Taylor Series Expansion for $\tan x$

Question

I'm trying to determine the Taylor series expansion for $\tan x$:

I know that the $n$th derivative of the expansion must be the same as the $n$th derivative of the function.

Please help, I have no idea what to do. :'(

Answer 1

You may calculate derivative using $\tan=\sin/\cos$, as comment say. But you can also try the following strategy.

Let $\tan x=\sum_n a_nx^n$

you know that the derivative of $\tan x$ equals $1+(\tan x)^2$

So you know

$$\sum_n na_nx^{n-1}=1+(\sum_i a_ix^i)(\sum_i a_i x^i)=1+\sum_n(\sum_{k+m=n-1}a_ka_m)x^n$$

You also knonw that $\tan(-x)=-\tan x$, whence $a_n=0$ for $n$ even.

Using this you can arrange an inductive process to calculate the taylor expansion. (Namely, $a_{2k}=0, a_1=1, a_3=1/3, a_5=2/15...$ and so on)

See also http://en.wikipedia.org/wiki/Taylor_series and http://en.wikipedia.org/wiki/Bernoulli_numbers for the explicit result