MacLaurin Series of $\tan(x)$

Question

I am trying to answer the following:

Compute the MacLaurin series of $\tan(x)$.

I know this one is: $$\tan x=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}x^{2n-1}}{(2n)!}$$ And I know how to derive this formula. Indeed I simply express $\tan$ as a linear combination of $\cot(x)$ and $\cot(2x)$, for which we know the explicit formula $$\cot(x)=\sum_{n=0}^\infty \frac{(-1)^n2^{2n}B_{2n}x^{2n-1}}{(2n)!}$$ This formula is derived by writing $\cot$ in its exponential form and doing some algebra.

I know how to derive these formulas, but I do not understand what makes it the MacLaurin series for $\tan(x)$. Why couldn't they be any Taylor series centered somewhere else? And what even makes them a Taylor series, I only see it as a power series...

Thank you for you responses and help!

Answer 1

Because if we have a function $f\colon(a-r,a+r)\longrightarrow\Bbb R$ and a power series$$\sum_{n=0}^\infty a_n(x-a)^n\tag1$$centered at $a$ such that$$\bigl(\forall x\in(a-r,a+r)\bigr):f(x)=\sum_{n=0}^\infty a_n(x-a)^n,$$then automatically $f$ is a $C^\infty$ function and its Taylor series centered at $a$ is $(1)$.

For instance,$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n$$and therefore the Taylor series of $\frac1{1-x}$ centered at $0$ is $\sum_{n=0}^\infty x^n$.

Answer 2

In order to expand the function $\tan x$ at any point $x_0\in\bigl(-\frac\pi2,\frac\pi2\bigr)$, it is sufficient to give a general formula for the $n$th derivative of the function $\tan x$. Such a general formula can be alternatively given as follows. \begin{gather*} (\tan x)^{(n)}=-(\ln\cos x)^{(n+1)}\\ =-\sum_{k=1}^{n+1} \frac{(-1)^{k-1}(k-1)!}{(\cos x)^k} B_{n+1,k}\biggl(-\sin x,-\cos x,\sin x,\cos x,\dotsc, \cos\biggl[x+\frac{(n-k+2)\pi}{2}\biggr]\biggr)\\ =-\sum_{k=1}^{n+1} \frac{(-1)^{k-1}(k-1)!}{(\cos x)^k} \frac{(-1)^k(\cos x)^k}{k!}\sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^\ell}{(2\cos x)^\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{n+1} \cos\biggl[(2q-\ell)x+\frac{(n+1)\pi}2\biggr]\\ =\sum_{k=1}^{n+1} \frac{1}{k}\sum_{\ell=0}^k\binom{k}{\ell}\frac{(-1)^{\ell}}{(2\cos x)^\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{n+1} \cos\biggl[(2q-\ell)x+\frac{(n+1)\pi}2\biggr]. \end{gather*} To the best of my knowledge, there are several other ways to compute the general formula of the $n$th derivative of the function $\tan x$ in the following, but not limited to, references.

References

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