Calculate the power series expansion about $\pi/2$ of $g(z)=\tan(z/2)$.
Now calculate the expansion about $0$.
I'm having trouble doing this. I'm not even sure which is the best way to approach it, for instance I tried deriving and calculating the coefficients but that quickly proved too tedious. Should I try constructing $\tan(z)$ by using the power series of $\sin(z)$ and $\cos(z)$, then plugging in $z/2$ for $z$, or is there an easier way?
Best Answer
What we are looking for is the Taylor series at $z=\pi/2$ for $g(z)$. The Taylor series is defined as $$ f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(z-a)^n $$ Therefore, for $g(z)$, we have \begin{align} g(z) &= g(\pi/2) + g'(\pi/2)(z - \pi/2) + \frac{g''(\pi/2)}{2!}(z - \pi/2)^2 + \cdots\\ &= 1 + (z - \pi/2) + \frac{1}{2!}(z - \pi/2)^2 + \frac{2}{3!}(z - \pi/2)^2 + \cdots\\ &= \sum_{n=0}^{\infty}\frac{(z - \pi/2)^n}{n!}U_{n + 1} \end{align} where $U_n$ are the up down numbers. You only need the first four to figure out you need up down numbers. I checked the next 3 terms as well to verify. As a note, the up down numbers are $$ 1,1,1,2,5,16,61,272,1385,\text{look up rest} $$ In your series above, we start at $U_{n + 1}$ for $n\geq 0$. At $z = 0$, we get $g(0) = 0$, $g'(0) = 1/2$, etc. so we yield the Taylor series for $g(z)$ at $z = 0$.