As the title states, I'm trying to find the coefficients $a_0$, $a_1$ and $a_2$ of the power series $\sum_{n = 0}^\infty a_n z^n$ around 0 of \begin{align*} \frac{z + 1}{(z + 2) \cos z}, \end{align*} where we use the power series of $\cos z$ around 0 as well.
Now as far as I'm aware, this can be done pretty easily by using the complex derivatives $f^{(n)}$, however I've been tasked NOT to use any derivatives, so this method is out of the question. This has left my puzzled on what to do thus far, if any of you could provide a hint or a direction for a suitable method, I would be grateful.
What I've done so far:
- Obviously just replacing $\cos z$ by its power series does not work, since you cannot get the sum out of the numerator.
- Secondly, I tried replacing $\cos z$ by the usual $\frac{1}{2}(e^{iz} + e^{-iz})$, but that gives the same issue..
Thank you for any help you can provide!
Note that $\frac1{1+y}=1-y+y^2-y^3\pm\ldots$, so $$\frac1{2+z}=\frac12\cdot\frac1{1+z/2} =\frac12\left(1-\frac12z+\frac14z^2\pm\ldots \right)$$ $$ \frac1{\cos z}=\frac1{1-\frac12z^2\pm\ldots}=1+\frac12z^2\pm\ldots$$ and finally by putting enough trust into the well-behaving of converging power series $$ \begin{align}\frac{z+1}{(z+2)\cos z}&=(1+z)\cdot\left(\frac12-\frac14z+\frac18z^2\pm\ldots\right)\cdot\left(1+\frac12z^2\pm\ldots\right)\\&=\frac12+\frac14z+\frac18z^2+\ldots\end{align}$$ (it does not continue with $\frac1{16}z^3$)