left(\dfrac{e^{2ix}+1}{e^{2ix}-1}\right)\tag{2}$$

This is the tricky part, we need to make the $\dfrac{2ix}{e^{2ix}-1}$ appear to use substitution to obtain the power series for $\left(\dfrac{u}{e^{u}-1}\right)$ with $u=2ix$

$$i\left(\dfrac{e^{2ix}+1}{e^{2ix}-1}\right)=i\left(\dfrac{e^{2ix}-1+2}{e^{2ix}-1}\right)=i\left(1+\dfrac{2}{e^{2ix}-1}\right)=i+\dfrac{2i}{e^{2ix}-1}\tag{3}$$

We factor out $\dfrac{1}{x}$ from $\dfrac{2i}{e^{2ix}-1}$, and we obtain:

$$i+\dfrac{1}{x}\left(\dfrac{2ix}{e^{2ix}-1}\right)=i+\dfrac{1}{x}\left(\sum_{n=0}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)$$

Next, we need to "release" two terms from the sigma notation chunk above. Why do we need to release the terms and readjust the index $n=2$? Because we want to take out the first odd Bernoulli number $B_1=-\dfrac{1}{2}$. Then we can adjust the index to include all even (non-zero) Bernoulli numbers. The first 2 terms to be released is $1$ and $-ix$. We need to remember that the second term is eliminated, but the first term remained. So when we shift the index of the summation notation to $n=0$, we need to include the first term to avoid duplicating the terms.

$$i+\dfrac{1}{x}\left(\color{red}{1}-ix+\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)=i+\color{red}{\dfrac{1}{x}}-i+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)=\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)$$

Convert the summation index to $n=1$, and $B_n=B_{2n}$ and change all index of $n$ to $2n$ to include all even Bernoulli numbers, we have:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=1}}^{+\infty}\dfrac{B_{2n}(2ix)^{2n}}{(2n)!}\right)$$

Shift the summation index to $n=0$, we need to eliminate $\dfrac{1}{x}$ to avoid duplicating terms. In other words $\dfrac{1}{x}$ is "absorbed" into the sum. We have:

$$\dfrac{1}{x}\left(\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}(2ix)^{2n}}{(2n)!}\right)=\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}2^{2n}(i^{2})^nx^{2n-1}}{(2n)!}$$

Finally, $(i^2)^n=(-1)^n$, so we arrived at the final conclusion of $\cot(x)$

$$\boxed{\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}}$$

Deriving power series for $\coth(x)$

We have proved here that $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ and $\cosh(x)=\dfrac{e^{x}+e^{-x}}{2}$. By definition:

$$\coth(x)=\dfrac{\cosh(x)}{\sinh(x)}=\dfrac{e^{x}+e^{-x}}{2}\cdot\dfrac{2}{e^{x}-e^{-ix}}=\left(\dfrac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right)\tag{1}$$

Multiply the numerator and denominator for $e^{x}$, we have:

$$\dfrac{e^{2x}+1}{e^{2x}-1}\tag{2}$$

This is the tricky part, we need to make the $\dfrac{2x}{e^{2x}-1}$ appear to use substitution to obtain the power series for $\left(\dfrac{u}{e^{u}-1}\right)$ with $u=2x$

$$\dfrac{e^{2x}+1}{e^{2ix}-1}=\dfrac{e^{2x}-1+2}{e^{2x}-1}=1+\dfrac{2}{e^{2x}-1}\tag{3}$$

We factor out $\dfrac{1}{x}$ from $\dfrac{2}{e^{2x}-1}$, and we obtain:

$$1+\dfrac{1}{x}\left(\dfrac{2x}{e^{2x}-1}\right)=1+\dfrac{1}{x}\left(\sum_{n=0}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)$$

Similarly, as above, we have

$$1+\dfrac{1}{x}\left(\color{red}{1}-x+\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)=1+\color{red}{\dfrac{1}{x}}-1+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)=\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)$$

Convert the summation index to $n=1$, and $B_n=B_{2n}$ and change all index of $n$ to $2n$ to include all even Bernoulli numbers, we have:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=1}}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{(2n)!}\right)$$

Shift the summation index to $n=0$, we need to eliminate $\dfrac{1}{x}$ to avoid duplicating terms. In other words $\dfrac{1}{x}$ is "absorbed" into the sum. We have:

$$\dfrac{1}{x}\left(\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{(2n)!}\right)=\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

We arrived at the final conclusion of $\coth(x)$

$$\boxed{\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}}$$

Deriving power series for $\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)$

Why are we interested in the series expansion of this function? Because it is the generating function of even Bernoulli numbers. Since we have derived the series representation for $\cot(x)$ and $\coth(x)$, we can go on to easily produce the series for these functions.

$\cot(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\implies\cot\left(\dfrac{x}{2}\right)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}\left(\dfrac{x}{2}\right)^{2n-1}}{(2n)!}$

$$=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{2^{2n-1}(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2x^{2n-1}}{(2n)!}$$

$$\implies\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)=\boxed{\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}x^{2n}}{(2n)!}}$$

Deriving power series for $\dfrac{x}{2}\coth\left(\dfrac{x}{2}\right)$

We can repeat the same process as above, utilizing the series of $\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)$ and obtain the following result:

$$\boxed{\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}x^{2n}}{(2n)!}}$$