I found the following formula here: Taylor Series of $\tan x$.
Taylor series of $\tan x$:
$$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n – 1} $$.
I do not understand what the Bernoulli number means in this formula. My problems are the following ones:
I do not know anything about Bernoulli numbers. I do not know how to expand the Bernoulli number in the formula you see further up. What expansion will be the final result? How did you arrive there? Why did you arrive there? How did you know that you should use the specific Bernoulli-related terms that you used in your answer? What is the purpose of the Bernoulli number in this formula?
A great deal of thanks.
Best Answer
The exponential generating function for the Bernoulli Numbers is $$ \sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac{x}{e^x-1}\tag{1} $$ The even part of $(1)$ is $$ \sum_{n=0}^\infty\frac{B_{2n}x^{2n}}{(2n)!}=\frac x2\coth\left(\frac x2\right)\tag{2} $$ Since $\cot(x)=i\coth(ix)$, by substituting $x\mapsto ix$, we get $$ \begin{align} \sum_{n=0}^\infty\frac{(-1)^nB_{2n}x^{2n}}{(2n)!} &=\frac{ix}2\coth\left(\frac{ix}2\right)\\ &=\frac x2\cot\left(\frac x2\right)\tag{3} \end{align} $$ Therefore, $$ \cot(x)=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4} $$ Since $\tan(x)=\cot(x)-2\cot(2x)$, we get $$ \begin{align} \tan(x) &=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}\left(x^{2n-1}-2\cdot2^{2n-1}x^{2n-1}\right)}{(2n)!}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}B_{2n}\left(2^{2n}-1\right)2^{2n}x^{2n-1}}{(2n)!}\tag{5} \end{align} $$