I feel a bit daft asking, but I'm desperately trying to find the general expression for the Taylor Series of $\tan^2(x)$ at $x = 0$.
As given here we see the Taylor Series for $\tan(x)$:
$$ \tan(x) = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} x^{2n-1}$$
As such:
\begin{align}
\tan^{2}(x) &= \left[\sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} x^{2n-1}\right]^{2} \\
&= \left[\sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} x^{2n-1}\right]\left[\sum^{\infty}_{m=1} \frac{B_{2m} (-4)^m \left(1-4^m\right)}{(2m)!} x^{2m-1}\right] \\
&= \sum^{\infty}_{n=1}\sum^{\infty}_{m=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} x^{2n-1}\frac{B_{2m} (-4)^m \left(1-4^m\right)}{(2m)!} x^{2m-1} \\
&= \sum^{\infty}_{n=1}\sum^{\infty}_{m=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)}{(2n)!} \frac{B_{2m} (-4)^m \left(1-4^m\right)}{(2m)!} x^{2n + 2m-2}
\end{align}
I am lost however in trying to find $a_{n}$ such that,
$$\tan^{2}(x) = \sum_{n} a_{n} x^{n} $$
Does anyone have any pointers on how to approach this problem?
Also, is it easier to work with $\sec^2(x)$ instead?
Best Answer
If $\tan x=\sum_{n=0}^{\infty}a_nx^{n}$, and we take an interval
$-\pi/2<-r<0<r<\pi/2$ then for $x\in (-r,r)$, we have
$1+\tan^2x=\sec^2x=\frac{d\tan x}{dx}\Rightarrow$
$1+\tan^2x=\sum_{n=1}^{\infty}na_nx^{n-1}\Rightarrow$
$\tan^2x=\sum_{n=1}^{\infty}na_nx^{n-1}-1$
and so you can get the Taylor coefficients of $\tan^2$ from those of $\tan.$