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I am trying to 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!

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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 $\sum_{n=0}^\infty x^n$.

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